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niverse. 4.1 Gravitational Forces due to Earth 4.2 Newton’s Law of Universal Gravitation 4.3 Relating Gravitational Field Strength to Gravitational Force Unit Themes and Emphases • Change and Systems • Social and Environmental Contexts • Problem-Solving Skills Focussing Questions In this study of dynamics and gravitation, you will investigate different types of forces and how they change the motion of objects and affect the design of various technological systems. As you study this unit, consider these questions: • How does an understanding of forces help humans interact with their environment? • How do the principles of dynamics affect mechanical and other systems? • What role does gravity play in the universe? Unit Project Tire Design, Stopping Distance, and Vehicle Mass • By the time you complete this unit, you will have the skills to evaluate how tire treads, road surfaces, and vehicle mass affect stopping distances. You will need to consider human reaction times and the amount of moisture on road surfaces to investigate this problem. Unit II Dynamics 123 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 124 C H A P T E R Forces can change velocity. 3 Key Concepts In this chapter, you will learn about: vector addition Newton’s laws of motion static and kinetic friction Learning Outcomes When you have completed this chapter, you will be able to: Knowledge explain that a non-zero net force causes a change in velocity calculate the net force apply Newton’s three laws to solve motion problems explain static and kinetic friction Science, Technology, and Society explain that the goal of technology is to provide solutions to practical problems explain that science and technology develop to meet societal needs explain that science develops through experimentation Screeching tires on the road and the sound of metal and fibreglass being crushed are familiar sounds of a vehicle collision. Depending on the presence of airbags and the correct use of seat belts and headrests, a motorist may suffer serious injury. In order to design these safety devices, engineers must understand what forces are and how forces affect the motion of an object. When a driver suddenly applies the brakes, the seat belts of all occupants lock. If the vehicle collides head-on with another vehicle, airbags may become deployed. Both seat belts and airbags are designed to stop the forward motion of motorists during a head-on collision (Figure 3.1). Motorists in a stationary vehicle that is rear-ended also experience forces. The car seats move forward quickly, taking the lower part of each person’s body with it. But each person’s head stays in the same place until yanked forward by the neck. It is this sudden yank that causes whiplash. Adjustable headrests are designed to prevent whiplash by supporting the head of each motorist. In this chapter, you will investigate how forces affect motion and how to explain and predict the motion of an object using Newton’s three laws. Figure 3.1 To design cars with better safety features, accident researchers use dummies to investigate the results of high-speed collisions. 124 Unit II 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 125 3-1 QuickLab 3-1 QuickLab Accelerating a Cart Problem If you pull an object with a force, how do force and mass affect the acceleration of the object? loaded cart spring scale Materials dynamics cart with hook two 200-g standard masses one 1-kg standard mass spring scale (0–5 N) smooth, flat surface (about 1.5 m long) Procedure 1 Place the 200-g standard masses on the cart and attach the spring scale to the hook on the cart. Figure 3.2 Questions 1. Why do you think it was difficult to apply a constant force when pulling the cart each time? 2. Describe how the acceleration of the cart changed from what it was in step 2 when (a) you used the 1-kg standard mass instead of the 2 Pull the spring scale so that the cart starts to 200-g masses, accelerate forward (Figure 3.2). Make sure that the force reading on the spring scale is 2 N and that the force remains as constant as possible while pulling the cart. Observe the acceleration of the cart. 3 Replace the 200-g masses on the cart with the 1-kg standard mass. Then pull the cart, applying the same force you used in step 2. Observe the acceleration of the cart. 4 Remove all the objects from the cart. Then pull the cart, applying the same force you used in step 2. Observe the acceleration of the cart. 5 Repeat step 4 but this time pull with a force of 1 N. (b) you removed all the objects from the cart, (c) you decreased the pulling force to 1 N, and (d) you increased the pulling force to 3 N. 3. What force was required to start the cart moving in step 7? 4. Suppose, instead of hooking a spring scale to the cart in steps 2 to 4, you gave the cart a push of the same magnitude each time. (a) Which cart would you expect to travel the farthest distance? (b) Which cart would you expect to slow down sooner? 6 Repeat step 4 but this time pull with a force of 3 N. (c) What force do you think makes the cart 7 Repeat step 4 but now only pull with just enough force to start the cart moving. Measure the force reading on the spring scale. eventually come to a stop? Think About It 1. Describe the motion of a large rocket during liftoff using the concept of force. Include diagrams in your explanation. 2. Is a plane during takeoff accelerating or moving with constant velocity? Explain in words and with diagrams. Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 3 Forces can change velocity. 125 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 126 info BIT Statics is a branch of dynamics that deals with the forces acting on stationary objects. Architecture is primarily a practical application of statics. dynamics: branch of mechanics dealing with the cause of motion 3.1 The Nature of Force The Petronas Twin Towers in Kuala Lumpur, Malaysia, are currently the world’s tallest twin towers. Including the spire on top, each tower measures 452 m above street level. To allow for easier movement of people within the building, architects designed a bridge to link each tower at the 41st floor. What is interesting is that this bridge is not stationary. In order for the bridge to not collapse, it must move with the towers as they sway in the wind (Figure 3.3). In Unit I, you learned that kinematics describes the motion of an object without considering the cause. When designing a structure, the kinematics quantities that an architect considers are displacement, velocity, and acceleration. But to predict how and explain why a structure moves, an architect must understand dynamics. Dynamics deals with the effects of forces on objects. Structures such as bridges and buildings are required to either remain stationary or move in appropriate ways, depending on the design, so that they are safe to use. Architects must determine all the forces that act at critical points of the structure. If the forces along a particular direction do not balance, acceleration will occur. Before you can predict or explain the motion of an object, it is important to first understand what a force is and how to measure and calculate the sum of all forces acting on an object. Figure 3.3 The design of tall buildings involves understanding forces. Towering buildings are susceptible to movement from the wind. 126 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 127 Force Is a Vector Quantity You experience a force when you push or pull an object. A push or a pull can have different magnitudes and can be in different directions. For this reason, force is a vector quantity. In general, any force acting on an object can change the shape and/or velocity of the object (Figure 3.4). If you want to deform an object yet keep it stationary, at least two forces must be present. (a) force exerted by hand on ball (b) force exerted by hand on ball force exerted by hand on ball (c) force: a quantity measuring a push or a pull on an object force exerted by table on ball distorting starting stopping (d) force exerted by hand on ball (e) force exerted by hand on ball (f) force exerted by hand on ball info BIT One newton is roughly equal to the magnitude of the weight of a medium-sized apple or two golf balls. speeding up slowing down changing the direction of motion Figure 3.4 Different forces acting on a ball change either the shape or the motion of the ball. (a) The deformation of the ball is caused by both the hand and the table applying opposing forces on the ball. (b)–(f) The motion of the ball is changed, depending on the magnitude and the direction of the force applied by the hand. The symbol of force is F and the SI unit for force is the newton (N), named in honour of physicist Isaac Newton (1642–1727). One newton is equal to one kilogram-metre per second squared (1 kgm/s2), which is the force required to move a 1-kg object with an acceleration of 1 m/s2. The direction of a force is described using reference coordinates that you choose for a particular situation. You may use [forward] or [backward], compass directions, or polar coordinates. When stating directions using polar coordinates, measure angles counterclockwise from the positive x-axis (Figure 3.5). (a) y 1 0 N 30.0° x (b) y II III 330° 10 N I x IV Figure 3.5 Two vectors of the same magnitude but with different directions. (a) 10 N [30] (b) 10 N [330] Chapter 3 Forces can change velocity. 127 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 128 Measuring Force One way you could measure forces involves using a calibrated spring scale. To measure the force of gravity acting on an object, attach the object to the end of a vertical spring and observe the stretch of the spring. The weight of an object is the force of gravity acting on the object. The symbol g. of weight is F When the spring stops stretching, the gravit |
ational and elastic forces acting on the object balance each other (Figure 3.6). At this point, the elastic force is equal in magnitude to the weight of the object. So you can determine the magnitude of the weight of an object by reading the pointer position on a calibrated spring scale once the spring stops stretching. pointer elastic force vector Find out the relationship between the stretch of a spring and the weight of an object by doing 3-2 QuickLab. object 10 6 N gravitational force vector 6 N Figure 3.6 A spring scale is one type of instrument that can be used to measure forces. 3-2 QuickLab 3-2 QuickLab Measuring Force Using a Spring Scale Problem How is the amount of stretch of a calibrated spring related to the magnitude of the force acting on an object? 3 Hang additional objects from the spring, up to a total mass of 1000 g. Each time, record the mass and the magnitudes of the corresponding gravitational and elastic forces. Materials set of standard masses with hooks spring scale (0–10 N) Procedure 1 Hold the spring scale vertically and make sure the pointer reads zero when nothing is attached. 2 Gently suspend a 100-g standard mass from the spring. Use a table to record the mass and the magnitudes of the gravitational and elastic forces acting on the object. Questions 1. What was the reading on the spring scale when the 100-g mass was attached? 2. What happened to the stretch of the spring when the mass of the object attached to the spring scale (a) doubled? (b) tripled? (c) changed by a factor of 10? 3. Why is a spring scale ideal for measuring force? 128 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 129 Representing Forces Using Free-Body Diagrams A free-body diagram is a powerful tool that can be used to analyze situations involving forces. This diagram is a sketch that shows the object by itself, isolated from all others with which it may be interacting. Only the force vectors exerted on the object are included and, in this physics course, the vectors are drawn with their tails meeting at the centre of the object (Figure 3.7). However, it does not necessarily mean that the centre of the object is where the forces act. When drawing a free-body diagram, it is important to show the reference coordinates that apply to the situation in a given problem. Remember to always include which directions you will choose as positive. Figure 3.8 shows the steps for drawing free-body diagrams. free-body diagram: vector diagram of an object in isolation showing all the forces acting on it Fs elastic force exerted by spring on mass + up down Fg gravitational force exerted by Earth on mass Identify the object for the free-body diagram Sketch the object in isolation with a dot at its centre Identify all the forces acting on the object Include in free-body diagram Choose appropriate reference coordinates and include which directions are positive Forces on other objects? Ignore Figure 3.7 The free-body diagram for the object in Figure 3.6 that is suspended from the spring scale. The spring scale is not included here because it is not the object being studied. N, S, E, W up, down, forward, backward x, y If possible, choose an appropriate scale Draw forces proportionally extending out from the dot at the centre of the object Figure 3.8 Flowchart summarizing the steps for drawing a free-body diagram Chapter 3 Forces can change velocity. 129 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 130 normal force: force on an object that is perpendicular to a common contact surface Some Types of Forces There are different types of forces and scientists distinguish among them by giving these forces special names. When an object is in contact with another, the objects will have a common surface of contact, and the two objects will exert a normal force on each other. The normal force, F N, is a force that is perpendicular to this common surface. Depending on the situation, another force called friction, F f, may be present, and this force acts parallel to the common surface. The adjective “normal” simply means perpendicular. Figure 3.9 (a) shows a book at rest on a level table. The normal force exerted by the table on the book is represented by the vector directed upward. If the table top were slanted and smooth as in Figure 3.9 (b), the normal force acting on the book would not be directed vertically upward. Instead, it would be slanted, but always perpendicular to the contact surface. (a) (b) FN Fg FN Ff Fg θ Figure 3.9 Forces acting on (a) a stationary book on a level table and on (b) a book accelerating down a smooth, slanted table. app, if, say, a A stationary object may experience an applied force, F person pushes against the object (Figure 3.10). In this case, the force of friction acting on the object will oppose the direction of impending motion. FN Ff Fapp Fg Figure 3.10 The forces acting on a stationary box Example 3.1 demonstrates how to draw a free-body diagram for a car experiencing different types of forces. In this situation, the normal force acting on the car is equal in magnitude to the weight of the car. 130 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 131 Example 3.1 g, of 10 000 N [down] is coasting on a level road. A car with a weight, F N, of 10 000 N [up], a force of air The car experiences a normal force, F f, exerted air, of 2500 N [backward], and a force of friction, F resistance, F by the road on the tires of 500 N [backward]. Draw a free-body diagram for this situation. Analysis and Solution While the car is coasting, there is no forward force acting on the car. The free-body diagram shows four forces (Figure 3.11). 2000 N up down backward forward FN Ff Fair Fg Figure 3.11 Practice Problems 1. The driver in Example 3.1 sees a pedestrian and steps on the brakes. The force of air resistance is 2500 N [backward]. With the brakes engaged, the force of friction exerted on the car is 5000 N [backward]. Draw a free-body diagram for this situation. 2. A car moving at constant velocity starts to speed up. The weight of the car is 12 000 N [down]. The force of air resistance is 3600 N [backward]. With the engine engaged, the force of friction exerted by the road on the tires is 7200 N [forward]. Draw a free-body diagram for this situation. Answers 1. and 2. See page 898. Using Free-Body Diagrams to Find Net Force Free-body diagrams are very useful when you need to calculate the net force, F net, on an object. The net force is a vector sum of all the forces acting simultaneously on an object. The force vectors can be added using either a scale vector diagram or using components. net force: vector sum of all the forces acting simultaneously on an object Concept Check Can the net force on an object ever equal zero? Explain using an example and a free-body diagram. Chapter 3 Forces can change velocity. 131 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 132 e SIM Learn how to use free-body diagrams to find the net force on an object. Follow the eSim links at www.pearsoned.ca/ school/physicssource. Adding Collinear Forces Vectors that are parallel are collinear, even if they have opposite directions. Example 3.2 demonstrates how to find the net force on an object given two collinear forces. In this example, a canoe is dragged using two ropes. The magnitude of the force F T exerted by a rope on an object at the point where the rope is attached to the object is called the tension in the rope. In this physics course, there are a few assumptions that you need to make about ropes or cables to simplify calculations. These assumptions and the corresponding inferences are listed in Table 3.1. Note that a “light” object means that it has negligible mass. Table 3.1 Assumptions about Ropes or Cables Assumption Inference The mass of the rope is negligible. The rope has a negligible thickness. The tension is uniform throughout the length of the rope. F T acts parallel to the rope and is directed away from the object to which the rope is attached. The rope is taut and does not stretch. Any objects attached to the rope will have the same magnitude of acceleration as the rope. Example 3.2 Two people, A and B, are dragging a canoe out of a lake onto a beach using light ropes (Figure 3.12). Each person applies a force of 60.0 N [forward] on the rope. The force of friction exerted by the beach on the canoe is 85.0 N [backward]. Starting with a free-body diagram, calculate the net force on the canoe. forward backward Practice Problems 1. Two dogs, A and B, are pulling a sled across a horizontal, snowy surface. Dog A exerts a force of 200 N [forward] and dog B a force of 150 N [forward]. The force of friction exerted by the snow on the sled is 60 N [backward]. The driver attempts to slow down the sled by pulling on it with a force of 100 N [backward]. Starting with a free-body diagram, calculate the net force on the sled. 132 Unit II Dynamics Given F F T1 T2 F f 60.0 N [forward] 60.0 N [forward] 85.0 N [backward] Figure 3.12 Required net force on canoe (F net) Analysis and Solution Draw a free-body diagram for the canoe (Figure 3.13). backward forward FT1 FT2 Ff Figure 3.13 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 133 backward forward FT1 Fnet Figure 3.14 FT2 Ff Add the force vectors shown in the vector addition diagram (Figure 3.14). F f F Ff F T2 FT2 F T1 FT1 60.0 N 60.0 N (85.0 N) 60.0 N 60.0 N 85.0 N 35.0 N 35.0 N [forward] net Fnet F net Paraphrase The net force on the canoe is 35.0 N [forward]. Practice Problems 2. In a tractor pull, four tractors are connected by strong chains to a heavy load. The load is initially at rest. Tractors A and B pull with forces of 5000 N [E] and 4000 N [E] respectively. Tractors C and D pull with forces of 4500 N [W] and 3500 N [W] respectively. The magnitude of the force of friction exerted by the ground on the load is 1000 N. (a) Starting with a free-body diagram, calculate the net force on the load. (b) If the load is initially at rest, will it start moving? Explain. Answers 1. |
190 N [forward] 2. (a) 0 N, (b) no Adding Non-Collinear Forces Example 3.3 demonstrates how to find the net force on an object if the forces acting on it are neither parallel nor perpendicular. By observing the relationship between the components of the force vectors, you can greatly simplify the calculations. Example 3.3 Refer to Example 3.2 on page 132. Person A thinks that if A and B each pull a rope forming an angle of 20.0° with the bow, the net force on the canoe will be greater than in Example 3.2 (Figure 3.15). The canoe is being dragged along the beach using ropes that are parallel to the surface of the beach. Starting with a free-body diagram, calculate the net force on the canoe. Is person A’s thinking correct? 20.0° θ 1 20.0° θ 2 Figure 3.15 Given F 60.0 N [along rope] 85.0 N [backward] T1 F f F T2 1 60.0 N [along rope] 2 20.0 y Required net force on canoe (F net) Analysis and Solution Draw a free-body diagram for the canoe (Figure 3.16). x Ff Figure 3.16 FT1 20.0° 20.0° FT2 Chapter 3 Forces can change velocity. 133 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 134 Practice Problems 1. Refer to Example 3.3. Suppose person A pulls a rope forming an angle of 40.0 with the bow and person B pulls a rope forming an angle of 20.0 with the bow. Each person applies a force of 60.0 N on the rope. The canoe and ropes are parallel to the surface of the beach. If the canoe is being dragged across a horizontal, frictionless surface, calculate the net force on the canoe. 2. Two people, A and B, are dragging a sled on a horizontal, icy surface with two light ropes. Person A applies a force of 65.0 N [30.0] on one rope. Person B applies a force of 70.0 N [300] on the other rope. The force of friction on the sled is negligible and the ropes are parallel to the icy surface. Calculate the net force on the sled. Answers 1. 1.04 102 N [10.0] 2. 95.5 N [343] Separate all forces into x and y components. T1 Vector F F F T2 f x component y component (60.0 N)(cos 20.0) (60.0 N)(cos 20.0) 85.0 N (60.0 N)(sin 20.0) (60.0 N)(sin 20.0) 0 From the chart, FT1y F nety Fnety F T1y FT1y So y FT2y. F T2y FT2y 0 N FT1 20.0° Fnet 20.0° FT2 Ff x Figure 3.17 Figure 3.17 Add the x components of all force vectors in the vector addition diagram (Figure 3.17). netx Fnetx F f Ff F T2x FT2x x direction F F T1x FT1x (60.0 N)(cos 20.0) (60.0 N)(cos 20.0) (85.0 N) (60.0 N)(cos 20.0) (60.0 N)(cos 20.0) 85.0 N 27.8 N 27.8 N [0] F net Paraphrase The net force is 27.8 N [0°]. Since the net force in Example 3.3 is less than that in Example 3.2, person A’s thinking is incorrect. Applying Free-Body Diagrams to Objects in Equilibrium At the beginning of this section, you learned that architects consider the net force acting at critical points of a building or bridge in order to prevent structure failure. Example 3.4 demonstrates how free-body diagrams and the concept of net force apply to a stationary object. Stationary objects are examples of objects at equilibrium because the net force acting on them equals zero. Example 3.4 A store sign that experiences a downward gravitational force of 245 N is suspended as shown in Figure 3.18. T1 and F Calculate the forces F exerted at the point at which the sign is suspended. T2 Figure 3.18 θ 1 55.0° FT1 θ 2 FT2 134 Unit II Dynamics 03-Phys20-Chap03.qxd 7/25/08 8:17 AM Page 135 Given F g 245 N [down] Required T2) T1 and F forces (F 1 55.0° 2 90.0° Analysis and Solution Draw a free-body diagram for the sign (Figure 3.19). Resolve all forces into x and y components. T1 Vector F F F T2 g x component FT1 cos 55.0 FT2 0 y component FT1 sin 55.0 0 Fg Since the sign is at equilibrium, the net force in both the x and y directions is zero. Fnetx Fnety 0 N Add the x and y components of all force vectors separately. x direction F netx Fnetx F T1x FT1x F T2x FT2x 0 FT1 cos 55.0 F T2 FT2 FT1 cos 55.0 y direction F F T1y FT1y nety Fnety F g Fg 0 FT1 sin 55.0 ( Fg) 0 FT1 Fg sin 55.0 FT1 Fg sin 55.0 245 N sin 55.0 kgm s2 299.1 299 N Substitute FT1 into the equation for FT2. FT2 F T2 FT1 cos 55.0 (299.1 N)(cos 55.0) 171.6 N 172 N [0] From Figure 3.19, the direction of F T1 measured counterclockwise from the positive x-axis is 180 55.0 125 . F T1 299 N [125] Paraphrase T2 is 172 N [0]. T1 is 299 N [125] and F F 55.0° FT1 55.0° y x FT2 Fg Figure 3.19 Practice Problems 1. If the sign in Example 3.4 had half the weight, how would the forces T2 compare? T1 and F F 2. Suppose the sign in Example 3.4 is suspended as shown in Figure 3.20. Calculate the forces FF T1 T2. and FF θ 1 40.0° FT1 θ 2 FT2 Figure 3.20 3. Refer to the solutions to Example 3.4 and Practice Problem 2 above. (a) As 1 decreases, what happens T2? T1 and F to F (b) Explain why 1 can never equal zero. Answers 1. directions of F T1 and F T2 would remain the same as before, but the respective magnitudes would be half 3.81 102 N [140°] 2.92 102 N [0°] T1 2. F F 3. (a) F T2 T1 and F (b) magnitude of F T2 increase in value T1y must always equal Fg Chapter 3 Forces can change velocity. 135 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 136 3.1 Check and Reflect 3.1 Check and Reflect Knowledge 1. (a) Explain what a force is, and state the SI unit of force. (b) Why is force a dynamics quantity and not a kinematics quantity? Applications 2. Sketch a free-body diagram for (a) a bicycle moving west on a level road with decreasing speed (b) a ball experiencing forces of 45 N [12.0], 60 N [100], and 80 N [280] simultaneously 3. The total weight of a biker and her motorbike is 1800 N [down]. With the engine engaged, the force of friction exerted by the road on the tires is 500 N [forward]. The air resistance acting on the biker and bike is 200 N [backward]. The normal force exerted by the road on the biker and bike is 1800 N [up]. (a) Consider the biker and bike as a single object. Draw a free-body diagram for this object. (b) Calculate the net force. 4. If two forces act on an object, state the angle between these forces that will result in the net force given below. Explain using sketches. (a) maximum net force (b) minimum net force 5. Two people, A and B, are pulling on a tree with ropes while person C is cutting the tree down. Person A applies a force of 80.0 N [45.0] on one rope. Person B applies a force of 90.0 N [345] on the other rope. Calculate the net force on the tree. 6. Three forces act simultaneously on an 2 is 80 N [115], 1 is 65 N [30.0], F object: F 3 is 105 N [235]. Calculate the net and F force acting on the object. Extensions 7. A sign that experiences a downward gravitational force of 245 N is suspended, as shown below. Calculate the forces F T2. and F T1 θ 1 55.0° θ 2 55.0° FT1 FT2 8. The blanket toss is a centuries-old hunting technique that the Inuit used to find herds of caribou. During the toss, several people would hold a hide taut while the hunter would jump up and down, much like on a trampoline, increasing the jump height each time. At the top of the jump, the hunter would rotate 360 looking for the herd. Draw a free-body diagram for a hunter of weight 700 N [down] while (a) standing at rest on the taut hide just before a jump (b) at the maximum jump height 9. Construct a flowchart to summarize how to add two or more non-collinear forces using components. Refer to Figure 3.8 on page 129 or Student References 4: Using Graphic Organizers on page 869. e TEST To check your understanding of forces, follow the eTest links at www.pearsoned.ca/school/ physicssource. 136 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 137 info BIT The skeleton is an Olympic sledding sport believed to have originated in St. Moritz, Switzerland. 3.2 Newton’s First Law At the 2006 Winter Olympics in Turin, Italy, Duff Gibson charged head-first down a 1.4-km icy track with 19 challenging curves. He reached speeds well over 125 km/h and ended up clinching the gold medal in the men’s skeleton event (Figure 3.21). Gibson’s success had a lot to do with understanding the physics of motion. Imagine an ideal situation in which no friction acts on the sled and no air resistance acts on the athlete. Scientist Galileo Galilei (1564–1642) thought that an object moving on a level surface would continue moving forever at constant speed and in the same direction if no external force acts on the object. If the object is initially stationary, then it will remain stationary, provided no external force acts on the object. In the real world, friction and air resistance are external forces that act on all moving objects. So an object that is in motion will eventually slow down to a stop, unless another force acts to compensate for both friction and air resistance. Galileo recognized the existence of friction, so he used thought experiments, as well as experiments with controlled variables, to understand motion. Thought experiments are theoretical, idealized situations that can be imagined but cannot be created in the real world. Figure 3.21 In the 2006 Winter Olympics, Calgary residents Duff Gibson (shown in photo) and Jeff Pain competed in the men’s skeleton. Gibson edged Pain by 0.26 s to win the gold medal. Pain won the silver medal. Chapter 3 Forces can change velocity. 137 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 138 inertia: property of an object that resists acceleration The Concept of Inertia Since ancient times, many thinkers attempted to understand how and why objects moved. But it took thousands of years before satisfactory explanations were developed that accounted for actual observations. A major stumbling block was not identifying friction as a force that exists in the real world. In his study of motion, Galileo realized that every object has inertia, a property that resists acceleration. A stationary curling stone on ice requires a force to start the stone moving. Once it is moving, the curling stone requires a force to stop it. To better understand the concept of inertia, try the challenges in 3-3 QuickLab. Concept Check Compare the inertia of an astronaut on Earth’s surface, in o |
rbit around Earth, and in outer space. Can an object ever have an inertia of zero? Explain. 3-3 QuickLab 3-3 QuickLab Challenges with Inertia Problem What role does inertia play in each of these challenges? 3 Set up a stack of loonies and the ruler as shown in Figure 3.22 (c). Use the ruler to remove the loonie at the very bottom without toppling the stack. Materials glass tumbler small, stiff piece of cardboard several loonies empty soft-drink bottle plastic hoop (about 2 cm wide, cut from a large plastic bottle) small crayon with flat ends ruler (thinner than thickness of one loonie) Procedure 1 Set up the tumbler, piece of cardboard, and loonie as shown in Figure 3.22 (a). Remove the cardboard so the loonie falls into the tumbler. 2 Set up the soft-drink bottle, plastic hoop, and a crayon as shown in Figure 3.22 (b). Remove the hoop so the crayon falls into the bottle. Caution: Keep your eyes well above the coin stack. Questions 1. (a) What method did you use to successfully perform each step in the procedure? (b) Apply the concept of inertia to explain why each of your procedures was effective. (a) (b) (c) Figure 3.22 138 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 139 Newton’s First Law and Its Applications Newton modified and extended Galileo’s ideas about inertia in a law, called Newton’s first law of motion (Figure 3.23). (a) Fnet 0 An object will continue either being at rest or moving at constant velocity unless acted upon by an external non-zero net force. v 0 If F net 0, then v 0. So if you want to change the motion of an object, a non-zero net (b) Fnet 0 force must act on the object. Concept Check The Voyager 1 and 2 space probes are in interstellar space. If the speed of Voyager 1 is 17 km/s and no external force acts on the probe, describe the motion of the probe (Figure 3.24). v constant Figure 3.23 If the net force on an object is zero, (a) a stationary object will remain at rest, and (b) an object in motion will continue moving at constant speed in the same direction. Figure 3.24 The Voyager planetary mission is NASA’s most successful in terms of the number of scientific discoveries. Newton’s First Law and Sliding on Ice Many winter sports involve a person sliding on ice. In the case of the skeleton event in the Winter Olympics, an athlete uses a sled to slide along a bobsled track. In hockey, a player uses skates to glide across the icy surface of a rink. Suppose a person on a sled is sliding along a horizontal, icy surface. If no external force acts on the person-sled system, then according to Newton’s first law, the person would maintain the same speed. In fact, the person would not stop at all (Figure 3.25). In real life, the external forces of friction and air resistance act on all moving objects. So the system would eventually come to a stop. FN Fg Figure 3.25 Free-body diagram of a person-sled system sliding on a horizontal, frictionless surface Chapter 3 Forces can change velocity. 139 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 140 Newton’s First Law and Vehicle Safety Devices When you are in a moving car, you can feel the effects of your own inertia. If the car accelerates forward, you feel as if your body is being pushed back against the seat, because your body resists the increase in speed. If the car turns a corner, you feel as if your body is being pushed against the door, because your body resists the change in the direction of motion. If the car stops suddenly, you feel as if your body is being pushed forward, because your body resists the decrease in speed (Figure 3.26). (a) Inertia of motorist makes her feel like she is being pushed backward. direction of motion (c) Inertia of motorist makes her feel like she is being thrown forward. direction of acceleration of vehicle (b) Inertia of motorist makes her feel like she wants to continue moving in a straight line. direction of acceleration of vehicle Figure 3.26 The inertia of a motorist resists changes in the motion of a vehicle. (a) The vehicle is speeding up, (b) the vehicle is changing direction, and (c) the vehicle is stopping suddenly. When a car is rear-ended, a motorist’s body moves forward suddenly as the car seat moves forward. However, the motorist’s head resists moving forward. A properly adjusted headrest can minimize or prevent whiplash, an injury resulting from the rapid forward accelerations in a rear-end collision (Figure 3.27). Research shows that properly adjusted headrests can reduce the risk of whiplash-related injuries by as much as 40%. A poorly adjusted headrest, however, can actually worsen the effects of a rear-end collision on the neck and spine. When a car is involved in a head-on collision, the motorist continues to move forward. When a seat belt is worn properly, the forward motion of a motorist is safely restricted. If a head-on collision is violent enough, sodium azide undergoes a rapid chemical reaction to produce non-toxic nitrogen gas, which inflates an airbag. The inflated airbag provides a protective cushion to slow down the head and body of a motorist (Figure 3.28). 5–10 cm Figure 3.27 The ideal position for a headrest Figure 3.28 Airbag systems in vehicles are designed to deploy during vehicle collisions. 140 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 141 3-4 Decision-Making Analysis 3-4 Decision-Making Analysis The Airbag Debate The Issue Front airbags were introduced in the 1990s to help prevent injury to motorists, especially during head-on collisions. Side airbags can also help. Yet front airbags have also been the cause of serious injury, even death. Furthermore, airbags add to the cost of a vehicle. Airbag advocates want both front and side airbags installed, better airbags, and greater control over their operation. Opponents want airbags removed from cars altogether. Background Information Airbags are connected to sensors that detect sudden changes in acceleration. The process of triggering and inflating an airbag occurs in about 40 ms. It is in that instant that arms and legs have been broken and children have been killed by the impact of a rapidly inflating airbag. Tragically, some of these deaths occurred during minor car accidents. Manufacturers have placed on/off switches for airbags on some vehicles, and some engineers are now developing “smart” airbags, which can detect the size of a motorist and the distance that person is sitting from an airbag. This information can then be used to adjust the speed at which the airbag inflates. Required Skills Defining the issue Developing assessment criteria Researching the issue Analyzing data and information Proposing a course of action Justifying the course of action Evaluating the decision Analysis 1. Identify the different stakeholders involved in the airbag controversy. 2. Research the development and safety history of airbags in cars. Research both front and side airbags. Consider head, torso, and knee airbags. Analyze your results, and identify any trends in your data. 3. Propose a solution to this issue, based on the trends you identified. 4. Propose possible changes to current airbag design that could address the issues of safety and cost. 5. Plan a class debate to argue the pros and cons of airbag use. Identify five stakeholders to represent each side in the debate. Support your position with research. Participants will be assessed on their research, organizational skills, debating skills, and attitudes toward learning. Concept Check Use Newton’s first law to explain why (a) steel barriers usually separate the cab of a truck from the load (Figure 3.29), (b) trucks carrying tall loads navigate corners slowly, and (c) customers who order take-out drinks are provided with lids. e TECH Explore the motion of an object that experiences a net force of zero. Follow the eTECH links at www.pearsoned.ca/school/ physicssource. Figure 3.29 Chapter 3 Forces can change velocity. 141 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 142 3.2 Check and Reflect 3.2 Check and Reflect Knowledge 6. Imagine you are the hockey coach for a 1. In your own words, state Newton’s first law. 2. Give two examples, other than those in the text, that illustrate the property of inertia for both a stationary and a moving object. 3. Use Newton’s first law to describe the motion of (a) a car that attempts to go around an icy curve too quickly, and (b) a lacrosse ball after leaving the lacrosse stick. 4. Apply Newton’s first law and the concept of inertia to each of these situations. team of 10-year-olds. At a hockey practice, you ask the players to skate across the ice along the blue line (the line closest to the net), and shoot the puck into the empty net. Most of the shots miss the net. The faster the children skate, the more they miss. Newton’s first law would help the players understand the problem, but a technical explanation might confuse them. (a) Create an explanation that would make sense to the 10-year-olds. (b) With the aid of a diagram, design a drill for the team that would help the players score in this type of situation. (a) How could you remove the newspaper without toppling the plastic beaker? Extensions 7. Research why parents use booster seats for young children using information from Safe Kids Canada. Summarize the “seat belt test” that determines whether a child is big enough to wear a seat belt without a booster seat. Begin your search at www.pearsoned.ca/school/physicssource. 8. During a sudden stop or if a motorist tries to adjust a seat belt suddenly, the seat belt locks into position. Research why seat belts lock. Write a brief report, including a diagram, of your findings. Begin your search at www.pearsoned.ca/school/physicssource. 9. Make a web diagram to summarize concepts and ideas associated with Newton’s first law. Label the oval in the middle as “Newton’s first law.” Cluster your words or ideas in other ovals around it. Connect these ovals to the central one and one another, where appropriate, with lines. S |
ee Student References 4: Using Graphic Organizers on page 869 for an example. e TEST To check your understanding of inertia and Newton’s first law, follow the eTest links at www.pearsoned.ca/school/physicssource. (b) While moving at constant speed on a level, snowy surface, a snowmobiler throws a ball vertically upward. If the snowmobile continues moving at constant velocity, the ball returns to the driver. Why does the ball land ahead of the driver if the snowmobile stops? Assume that the air resistance acting on the ball is negligible. x y vs vby vbx Applications 5. Design an experiment using an air puck on an air table or spark air table to verify Newton’s first law. Report your findings. Caution: A shock from a spark air table can be dangerous. 142 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 143 3.3 Newton’s Second Law If a speed skater wants to win a championship race, the cardiovascular system and leg muscles of the athlete must be in peak condition. The athlete must also know how to effectively apply forces to propel the body forward. World-class speed skaters such as Cindy Klassen know that maximizing the forward acceleration requires understanding the relationship among force, acceleration, and mass (Figure 3.30). Newton spent many years of his life trying to understand the motion of objects. After many experiments and carefully analyzing the ideas of Galileo and others, Newton eventually found a simple mathematical relationship that models the motion of an object. This relationship, known as Newton’s second law, relates the net force acting on an object, the acceleration of the object, and its mass. Begin by doing 3-5 Inquiry Lab to find the relationship between the acceleration of an object and the net force acting on it. info BIT Cindy Klassen won a total of five medals during the 2006 Winter Olympics, a Canadian record, and is currently Canada’s most decorated Olympian with six medals. Figure 3.30 Cindy Klassen, originally from Winnipeg but now a Calgary resident, won the gold medal in the 1500-m speed skating event in the 2006 Winter Olympics in Turin, Italy. Chapter 3 Forces can change velocity. 143 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 144 3-5 Inquiry Lab 3-5 Inquiry Lab Relating Acceleration and Net Force Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 1 2 it a (t)2 relates The kinematics equation d v displacement d, initial velocity v and acceleration aa. If v dd 1 a (t)2. If you solve for acceleration, you get 2 2d a . Remember to use the scalar form of this equation ( )2 t i 0, the equation simplifies to i, time interval t, when solving for acceleration. Question How is the acceleration of an object related to the net force acting on the object? Hypothesis State a hypothesis relating acceleration and net force. Remember to write an “if/then” statement. Variables The variables involved in this lab are the mass of the system, the applied force acting on the system, friction acting on the system, time interval, the distance the system travels, and the acceleration of the system. Read the procedure and identify the controlled, manipulated, and responding variable(s). Materials and Equipment C-clamp dynamics cart three 100-g standard masses pulley smooth, flat surface (about 1.5 m) string (about 2 m) recording tape recording timer with power supply metre-stick masking tape graph paper e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. 144 Unit II Dynamics Procedure 1 Copy Tables 3.2 and 3.3 from page 145 into your notebook. 2 Measure the mass of the cart. Record the value in Table 3.2. 3 Set up the recording timer, pulley, and cart loaded with three 100-g standard masses on a lab bench (Figure 3.31). Make a loop at each end of the string. Hook one loop to the end of the cart and let the other loop hang down over the pulley. recording tape recording timer 100 g 100 g string pulley 100 g 1 N Figure 3.31 Caution: Position a catcher person near the edge of the lab bench. Do not let the cart or hanging objects fall to the ground. 4 Attach a length of recording tape to the cart and thread it through the timer. 5 While holding the cart still, transfer one 100-g mass from the cart to the loop of string over the pulley and start the timer. When you release the cart, the hanging 100-g mass will exert a force of about 1 N on the system. Release the cart but stop it before it hits the pulley. Label the tape “trial 1; magnitude of F app 1 N.” 6 Repeat steps 4 and 5 using the same cart but this time transfer another 100-g mass from the cart to the first hanging object. Label the tape “trial 2; magnitude of F app = 2 N.” By transferring objects from the cart to the end of the string hanging over the pulley, the mass of the system remains constant but the net force acting on the system varies. 7 Repeat steps 4 and 5 using the same cart but this time transfer another 100-g mass from the cart to the two hanging masses. Label the tape “trial 3; magnitude of F app = 3 N.” 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 145 Analysis 1. Calculate the mass of the system, mT. Record the value in Table 3.2. 2. Using the tape labelled “trial 3,” label the dot at the start t 0 and mark off a convenient time interval. If the 1 s, a time interval of 30 dot period of the timer is 0 6 1 s 0.5 s). Record spaces represents 0.5 s (30 0 6 the time interval in Table 3.2. 3. Measure the distance the system travelled during this time interval. Record this value in Table 3.2. 2 d to calculate the magnitude 4. Use the equation a ( )2 t of the acceleration of the system. Record the value in Tables 3.2 and 3.3. 5. Using the same time interval, repeat questions 3 and 4 for the tapes labelled “trial 1” and “trial 2.” 6. Why is it a good idea to choose the time interval using the tape labelled “trial 3”? 7. Plot a graph of the magnitude of the acceleration vs. the magnitude of the applied force (Table 3.3). 8. (a) Describe the graph you drew in question 7. (b) Where does the graph intersect the x-axis? Why? What conditions would have to be present for it to pass through the origin? (c) For each trial, subtract the x-intercept from the applied force to find the net force. Record the values in Table 3.3. Then plot a graph of the magnitude of the acceleration vs. the magnitude of the net force. 9. When the magnitude of the net force acting on the system is doubled, what happens to the magnitude of the acceleration of the system? 10. What is the relationship between the magnitude of the acceleration and the magnitude of the net force? Write this relationship as a proportionality statement. Does this relationship agree with your hypothesis? Table 3.2 Mass, Time, Distance, and Acceleration Trial Mass of Cart mc (kg) Mass of Load on Cart ml (kg) Mass of Load Hanging over Pulley mh (kg) mT 1 2 3 0.200 0.100 0 0.100 0.200 0.300 Table 3.3 Force and Acceleration Total Mass mc ml (kg) mh Time Interval t (s) Distance d (m) Magnitude of a (m/s2) Trial 1 2 3 Magnitude of F app Acting Magnitude of F net Acting on System (N) on System (N) 1 2 3 Magnitude of a of System (m/s2) Chapter 3 Forces can change velocity. 145 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 146 FN Fair Ff Fg Fapp Fapp Fnet Ff Fair Fnet Fapp Fair Ff Relating Acceleration and Net Force For the system in 3-5 Inquiry Lab, you discovered that there is a linear relationship between acceleration and net force. This relationship can be written as a proportionality statement: Horizontal Forces a Fnet This relationship applies to speed skating. In the short track relay event, a speed skater pushes the next teammate forward onto the track when it is the teammate’s turn to start skating. While the teammate is being pushed, the horizontal forces acting on the skater are the applied push force, friction, and air resistance (Figure 3.32). As long as the applied push force is greater in magnitude than the sum of the force of friction acting on the skates and the air resistance acting on the skater’s body, the net force on the teammate acts forward. Figure 3.32 (left) Free-body diagram showing the forces acting on a speed skater being pushed by a teammate in the short track relay event; (right) vector addition diagram for the horizontal forces. The harder the forward push, the greater will be the forward net force on the teammate (Figure 3.33). So the acceleration of the teammate will be greater. Note that the acceleration is in the same direction as the net force. Find out the relationship between the acceleration of an object and its mass by doing 3-6 Design a Lab. m constant a a Fnet Fnet Concept Check What is the difference between a net force and an applied force? Can a net force ever equal an applied force? Explain using an example and a free-body diagram. Figure 3.33 For the same mass, a greater net force results in a greater acceleration. 146 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 147 3-6 Design a Lab 3-6 Design a Lab Relating Acceleration and Mass In this lab, you will investigate the relationship between acceleration and mass when the net force acting on the system is constant. The Question How is the acceleration of an object related to the mass of the object? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Design and Conduct Your Investigation • State a hypothesis relating acceleration and mass. • Then use the set-up in Figure 3.31 on page 144 to design an experiment. List the materials you will use as well as a detailed procedure. Use the procedure and questions in 3-5 Inquiry Lab to help you. • Plot a graph of the magnitude of the acceleration vs. the mass of the system. Then plot a graph of the magnitude of the acceleration vs. the reciprocal of the mass of the system. • Analyze your data and form conclusions. How well did your results agree with your hypothesis? e TECH Explore how the net force on an object and its mass affect |
its acceleration. Follow the eTech links at www.pearsoned.ca/ school/physicssource. Relating Acceleration and Mass In 3-6 Design a Lab, you discovered that the relationship between acceleration and mass is non-linear. But if you plot acceleration as a function of the reciprocal of mass, you get a straight line. This shows that there is a linear relationship between acceleration and the reciprocal of mass. This relationship can be written as a proportionality statement: 1 a m In speed skating, evidence of this relationship is the different accelerations that two athletes of different mass have. Suppose athlete A has a mass of 60 kg and athlete B a mass of 90 kg. If the net force acting on A and B is the same, you would expect A to have a greater acceleration than B (Figure 3.34). This observation makes sense in terms of inertia, because the inertia of B resists the change in motion more so than the inertia of A does. In fact, you observed this relationship in 3-1 QuickLab when you compared the acceleration of an empty cart and a cart loaded with a 1-kg standard mass. m Fnet constant m a a A B Figure 3.34 For the same net force, a more massive person has a smaller acceleration than a less massive one does. Chapter 3 Forces can change velocity. 147 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 148 Newton’s Second Law and Inertial Mass 1 can be combined The proportionality statements a Fnet and a m F F net where k is the proportionality net or a k into one statement, a m m constant. Since 1 N is defined as the net force required to accelerate a 1-kg object at 1 m/s2, k is equal to 1. So F net. the equation can be written as a m This mathematical relationship is Newton’s second law. When an external non-zero net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object. The equation for Newton’s second law is usually written with F net on the left side: F net ma The Concept of Inertial Mass All objects have mass, so all objects have inertia. From experience, it is more difficult to accelerate a curling stone than to accelerate a hockey puck (Figure 3.35). This means that the inertia of an object is related to its mass. The greater the mass of the object, the greater its inertia. The mass of an object in Newton’s second law is determined by finding the ratio of a known net force acting on an object to the acceleration of the object. In other words, the mass is a measure of the inertia of an object. Because of this relationship, the mass in Newton’s second law is called inertial mass, which indicates how the mass is measured. m Fnet a a constant Fnet m m Fnet Figure 3.35 If the acceleration of the curling stone and the hockey puck is the same, F net on the curling stone would be 95 times greater than F inertial mass of the curling stone is that much greater than the hockey puck. net on the hockey puck because the Concept Check What happens to the acceleration of an object if (a) the mass and net force both decrease by a factor of 4? (b) the mass and net force both increase by a factor of 4? (c) the mass increases by a factor of 4, but the net force decreases by the same factor? (d) the mass decreases by a factor of 4, and the net force is zero? e MATH Use technology to explore the relationship among Fnet, m, and a in Newton’s second law. Follow the eMath links at www.pearsoned.ca/school/ physicssource to download sample data. inertial mass: mass measurement based on the ratio of a known net force on an object to the acceleration of the object 148 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 149 Applying Newton’s Second Law to Horizontal Motion Example 3.5 demonstrates how to use Newton’s second law to predict the average acceleration of a lacrosse ball. In this situation, air resistance is assumed to be negligible to simplify the problem. Example 3.5 A lacrosse player exerts an average net horizontal force of 2.8 N [forward] on a 0.14-kg lacrosse ball while running with it in the net of his stick (Figure 3.36). Calculate the average horizontal acceleration of the ball while in contact with the lacrosse net. Given F 2.8 N [forward] net m 0.14 kg up down forward backward Figure 3.36 Required average horizontal acceleration of ball (a) Analysis and Solution The ball is not accelerating up or down. So in the vertical direction, F In the horizontal direction, the acceleration of the ball is in the direction of the net force. So use the scalar form of Newton’s second law. 0 N. net Fnet ma F net a m N .8 2 0 4 kg .1 m 2.8 kg s2 0.14 kg 20 m/s2 a 20 m/s2 [forward] Practice Problems 1. The net force acting on a 6.0-kg grocery cart is 12 N [left]. Calculate the acceleration of the cart. 2. A net force of 34 N [forward] acts on a curling stone causing it to accelerate at 1.8 m/s2 [forward] on a frictionless icy surface. Calculate the mass of the curling stone. Answers 1. 2.0 m/s2 [left] 2. 19 kg Paraphrase The average horizontal acceleration of the lacrosse ball is 20 m/s2 [forward]. In Example 3.6, a free-body diagram is used to first help determine the net force acting on a canoe. Then Newton’s second law is applied to predict the average acceleration of the canoe. Chapter 3 Forces can change velocity. 149 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 150 Example 3.6 Two athletes on a team, A and B, are practising to compete in a canoe race (Figure 3.37). Athlete A has a mass of 70 kg, B a mass of 75 kg, and the canoe a mass of 20 kg. Athlete A can exert an average force of 400 N [forward] and B an average force of 420 N [forward] on the canoe using the paddles. During paddling, the magnitude of the water resistance on the canoe is 380 N. Calculate the initial acceleration of the canoe. up down forward backward Practice Problem 1. In the men’s four-man bobsled event in the Winter Olympics, the maximum mass of a bobsled with two riders, a pilot, and a brakeman is 630 kg (Figure 3.39). Figure 3.37 Given mA F A F f 70 kg 400 N [forward] 380 N [backward] mB F B 75 kg 420 N [forward] mc 20 kg Required initial acceleration of canoe (a) Analysis and Solution The canoe and athletes are a system because they move together as a unit. Find the total mass of the system. mT mc mB mA 70 kg 75 kg 20 kg 165 kg Draw a free-body diagram for the system (Figure 3.38). Figure 3.39 During a practice run, riders A and B exert average forces of 1220 N and 1200 N [forward] respectively to accelerate a bobsled of mass 255 kg, a pilot of mass 98 kg, and a brakeman of mass 97 kg. Then they jump in for the challenging ride down a 1300-m course. During the pushing, the magnitude of the force of friction acting on the bobsled is 430 N. Calculate the average acceleration of the bobsled, pilot, and brakeman. up down forward FN FA backward Ff FB Fg Figure 3.38 FA FB Ff Fneth Answer 1. 4.4 m/s2 [forward] 150 Unit II Dynamics The system is not accelerating up or down. So in the vertical direction, Fnetv Write equations to find the net force on the system in both the horizontal and vertical directions. 0 N. 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 151 neth Fneth horizontal direction F FF f Ff F B FB F A FA 400 N 420 N (380 N) 400 N 420 N 380 N 440 N FF g vertical direction FF F N 0 netv Fnetv Calculations in the vertical direction are not required in this problem. Apply Newton’s second law to the horizontal direction. Fneth a mTa Fnet h mT 4 N 0 4 5 g k 6 1 440 kg m 2 s 165 kg 2.7 m/s2 a 2.7 m/s2 [forward] Paraphrase The canoe will have an initial acceleration of 2.7 m/s2 [forward]. Applying Newton’s Second Law to Vertical Motion Example 3.7 demonstrates how to apply Newton’s second law to determine the vertical acceleration of a person riding an elevator. To determine the net force on the elevator, use a free-body diagram. Example 3.7 A person and an elevator have a combined mass of 6.00 102 kg (Figure 3.40). The elevator cable exerts a force of 6.50 103 N [up] on the elevator. Find the acceleration of the person. Given 6.00 102 kg mT 6.50 103 N [up] F T g 9.81 m/s2 [down] Required acceleration of person (a ) up down Figure 3.40 Analysis and Solution Draw a free-body diagram for the person-elevator system [Figure 3.41 (a)]. The system is not accelerating left or right. 0 N. So in the horizontal direction, F net Since the person is standing on the elevator floor, both the person and the elevator will have the same vertical acceleration. up down FT Fg Figure 3.41 (a) Chapter 3 Forces can change velocity. 151 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 152 For the vertical direction, write an equation to find the net force on the system [Figure 3.41 (b)]. F T F g F net Practice Problems 1. The person in Example 3.7 rides the same elevator when the elevator cable exerts a force of 5.20 103 N [up] on the elevator. Find the acceleration of the person. 2. An electric chain hoist in a garage exerts a force of 2.85 103 N [up] on an engine to remove it from a car. The acceleration of the engine is 1.50 m/s2 [up]. What is the mass of the engine? Answers 1. 1.14 m/s2 [down] 2. 252 kg Apply Newton’s second law. ma F F T g ma FT Fg mg ma FT 9.81 m/s2.50 103 kg m 2 s 6.00 102 kg 1.02 m/s2 a 1.02 m/s2 [up] FT Fg Fnet 9.81 m/s2 Figure 3.41 (b) e WEB Air resistance is the frictional force that acts on all objects falling under the influence of gravity. Research how this force affects the maximum speed that an object reaches during its fall. Write a brief summary of your findings. Begin your search at www.pearsoned.ca/ school/physicssource. F f F g up down Figure 3.42 152 Unit II Dynamics Paraphrase The acceleration of the person is 1.02 m/s2 [up]. In Example 3.8, the force of gravity causes a skydiver to accelerate downward. Since the only motion under consideration is that of the skydiver and the direction of motion is down, it is convenient to choose down to be positive. Example 3.8 A skydiver is j |
umping out of an airplane. During the first few seconds of one jump, the parachute is unopened, and the magnitude of the air resistance acting on the skydiver is 250 N. The acceleration of the skydiver during this time is 5.96 m/s2 [down]. Calculate the mass of the skydiver. Given F f g 9.81 m/s2 [down] 250 N [up] Required mass of skydiver (m) a 5.96 m/s2 [down] Analysis and Solution Draw a free-body diagram for the skydiver (Figure 3.42). The skydiver is not accelerating left or right. So in the horizontal direction, F net For the vertical direction, write an equation to find the net force on the skydiver (Figure 3.43). F net 0 N. F g F f Fnet Fg Ff Figure 3.43 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 153 Apply Newton’s second law. maa F FF f g ma Fg Ff ma mg (250 N) mg 250 N 250 N mg ma m(g a) 250 N g a m 250 N 9.81 m/s2 5.96 m/s2 250 kgm s2 3.85 m s2 64.9 kg Paraphrase The mass of the skydiver is 64.9 kg. Practice Problems 1. A 55-kg female bungee jumper fastens one end of the cord (made of elastic material) to her ankle and the other end to a bridge. Then she jumps off the bridge. As the cord is stretching, it exerts an elastic force directed up on her. Calculate her acceleration at the instant the cord exerts an elastic force of 825 N [up] on her. 2. During a bungee jump, the velocity of the 55-kg woman at the lowest point is zero and the cord stretches to its maximum. (a) Compare the direction of her acceleration at the lowest point of the jump to the part of the jump where she is accelerating due to gravity. (b) At this point, what is the direction of her acceleration? Applying Newton’s Second Law to Two-Body Systems When two objects are connected by a light rope as in Example 3.9, applying a force on one of the objects will cause both objects to accelerate at the same rate and in the same direction. In other words, the applied force can be thought to act on a single object whose mass is equivalent to the total mass. Answers 1. 5.2 m/s2 [up] 2. (b) up Example 3.9 Two blocks of identical material are connected by a light rope on a level surface (Figure 3.44). An applied force of 55 N [right] causes the blocks to accelerate. While in motion, the magnitude of the force of friction on the block system is 44.1 N. Calculate the acceleration of the blocks. Given mA mB F app F f 20 kg 10 kg 55 N [right] 44.1 N [left] 20 kg 10 kg Figure 3.44 Required acceleration (a) Analysis and Solution The two blocks move together as a unit with the same acceleration. So consider the blocks to be a single object. Find the total mass of both blocks. mA mB 20 kg 10 kg 30 kg mT Practice Problems 1. Two buckets of nails are hung one above the other and are pulled up to a roof by a rope. Each bucket has a mass of 5.0 kg. The tension in the rope connecting the buckets is 60 N. Calculate the acceleration of the buckets. 2. Refer to Example 3.9. The force of friction on the 10-kg block has a magnitude of 14.7 N. (a) Calculate the tension in the rope connecting the two blocks. (b) Calculate the tension in the rope between the hand and the 10-kg block. Answers 1. 2.2 m/s2 [up] 2. (a) 37 N (b) 55 N Chapter 3 Forces can change velocity. 153 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 154 Draw a free-body diagram for this single object (Figure 3.45). The single object is not accelerating up or down. 0 N. So in the vertical direction, F net Write equations to find the net force on the single object in both the horizontal and vertical directions. horizontal direction F FF F f app neth Apply Newton’s second law. mTaa F F f mTa Fapp Ff Ff Fap p a m T app 55 N (44.1 N) 30 kg 0.36 m/s2 a 0.36 m/s2 [right] vertical direction F F N 0 F g netv Fnetv FN up right left down Calculations in the vertical direction are not required in this problem. Ff mT Fapp Fg Figure 3.45 Fapp Ff Fneth e SIM Apply Newton’s second law to determine the motion of two blocks connected by a string over a pulley. Follow the eSim links at www.pearsoned.ca/ school/physicssource. Paraphrase The acceleration of the blocks is 0.36 m/s2 [right]. Applying Newton’s Second Law to a Single Pulley System In Example 3.10, two objects are attached by a rope over a pulley. The objects, the rope, and the pulley form a system. You can assume that the rope has a negligible mass and thickness, and the rope does not stretch or break. To simplify calculations in this physics course, you need to also assume that a pulley has negligible mass and has no frictional forces acting on its axle(s). In Example 3.10, the external forces on the system are the gravitational forces acting on the hanging objects. The internal forces on the system are the forces along the string that pull on each object. The magnitude of both the internal and external forces acting on the system are not affected by the pulley. The pulley simply redirects the forces along the string that pulls on each object. Example 3.10 Two objects, A and B, are connected by a light rope over a light, frictionless pulley (Figure 3.46). A has a mass of 25 kg and B a mass of 35 kg. Determine the motion of each object once the objects are released. Given mA 25 kg g 9.81 m/s2 [down] mB 35 kg Required acceleration of each object (a A and a B) 25 kg 35 kg Figure 3.46 154 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 155 Analysis and Solution The difference in mass between objects A and B will provide the net force that will accelerate both objects. Since mB expect mB to accelerate down while mA accelerates up. The rope has a negligible mass. So the tension in the rope is the same on both sides of the pulley. The rope does not stretch. So the magnitude of aa Find the total mass of both objects. A is equal to the magnitude of a mA, you would B. mT mB mA 25 kg 35 kg 60 kg Choose an equivalent system in terms of mT to analyze the motion [Figure 3.47 (a)]. left right FA mT FB Figure 3.47 (a) F net Fnet F A is equal to the gravitational force acting on mA, and F B is equal to the gravitational force acting on mB. Apply Newton’s second law to find the net force acting on mT [Figure 3.47 (b)]. F F B A FA FB mAg mBg mA)g (mB (35 kg 25 kg)(9.81 m/s2) m 98.1 kg 2 s 98.1 N Figure 3.47 (b) Fnet FA FB Use the scalar form of Newton’s second law to calculate the magnitude of the acceleration. Fnet mTa a F et n m T .1 N 8 9 g k 0 6 98.1 kg m 2 s 60 kg 1.6 m/s2 1.6 m/s2 [up] and a a A 1.6 m/s2 [down] B Paraphrase Object A will have an acceleration of 1.6 m/s2 [up] and object B will have an acceleration of 1.6 m/s2 [down]. Practice Problems 1. Determine the acceleration of the system shown in Example 3.10 for each situation below. State the direction of motion for each object. Express your answer in terms of g. (a) mA (b) mA (c) mA mB 1 3 2mB mB 2. Use the result of Example 3.10 and a free-body diagram to calculate the tension in the rope. 3. Draw a free-body diagram for each object in Example 3.10. Answers 1 g, mA moves up, mB moves down 1. (a) a 2 1 g, mA moves down, mB moves up (b) a 3 (c) a 0, neither mass moves 2. 2.9 102 N 3. FT a 25 kg FT 35 kg a Fg Fg Chapter 3 Forces can change velocity. 155 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 156 Applying Newton’s Second Law to a Two-Pulley System In Example 3.11, the system is made up of three objects (A, B, and C). As in Example 3.10, the difference in weight between objects B and C will provide the net force that will accelerate the system. Example 3.11 A 20-kg truck tire (object A) is lying on a horizontal, frictionless surface. The tire is attached to two light ropes that pass over light, frictionless pulleys to hanging pails B and C (Figure 3.48). Pail B has a mass of 8.0 kg and C a mass of 6.0 kg. Calculate the magnitude of the acceleration of the system. tire (A) pail B pail C Figure 3.48 Given mA 20 kg g 9.81 m/s2 [down] mB 8.0 kg mC 6.0 kg Required magnitude of the acceleration of the system (a) Analysis and Solution Since mB accelerates up. Since object A will accelerate left, choose left to be positive. mC, you would expect mB to accelerate down while mC The rope has a negligible mass and the rope does not stretch. So the magnitude of a equal to a C. Find the total mass of the system. A is equal to the magnitude of a B, which is also mT mC mB mA 20 kg 8.0 kg 6.0 kg 34 kg Choose an equivalent system in terms of mT to analyze the motion (Figure 3.49). left right Figure 3.49 FB mT FC 156 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 157 F C is equal to B is equal to the gravitational force acting on mB, and F the gravitational force acting on mC. left right FB FC Fnet Figure 3.50 C Fnet Apply Newton’s second law to find the net force acting on mT (Figure 3.50). F F F B net FB FC mBg mCg mC)g (mB (8.0 kg 6.0 kg)(9.81 m/s2) m 19.6 kg 2 s 19.6 N Use the scalar form of Newton’s second law to calculate the magnitude of the acceleration. Fnet a mTa Fnet mT 19.6N 34 kg 19.6 kg m 2 s 34 kg 0.58 m/s2 Paraphrase The system will have an acceleration of magnitude 0.58 m/s2. Practice Problems 1. Calculate the acceleration of the tire in Example 3.11 if the mass of pail B is increased to 12 kg, without changing the mass of pail C. 2. If the tire in Example 3.11 is replaced by a car tire of mass 15 kg, calculate the acceleration of each object. Answers 1. 1.5 m/s2 [left] 2. (A) 0.68 m/s2 [left], (B) 0.68 m/s2 [down], (C) 0.68 m/s2 [up] Chapter 3 Forces can change velocity. 157 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 158 3.3 Check and Reflect 3.3 Check and Reflect Knowledge 1. In your own words, state Newton’s second law. app acting on an object 2. An applied force F of constant mass causes the object to accelerate. Sketch graphs to show the relationship between a and Fapp when friction is (a) present, and (b) absent. Refer to Student References 5.1: Graphing Techniques on pp. 872–873. 3. Sketch a graph to show the relationship between the magnitude of acceleration and mass for constant net force. 4. Explain why vehicles with more powerful engines are able to accelerate faster. Application |
s 5. A dolphin experiences a force of 320 N [up] when it jumps out of the water. The acceleration of the dolphin is 2.6 m/s2 [up]. 8. Two boxes, A and B, are touching each other and are at rest on a horizontal, frictionless surface. Box A has a mass of 25 kg and box B a mass of 15 kg. A person applies a force of 30 N [right] on box A which, in turn, pushes on box B. Calculate the acceleration of the boxes. 9. A 4.0-kg oak block on a horizontal, rough oak surface is attached by a light string that passes over a light, frictionless pulley to a hanging 2.0-kg object. The magnitude of the force of friction on the 4.0-kg block is 11.8 N. 4.0 kg 2.0 kg (a) Calculate the acceleration of the system. (b) Calculate the tension in the string. (a) Calculate the mass of the dolphin. Extension (b) What would be the acceleration of the dolphin if it had the same strength but half the mass? 6. An ice hut used for winter fishing is resting on a level patch of snow. The mass of the hut is 80 kg. A wind exerts a horizontal force of 205 N on the hut, and causes it to accelerate. While in motion, the magnitude of the force of friction acting on the hut is 196 N. What is the acceleration of the hut? 7. Suppose the only horizontal forces acting on a 20-N object on a smooth table are 36 N [45] and 60 N [125]. (a) What is the net force acting on the object? (b) Calculate the acceleration of the object. 10. Summarize concepts and ideas associated with Newton’s second law using a graphic organizer of your choice. See Student References 4: Using Graphic Organizers on pp. 869–871 for examples of different graphic organizers. Make sure that the concepts and ideas are clearly presented and are linked appropriately. e TEST To check your understanding of Newton’s second law, follow the eTest links at www.pearsoned.ca/school/physicssource. 158 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 159 3.4 Newton’s Third Law Volleyball is a sport that involves teamwork and players knowing how to apply forces to the ball to redirect it. When the velocity of the ball is large, a player will usually bump the ball to slow it down so that another player can redirect it over the net (Figure 3.51). At the instant the player bumps the ball, the ball exerts a large force on the player’s arms, often causing sore arms. Immediately after the interaction, the ball bounces upward. To explain the motion of each object during and after this interaction requires an understanding of Newton’s third law. Newton’s first two laws describe the motion of an object or a system of objects in isolation. But to describe the motion of objects that are interacting, it is important to examine how the force exerted by one object on another results in a change of motion for both objects. Find out what happens when two initially stationary carts interact by doing 3-7 QuickLab. info BIT In order to walk, you must apply a force backward on the ground with one foot. The ground then pushes forward on that foot. Figure 3.51 Conrad Leinemann of Kelowna, British Columbia, bumps the ball while teammate Jody Holden of Shelburne, Nova Scotia, watches during the beach volleyball competition at the 1999 Pan Am Games in Winnipeg, Manitoba. Chapter 3 Forces can change velocity. 159 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 160 3-7 QuickLab 3-7 QuickLab Exploding Carts Problem If a stationary cart exerts a net force on another identical cart, what will be the motion of both carts after the interaction? Materials dynamics cart with spring dynamics cart without spring 500-g standard mass Procedure 1 Note the position of the spring on the one cart, and standard mass barrier C-clamp cocked spring spring release Figure 3.52 Questions 1. What did you observe when you released the spring when the cart was initially at rest and not touching the other cart? how to cock and release the spring. 2. (a) What did you observe when you released the 2 Cock the spring and place the cart on the table. Release the spring. CAUTION: Do not cock the spring unless it is safely attached to the cart. Do not point the spring at anyone when releasing it. 3 Repeat step 2, this time making the cart with the spring touch the second cart (Figure 3.52). Release the spring. 4 Repeat step 3 but add a 500-g standard mass to one of the carts before releasing the spring. spring when one cart was touching the other cart? (b) What evidence do you have that two forces were present? (c) What evidence do you have that a force was exerted on each cart? (d) How do the magnitudes and directions of the two forces compare? 3. Compare and contrast the results from steps 3 and 4. action force: force initiated by object A on object B reaction force: force exerted by object B on object A Forces Always Exist in Pairs When two objects interact, two forces will always be involved. One force is the action force and the other is the reaction force. The important points to remember are that the reaction force always acts on a different object than the action force, and that the reaction force acts in the opposite direction. Concept Check Is it possible to have an action force without a reaction force? 160 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 161 Newton’s Third Law and Its Applications Newton found that the reaction force is equal in magnitude to the action force, but opposite in direction. This relationship is called Newton’s third law of motion. If object A exerts a force on object B, then B exerts a force on A that is equal in magnitude and opposite in direction. F A on B F B on A Some people state Newton’s third law as “for every action force, there is an equal and opposite reaction force.” However, remembering Newton’s third law this way does not emphasize that the action and reaction forces are acting on different objects (Figure 3.53). PHYSICS INSIGHT In order to show action-reaction forces, you must draw two free-body diagrams, one for each object. Faction force exerted by student on ground Freaction force exerted by ground on student Figure 3.53 The action force is the backward force that the student exerts on the ground. The reaction force is the forward force that the ground exerts on the student. Only the action-reaction pair are shown here for simplicity. Concept Check If the action force is equal in magnitude to the reaction force, how can there ever be an acceleration? Explain using an example and free-body diagrams. Action-Reaction Forces Acting on Objects in Contact Let’s revisit the scenario of the volleyball player bumping the ball. At the instant that both the ball and the player’s arms are in contact, the action force is the upward force that the player exerts on the ball. The reaction force is the downward force that the ball exerts on the player’s arms. During the collision, the ball accelerates upward and the player’s arms accelerate downward (Figure 3.54). e TECH Explore how a stranded astronaut can return to a spacecraft by throwing away tools. Follow the eTech links at www.pearsoned.ca/school/ physicssource. F action force exerted by player on ball force exerted by ball on player F reaction Figure 3.54 The action-reaction forces at collision time when a volleyball player bumps the ball Chapter 3 Forces can change velocity. 161 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 162 A similar reasoning applies when a baseball bat strikes a baseball. The action force is the forward force that the bat exerts on the ball. The reaction force is the backward force that the ball exerts on the bat. During the collision, the ball accelerates forward and the bat slows down as it accelerates backward (Figure 3.55). Freaction force exerted by ball on bat Faction force exerted by bat on ball e WEB Fire hoses and extinguishers are difficult to control because their contents exit at high speed as they are redirected when putting out a fire. Research the operation and safe use of fire extinguishers. How do Newton’s three laws apply to fire extinguishers? Interview an experienced firefighter. Write a brief summary of your findings. Begin your search at www.pearsoned.ca/school/ physicssource. Figure 3.55 The action-reaction forces at collision time when a baseball bat strikes a baseball Action-Reaction Forces Acting on Objects Not in Contact Sometimes an object can exert a force on another without actually touching the other object. This situation occurs when an object falls toward Earth’s surface, or when a magnet is brought close to an iron nail. Action-reaction forces still exist in these interactions. When an apple falls toward the ground, the action force is the force of gravity that Earth exerts on the apple. The falling apple, in turn, exerts a reaction force upward on Earth. So while the apple is accelerating down, Earth is accelerating up (Figure 3.56). You see the acceleration of the apple but not of Earth because the inertial mass of the apple is far less than that of Earth. In fact, Earth does accelerate but at a negligible rate because the magnitude of the acceleration is inversely proportional to mass. force exerted by Earth on apple Faction force exerted by apple on Earth Freaction Figure 3.56 The action-reaction forces when an apple falls toward Earth’s surface 162 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 163 When a magnet is brought close to an iron nail, the action force is the magnetic force that the magnet exerts on the nail. The reaction force is the force that the nail exerts on the magnet. So the nail accelerates toward the magnet, and at the same time the magnet is accelerating toward the nail (Figure 3.57). Investigate the validity of Newton’s third law by doing 3-8 QuickLab. Freaction Faction N force exerted by nail on magnet force exerted by magnet on nail S Figure 3.57 The action-reaction forces when a magnet is brought close to an iron nail 3-8 QuickLab 3-8 QuickLab Skateboard Interactions Problem How does Newton’s third law apply to interactions involving skateboards? Mate |
rials two skateboards CAUTION: Wear a helmet and knee pads when doing this activity. Procedure 1 Choose a partner with a mass about the same as yours. 2 Sit on skateboards on a hard, level surface with your feet toward one another and touching (Figure 3.58). Figure 3.58 3 Give your partner a gentle push with your feet. Observe what happens to both skateboards. 4 Repeat steps 2 and 3 but this time, give your partner a harder push. Observe what happens to both skateboards. 5 Repeat steps 2 and 3 but this time, have you and your partner push simultaneously. Observe what happens to both skateboards. 6 Choose a partner with a significantly different mass than yours. 7 Repeat steps 2 to 5 with your new partner. 8 Sit on a skateboard near a wall. Then push against the wall. Observe the motion of your skateboard. Questions 1. Describe the motion of each skateboard when (a) you pushed a partner of equal mass, and (b) you pushed a partner of significantly different mass. 2. Compare and contrast the results from steps 4 and 5. 3. What happened to your skateboard when you pushed against the wall? 4. Explain each interaction in this activity using Newton’s laws. Draw a sketch showing the action-reaction forces in each situation. Chapter 3 Forces can change velocity. 163 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 164 Applying Newton’s Third Law to Situations Involving Frictionless Surfaces In Example 3.12, an applied force acts on box A, causing all three boxes to accelerate. Newton’s third law is used to calculate the force that box C exerts on box B. Example 3.12 Three boxes, A, B, and C, are positioned next to each other on a horizontal, frictionless surface (Figure 3.59). An applied force acting on box A causes all the boxes to accelerate at 1.5 m/s2 [right]. Calculate the force exerted by box C on box B. 10 kg 8.0 kg A B 5.0 kg C Figure 3.59 Practice Problems 1. For the situation in Example 3.12, calculate the force that box B exerts on box A. 2. For the situation in Example 3.9 Practice Problem 1 on page 153, calculate the applied force needed to lift both buckets up. Answers 1. 23 N [left] 2. 1.2 102 N [up] Given mA 8.0 kg a 1.5 m/s2 [right] mB 10 kg mC 5.0 kg Required C on B) force exerted by box C on box B (F FN Analysis and Solution Draw a free-body diagram for box C (Figure 3.60). Box C is not accelerating up or down. 0 N. So in the vertical direction, F net Write equations to find the net force on box C in both the horizontal and vertical directions. up left down right horizontal direction F neth Fneth F B on C FB on C F g vertical direction FF F netv N 0 Fnetv Calculations in the vertical direction are not required in this problem. FB on C mC Apply Newton’s second law. FB on C F B on C m s2 mCa (5.0 kg)1.5 m 7.5 kg 2 s 7.5 N 7.5 N [right] Apply Newton’s third law. F C on B F B on C 7.5 N [left] Fg Paraphrase The force exerted by box C on box B is 7.5 N [left]. Figure 3.60 164 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 165 Applying Newton’s Third Law to Situations Involving Friction In Example 3.13, a rough surface exerts a force of friction on two boxes. Newton’s third law is used to calculate the force exerted by box B on box A in this situation. Example 3.13 Two boxes, A and B, of identical material but different mass are placed next to each other on a horizontal, rough surface (Figure 3.61). An applied force acting on box A causes both boxes to accelerate at 2.6 m/s2 [right]. If the magnitude of the force of friction on box B is 28.3 N, calculate the force exerted by box B on box A. Given mA F f on B 6.5 kg 28.3 N [left] mB 8.5 kg 6.5 kg 8.5 kg A B Figure 3.61 FN left up down right a 2.6 m/s2 [right] Ff on B mB FA on B Required B on A) force exerted by box B on box A (F Analysis and Solution Draw a free-body diagram for box B (Figure 3.62). Box B is not accelerating up or down. 0 N. So in the vertical direction, F net Write equations to find the net force on box B in both the horizontal and vertical directions. horizontal direction FF neth Fneth FF A on B FA on B FF f on B Ff on B F g vertical direction FF F netv N 0 Fnetv Calculations in the vertical direction are not required in this problem. Apply Newton’s second law. mBa FA on B Ff on B FA on B F A on B mBa Ff on B (8.5 kg)(2.6 m/s2) (28.3 N) (8.5 kg)(2.6 m/s2) 28.3 N 50 N 50 N [right] Apply Newton’s third law. F B on A F A on B 50 N [left] Paraphrase The force exerted by box B on box A is 50 N [left]. FA on B Fg Figure 3.62 Ff on B Fneth Practice Problem 1. To minimize the environmental impact of building a road through a forest, a logger uses a team of horses to drag two logs, A and B, from the cutting location to a nearby road. A light chain connects log A with a mass of 150 kg to the horses’ harness. Log B with a mass of 250 kg is connected to log A by another light chain. (a) The horses can pull with a combined force of 2600 N. The ground exerts a force of friction of magnitude 2400 N on the logs. Calculate the acceleration of the logs. (b) If the force of friction on log A is 900 N, calculate the force exerted by log B on log A. Answer 1. (a) 0.500 m/s2 [forward] (b) 1.63 103 N [backward] Chapter 3 Forces can change velocity. 165 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 166 Applying Newton’s Second and Third Laws to Propeller Aircraft The acceleration of many devices such as propeller aircraft can be controlled in midair. To explain how these machines accelerate involves applying Newton’s second and third laws. A propeller airplane can move through air because as the propeller rotates, it exerts an action force on the air, pushing the air backward. According to Newton’s third law, the air, in turn, exerts a reaction force on the propeller, pushing the airplane forward (Figure 3.63). Propeller blades are slanted so that they scoop new air molecules during each revolution. The faster a propeller turns, the greater is the mass of air accelerated backward and, by Newton’s second law, the force exerted by the air on the propeller increases. Faction force exerted by propeller on air Freaction force exerted by air on propeller Figure 3.63 The action-reaction forces when a propeller airplane is in flight THEN, NOW, AND FUTURE Wallace Rupert Turnbull (1870–1954) Wallace Rupert Turnbull was an aeronautical engineer interested in finding ways to make aircraft wings stable (Figure 3.64). In 1902, he built the first wind tunnel in Canada at Rothesay, New Brunswick, for his experiments on propeller design. In 1909, Turnbull was awarded a bronze medal from the Royal Aeronautical Society for his research on efficient propeller design. One of his major inventions was the variable-pitch propeller, which is still used on aircraft today. During takeoff, the angle of the blades is adjusted to scoop more air. Air moving at a high speed backward gives a plane thrust, which causes the plane to accelerate forward. Once a plane maintains a constant altitude, the blade angle, or pitch, is decreased, reducing fuel consumption. This allows greater payloads to be carried efficiently and safely through the sky. By 1925, Turnbull had perfected a propeller that used an electric motor to change its pitch. In 1927, the Canadian Air Force successfully tested the propeller at Borden, Ontario. Turnbull was later inducted into the Canadian Aviation Hall of Fame in 1977. Questions 1. Research the forces that act on airplanes in flight. Define these forces and compare them to forces already discussed in this chapter. 2. Explain how and where the forces on an airplane act to cause changes in its horizontal motion. Use Newton’s laws and diagrams to support your explanations. Figure 3.64 Canadian inventor Wallace Rupert Turnbull 166 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 167 Applying Newton’s Third Law to Rockets The motion of rockets is a little different from that of propeller airplanes because a rocket does not have propellers that scoop air molecules. In fact, a rocket can accelerate in outer space where there is a vacuum. When a rocket engine is engaged, the highly combustible fuel burns at a tremendous rate. The action force of the exhaust gas leaving the rocket, according to Newton’s third law, causes a reaction force that pushes against the rocket. It is the action force of the exhaust gas being directed backward that accelerates the rocket forward (Figure 3.65). That is why a rocket can accelerate in outer space. Test out Newton’s third law with a toy rocket by doing 3-9 Design a Lab. Faction force exerted by rocket on exhaust gas Freaction force exerted by exhaust gas on rocket Figure 3.65 The action-reaction forces when a rocket is in flight 3-9 Design a Lab 3-9 Design a Lab Motion of a Toy Rocket Figure 3.66 shows a toy rocket partially filled with water about to be released from an air pump. The pump is used to add pressurized air into the rocket. air under pressure air pump water rocket release Figure 3.66 The Question What effect does increasing each of these quantities have on the motion of the rocket? • the amount of water inside the rocket • the air pressure inside the rocket Design and Conduct Your Investigation • State a hypothesis. Then design and conduct an experiment to test your hypothesis. Be sure to identify all variables and to control the appropriate ones. Caution: Never point the rocket at anyone. Perform this activity outside. • Compare the direction of motion of the water and the rocket when the rocket is released. • Explain the motion of the rocket, water, and air in terms of Newton’s third law. Include sketches showing at least three action-reaction pairs of forces. • How well did your results agree with your hypothesis? Chapter 3 Forces can change velocity. 167 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 168 3.4 Check and Reflect 3.4 Check and Reflect Knowledge 1. In your own words, state Newton’s third law. 2. Explain why (a) a swimmer at the edge of a pool pushes backward on the wall in order to move forward, and (b) |
when a person in a canoe throws a package onto the shore, the canoe moves away from shore. 3. No matter how powerful a car engine is, a car cannot accelerate on an icy surface. Use Newton’s third law and Figure 3.53 on page 161 to explain why. 4. State and sketch the action-reaction forces in each situation. (a) Water pushes sideways with a force of 600 N on the centreboard of a sailboat. (b) An object hanging at the end of a spring exerts a force of 30 N [down] on the spring. Applications 5. An object is resting on a level table. Are the normal force and the gravitational force acting on the object action-reaction forces? Explain your reasoning. 6. A vehicle pushes a car of lesser mass from rest, causing the car to accelerate on a rough dirt road. Sketch all the actionreaction forces in this situation. 7. Suppose you apply a force of 10 N to one spring scale. What is the reading on the other spring scale? What is the force exerted by the anchored spring scale on the wall? 8. Blocks X and Y are attached to each other by a light rope and can slide along a horizontal, frictionless surface. Block X has a mass of 10 kg and block Y a mass of 5.0 kg. An applied force of 36 N [right] acts on block X. Y X Fapp (a) Calculate the action-reaction forces the blocks exert on each other. (b) Suppose the magnitudes of the force of friction on blocks X and Y are 8.0 N and 4.0 N respectively. Calculate the action-reaction forces the blocks exert on each other. 9. A rectangular juice box has two holes punched near the bottom corners on opposite sides, and another hole at the top. The box is hung from a rigid support with a string. Predict what will happen if the box is filled with water through the top hole and the holes at the bottom are open. Use Newton’s third law to explain your answer. Test your prediction. Cover the holes at the bottom with tape before filling the box with water. Then remove the tape to let the water out and observe the motion of the box. string hole water juice box holes ? F 10 N water stream e TEST To check your understanding of Newton’s third law, follow the eTest links at www.pearsoned.ca/school/physicssource. 168 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 169 3.5 Friction Affects Motion Throughout this chapter, you encountered friction in all the lab activities and when solving several problems. Friction is a force that is present in almost all real-life situations. In some cases, friction is desirable while in other cases, friction reduces the effectiveness of mechanical systems. Without friction, you would not be able to walk. The wheels on a vehicle would have no traction on a road surface and the vehicle would not be able to move forward or backward. Parachutists would not be able to land safely (Figure 3.67). On the other hand, friction causes mechanical parts to seize and wear out, and mechanical energy to be converted to heat. For example, snowmobiles cannot move for long distances over bare ice. Instead, snowmobilers must detour periodically through snow to cool the moving parts not in contact with the ice. To determine the direction of the force of friction acting on an object, you need to first imagine the direction in which the object would move if there were no friction. The force of friction, then, opposes motion in that direction. info BIT Olympic cyclists now wear slipperier-than-skin suits with seams sown out of the airflow to reduce friction and improve race times by as much as 3 s. friction: force that opposes either the motion of an object or the direction the object would be moving in if there were no friction Figure 3.67 When a person falls in midair, the air resistance that acts on a parachute slows the fall. In this case, friction allows a parachutist to land without injury. Chapter 3 Forces can change velocity. 169 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 170 In a sport such as curling, friction affects how far the stone will travel along the ice. Sweeping the ice in front of a moving stone reduces the force of friction acting on the stone (Figure 3.68). The result is that the stone slides farther. To better understand how the nature of a contact surface affects the force of friction acting on an object, do 3-10 QuickLab. Figure 3.68 Brad Gushue, from St. John’s, Newfoundland, and his team won the gold medal in men’s curling at the 2006 Winter Olympics in Turin, Italy. 3-10 QuickLab 3-10 QuickLab Friction Acting on a Loonie Problem What factors affect the ability of a loonie to start sliding? 4 Use a piece of tape to fasten the fine sandpaper on the textbook, sandy-side facing up. Repeat step 3. Materials two loonies textbook tape protractor coarse, medium, and fine sandpaper: a 10 cm 25 cm piece of each Procedure 1 Read the procedure and design a chart to record your results. 2 Place your textbook flat on a lab bench and place a loonie at one end of the book. 5 Repeat step 4 for the medium sandpaper and then for the coarse sandpaper. Carefully remove and save the sandpaper. 6 Repeat steps 2 and 3 but this time increase the mass (not the surface area) by stacking one loonie on top of the other. Use a piece of tape between the two loonies to fasten them together. Questions 1. How consistent were your results for each trial? 2. Explain how the angle needed to start the loonie sliding down the incline was affected by 3 Slowly raise this end of the textbook until the loonie starts to slide down the incline (Figure 3.69). • • the roughness of the contact surface the mass of the coins (number of stacked coins) in contact with the contact surface Use the protractor to measure the angle the textbook makes with the lab bench when the loonie first starts to slide. Repeat this step several times, and find the average of the angles. 170 Unit II Dynamics 3. Identify the controlled, manipulated, and responding variables in this activity. Figure 3.69 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 171 Static Friction In 3-10 QuickLab, you discovered that the force of friction depends on the nature of the two surfaces in contact. If you drag an object on a smooth surface, the force of friction acting on the object is less than if you drag it on a rough or bumpy surface. If you drag a smooth block and a rough block on the same surface, the force of friction acting on each block will be different. Although there are different types of friction, the force of friction that acts on objects sliding across another surface is the main focus in this section. e SIM Learn how friction is created and how it affects the net force on an object. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Suppose an object A (the desk) is in contact with another object B (the floor) as in Figure 3.70. The contact surface would be the horizontal surface at the bottom of each leg of the desk. app, such that F Now suppose that a force acts on the desk, say F app has a vertical component as well as a horizontal component. If the desk remains at rest, even though F app acts on it, then the net force on the desk is zero, F net 0 N. y x vd 0 Fappx θ Fapp Fappy Ffstatic Figure 3.70 An applied force F floor exerts a force of static friction on the bottom of each leg of the desk. app is acting on the desk at a downward angle . The In the x direction, Fnetx 0 N, which means that F appx must be balanced by another force. This balancing force is the force of static friction, F fstatic. The equation for the net force acting on the desk in the x direction would then be F F F netx fstatic appx Ffstatic Fappx Fnetx 0 Fappcos Ffstatic Fappcos Ffstatic So the direction of FF fstatic opposes the x component of the applied force acting on the desk. static friction: force exerted on an object at rest that prevents the object from sliding Chapter 3 Forces can change velocity. 171 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 172 The Magnitude of Static Friction An important point about static friction is that its magnitude does not have a fixed value. Instead, it varies from zero to some maximum value. This maximum value is reached at the instant the object starts to move. If you push on a table with a force of ever-increasing magnitude, you will notice that the table remains at rest until you exceed a critical value. Because of Newton’s second law, the magnitude of the force of static friction must increase as the applied force on the table increases, if the forces are to remain balanced. Static Friction on a Horizontal Surface Suppose the applied force acting on the desk in Figure 3.70 on page 171 is given. Example 3.14 demonstrates how to calculate the force of static friction by using a free-body diagram to help write the equation for the net force on the desk. Since F app acts at an angle to the surface of the desk, it is convenient to use Cartesian axes to solve this problem. Example 3.14 The magnitude of the applied force in Figure 3.71 is 165 N and 30.0. If the desk remains stationary, calculate the force of static friction acting on the desk. Given magnitude of F app 30.0 165 N Required fstatic) force of static friction (F y x x θ 30.0° Fapp Figure 3.71 Practice Problem 1. A mountain climber stops during the ascent of a mountain (Figure 3.73). Sketch all the forces acting on the climber, and where those forces are acting. Figure 3.73 172 Unit II Dynamics Analysis and Solution Draw a free-body diagram for the desk (Figure 3.72). Fappx Ffstatic 0 Fnetx Fappy 0 Fnety FN Fg y x FN Ffstatic Fappx θ 30.0° Fappy Fapp Fg Figure 3.72 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 173 0 N in both Since the desk is not accelerating, F net the x and y directions. Write equations for the net force on the desk in both directions. x direction F FF netx appx Fappx Fnetx 0 Fappx y direction FF F nety N 0 Fnety Calculations in the y direction are not required in this problem. FF fstatic Ffstatic Ffstatic FF appy F g Ffstatic Fappx (165 N)(cos ) (165 N)(cos 30.0) 143 N F fstatic prevents the desk from sliding in the x directio |
n. The negative value for Ffstatic indicates that the direction of FF fstatic is along the negative x-axis or [180]. F fstatic 143 N [180] Paraphrase The force of static friction acting on the desk is 143 N [180]. Answer 1. FT Ffstatic FN FN Ffstatic Fg Static Friction on an Incline If an object is at rest on an incline, the net force acting on the object 0 N. Let’s first examine the forces acting on the object is zero, F net (Figure 3.74). FN Ffstatic Figure 3.74 (left) Free-body diagram for an object at rest on an incline; (below) vector addition diagrams for the and forces Fg θ Fg Fg θ Fg Forces Forces Fnet Fg Ffstatic Fnet 0 Ffstatic Fg Fg Fnet FN Fnet 0 FN Chapter 3 Forces can change velocity. 173 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 174 When working with inclines, it is easier to rotate the reference coordinates so that motion along the incline is described as either uphill or downhill. This means that only the gravitational force needs to be resolved into components, one parallel to the incline F g and one perpendicular to the incline F g . Usually, uphill is chosen to be positive unless the object is accelerating downhill. In Figure 3.74 on page 173, if there were no friction, the component g would cause the object to accelerate down the incline. So for the object F fstatic) must be acting up the incline. to remain at rest, a balancing force (FF The equation for the net force acting on the object parallel to the incline would then be g F net F F fstatic Fnet Fg Ffstatic 0 Fg Ffstatic Ffstatic Fg g requires using the geometry of To determine the expression for F g is 90.0 . Since a triangle. In Figure 3.75, the angle between F g and FF g is . g is 90.0, the angle between FF g and F g and F the angle between F 90° θ θ Fg Fg θ Fg Figure 3.75 Diagram for an object at rest on an incline showing only the force of gravity vector resolved into components Since the object is not accelerating perpendicular to the incline, the equation for the net force acting on the object in this direction is net FF F N Fnet FN 0 FN FN Fg F g Fg Fg In 3-10 QuickLab, the sandpaper exerted a force of static friction on the loonie, preventing the coin from sliding down the incline. Example 3.15 demonstrates how to calculate the force of static friction acting on a loonie at rest on an incline of 25.0. 174 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 175 Example 3.15 A loonie with a mass of 7.0 g is at rest on an incline of 25.0 (Figure 3.76). Calculate the force of static friction acting on the loonie. Given m 7.0 g 7.0 103 kg g 9.81 m/s2 Required fstatic) force of static friction (F 25.0 loonie θ 25.0° Figure 3.76 Analysis and Solution Draw a free-body diagram for the loonie (Figure 3.77). p u FN p u h ill h ill o w n d o w n d Ffstatic Fg Fg θ 25.0° θ 25.0° Fg Fg Ffstatic Fnet 0 Figure 3.77 Practice Problems 1. A loonie of mass 7.0 g is taped on top of a toonie of mass 7.3 g and the combination stays at rest on an incline of 30.0. Calculate the force of static friction acting on the face of the toonie in contact with the incline. 2. A loonie of mass 7.00 g is placed on the surface of a rough book. A force of static friction of magnitude 4.40 102 N acts on the coin. Calculate the maximum angle at which the book can be inclined before the loonie begins to slide. Answers 1. 7.0 102 N [uphill] 2. 39.8 0 N both Since the loonie is not accelerating, F net parallel and perpendicular to the incline. Write equations to find the net force on the loonie in both directions. direction net FF FF N Fnet 0 Calculations in the direction are not required in this problem. direction g FF net F FF fstatic Fnet Fg Ffstatic Fnet Fg Ffstatic Now, Fg mg sin So, Ffstatic F g 0 (mg sin ) mg sin (7.0 103 kg)(9.81 m/s2)(sin 25.0) 2.9 102 N F fstatic prevents the loonie from sliding downhill. The positive value for Ffstatic indicates that the direction of F fstatic is uphill. 2.9 102 N [uphill] F fstatic Paraphrase The force of static friction acting on the loonie is 2.9 102 N [uphill]. Chapter 3 Forces can change velocity. 175 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 176 kinetic friction: force exerted on an object in motion that opposes the motion of the object as it slides on another object Kinetic Friction Suppose you apply a force to the desk in Figure 3.78 and the desk starts to slide across the floor at constant velocity. In this situation, the force of static friction is not able to balance the applied force, so motion occurs. Now the floor will exert a force of friction on the desk that opposes the direction of motion of the desk. This force is the force f kinetic. of kinetic friction, F Kinetic friction is present any time an object is sliding on another, whether or not another force acts on the sliding object. If you stop pushing the desk once it is in motion, the desk will coast and eventually stop. While the desk is sliding, the floor exerts a force of kinetic friction on the desk. This frictional force is directed backward, and causes the desk to eventually come to a stop. y x vd constant Fappx θ Fapp Fappy Ffkinetic Figure 3.78 The applied force F desk, causing the desk to slide. While the desk is in motion, the floor exerts a force of kinetic friction that opposes the motion of the desk. app overcomes the force of static friction acting on the The Direction of Kinetic Friction on an Incline If an object is on an incline and the object begins to slide, the surface of the incline exerts a force of kinetic friction on the object that opposes its motion. Whether the object is accelerating uphill or downhill, F net 0 N parallel to the incline. 176 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 177 Accelerating Down an Incline Let’s first consider the case where an object accelerates downhill g causes the object to accelerate (Figure 3.79). In this situation, F downhill. The force of kinetic friction acts to oppose the motion of the fkinetic is uphill as shown below. object. So F a FN Ffkinetic Fg Fg θ Fg θ Figure 3.79 (left) Free-body diagram for an object accelerating downhill; (below) vector addition diagrams for the and forces Forces Forces Fg Fnet Ffkinetic Fg Ffkinetic Fnet Fnet 0 Fg FN Fg Fnet Fnet FN 0 The equation for the net force acting on the object parallel to the incline is g F net F F fkinetic net becomes If you apply Newton’s second law, the equation for F ma F g F fkinetic ma Fg Ffkinetic fkinetic acts uphill. For the object to In Figure 3.79, F g acts downhill and F accelerate downhill, the net force on the object, F net , is directed downhill. So the magnitude of F fkinetic. g must be greater than the magnitude of F Since the object is not accelerating perpendicular to the incline, the net force acting on the object in this direction is zero. The equation for the net force on the object in the perpendicular direction is net F F N Fnet FN 0 FN FN Fg F g Fg Fg Chapter 3 Forces can change velocity. 177 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 178 Accelerating Up an Incline If an object is accelerating uphill, the force of kinetic friction acts app, g also acts downhill. A force, F downhill to oppose the motion. F must act uphill on the object that is great enough to overcome both F fkinetic g (Figure 3.80). and F Fapp Forces a FN Ffkinetic Fg Fg θ Fg θ Fapp Fg Ffkinetic Fg Ffkinetic Fapp Fnet Fnet ≠ 0 Fnet Fg FN Forces Fg Fnet FN Fnet 0 Figure 3.80 (left) Free-body diagram for an object accelerating uphill; (right) vector addition diagrams for the and forces The equation for the net force acting on the object parallel to the incline is net F F app g F F fkinetic net becomes If you apply Newton’s second law, the equation for F ma F app ma Fapp g F F fkinetic Fg Ffkinetic app acts uphill. For fkinetic act downhill and F g and F In Figure 3.80, both F net is directed uphill. So the magnitude the object to accelerate uphill, F app must be greater than the sum of the magnitudes of F fkinetic. g and F of F Since the object is not accelerating perpendicular to the incline, the net force acting on the object in this direction is zero. The equation for the net force on the object in the perpendicular direction is net F F N Fnet FN 0 FN FF g Fg Fg FN Fg Concept Check What is the angle between the normal force and the force of friction? Is this angle always the same size? Explain your reasoning. 178 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 179 Example 3.16 demonstrates how to calculate the acceleration of a block sliding down an incline. Since the direction of motion of the block is downhill, it is convenient to choose downhill to be positive. Example 3.16 A 3.5-kg block is sliding down an incline of 15.0 (Figure 3.81). The surface of the incline exerts a force of kinetic friction of magnitude 3.9 N on the block. Calculate the acceleration of the block. 3.5 kg 15.0° Figure 3.81 Given m 3.5 kg magnitude of F fkinetic 3.9 N Required acceleration of block (a ) 15.0 g 9.81 m/s2 Analysis and Solution Draw a free-body diagram for the block (Figure 3.82). Since the block is accelerating downhill, F net to the incline, but F net 0 N perpendicular to the incline. 0 N parallel F Write equations to find the net force on the block in both directions. direction net F F N Fnet 0 Calculations in the direction are not required in this problem. direction g F net F F fkinetic Fnet Fg Ffkinetic ma Fg Ffkinetic Now, Fg mg sin So, ma mg sin (3.9 N) mg sin 3.9 N g FN Fg Ff kinetic θ 15.0° θ 15.0° Fg Fg Fg Ff kinetic Fnet 9 N 3. a g sin m 9 N . 3 (9.81 m/s2)(sin 15.0) g k 5 . 3 1.4 m/s2 Figure 3.82 The positive value for a indicates that the direction of a is downhill. a 1.4 m/s2 [downhill] Paraphrase The acceleration of the block is 1.4 m/s2 [downhill]. Practice Problems 1. Determine the acceleration of the block in Example 3.16 if friction is not present. 2. A 55.0-kg skier is accelerating down a 35.0 slope. The magnitude of the skier’s acceleration is 4.41 m/s2. Calculate the force of kinetic friction that |
the snowy surface exerts on the skis. Answers 1. 2.5 m/s2 [downhill] 2. 66.9 N [uphill] Chapter 3 Forces can change velocity. 179 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 180 Comparing the Magnitudes of Static and Kinetic Friction The magnitude of the force of kinetic friction is never greater than the maximum magnitude of the force of static friction. Often, the magni- fstatic. fkinetic is less than the magnitude of F tude of F Figure 3.83 shows a graph of a situation where a person is applying very little force to an object during the first 2 s. Then the person begins to push harder, and at t 4 s, the object starts to move. The graph does not provide any information about the applied force after 4 s ( Ff Magnitude of the Force of Friction vs. Time static friction maximum value of static friction kinetic friction 0 4 8 12 16 20 Time t (s) Figure 3.83 The force of static friction increases up to a maximum value. Concept Check Explain why it makes sense that the magnitude of the force of kinetic friction does not exceed the maximum magnitude of the force of static friction. e WEB Leonardo da Vinci was as creative in science as he was in art. Research some of da Vinci’s scientific ideas. Write a brief report of your findings, including diagrams where appropriate. Begin your search at www.pearsoned.ca/ school/physicssource. Determining the Magnitude of Frictional Forces Leonardo da Vinci (1452–1519) was one of the first people to experimentally determine two important relationships about friction. He discovered that for hard contact surfaces, the force of friction does not depend on the contact surface area. If you push a heavy box across the floor, the force of friction acting on the box is the same whether you push it on its bottom or on its side [Figure 3.84 (a) and (b)]. Da Vinci also discovered that the force of friction acting on an object depends on the normal force acting on that object. Find out what this relationship is by doing 3-11 Inquiry Lab. (a) (b) Figure 3.84 The force of friction acting on the box in each of these pictures is the same. For hard contact surfaces, the force of friction does not depend on contact surface area. 180 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 181 3-11 Inquiry Lab 3-11 Inquiry Lab Relating Static Friction and the Normal Force Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question What is the relationship between the maximum magnitude of the force of static friction and the magnitude of the normal force acting on an object? Hypothesis State a hypothesis relating the magnitude of F and the magnitude of F N. Write an “if/then” statement. fstatic Variables Read the procedure and identify the controlled, manipulated, and responding variable(s). Materials and Equipment balance wooden block with different face areas and a hook horizontal board spring scale, calibrated in newtons set of standard masses Procedure 1 Read the steps of the procedure and design a chart to record your results. 2 Measure the mass of the block using the balance. 3 Place the largest face of the block on the horizontal board. Attach the spring scale to the block. Pull with an ever-increasing horizontal force until the block just starts to move. Record this force, which is the maximum magnitude of the force of static friction. 4 Increase the mass of the block system by placing a standard mass on the upper surface. Record the total mass of the block with the standard mass. Use the spring scale to determine the maximum magnitude of the force of static friction for this system (Figure 3.85). Figure 3.85 5 Repeat step 4 three more times, increasing the added mass each time until you have five different masses and five corresponding maximum magnitudes of static friction. 6 Calculate the magnitude of the weight corresponding to each mass system. Record the magnitude of the normal force. 7 (a) Graph the maximum magnitude of the force of static friction as a function of the magnitude of the normal force. (b) Draw the line of best fit and calculate the slope of the graph. Analysis 1. Describe the graph you drew in step 7. 2. As the magnitude of the normal force acting on the mass system increased, what happened to the maximum magnitude of the force of static friction? 3. What is the relationship between the maximum magnitude of the force of static friction and the magnitude of the normal force? Write this as a proportionality statement. Does this relationship agree with your hypothesis? 4. On a level surface, how does the magnitude of the weight of an object affect the magnitude of the normal force and the maximum magnitude of the force of static friction? 5. Explain why adding a bag of sand to the trunk of a rear-wheel-drive car increases its traction. 6. Design and conduct an experiment to verify that contact surface area does not affect the maximum magnitude of the force of static friction for a sliding object. Identify the controlled, manipulated, and responding variables. Analyze your data and form conclusions. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Chapter 3 Forces can change velocity. 181 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 182 Project LINK How will the force of static friction acting on each vehicle in the Unit II Project on page 232 affect the stopping distance? How will the types of treads of the tires affect the force of static friction? coefficient of static friction: proportionality constant relating (Ffstatic)max and FN Coefficient of Static Friction In 3-11 Inquiry Lab, you found that the maximum magnitude of the force of static friction is directly proportional to the magnitude of the normal force. This proportionality can be written mathematically: (Ffstatic)max FN As an equation, the relationship is (Ffstatic)max sFN where s is a proportionality constant called the coefficient of static friction. Since the magnitude of the force of static friction can be anywhere from zero to some maximum value just before motion occurs, the general equation for the magnitude of the force of static friction must have an inequality sign. Ffstatic sFN for static friction Coefficient of Kinetic Friction Find out how the force of kinetic friction acting on an object is related to the normal force on that object by doing 3-12 Design a Lab. 3-12 Design a Lab 3-12 Design a Lab Relating Kinetic Friction and the Normal Force In this lab, you will investigate the relationship between the force of kinetic friction acting on an object and the normal force acting on that object. The Question What is the relationship between the magnitude of the force of kinetic friction and the magnitude of the normal force acting on an object? Design and Conduct Your Investigation • State a hypothesis relating the magnitudes of F • Then use the set-up in Figure 3.85 on page 181 to design an experiment. List the materials you will use as well as a detailed procedure. You will need to place objects of different mass on the block for each trial. and F N. fkinetic • For each trial, measure the force that must be applied to keep the block system moving at constant velocity. Then calculate the magnitude of the normal force. • Plot a graph of Ffkinetic • Analyze your data and form conclusions. as a function of FN. How well did your results agree with your hypothesis? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. From 3-12 Design a Lab, just as with static friction, the magnitude of kinetic friction is directly proportional to the magnitude of the normal force. This proportionality can be written mathematically: Ffkinetic FN 182 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 183 As an equation, the relationship is kFN for kinetic friction Ffkinetic where k is a proportionality constant called the coefficient of kinetic friction. The force of kinetic friction has only one value, unlike the force of static friction which varies from zero to some maximum value. So the equation for the force of kinetic friction has an equal sign, not an inequality as does the equation for the force of static friction. Characteristics of Frictional Forces and Coefficients of Friction There are a few important points to keep in mind about the force of friction and the variables that affect its magnitude: • The equations for static friction and kinetic friction are not fundamental laws. Instead, they are approximations of experimental results. coefficient of kinetic friction: proportionality constant relating Ffkinetic and FN • The equations (Ffstatic)max kFN cannot be written sFN and Ffkinetic N are perpendicular f and F as vector equations because the vectors F to each other. s and • Both • For a given pair of surfaces, the coefficient of static friction is usually k are proportionality constants that have no units. greater than the coefficient of kinetic friction. • The coefficients of friction depend on the materials forming the contact surface, how smooth or rough a surface is, whether the surface is wet or dry, the temperature of the two contact surfaces, and other factors. Table 3.4 lists coefficients of friction between pairs of materials. Table 3.4 Approximate Coefficients of Friction for Some Materials Material Coefficient of Static Friction s Coefficient of Kinetic Friction k Copper on copper Steel on dry steel Steel on greased steel Dry oak on dry oak Rubber tire on dry asphalt Rubber tire on wet asphalt Rubber tire on dry concrete Rubber tire on wet concrete Rubber tire on ice Curling stone on ice Teflon™ on Teflon™ Waxed hickory skis on dry snow Waxed hickory skis on wet snow Synovial fluid on joint 1.6 0.41 0.15 0.5 1.2 0.6 1.0 0.7 0.006 0.003 0.04 0.06 0.20 0.01 1.0 0.38 0.09 0.3 0.8 0.5 0.7 0.5 0.005 0.002 0.04 0.04 0.14 0.01 Chapter 3 Forces can change velocity. 183 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 184 How Friction Affects Motion Movable joints in the human body, s |
uch as elbows, knees, and hips, have membranes that produce a lubricating fluid called synovial fluid. Among other factors, the amount of synovial fluid and the smoothness of adjacent bone surfaces affect the coefficients of friction in synovial joints (Figure 3.86). The movement of synovial joints is very complicated because various biological processes are involved. In diseases such as arthritis, physical changes in joints and/or the presence of too much or too little synovial fluid affect the coefficients of friction. This, in turn, results in limited and painful movement. Figure 3.86 The amount of synovial fluid present depends on the need for a joint to move in a particular direction. info BIT Cars with wide tires experience no more friction than if the cars had narrow tires. Wider tires simply spread the weight of a vehicle over a greater surface area. This reduced pressure on the road reduces heating and tire wear. The effect of temperature on the coefficients of friction plays a role in drag racing. Drag racers often warm the tires on their cars by driving for a while. Tires that are warm stick to a racing track better than cooler tires. This increased coefficient of static friction increases traction and improves the acceleration of the car. The amount of moisture on a road surface, the temperature of the road surface and tires, and the type of tire treads are some factors that determine if a vehicle will skid. For a given tire, the coefficients of static and kinetic friction are greater on a dry road than if the same road is wet. The result is that vehicles are less likely to skid on a dry road than on a wet road. Tire treads and road surfaces also affect the force of friction acting on a vehicle (Figure 3.87). A ribbed tire increases friction acting sideways which helps a driver steer better. A lug tread provides more traction than a ribbed tire. Slicks, the tires on drag racing cars, have no treads at all to increase the surface area of the tire in contact with the racing track to better dissipate heat. (a) (b) (c) Figure 3.87 Different types of tires: (a) a ribbed tire with chains on it for better traction on snowy and icy surfaces, (b) a lug tread, and (c) slicks on a racing car Example 3.17 demonstrates how to use the coefficients of friction in Table 3.4 on page 183 to calculate the mass of a sled. Since the sled is at rest, the snowy surface exerts a force of static friction on the sled. 184 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 185 Example 3.17 A sled with waxed hickory runners rests on a horizontal, dry snowy surface (Figure 3.88). Calculate the mass of the sled if the maximum force that can be applied to the sled before it starts moving is 46 N [forward]. Refer to Table 3.4 on page 183. up down forward backward Figure 3.88 Given F app s 46 N [forward] 0.06 from Table 3.4 (waxed hickory skis on dry snow) g 9.81 m/s2 [down] up Required mass of sled (m) backward down Analysis and Solution Draw a free-body diagram for the sled (Figure 3.89). Since the sled is not accelerating, F net in both the horizontal and vertical directions. Write equations to find the net force on the sled in both directions. 0 N FF neth Fneth horizontal direction F app Fapp 0 Fapp Fapp Fapp sFN FF fstatic Ffstatic Ffstatic ( Fapp sFN sFN) vertical direction F F netv N FN Fnetv 0 FN FN mg FF g Fg (mg) mg FN mg into the equation for Fapp. Fapp Substitute FN smg Fapp sg m 46 N m (0.06)9.81 s2 46 kgm s2 (0.06)9.81 m s2 8 101 kg Paraphrase The mass of the sled is 8 101 kg. Fapp Ffstatic 0 Fneth FN forward Ffstatic Fapp Fg Figure 3.89 Practice Problems 1. An applied force of 24 N [forward] causes a steel block to start moving across a horizontal, greased steel surface. Calculate the mass of the block. Refer to Table 3.4 on page 183. 2. Suppose the sled in Example 3.17 is resting on a horizontal, wet snowy surface. Would the sled move if the applied force is 125 N? Explain. Refer to Table 3.4 on page 183. Answers 1. 16 kg 2. no, F fstatic > F app In Example 3.18, a toboggan is initially at rest on a snowy hill. By knowing only the angle of the incline, it is possible to determine the coefficient of static friction for the toboggan on the hill. Chapter 3 Forces can change velocity. 185 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 186 Example 3.18 A 50-kg toboggan is on a snowy hill. If the hill forms an angle of at least 20.0 with the horizontal, the toboggan just begins to slide downhill (Figure 3.90). Calculate the coefficient of static friction for the toboggan on the snow. Practice Problems 1. Calculate the coefficient of static friction if the toboggan in Example 3.18 is 20 kg and the hill forms an angle of 30.0 with the horizontal. 2. An 80-kg skier on a slushy surface starts moving down a hill forming an angle of at least 25.0 with the horizontal. (a) Determine the coefficient of static friction. (b) Calculate the maximum force of static friction on the skier. Given m 50 kg 20.0 Required coefficient of static friction ( s) Analysis and Solution Draw a free-body diagram for the toboggan [Figure 3.91 (a)]. When the angle of the incline is just enough for the toboggan to start moving, the surface of the incline is exerting the maximum magnitude of the force of static friction on the toboggan. u p p u h ill h ill o w n d o w n d θ 20.0° Figure 3.90 g 9.81 m/s2 u p p u h ill h ill o w n d o w n d Ffstatic FN Fg θ 20.0° θ 20.0° Fg Fg Answers 1. 0.58 2. (a) 0.47 (b) 3.3 102 N [uphill] info BIT The trigonometric function tan can be expressed in terms of sin and cos . in s tan s o c net 0 N in both the parallel and perpendicular Just before the toboggan begins to slide, F directions to the incline. Write equations to find the net force on the toboggan in both directions [Figure 3.91 (b)]. Figure 3.91 (a) direction net F F N Fnet FN 0 FN FN Fg F g Fg Fg direction g F net F F fstatic Fnet Fg Ffstatic 0 Fg Ffstatic Fg Ffstatic Fg mg sin Ffstatic Now, Fg mg cos So, FN (mg cos ) mg cos (mg sin ) mg sin mg sin mg cos into the last equation for the direction. sFN Figure 3.91 (b) Substitute FN Fg Ffstatic Fnet 0 s(mgcos ) mgsin s cos sin in s c s o tan tan 20.0 0.36 s Paraphrase The coefficient of static friction for the toboggan on the snow is 0.36. Note that angle of the hill. s does not depend on the mass of the toboggan, only on the 186 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 187 Kinetic Friction Applies to Skidding Tires When the tires of a vehicle lock or if the tires skid on a road surface, the tires no longer rotate. Instead, the tires slide along the road surface. At the area where the tire and the road are in contact, the road surface exerts a force of kinetic friction directed backward on the tire (Figure 3.92). e TECH Explore how the initial velocity of a skidding car and its mass affect the braking distance. Follow the eTech links at www.pearsoned.ca/school/ physicssource. Ffkinetic Ffkinetic Figure 3.92 Diagram showing the force of kinetic friction acting on the tires of a skidding car Safety features on vehicles such as anti-lock braking systems are designed to prevent the wheels of a vehicle from locking when a driver steps on the brakes. If the wheels lock, the tires no longer rotate on the road surface and the vehicle ends up skidding. As long as the wheels continue to turn, the road surface exerts a force of static friction on the tires. Anti-lock braking systems maximize the force of static friction acting on the tires, allowing the driver of a vehicle to come to a more controlled stop. In Example 3.19, a lift truck is skidding on a concrete surface. Since the wheels are not rotating, the concrete surface is exerting a force of kinetic friction on the tires. e WEB Research how anti-lock braking systems work, and identify the strengths and weaknesses. Interview a car salesperson and/or an owner. Write a brief report of your findings, including diagrams where appropriate. Begin your search at www.pearsoned.ca/school/ physicssource. Example 3.19 A 1640-kg lift truck with rubber tires is skidding on wet concrete with all four wheels locked (Figure 3.93). Calculate the acceleration of the truck. Refer to Table 3.4 on page 183. backward up down forward Given m 1640 kg k g 9.81 m/s2 [down] 0.5 from Table 3.4 (rubber on wet concrete) Required acceleration of lift truck (a) Analysis and Solution Draw a free-body diagram for the lift truck (Figure 3.94). Since the lift truck is accelerating forward, F net direction, but F net vertical direction. Write equations to find the net force on the lift truck in both directions. 0 N in the horizontal 0 N in the Figure 3.93 up backward FN down forward Ffkinetic Fg FN Fg 0 Fnetv Figure 3.94 Chapter 3 Forces can change velocity. 187 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 188 Practice Problems 1. An applied force of 450 N [forward] is needed to drag a 1000-kg crate at constant speed across a horizontal, rough floor. Calculate the coefficient of kinetic friction for the crate on the floor. 2. Calculate the force of kinetic friction if the truck in Example 3.19 is skidding downhill at constant speed on a hill forming an angle of 15.0 with the horizontal. Answers 1. 4.59 102 2. 4.16 103 N [uphill] horizontal direction F F fkinetic neth Ffkinetic Fneth ma Ffkinetic kFN vertical direction FF F netv N FN Fnetv 0 FN FN mg F g Fg (mg) mg FN mg into the equation for Ffkinetic. Substitute FN kmg ma a kg (0.5)9.81 5 m/s2 m s2 The negative value for a indicates that the direction of a backward. is a 5 m/s2 [backward] Paraphrase The acceleration of the truck is 5 m/s2 [backward]. Example 3.20 involves a snowmobile accelerating uphill while towing a sled. Since the motion of the sled is uphill, it is convenient to choose uphill to be positive. Example 3.20 A person wants to drag a 40-kg sled with a snowmobile up a snowy hill forming an angle of 25.0 (Figure 3.95). The coefficient of kinetic friction for the sled on the snow is 0.04. Calculate the force of the s |
nowmobile on the sled if the sled accelerates at 2.5 m/s2 [uphill]. Figure 3.95 p u p u h ill h ill o w n d o w n d θ 25.0° Given m 40 kg g 9.81 m/s2 25.0 a 2.5 m/s2 [uphill] k 0.04 Required app) applied force on sled (F 188 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 189 Analysis and Solution Draw a free-body diagram for the sled (Figure 3.96). FN Ffkinetic Figure 3.96 p u p u h ill h ill o w n d o w n d Fg θ 25.0° Fapp θ 25.0° Fg Fapp Fg Ffkinetic Fg Fnet Since the sled is accelerating uphill, 0 N parallel to the incline, but F 0 N perpendicular to the incline. F net net Write equations to find the net force on the sled in both directions. direction net FF F N Fnet FN 0 FN FN direction net FF F app Fnet Fapp ma Fapp Fapp F g Fg Fg Fg FF g F fkinetic Fg Ffkinetic Fg Ffkinetic ma Fg Ffkinetic Now, Fg mg cos So, FN (mg cos ) mg cos Also, Fg mg sin and kFN Ffkinetic Fapp ma (mg sin ) kFN) ( ma mg sin kFN Substitute FN mg cos into the equation for Fapp. Fapp ma mg sin ma mg(sin kmg cos kcos ) Practice Problems 1. A roofer is shingling a roof that rises 1.0 m vertically for every 2.0 m horizontally. The roofer is pulling one bundle of shingles (A) with a rope up the roof. Another rope connects bundle A to bundle B farther down the roof (Figure 3.97). mA 15 kg mB 15 kg 10 m 20 m Figure 3.97 Each of the two bundles of shingles has a mass of 15 kg. The coefficient of kinetic friction for the bundles on plywood sheeting is 0.50. (a) What force must the roofer exert up the roof to drag the bundles at constant speed? (b) Calculate the force exerted by bundle A on bundle B. (c) What total force would the roofer have to exert to accelerate both bundles at 2.0 m/s2 [up roof]? Answers 1. (a) 2.6 102 N [up roof] (b) 1.3 102 N [up roof] (c) 3.2 102 N [up roof] (40 kg)(2.5 m/s2) (40 kg)(9.81 m/s2) [(sin 25.0) (0.04)(cos 25.0)] 3 102 N app is uphill. The positive value for Fapp indicates that the direction of F F app 3 102 N [uphill] Paraphrase The snowmobile must apply a force of 3 102 N [uphill]. Chapter 3 Forces can change velocity. 189 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 190 3.5 Check and Reflect 3.5 Check and Reflect Knowledge 1. In your own words, define friction. 2. What are some situations where friction is so small that it could be neglected? 3. Distinguish between static friction and kinetic friction. Applications 4. A pair of skis weigh 15 N [down]. Calculate the difference in the maximum force of static friction for the skis on a wet and dry snowy, horizontal surface. Refer to Table 3.4 on page 183. 5. A force of 31 N [forward] is needed to start an 8.0-kg steel slider moving along a horizontal steel rail. What is the coefficient of static friction? 6. A biker and his motorcycle have a weight of 2350 N [down]. Calculate the force of kinetic friction for the rubber tires and dry concrete if the motorcycle skids. Refer to Table 3.4 on page 183. 7. A 15-kg box is resting on a hill forming an angle with the horizontal. The coefficient of static friction for the box on the surface is 0.45. Calculate the maximum angle of the incline just before the box starts to move. 8. The coefficient of static friction for a wheelchair with its brakes engaged on a conveyor-type ramp is 0.10. The average mass of a person including the wheelchair is 85 kg. Determine if a ramp of 8.0° with the horizontal will prevent motion. 9. A truck loaded with a crate of mass m is at rest on an incline forming an angle of 10.0 with the horizontal. The coefficient of static friction for the crate on the truck bed is 0.30. Find the maximum possible acceleration uphill for the truck before the crate begins to slip backward. 190 Unit II Dynamics 10. A loaded dogsled has a mass of 400 kg and is being pulled across a horizontal, packed snow surface at a velocity of 4.0 m/s [N]. Suddenly, the harness separates from the sled. If the coefficient of kinetic friction for the sled on the snow is 0.0500, how far will the sled coast before stopping? Extensions 11. A warehouse employee applies a force of 120 N [12.0] to accelerate a 35-kg wooden crate from rest across a wooden floor. The coefficient of kinetic friction for the crate on the floor is 0.30. How much time elapses from the time the employee starts to move the crate until it is moving at 1.2 m/s [0°]? 12. Make a Venn diagram to summarize the similarities and differences between static and kinetic friction. See Student References 4: Using Graphic Organizers on page 869 for an example. 13. Research how the type of tread on a tire affects the coefficients of static friction and kinetic friction given the same road surface. Find out what hydroplaning is and how tires are designed to minimize this problem. Write a brief report of your findings, including diagrams where appropriate. Begin your search at www.pearsoned.ca/school/physicssource. 14. Design an experiment to determine the coefficients of static and kinetic friction for a curling stone on an icy surface. Perform the experiment at a local arena or club. Ask the icemaker to change the temperature of the ice, and repeat the experiment to determine if there is a difference in your values. Write a brief report of your findings. e TEST To check your understanding of friction and inclines, follow the eTest links at www.pearsoned.ca/school/physicssource. 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 191 CHAPTER 3 SUMMARY Key Terms and Concepts dynamics force free-body diagram normal force net force inertia inertial mass action force Key Equations Newton’s first law: F net Newton’s second law: F net F A on B Newton’s third law: 0 when v 0 ma F B on A reaction force friction static friction kinetic friction coefficient of static friction coefficient of kinetic friction Static friction: Ffstatic sFN Kinetic friction: Ffkinetic kFN Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. Dynamics involves forces free-body diagrams vector sums Newton’s first law Newton’s second law Newton’s third law are are measured in show all the forces is adding the involves inertia states that states that states that sliding friction two kinds kinetic equations so they have acting on determines the when magnitude and direction on Figure 3.98 Chapter 3 Forces can change velocity. 191 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 192 CHAPTER 3 REVIEW Knowledge 1. (3.1, 3.3) Two people, A and B, are pushing a stalled 2000-kg truck along a level road. Person A exerts a force of 300 N [E]. Person B exerts a force of 350 N [E]. The magnitude of the force of friction on the truck is 550 N. Calculate the acceleration of the truck. 2. (3.2) Use a free-body diagram and Newton’s first law to explain the motion of (a) a figure skater during a glide, and (b) a hockey puck during a cross-ice pass. Assume ice is frictionless. 3. (3.4) A transport truck pulls a trailer with a force of 1850 N [E]. What force does the trailer exert on the transport truck? 4. (3.5) An inexperienced driver, stuck in snow, tends to spin the car tires to increase the force of friction exerted by the snow on the tires. What advice would you give to the driver? Why? Applications 5. A device used to treat a leg injury is shown below. The pulley is attached to the foot, and the weight of the 3.0-kg object provides a tension force to each side of the pulley. The pulley is at rest to the pulley, because the foot applies a force F T2 in T1 and FF which is balanced by the forces FF the rope. The weight of the leg and foot is supported by the pillow. (a) Using a free-body diagram for the pulley, determine the force FF . (b) What will happen to the magnitude of F T2 decreases? Why? T1 and FF the angle between FF if support leg harness foot pulley F T1 F 45.0º 45.0º F T2 pillow 3.0 kg 192 Unit II Dynamics 6. Refer to Example 3.6 Practice Problem 1 on page 150. In a second practice run, the initial acceleration of the bobsled, pilot, and brakeman is 4.4 m/s2 [forward]. Rider A exerts an average force of magnitude 1200 N on the bobsled, and the force of friction decreases to 400 N. What average force does rider B exert? 7. During its ascent, a loaded jet of mass 4.0 105 kg is flying at constant velocity 20.0 above the horizontal. The engines of the plane provide a of 4.60 106 N [forward] to provide the thrust T [perpendicular to wings]. The air lift force L opposes the motion of the jet. resistance R Determine the magnitudes of L and R . L T 20.0° Fg R 8. Suppose the force of kinetic friction on a sliding block of mass m is 2.5 N [backward]. What is the force of kinetic friction on the block if another block of mass 2m is placed on its upper surface? 9. A 1385-kg pickup truck hitched to a 453-kg trailer accelerates along a level road from a stoplight at 0.75 m/s2 [forward]. Ignore friction and air resistance. Calculate (a) the tension in the hitch, (b) the force of friction exerted by the road on the pickup truck to propel it forward, and (c) the force the trailer exerts on the pickup truck. 10. Two curlers, A and B, have masses of 50 kg and 80 kg respectively. Both players are standing on a carpet with shoes having Teflon™ sliders. The carpet exerts a force of friction of 24.5 N [E] on player A and a force of friction of 39.2 N [W] on player B. Player A pushes player B with a force of 60 N [E]. (a) Calculate the net force acting on each player. (b) Calculate the acceleration of each player. 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 193 11. A force of 15 N [S] moves a case of soft drinks weighing 40 N [down] across a level counter at constant velocity. Calculate the coefficient of kinetic friction for the case on the counter. 12. A 1450-kg car is towing a trailer of mass 454 kg. The force of air resistance on both vehicles is 7471 N [backward]. If the acceleration of both vehicles is 0.225 m/s2, what is the coefficient of static friction for the wheels on the ground? 13. Two bags of potatoes, m1 6 |
0 kg and m2 40 kg, are connected by a light rope that passes over a light, frictionless pulley. The pulley is suspended from the ceiling using a light spring scale. (a) What is the reading on the scale if the pulley is prevented from turning? (b) Draw a free-body diagram for each bag when the pulley is released. (i) Calculate the acceleration of the system. (ii) Calculate the tension in the rope. (c) What is the reading on the scale when the bags are accelerating? (d) Explain the difference between your answers in parts (a) and (c). spring scale 10 pulley m1 60 kg POTATOES POTATOES m2 40 kg POTATOES m1 60 kg 14. A drag racing car initially at rest can reach a speed of 320 km/h in 6.50 s. The wheels of the car can exert an average horizontal force of 1.52 104 N [backward] on the pavement. If the force of air resistance on the car is 5.2 103 N [backward], what is the mass of the car? 15. A tractor and tow truck have rubber tires on wet concrete. The tow truck drags the tractor at constant velocity while its brakes are locked. If the tow truck exerts a horizontal force of 1.0 104 N on the tractor, determine the mass of the tractor. Refer to Table 3.4 on page 183. 16. Create a problem involving an object of mass m on an incline of angle . Write a complete solution, including an explanation of how to resolve the gravitational force vector into components. 17. The table below shows some coefficients of static and kinetic friction ( tires in contact with various road surfaces. s and k) for rubber Coefficient Dry Concrete Wet Concrete Dry Asphalt Wet Asphalt s k 1.0 0.7 0.7 0.5 1.2 0.6 0.6 0.5 (a) Which road surface exerts more static friction on a rubber tire, dry concrete or dry asphalt? Explain. (b) On which surface does a car slide more easily, on wet concrete or on wet asphalt? Why? (c) On which surface will a moving car begin to slide more easily, on dry concrete or on dry asphalt? Why? (d) On which surface will a car with locked brakes slide a shorter distance, on dry concrete or on dry asphalt? Explain. Extensions 18. An 80-kg baseball player slides onto third base. The coefficient of kinetic friction for the player on the ground is 0.70. His speed at the start of the slide is 8.23 m/s. (a) Calculate his acceleration during the slide. (b) For how long does he slide until he stops? (c) Show that the time it takes the player to come to a stop is given by the equation t vi kg . Consolidate Your Understanding 19. Write a paragraph explaining the similarities and differences among Newton’s three laws. Include an example that involves all three laws and explain how each law applies. Use the example to teach the laws to a student who has not studied dynamics. 20. Write a paragraph describing the differences between static and kinetic friction, and between the coefficients of static and kinetic friction. Include an example with a free-body diagram for each type of friction. Think About It Review your answers to the Think About It questions on page 125. How would you answer each question now? e TEST To check your understanding of forces and Newton’s laws of motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 3 Forces can change velocity. 193 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 194 Gravity extends throughout the universe. Skydiving, hang-gliding, bungee jumping, and hot-air ballooning are just a few activities that take advantage of gravitational forces for a thrill (Figure 4.1). Gravitational force attracts all objects in the universe. It holds you to Earth, and Earth in its orbit around the Sun. In 1665, Isaac Newton began his study of gravity when he attempted to understand why the Moon orbits Earth. His theories led him to an understanding of the motion of planets and their moons in the solar system. Several centuries later, these theories led to the launch of satellites and the success of various space missions such as Mariner and Voyager. Gravity is one of the four basic forces of nature, called fundamental forces, that physicists think underlie all interactions in the universe. These forces are the gravitational force, the electromagnetic force, the weak nuclear force, and the strong nuclear force. In this chapter, you will investigate how gravity affects the motion of objects on Earth and on other planets, and how it affects the motion of satellites orbiting Earth. Figure 4.1 Understanding how forces and gravity affect motion determines how successful the design of a hot-air balloon will be and the best way to navigate it. C H A P T E R 4 Key Concepts In this chapter, you will learn about: gravitational force Newton’s law of universal gravitation gravitational field Learning Outcomes When you have completed this chapter, you will be able to: Knowledge identify gravity as a fundamental force in nature describe Newton’s law of universal gravitation explain the Cavendish experiment define and apply the concept of a gravitational field compare gravitational field strength and acceleration due to gravity predict the weight of objects on different planets Science, Technology, and Society explain that concepts, models, and theories help interpret observations and make predictions 194 Unit II 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 195 4-1 QuickLab 4-1 QuickLab Falling Coins Problem Suppose you drop two coins of different shapes and sizes from the same height. How do the rates of both coins falling compare? Materials variety of coins (penny, nickel, dime, quarter, loonie, and toonie) ruler Styrofoam™ disk (size of a loonie) Procedure 1 Choose any two different coins and place them at the edge of a table above an uncarpeted floor. 2 Using a ruler, push the coins off the table so they leave at the same time (Figure 4.2). 3 Listen carefully for the sounds of the coins as they hit the floor. 4 Repeat this activity with different combinations of two coins, including the loonie with the Styrofoam™ disk. Record your observations. Figure 4.2 Questions 1. When the coins landed, how many sounds did you hear? Did all combinations of two coins give the same result? Explain. 2. If all the coins fall at the same rate, how many sounds would you expect to hear when they land? 3. How would the results compare if two coins were released at the same time from a greater height, such as 10 m? 4. How did the average acceleration of the loonie differ from that of the Styrofoam™ disk? Explain why. Think About It 1. (a) What factors affect the weight of an astronaut during a rocket flight? (b) How does the astronaut’s weight change? 2. What would be the motion of Earth if the Sun’s gravity were zero? Assume that no other celestial bodies affect Earth. Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 4 Gravity extends throughout the universe. 195 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 196 info BIT Gravitational force is an attraction force only. There is no such thing as a repulsive gravitational force. gravitational force: attractive force between any two objects due to their masses 4.1 Gravitational Forces due to Earth One of Newton’s great achievements was to identify the force that causes objects to fall near Earth’s surface as the same force that causes the Moon to orbit Earth. He called this force “gravity,” and he reasoned that this force is present throughout the universe. g, is the force that attracts any two objects Gravitational force, F together. Although this force is the weakest fundamental force, you can feel its effect when you interact with an object of very large mass such as Earth. When you slide down a waterslide, you can feel the gravitational force exerted by Earth pulling you downward toward the bottom of the slide (Figure 4.3). But if you want to feel the gravitational force exerted by the person sitting next to you, you will not be able to sense anything because the magnitude of the force is so small. Figure 4.3 The attractive force between Earth and you is far greater than that between you and another person coming down a waterslide. 196 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 197 Gravitational force is a force that always exists in pairs. This is another example of Newton’s third law. If Earth exerts a gravitational force of magnitude 700 N on you, then you exert a gravitational force of magnitude 700 N on Earth. Earth attracts you and you attract Earth. The force you exert on Earth has a negligible effect because Earth’s mass (5.97 1024 kg) is huge in comparison to yours. However, the gravitational force that Earth exerts on you causes a noticeable acceleration because of your relatively small mass. Concept Check Which diagram best represents the gravitational force acting on you and on Earth (Figure 4.4)? Explain your reasoning. (a) (b) (c) (d) Figure 4.4 The Concept of Weight In a vacuum, all objects near Earth’s surface will fall with the same acceleration, no matter what the objects consist of or what their masses are. The only force acting on a falling object in a vacuum is the gravitational force exerted by Earth on the object (Figure 4.5). Suppose you analyze this situation using a free-body diagram and Newton’s second law. Fg Figure 4.5 Free-body diagram for a falling object in a vacuum Chapter 4 Gravity extends throughout the universe. 197 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 198 Fs 2 6 10 14 18 0 4 8 12 16 20 The equation for the net force acting on the falling object is F net F g ma F g Since the object is accelerating due to gravity, a g. So the equation for the net force becomes mg F g or F g mg The equation F mg is valid in general, because the gravitational force acting on an object is the same, whether or not the object is at rest or is moving. This equation relates the gravitational force acting on an object, the so-called weight of the object, to its mass. g Fg Figure 4.6 Diagram sh |
owing the forces acting on an object that is suspended from a spring scale weight: gravitational force exerted on an object by a celestial body 4-2 QuickLab 4-2 QuickLab One way to measure the magnitude of the weight of an object directly involves using a spring scale (Figure 4.6). When the object stops moving at the end of the spring, Earth exerts a downward gravitational force on the object while the spring exerts an upward elastic force of equal magnitude on the object. Find out what the relationship is between mass and gravitational force in the vicinity of your school by doing 4-2 QuickLab. Relating Mass and Weight Problem What is the relationship between the mass of an object and the local value of the gravitational force exerted on that object? Materials set of standard masses with hooks spring scale (010 N) graph paper Procedure 1 Design a procedure to determine the gravitational force acting on a set of standard masses (Figure 4.7). 2 Use a table to record the magnitude of the gravitational force and mass. Add another column in the table for the ratio of Fg to m. 3 Calculate the ratio of Fg to m for each standard mass. Calculate the average ratio for all masses. Include units 10 Figure 4.7 198 Unit II Dynamics 4 Plot a graph of Fg vs. m. Draw a line of best fit through the data points. Calculate the slope of the graph. Questions 1. What does the ratio of Fg to m represent? How constant is this value? 2. Describe the graph of Fg vs. m. How does the slope compare to the average ratio calculated in step 3? 3. Write an equation relating Fg and m. Use the symbol g for the proportionality constant. 4. (a) The Moon exerts a gravitational force that is 1 that exerted by Earth. If you did this about 6 activity on the Moon using an appropriate spring scale and the same standard masses, would • the graph be a straight line? • the slope be the same as before? • the line go through the origin? • the proportionality constant g be the same as before? (b) Why would a 05 N spring scale be more ideal to use on the Moon, rather than a 010 N spring scale? 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 199 Gravitational Mass mg is determined by finding the ratio of The mass in the equation F g the gravitational force acting on an object to the acceleration due to g and g are in the same direction, the scalar form of the gravity. Since F equation may be used: Fg mg m Fg g A practical way to measure this mass involves using a balance. In Figure 4.8, an object of unknown mass (A) is placed on one pan and standard masses (B) are added to the other pan until both pans balance. This method involves comparing the weights of two objects: one unknown and the other known. Mass measured using the concept of weight is called gravitational mass. gravitational force exerted on standard masses gravitational force exerted on an unknown mass gravitational mass: mass measurement based on comparing the known weight of one object to the unknown weight of another object Figure 4.8 When both pans are balanced, the gravitational force acting on the unknown mass is equal to the gravitational force acting on the standard masses. If the balance in Figure 4.8 were moved to the Moon, the process of determining the gravitational mass of object A would be the same. However, the weight of A and B would be different from that at Earth’s surface because the acceleration due to gravity at the Moon’s surface, gMoon, is 1.62 m/s2 compared to 9.81 m/s2, the average value at Earth’s surface. e SIM Explore how mass and weight are measured. Follow the eSim links at www.pearsoned.ca/school/ physicssource. FgA FgB mAgMoon mBgMoon mA mB But since both objects A and B experience the same value of gMoon, mB. So the gravitational mass of an object is the same whether the mA object is on Earth, the Moon, or anywhere else in the universe. Both gravitational mass and inertial mass are properties of an object that do not depend on the location of the object. Is Inertial Mass the Same as Gravitational Mass? You can determine the mass of an object by using either the concept of inertia or weight. Experiments since Newton’s day have shown that for any object, the numerical value of its inertial mass is equal to its gravitational mass. Later, Albert Einstein (1879–1955) showed that inertial mass is actually equivalent to gravitational mass. So it does not matter whether you determine the mass of an object using inertia or weight, because the numerical value of the mass is the same. Chapter 4 Gravity extends throughout the universe. 199 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 200 action-at-a-distance force: force that acts even if the objects involved are not touching field: three-dimensional region of influence gravitational field: region of influence surrounding any object that has mass Describing Gravitational Force as a Field Gravitational force is an example of a force that acts on objects whether or not they actually touch each other, even if the objects are in a vacuum. These forces are referred to as action-at-a-distance forces. In the 1800s, physicists introduced the concept of a field to explain action-at-a-distance forces. You encountered some fields in previous science courses when you worked with magnets. Imagine you are moving the north pole of a magnet close to the north pole of a fixed magnet. As you move the magnet closer to the fixed magnet, you can feel an increasing resistance. Somehow, the fixed magnet has created a region of influence in the space surrounding it. Physicists refer to a three-dimensional region where there is some type of an influence, whether it is an attraction or a repulsion, on a suitable object as a field. Since every object exerts a gravitational force in three dimensions, it influences the space around it (Figure 4.9). This region of influence is a gravitational field, and it is through this region that two objects interact. gravitational field of Earth gravitational field of the Moon Figure 4.9 This figure shows a two-dimensional representation of Earth’s and the Moon’s gravitational field. A gravitational field is three-dimensional and is directed toward the centre of the object. 200 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 201 To determine the magnitude and direction of a gravitational field created by an object, you could use a test mass mtest. At different locations around the object, this test mass will experience a gravitational force that has a certain magnitude and direction. The direction of the gravitational force will be directed toward the centre of the object. Gravitational field strength is defined as the gravitational force per F unit mass, g g . If you release the test mass, it will accelerate toward m te st the object with an acceleration equal to g. Figure 4.10 shows how the magnitude of Earth’s gravitational field strength changes as a test mass is moved farther away from Earth’s centre. The farther the test mass is moved, the more significant is the decrease in g. In fact, the graph in Figure 4.10 shows an inverse square relationship: g 1 r 2 Magnitude of Gravitational Field Strength vs. Distance from Earth’s Centre gravitational field strength: gravitational force per unit mass at a specific location 10.0 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 Distance from Earth’s centre r (Earth radii) Figure 4.10 The magnitude of the gravitational field strength as a function of distance from Earth’s centre Since force is measured in newtons and mass in kilograms, the units of gravitational field strength are newtons per kilogram, or N/kg. The ratio you determined in 4-2 QuickLab was the gravitational field strength at the vicinity of your school. Concept Check What happens to the magnitude of the gravitational field strength if (a) r decreases by a factor of four? (b) r increases by a factor of two? (c) mtest doubles? (d) mtest is halved? Chapter 4 Gravity extends throughout the universe. 201 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 202 4.1 Check and Reflect 4.1 Check and Reflect Knowledge 1. Distinguish between mass and weight. Explain using an example and a diagram. 10. How could you distinguish between a 5.0-kg medicine ball and a basketball in outer space without looking at both objects? 2. Distinguish between inertial mass and Extensions gravitational mass. 3. In your own words, define gravitational acceleration and gravitational field strength. State the units and symbol for each quantity. 11. Visit a local fitness gymnasium. Find out how athletes use gravitational and elastic forces to improve their fitness. Is friction a help or a hindrance? Write a brief report of your findings. 4. In your own words, explain the concept of a gravitational field. Include an example of how a gravitational field affects another object. 12. List some occupations that might require a knowledge of gravitational field strength. Briefly explain how gravitational field strength applies to these occupations. 5. Why do physicists use the concept of 13. Complete the gathering grid below to summarize the similarities and differences among gravitational mass, inertial mass, and gravitational force. Gravitational Mass Inertial Mass Gravitational Force Definition SI Unit Measuring Instrument(s) How the Quantity Is Measured Factors It Depends On Variability with Location e TEST To check your understanding of gravitational force, weight, mass, and gravitational field strength, follow the eTest links at www.pearsoned.ca/school/physicssource. a field to describe gravity? 6. In a vacuum, a feather and a bowling ball are released from rest at the same time from the same height. Compare the time it takes for each object to fall. Explain your answer. Applications 7. The Moon exerts a gravitational force that is about 1 6 that exerted by Earth. Explain why the mass of an object measured on the Moon using a balance is the same as if the object were on Earth’s surface. 8. Describe a situation where measuring the inertial mass of an |
object is easier than measuring its gravitational mass. 9. The table below shows the magnitude of the gravitational force on objects of different mass in Banff, Alberta. Mass (kg) 0 1.50 3.00 4.50 6.00 7.50 10.0 Magnitude of 0 14.7 29.4 44.1 58.9 73.6 98.1 Gravitational Force (N) (a) Graph the data. (b) Calculate the slope of the line. (c) What does the slope represent? 202 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 203 info BIT Gravity is the dominant force throughout the universe. This force is attractive and has an infinite range. 4.2 Newton’s Law of Universal Gravitation Gravity affects all masses in the universe. No matter where you are on Earth or in outer space, you exert a gravitational force on an object and an object exerts a gravitational force, of equal magnitude but opposite direction, on you. Because gravitational force acts over any distance, the range of its effect is infinite. Near Earth’s surface, the magnitude of the gravitational force exerted mBg. by Earth (object A) on object B is given by the equation FA on B But object B also exerts a gravitational force of equal magnitude on mAg. Newton hypothesized that, given two objects A Earth, FB on A and B, the magnitude of the gravitational force exerted by one object on the other is directly proportional to the product of both masses: Fg mAmB Figure 4.11 shows the magnitude of the gravitational force acting on an object at Earth’s surface (rEarth), one Earth radius above Earth (2rEarth), and two Earth radii above Earth (3rEarth). If the separation distance from Earth’s centre to the centre of the object doubles, Fg decreases 1 of its original value. If the separation distance from Earth’s centre to to 4 1 of its original value. the centre of the object triples, Fg decreases to 9 1 , Fg is inversely proportional to the square of the separation So, just as g 2 r distance (Figure 4.12): Fg 1 2 r Magnitude of Gravitational Force vs. Distance from Earth’s Centre Earth Fg rEarth 1 4 Fg 1 9 Fg 3rEarth 2rEarth 10.0 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 Distance from Earth’s centre r (Earth radii) Figure 4.11 The magnitude of the gravitational force acting on an object some distance from Earth varies inversely with the square of the separation distance. Figure 4.12 The magnitude of the gravitational force acting on a 1.00-kg object as a function of distance from Earth’s centre Chapter 4 Gravity extends throughout the universe. 203 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 204 If you combine both proportionalities into one statement, you get mAmB r 2 GmAmB r 2 ship is Newton’s law of universal gravitation. (Figure 4.13). This mathematical relation- or Fg Fg Any two objects, A and B, in the universe exert gravitational forces of equal magnitude but opposite direction on each other. The forces are directed along the line joining the centres of both objects. The magnitude of the gravitational force is given by Fg GmAmB r 2 , where mA and mB are the masses of the two objects, r is the separation distance between the centres of both objects, and G is a constant called the universal gravitational constant. First mass Separation distance Second mass Magnitude of gravitational force Fg 2 Fg 6 Fg 4 Fg 1 4 Fg m 3 m 6 4 Fg Figure 4.13 The magnitude of the gravitational force is directly proportional to the product of the two masses, and inversely proportional to the square of the separation distance. Experiments have shown that the magnitude of the gravitational force acting on any pair of objects does not depend on the medium in which the objects are located (Figure 4.14). In other words, given two fish, the gravitational force acting on either fish will have the same magnitude if both fish are underwater or in midair. (a) (b) 1 m 1 m Figure 4.14 The magnitude of the gravitational force acting on either fish is the same whether both fish are (a) underwater or (b) above water. e SIM Explore the relationship among mA, mB, r, and Fg in Newton’s law of gravitation. Follow the eSim links at www.pearsoned.ca/school/ physicssource. 204 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 205 Concept Check Two identical stationary baseballs are separated by a distance r. What will happen to the magnitude of the gravitational force acting on either ball if (a) the mass of each ball doubles? (b) r is halved? (c) the mass of each ball is halved and r doubles? e WEB Edmund Halley, an associate of Newton, paid for the publication of some of Newton’s famous work. Find out about Edmund Halley and his contributions to Newton’s work. Begin your search at www.pearsoned.ca/school/ physicssource. Determining the Value of the Universal Gravitational Constant Although Newton found a mathematical relationship for gravitational force, he was unable to determine the value of G. In 1798, scientist Henry Cavendish (1731–1810) confirmed experimentally that Newton’s law of gravitation is valid, and determined the density of Earth. Cavendish’s experimental set-up was later used to determine the value of G. The magnitude of the gravitational force acting on most pairs of objects is very weak and the magnitude decreases significantly as the separation distance between the objects increases. However, if you use two light spheres (each of mass m) and two heavy spheres (each of mass M) that are very close to each other, it is possible to determine the magnitude of the gravitational force exerted by M on m. The trick is to use a device that can accurately measure the very small gravitational force. A modern torsion balance is a device that uses a sensitive fibre and a beam of light to measure very minute forces due to gravity, magnetic fields, or electric charges. Cavendish used a modified torsion balance invented by John Michell (1724–1793) to verify Newton’s law of gravitation. A modern torsion balance consists of a small, light, rigid rod with two identical, light, spheres (m) attached to each end (Figure 4.15). The rod is suspended horizontally by a thin fibre connected to the centre of the rod. A mirror is also attached to the fibre and rod so that when the rod turns, the mirror also turns by the same amount. The entire assembly is supported in an airtight chamber. The torsion balance initially experiences no net force and the spheres m are stationary. fibre support torsion fibre mirror m zero deflection screen laser M m rod M Figure 4.15 A modern torsion balance uses a laser beam to measure the amount of twist in the fibre. The most accurate value of G has been determined using such a device. torsion balance: device used to measure very small forces info BIT Charles de Coulomb invented the original torsion balance in 1777 to measure small magnetic forces and forces in fluids. However, John Michell independently invented the same type of device in 1784. Chapter 4 Gravity extends throughout the universe. 205 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 206 m Fg M on m Fg m on M M fibre direction of motion M Fg m on M Fg M on m m Figure 4.16 This figure shows the top view of spheres m and M from Figure 4.15. The original position of spheres m is shown with dashed lines. The gravitational force exerted by M on m causes m to rotate toward M. The greater the magnitude of the gravitational force, the greater the angle of rotation. When two identical, heavy, spheres (M) are moved close to spheres m, the gravitational force exerted by M on m causes m to rotate horizontally toward M. This rotation causes the fibre to twist slightly (Figure 4.16). As the fibre twists, the mirror attached to both the fibre and the rod turns through an angle in the horizontal plane. A beam of light reflected from the mirror becomes deflected as spheres m rotate. The amount of deflection is an indication of how much the spheres rotate. The greater the magnitude of the gravitational force, the more the fibre twists, and the greater the angle of rotation. By measuring the amount of deflection, the gravitational force exerted by M on m can be determined. Spheres M are then moved to a symmetrical position on the opposite side of m, and the procedure is repeated. Since the separation distance between m and M, the values of m and M, and the gravitational force can all be measured, it is possible to calculate G using Newton’s law of gravitation. Fg GmM r 2 Fgr 2 mM G G Fgr 2 mM info BIT Scientists use Newton’s law of universal gravitation to calculate the masses of planets and stars. The current accepted value of G to three significant digits is 6.67 1011 Nm2/kg2. In Example 4.1, Newton’s law of gravitation is used to show that a person weighs slightly less at the top of the mountain than at its base. Example 4.1 Mount Logan in the Yukon is 5959 m above sea level, and is the highest peak in Canada. Earth’s mass is 5.97 1024 kg and Earth’s equatorial radius is 6.38 106 m. What would be the difference in the magnitude of the weight of a 55.0-kg person at the top of the mountain as compared to at its base (Figure 4.17)? Assume that Earth’s equatorial radius is equal to the distance from Earth’s centre to sea level. Given mp 55.0 kg h 5959 m mEarth rEarth 5.97 1024 kg 6.38 106 m Required difference in magnitude of weight (Fg) Mt. Logan 5959 m Figure 4.17 206 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 207 Practice Problems 1. Two people, A and B, are sitting on a bench 0.60 m apart. Person A has a mass of 55 kg and person B a mass of 80 kg. Calculate the magnitude of the gravitational force exerted by B on A. 2. The mass of the Titanic was 4.6 107 kg. Suppose the magnitude of the gravitational force exerted by the Titanic on the fatal iceberg was 61 N when the separation distance was 100 m. What was the mass of the iceberg? Answers 1. 8.2 107 N 2. 2.0 108 kg Analysis and Solution Assume that the separation distance between the person at the base of the mountain and Earth is equal to Earth’s equatorial radius. Base of mountain: rB rEarth 6.38 106 m Top of mountain: rT 6.38 106 m 5959 m Th |
e person’s weight is equal to the gravitational force exerted by Earth on the person, and is directed toward Earth’s centre both at the base and at the top of the mountain. Calculate Fg at the base of the mountain using Newton’s law of gravitation. (Fg)B GmpmEarth (rB)2 2 m N (55.0 kg)(5.97 1024 kg) 6.67 1011 2 g k (6.38 106 m)2 538.049 N Calculate Fg at the top of the mountain using Newton’s law of gravitation. (Fg)T GmpmEarth (rT)2 2 m N (55.0 kg)(5.97 1024 kg) 6.67 1011 2 g k (6.38 106 m 5959 m)2 537.045 N The difference in the magnitude of the weight is equal to the difference in magnitude of both gravitational forces. Fg (Fg)T (Fg)B 538.049 N 537.045 N 1.00 N Paraphrase The difference in the magnitude of the person’s weight is 1.00 N. Using Proportionalities to Solve Gravitation Problems Example 4.2 demonstrates how to solve gravitation problems using proportionalities. This technique is useful if you are given how the separation distance and masses change from one situation to another. Chapter 4 Gravity extends throughout the universe. 207 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 208 Example 4.2 Object A exerts a gravitational force of magnitude 1.3 1010 N on object B. Determine the magnitude of the gravitational force if the separation distance is doubled, mA increases by 6 times, and mB is halved. Explain your reasoning. Practice Problem 1. Object A exerts a gravitational force of magnitude 5.9 1011 N on object B. For each situation, determine the magnitude of the gravitational force. Explain your reasoning. (a) the separation distance 4 increases to of its original 3 3 of its value, mA increases to 2 original value, and mB is halved (b) the separation distance decreases 1 of its original value, mA is to 6 5 halved, and mB increases to 4 of its original value Answer 1. (a) 2.5 1011 N (b) 1.3 109 N Analysis and Solution mAmB and Fg From Newton’s law of gravitation, Fg Figure 4.18 represents the situation of the problem. 1 r 2 . before after mA mA Fg 1.3 1010 N mB r Fg ? 2r Figure 4.18 Fg (6mA) mB 1 2 and Fg 1 (2r)2 (6) 1 2 mAmB 3mAmB 1 1 22 r 2 1 1 2 r 4 mB Calculate the factor change of Fg. 3 1 4 3 4 Calculate Fg. 3 4 3 4 Fg (1.3 1010 N) 9.8 1011 N The new magnitude of the gravitational force will be 9.8 1011 N. Using Superposition to Find the Net Gravitational Force on an Object Example 4.3 demonstrates how to calculate the gravitational force exerted by both the Moon and the Sun on Earth. A free-body diagram is used to determine the gravitational forces acting on Earth. The technique of adding the gravitational force due to each pair of objects (Earth and the Moon, and Earth and the Sun) to find the net gravitational force is called superposition. 208 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 209 Example 4.3 During a lunar eclipse, Earth, the Moon, and the Sun are aligned on the same plane as shown in Figure 4.19. Using the data in the chart below, calculate the net gravitational force exerted by both the Moon and the Sun on Earth. Celestial Body Mass* (kg) Mean Separation Distance from Earth* (m) Earth Earth’s Moon Sun 5.97 1024 7.35 1022 1.99 1030 — 3.84 108 1.50 1011 *Source: Jet Propulsion Laboratory, California Institute of Technology (See JPL link at www.pearsoned.ca/school/physicssource.) 3.84 108 m 1.50 1011 m Moon Earth Sun Figure 4.19 Given mEarth mMoon mSun 5.97 1024 kg 7.35 1022 kg 1.99 1030 kg rE to M rE to S 3.84 108 m 1.50 1011 m Required net gravitational force on Earth (F g) Analysis and Solution Draw a free-body diagram for Earth (Figure 4.20). Practice Problems 1. During a solar eclipse, Earth, the Moon, and the Sun are aligned on the same plane as shown in Figure 4.21. Calculate the net gravitational force exerted by both the Moon and the Sun on Earth. 1.50 1011 m 3.84 108 m Earth Moon Sun Figure 4.21 2. During the first quarter phase of the Moon, Earth, the Moon, and the Sun are positioned as shown in Figure 4.22. Calculate the net gravitational force exerted by both the Moon and the Sun on Earth. Top View of Earth, the Moon, and the Sun Moon y 3.84 108 m Earth x Sun toward Sun toward Moon Fg1 Figure 4.20 Fg1 Fg2 Fnet 1.50 1011 m Figure 4.22 Answers 1. 3.54 1022 N [toward Sun’s centre] 2. 3.52 1022 N [0.3] Fg2 Diagram is not to scale. Calculate Fg exerted by the Moon on Earth using Newton’s law of gravitation. (Fg)1 GmEarthmMoon (rE to M)2 m N (5.97 1024 kg)(7.35 1022 kg) 6.67 1011 2 g k (3.84 108 m)2 2 1.985 1020 N 1.985 1020 N [toward Moon’s centre] F g1 Chapter 4 Gravity extends throughout the universe. 209 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 210 Calculate Fg exerted by the Sun on Earth using Newton’s law of gravitation. (Fg)2 GmEarthmSun (rE to S)2 m N (5.97 1024 kg)(1.99 1030 kg) 6.67 1011 2 g k (1.50 1011 m)2 2 3.522 1022 N 3.522 1022 N [toward Sun’s centre] F g2 gnet Fgnet F g2 Fg2 Find the net gravitational force on Earth. F F g1 Fg1 1.985 1020 N 3.522 1022 N 3.50 1022 N 3.50 1022 N [toward Sun’s centre] F gnet Paraphrase The net gravitational force on Earth due to the Sun and Moon during a lunar eclipse is 3.50 1022 N [toward Sun’s centre]. The Role of Gravitational Force on Earth’s Tides Newton used the concept of gravitational force to account for Earth’s tides. Although he correctly identified the gravitational force exerted by the Moon and the Sun on Earth as the major cause, a complete understanding of tides must take into account other factors as well. The height of the tides varies depending on the location on Earth (Figure 4.23). In the middle of the Pacific Ocean, the difference between high and low tides is about 0.5 m. But along the coastline of the continents, the difference may be considerably greater. Figure 4.23 The tides in the Bay of Fundy are the highest in the world. In some locations, the water level rises up to 18 m between low and high tides. 210 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 211 Some factors that affect tides are • the shape of the coastline • the topography of the ocean floor near the coastline • friction between Earth and the ocean water • Earth’s position in its orbit around the Sun • Earth’s rotation about its axis • the tilt of Earth’s axis • the alignment of Earth, the Moon, and the Sun First consider only the Moon’s influence on Earth. Since the Moon exerts a gravitational force on Earth, the Moon is in a sense pulling Earth closer to it. So the land mass and ocean water on Earth are all “falling” toward the Moon. In Figure 4.24, this gravitational force is greatest at side A, then decreases at the midpoints of A and B, and is least at side B, because the magnitude of the gravitational force varies inversely with the square of the separation distance. e WEB Newspapers in cities near an ocean, such as Halifax, often publish tidal charts listing the times of local high and low tides. Find an example of a tidal chart and suggest how different people would find this information useful. Begin your search at www.pearsoned.ca/school/ physicssource. 23.5° B Earth A Moon Figure 4.24 Earth experiences high tides on sides A and B at the same time. The vectors show the relative gravitational force exerted by the Moon on a test mass at various locations near Earth’s surface. The bulges at A and B are the high tides. The low tides occur at the midpoints of A and B. The bulge at B occurs because the land mass of Earth at B is pulled toward the Moon, leaving the ocean water behind. Next consider the fact that Earth rotates on its axis once every 24 h. As the bulges remain fixed relative to the Moon, Earth rotates underneath those bulges. So at a given location on Earth’s surface, a high tide is replaced by a low tide about 6 h later, followed again by a high tide about 6 h later, and so on. These time intervals are actually a bit longer than 6 h because the Moon is orbiting Earth every 27 1 3 days with respect to distant stars, and the Moon is taking the bulges along with it. Now consider that Earth is tilted on its axis. When the northern hemisphere is under the bulge at A, the southern hemisphere is under the bulge at B. So the high tides that are 12 h apart are not equally high, and low tides that are 12 h apart are not equally low. Example 4.4 demonstrates how to calculate the gravitational force exerted by the Moon on 1.0000 kg of water at A. info BIT Io, one of Jupiter’s moons, has tidal bulges of up to 100 m compared to typical tidal bulges of 1 m on Earth. Chapter 4 Gravity extends throughout the universe. 211 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 212 Example 4.4 Calculate the gravitational force exerted by the Moon on 1.0000 kg of water at A (Figure 4.25). Use G 6.672 59 1011 Nm2/kg2. Celestial Body Mass* (kg) Equatorial Radius* (m) Mean Separation Distance from Earth* (m) Earth 5.9742 1024 6.3781 106 — Earth’s Moon 7.3483 1022 1.7374 106 3.8440 108 *Source: Jet Propulsion Laboratory, California Institute of Technology (See JPL link at www.pearsoned.ca/school/physicssource.) Earth A water B rEarth 3.8440 108 m Moon Figure 4.25 Given mw mEarth mMoon rE to M 1.0000 kg 5.9742 1024 kg 7.3483 1022 kg 3.8440 108 m rEarth rMoon 6.3781 106 m 1.7374 106 m Required g) gravitational force exerted by Moon on water (F Analysis and Solution Draw a free-body diagram for the water showing only F g due to the Moon (Figure 4.26). away from Moon toward Moon Fg Figure 4.26 Practice Problem 1. Using the value of G given in Example 4.4, calculate the gravitational force exerted by the Moon on 1.0000 kg of water (a) at the midpoints of A and B, and (b) at B. Answer 1. (a) 3.3183 105 N [toward Moon’s centre] (b) 3.2109 105 N [toward Moon’s centre] 212 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 213 Find the separation distance between the water and the Moon. r rE to M rEarth 3.8440 108 m 6.3781 106 m 3.780 22 108 m Calculate Fg exerted by the Moon on the water using Newton’s law of gravitation. GmwmMoon r 2 Fg m N (1.0000 kg)(7.3483 1022 kg) 6.672 59 1011 2 g k (3.780 22 108 m)2 2 3.4312 105 N |
3.4312 105 N [toward Moon’s centre] F g Paraphrase The gravitational force exerted by the Moon on the water at A is 3.4312 105 N [toward Moon’s centre]. The Role of Gravitational Force on Interplanetary Travel Scientists who plan space missions take advantage of the gravitational force exerted by planets and other celestial bodies to change the speed and direction of spacecraft. Distances between celestial bodies are huge compared to distances on Earth. So a space probe leaving Earth to study Jupiter and Saturn and their moons would take many years to arrive there. Scientists have to calculate the position and velocity of all the celestial bodies that will affect the motion of the probe many years in advance. If several planets are moving in the same direction and their positions are aligned, a space probe launched from Earth can arrive at its destination many years sooner, provided the probe moves near as many of those planets as possible (Figure 4.27). info BIT The Voyager mission was intended to take advantage of a geometric alignment of Jupiter, Saturn, Uranus, and Neptune. This arrangement occurs approximately every 175 years. By using the concept of gravity assist, the flight time to Neptune was reduced from 30 to 12 years, and a minimum of onboard propellant on the spacecraft was required. Voyager 1 Voyager 1 Launch Launch Sept. 5, 1977 Sept. 5, 1977 Voyager 2 Voyager 2 Launch Launch Aug. 20, 1977 Aug. 20, 1977 Jupiter Mar. 5, 1979 Jupiter Mar. 5, 1979 Jupiter Jul. 9, 1979 Jupiter Jul. 9, 1979 Saturn Aug. 25, 1981 Saturn Aug. 25, 1981 Saturn Nov. 12, 1980 Saturn Nov. 12, 1980 Voyager 2 Neptune Aug. 25, 1989 Neptune Aug. 25, 1989 Uranus Jan. 24, 1986 Uranus Jan. 24, 1986 Voyager 1 Figure 4.27 Both Voyager spacecraft were launched from Cape Canaveral, Florida, in 1977. Voyager 1 had close encounters with Jupiter and Saturn, while Voyager 2 flew by all four of the gaseous planets in the solar system. Chapter 4 Gravity extends throughout the universe. 213 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 214 Each time the probe gets near enough to one of these planets, the gravitational field of the planet causes the path of the probe to curve (Figure 4.28). The planet deflects the space probe and, if the planet is moving in the same direction as the probe, the speed of the probe after its planetary encounter will increase. The use of the gravitational force exerted by celestial bodies to reduce interplanetary travel times is called gravity assist. info BIT Voyager 1 identified nine active volcanoes on Io, one of Jupiter’s moons. Up until that point, scientists knew of no other celestial body in the solar system, other than Earth, that has active volcanoes. According to Voyager 1’s instruments, the debris being ejected from Io’s volcanoes had a speed of 1.05 103 m/s compared to speeds of 50 m/s at Mount Etna on Earth. Figure 4.28 Voyager 1 passed close to Io, Ganymede, and Callisto, three of Jupiter’s moons. Jupiter has a total of 62 moons. Europa Ganymede Voyager 1 Jupiter Io Callisto THEN, NOW, AND FUTURE Small Steps Lead to New Models When Newton began his study of gravity in 1665, he was not the first person to tackle the challenge of explaining planetary motion. Ancient Greek astronomer Ptolemy (Claudius Ptolemaeus, 2nd century A.D.) and eventually Nicolaus Copernicus (1473–1543) proposed two different models of the solar system: one Earth-centred and the other Sun-centred. Later Johannes Kepler (1571–1630) developed three empirical laws describing planetary motion using astronomical data compiled by astronomer Tycho Brahe (1546–1601). Kepler’s laws confirmed that a Sun-centred system is the correct model, because it was possible to predict the correct position of plan- ets. However, Kepler was unable to explain why planets move. Newton was the first scientist to explain the motion of planets in terms of forces. By using his three laws of motion and his law of gravitation, Newton derived Kepler’s laws, providing further evidence of the validity of his force laws and of the model of a Sun-centred solar system. Newton was able to develop his law of gravitation because many scientists before him developed theories and made observations about planetary motion. Newton’s laws and the concept of gravity could describe the motion of objects on Earth and throughout the universe. This was a tremendous breakthrough because up until that time, scientists were unable to predict or explain motion. While Newton’s model can still be used today for most everyday situations, scientists have further modified it. The process of developing new models and theories has helped scientists tackle questions about the universe in ways Newton could never have imagined. Questions 1. Research the scientific developments that led to Newton’s law of gravitation. 2. What are some benefits of developing new scientific models and theories? 214 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 215 4.2 Check and Reflect 4.2 Check and Reflect Knowledge 1. Why is G called a “universal” constant? 2. Describe how a torsion balance can be used to measure the constant G. 3. Suppose Fg is the magnitude of the gravitational force between two people with a separation distance of 1.0 m. How would Fg change if (a) the separation distance became 2.0 m? (b) one person was joined by an equally massive friend while at this 2.0-m separation distance? Applications 4. The Moon has a mass of 7.35 1022 kg and its equatorial radius is 1.74 106 m. Earth’s mass is 5.97 1024 kg and its equatorial radius is 6.38 106 m. (a) Calculate the magnitude of the gravitational force exerted by (i) the Moon on a 100-kg astronaut standing on the Moon’s surface, and (ii) Earth on a 100-kg astronaut standing on Earth’s surface. (b) Explain why the values of Fg in part (a) are different. 5. Mars has two moons, Deimos and Phobos, each named after an attendant of the Roman war god Mars. Deimos has a mass of 2.38 1015 kg and its mean distance from Mars is 2.3 107 m. Phobos has a mass of 1.1 1016 kg and its mean distance from Mars is 9.4 106 m. (a) Without doing any calculations, predict which moon will exert a greater gravitational force on Mars. Explain your reasoning. (b) Check your prediction in part (a) by calculating the magnitude of the gravitational force exerted by each moon on Mars. Mars’ mass is 6.42 1023 kg. Show complete solutions. 6. Suppose the equatorial radius of Earth was the same as the Moon, but Earth’s mass remained the same. The Moon has an equatorial radius of 1.74 106 m. Earth’s mass is 5.97 1024 kg and its equatorial radius is 6.38 106 m. (a) Calculate the gravitational force that this hypothetical Earth would exert on a 1.00-kg object at its surface. (b) How does the answer in part (a) compare to the actual gravitational force exerted by Earth on this object? Extensions 7. Prepare a problem involving Newton’s law of gravitation for each situation. Work with a partner to solve each problem, and discuss the steps you use. (a) Choose the values of the two masses and the separation distance. (b) Use values of mA, mB, and r that are multiples of those in part (a). Use proportionalities to solve the problem. 8. During Newton’s time, scientists often • worked alone and contact with other scientists working on similar problems was difficult. • were knowledgeable in many different fields. Newton, for example, spent many years doing alchemy. (a) In paragraph form, assess the impacts that these factors might have had on science in Newton’s day. (b) In a paragraph, describe in what ways these factors are relevant to scientists today. e TEST To check your understanding of Newton’s law of gravitation, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 4 Gravity extends throughout the universe. 215 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 216 info BIT A g force is a force that causes an acceleration with a magnitude of some multiple of g. A force of 2g means the magnitude of the acceleration is 2 9.81 m/s2 19.6 m/s2. 4.3 Relating Gravitational Field Strength to Gravitational Force The acceleration due to gravity g near or on Earth’s surface is about 9.81 m/s2 [down]. But where does the value of 9.81 m/s2 come from? Consider the forces acting on a test mass mtest some distance above Earth’s surface, where Earth has a mass of Msource. The only force acting on mtest is the gravitational force exerted by Earth on the test mass (Figure 4.29). r Msource mtest Fg Figure 4.29 The gravitational force exerted by Earth on test mass mtest info BIT During a roller coaster ride, riders may experience a 4g change in acceleration between the top and bottom of a loop. This dramatic change in acceleration causes the thrill and occasional dizziness experienced by riders. The magnitude of F weight or using Newton’s law of gravitation. g can be evaluated two ways: using the concept of Weight Newton’s law of gravitation Fg mtestg Fg GmtestMsource r 2 Since the value of Fg is the same no matter which equation you use, set both equations equal to each other. mtest g Gmtest Msource r 2 g GMsource r 2 So no matter where the test mass is located in the universe, you can calculate the magnitude of the gravitational field strength (or gravitational acceleration) at any distance from a celestial body if you know the mass of the celestial body Msource and the separation distance between the centre of the test mass and the celestial body r. Find out the relationship between gravitational field strength and the acceleration due to gravity by doing 4-3 Design a Lab. 216 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 217 4-3 Design a Lab 4-3 Design a Lab Comparing Gravitational Field Strength to Gravitational Acceleration The Question What is the relationship between gravitational field strength and the local value of the gravitational acceleration? Design and Conduct Your Investigation State a hypothesis. Then design an experiment. Identify the controlled, manipula |
ted, and responding variables. Review the procedure in 4-2 QuickLab on page 198. List the materials you will use, as well as a detailed procedure. Check the procedure with your teacher and then do the investigation. Analyze your data and form a conclusion. How well did your results agree with your hypothesis? How Is Gravitational Field Strength Related to Gravitational Acceleration? To determine how gravitational field strength is related to gravitational acceleration, use the definition of a newton, 1 N 1 kgm/s2. Then substitute kilogram-metres per second squared for newtons in the equation for gravitational field strength: 1 N kg 1 m kg s2 kg m 1 s2 Metres per second squared are the units of acceleration. So in terms of units, gravitational field strength and gravitational acceleration are equivalent (Figure 4.30). Figure 4.30 Jennifer Heil of Spruce Grove, Alberta, won the gold medal in the women’s freestyle skiing moguls in the 2006 Winter Olympics in Turin, Italy. The gravitational field strength at the surface of the Moon is about 1 6 that at Earth’s surface. How would Jennifer’s jump on Earth compare with one on the Moon? Chapter 4 Gravity extends throughout the universe. 217 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 218 Calculating the Gravitational Acceleration of an Object on Two Celestial Bodies Example 4.5 demonstrates how to calculate the gravitational acceleration at the equator on Earth’s surface and that on the surface of the Moon. These two values are then compared to find the ratio of gEarth to gMoon. To solve the problem requires using data from Table 4.1, which shows the mass and equatorial radius of the Sun, the Moon, and each planet in the solar system. Table 4.1 Masses and Radii for Celestial Bodies in the Solar System* Celestial Body Sun Mercury Venus Earth Earth’s Moon Mars Jupiter Saturn Uranus Neptune Mass (kg) 1.99 1030 3.30 1023 4.87 1024 5.97 1024 7.35 1022 6.42 1023 1.90 1027 5.69 1026 8.68 1025 1.02 1026 Equatorial Radius (m) 6.96 108 2.44 106 6.05 106 6.38 106 1.74 106 3.40 106 7.15 107 6.03 107 2.56 107 2.48 107 *Source: Jet Propulsion Laboratory, California Institute of Technology (See JPL link at www.pearsoned.ca/school/physicssource.) Example 4.5 (a) Calculate the magnitude of the gravitational acceleration of an object at the equator on the surface of Earth and the Moon (Figure 4.31). Refer to Table 4.1 above. (b) Determine the ratio of gEarth to gMoon. How different would your weight be on the Moon? gEarth ? gMoon ? Moon Earth Figure 4.31 Given mEarth mMoon 5.97 1024 kg 7.35 1022 kg rEarth rMoon 6.38 106 m 1.74 106 m Required (a) magnitude of gravitational acceleration at equator on Earth and the Moon (gEarth and gMoon) (b) ratio of gEarth to gMoon e WEB Globular clusters are groups of about 1 000 000 stars that are bound together by gravity. Find out who discovered the first cluster and how many have been identified so far. Research the approximate size and shape of a globular cluster and the forces involved in its formation. Summarize your findings. Begin your search at www.pearsoned.ca/school/ physicssource. 218 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 219 Analysis and Solution (a) Use the equation g to calculate the GMsource r2 magnitude of the gravitational field strength on each celestial body. The magnitude of the gravitational acceleration is numerically equal to the magnitude of the gravitational field strength. Earth gEarth GmEarth (rEarth)2 m N (5.97 1024 kg) 6.67 1011 2 g k (6.38 106 m)2 2 9.783 N/kg 9.783 m/s2 The Moon gMoon GmMoon (rMoon)2 m N (7.35 1022 kg) 6.67 1011 2 g k (1.74 106 m)2 2 1.619 N/kg 1.619 m/s2 (b) Calculate the ratio of gEarth to gMoon. m 9.783 s2 m 1.619 s2 gEarth gMoon 6.04 Paraphrase (a) The magnitude of the gravitational acceleration at the equator on the surface of Earth is 9.78 m/s2 and of the Moon is 1.62 m/s2. Practice Problems 1. A satellite orbits Earth at a distance of 3rEarth above Earth’s surface. Use the data from Table 4.1 on page 218. (a) How many Earth radii is the satellite from Earth’s centre? (b) What is the magnitude of the gravitational acceleration of the satellite? 2. An 80.0-kg astronaut is in orbit 3.20 104 km from Earth’s centre. Use the data from Table 4.1 on page 218. (a) Calculate the magnitude of the gravitational field strength at the location of the astronaut. (b) What would be the magnitude of the gravitational field strength if the astronaut is orbiting the Moon with the same separation distance? 3. The highest satellites orbit Earth at a distance of about 6.6rEarth from Earth’s centre. What would be the gravitational force on a 70-kg astronaut at this location? Answers 1. (a) 4rEarth (b) 6.11 101 m/s2 2. (a) 3.89 101 N/kg (b) 4.79 103 N/kg 3. 16 N [toward Earth’s centre] (b) The ratio of gEarth to gMoon is 6.04. So your weight would be about 6 times less on the surface of the Moon than on Earth. Calculating the Weight of an Object on Mars The equation g GMsource r 2 can be used with F mg to calculate the weight of an object on any celestial body. In Example 4.6, the weight of a student on Mars is calculated. This quantity is then compared with the student’s weight on Earth’s surface. An interesting application of the variation in the weight of an object involves the Mars rover (Figure 4.32). The rover had a mass of about 175 kg, but on the surface of Mars, the rover weighed about 2.5 times less than on Earth’s surface. The rover was designed to avoid inclines greater than 30 but gEarth, the rover could travel farther up an because gMars incline on Mars than on Earth using the same battery charge. Figure 4.32 In full sunlight, a 140-W battery enabled the Mars rover to travel about 100 m per day on level ground with an average speed of 1.0 cm/s between charges. Chapter 4 Gravity extends throughout the universe. 219 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 220 Example 4.6 (a) What is the mass of a 60.0-kg student on Mars and on Earth? (b) What is the student’s weight at the equator on the surface of Mars and of Earth (Figure 4.33). Use the data from Table 4.1 on page 218. Fg ? Mars Given (a) ms (b) mMars mEarth 60.0 kg 6.42 1023 kg 5.97 1024 kg rMars rEarth 3.40 106 m 6.38 106 m Fg ? Earth Required (a) mass on Mars and on Earth (m) gEarth) gMars and F (b) weight on Mars and on Earth (F Figure 4.33 Analysis and Solution (a) Mass is a scalar quantity and does not depend on location. So the (b) Use the equation g student’s mass will be the same on Mars as on Earth. GMsource r 2 gravitational field strength on Mars and on Earth. to calculate the magnitude of the Practice Problems Use the data from Table 4.1, page 218, to answer the following questions. 1. What would be the weight of a 22.0-kg dog at the equator on Saturn’s surface? 2. (a) Do you think your skeleton could support your weight on Jupiter? (b) Compared to Earth, how much stronger would your bones need to be? 3. (a) What is the magnitude of the gravitational field strength at the equator on Uranus’ surface? (b) Compared to Earth, how would your weight change on Uranus? Answers 1. 230 N [toward Saturn’s centre] 2. (a) no (b) 2.53 times 3. (a) 8.83 N/kg, (b) 0.903 FgEarth Mars gMars GmMars (rMars)2 2 m N (6.42 1023 kg) 6.67 1011 2 g k (3.40 106 m)2 3.704 N/kg Earth gEarth GmEarth (rEarth)2 2 m N (5.97 1024 kg) 6.67 1011 2 g k (6.38 106 m)2 9.783 N/kg Since the direction of F celestial body, use the scalar equation Fg magnitude of the weight. g will be toward the centre of each mg to find the Mars FgMars msgMars Earth FgEarth msgEarth N (60.0 kg)3.704 g k 222 N N (60.0 kg)9.783 g k 587 N 220 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 221 Paraphrase and Verify (a) The mass of the student would be 60.0 kg on both Mars and Earth. (b) The weight of the student on Mars would be 222 N [toward Mars’ centre] and on Earth 587 N [toward Earth’s centre]. So the student would weigh about 2.6 times more on Earth than Mars. Different Values of Gravitational Field Strength on Earth For a long time, people thought that the magnitude of the gravitational field strength was constant at any location on Earth’s surface. However, scientists discovered that the value of g depends on both latitude and altitude. Latitude is the angular distance north or south of the equator. Altitude is the elevation of the ground above sea level. Figure 4.34 shows how the magnitude of the gravitational field strength at sea level varies with latitude. Measured Magnitude of Gravitational Field Strength at Sea Level vs. Latitude info BIT The value 9.81 N/kg is an average of the magnitude of the gravitational field strength at different locations on Earth’s surface. 9.84 9.83 9.82 9.81 9.80 9.79 9.78 20° 60° 80° 40° Latitude Equator Figure 4.34 Gravitational field strength at sea level as a function of latitude. At what location on Earth’s surface would you weigh the least? The most? P ole s The value of g increases as you move toward either the North or South Pole, because Earth is not a perfect sphere. It is flatter at the poles and it bulges out slightly at the equator. In fact, Earth’s radius is 21 km greater at the equator than at the poles. So an object at the equator is farther away from Earth’s centre than if the object were at the North Pole. Since g , the farther an object is from Earth’s centre, the smaller 1 r 2 the value of g will be. Other factors affect the value of g at Earth’s surface. The materials that make up Earth’s crust are not uniformly distributed. Some materials, such as gold, are more dense than materials such as zinc. Earth’s rotation about its axis also reduces the measured value of g, but the closer an object is to the North or South Pole, the less effect Earth’s rotation has on g. Concept Check Leo weighs 638 N [down] in Calgary, Alberta. What are some problems with Leo saying he weighs 638 N [down] anywhere on Earth? What property of matter would he be more accurate to state? Chapter 4 Gravity extends throughout the universe |
. 221 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 222 Gravity (mGal 20.5 20. 20.0 20.0 0 . 5 0 . 5 2 2 AA Applications of the Variation in g in Geology The variation in the value of g on Earth is used to detect the presence of minerals and oil. Geophysicists and geologists use sensitive instruments, called gravimeters, to detect small variations in g when they search for new deposits of ore or oil. Gold and silver deposits increase the value of g, while deposits of oil and natural gas decrease g. Figure 4.35 is an example of a map that shows different measured values of g as lines, where each line represents a specific value of g. 19.0 19.1 19.2 19.3 19.4 19.5 19.6 19.7 19.8 19.9 20.0 20.1 20.2 20.3 20.4 20.5 20.6 20.7 20.8 20.9 21.0 21.1 21.2 21.3 21.4 21.5 21.6 21.7 21.8 21.9 22.0 22.5 23.0 23.5 mGal Figure 4.35 A map showing the location of sulphide deposits in northern New Brunswick (shown in black) true weight: gravitational force acting on an object that has mass a 0 v constant FN Fg True Weight vs. Apparent Weight mg to calculate the weight of an object So far, you used the equation F g at any location in the universe. The gravitational force that you calculate g, of an object. with this equation is really called the true weight, F Suppose a student is standing on a scale calibrated in newtons in an elevator (Figure 4.36). If the elevator is at rest or is moving at constant velocity, the scale reads 600 N. Fg 600 N 300 200 100 400 0 500 600 700 Figure 4.36 The elevator and student are either at rest or moving at constant velocity. Using the free-body diagram for the student (Figure 4.37), the equation for the net force on the student is F N F F FN F g net 0 F g 0 Fg 0 mg FN N Figure 4.37 The free-body diagram for the student in Figure 4.36 FN mg 222 Unit II Dynamics a FN Fg Figure 4.39 The free-body diagram for the student in Figure 4.38 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 223 So when F is equal to the magnitude of the student’s weight. 0 N on the student, the magnitude of the normal force net Now suppose the elevator is accelerating up uniformly (Figure 4.38). In this situation, the scale reads 750 N. Fg 600 N 300 200 100 400 0 500 600 700 Figure 4.38 The elevator and student are accelerating up uniformly. N net To understand why the reading on the scale is different, draw the freebody diagram for the student (Figure 4.39) and write the equation for the net force: F F F g ma F F g N ma F F N ma mg m(a g) g The equation for F N is valid whether the student is accelerating up or down. In Figure 4.38, the student feels heavier than usual because the scale is pushing up on him with a force greater than mg. If the elevator is accelerating down uniformly, the scale reads m(a g ) but this time a and g are 525 N (Figure 4.40). As before, F in the same direction. So FN is less than mg, and the student feels lighter than usual. N Fg 600 N 300 200 100 400 0 500 600 700 Figure 4.40 The elevator and student are accelerating down uniformly. Chapter 4 Gravity extends throughout the universe. 223 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 224 apparent weight: negative of the normal force acting on an object The quantity F N is called the apparent weight, w, of an object. For the situations shown in Figures 4.38 and 4.40 on page 223, the equation for the apparent weight of the student is e SIM Calculate the true weight, normal force, and apparent weight of a person during an elevator ride. Follow the eSim links at www.pearsoned.ca/school/ physicssource. 224 Unit II Dynamics w F N m(a g) m(g a ) Example 4.7 demonstrates how to calculate the true weight and apparent weight of an astronaut in a rocket during liftoff on Earth’s surface. Example 4.7 A 100.0-kg astronaut in a spacesuit is standing on a scale in a rocket (Figure 4.41). The acceleration of the rocket is 19.6 m/s2 [up]. Calculate her true weight and apparent weight during liftoff on Earth. The acceleration due to gravity on Earth’s surface is 9.81 m/s2 [down]. Given m 100.0 kg a 19.6 m/s2 [up] g 9.81 m/s2 [down] up down Required true weight and apparent weight during liftoff (F g and w) 0 3000 1000 2000 Figure 4.41 Analysis and Solution Draw a free-body diagram and a vector addition diagram for the astronaut (Figure 4.42). up down FN Fg FN Fnet Fg Use the equation F g mg to find the astronaut’s true weight. Figure 4.42 F g mg (100.0 kg)9.81 9.81 102 N m s2 The astronaut is not accelerating left or right. So in the horizontal direction, F 0 N. net 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 225 For the vertical direction, write an equation to find the net force on the astronaut. F F F net N g g N N FN Apply Newton’s second law. ma F F g ma FF FF ma Fg m (9.81 102 N) (100.0 kg)19.6 s2 m 9.81 102 N (100.0 kg)19.6 s2 2.94 103 N 2.94 103 N [up] F Use the equation w F N Practice Problems 1. In Example 4.7, draw the free-body diagram for the scale during liftoff. 2. Suppose the rocket in Example 4.7 has an acceleration of 19.6 m/s2 [down] while it is near Earth’s surface. What will be the astronaut’s apparent weight and true weight? Answers 1. See page 898. 2. 9.79 102 N [up], 9.81 102 N [down] N to find the astronaut’s apparent weight. w F N (2.94 103 N) 2.94 103 N Paraphrase During liftoff, the astronaut’s true weight is 9.81 102 N [down] and her apparent weight is 2.94 103 N [down]. In Example 4.8, an astronaut is accelerating in deep space, a location in which the gravitational force acting on an object is not measurable. So in deep space, F 0. g Example 4.8 Refer to Example 4.7 on pages 224 and 225. What is the magnitude of the astronaut’s true weight and apparent weight if the rocket is in deep space? The magnitude of the acceleration of the rocket is 19.6 m/s2. Given m 100.0 kg magnitude of a 19.6 m/s2 g 0 m/s2 Required magnitude of true weight and apparent weight in deep space (F g and w) Analysis and Solution In deep space, the mass of the astronaut is still 100.0 kg, but g is negligible. F So g mg 0 N The astronaut is not accelerating left or right. So in the horizontal direction, F 0 N. net Practice Problems 1. An 80.0-kg astronaut is standing on a scale in a rocket leaving the surface of the Moon. The acceleration of the rocket is 12.8 m/s2 [up]. On the Moon, g 1.62 N/kg [down]. Calculate the magnitude of the true weight and apparent weight of the astronaut (a) during liftoff, and (b) if the rocket has the same acceleration in deep space. 2. A 60.0-kg astronaut is standing on a scale in a rocket about to land on the surface of Mars. The rocket slows down at 11.1 m/s2 while approaching Mars. Use the data from Table 4.1 on page 218. Calculate the true weight and apparent weight of the astronaut (a) as the rocket lands, and (b) if the rocket is accelerating at 7.38 m/s2 [up] when leaving Mars. Chapter 4 Gravity extends throughout the universe. 225 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 226 For the vertical direction, write an equation to find the net force on the astronaut. Refer to the free-body diagram in Figure 4.42 on page 224. Answers 1. (a) (F g) 1.30 102 N [down], (w) 1.15 103 N [down] F net b) (F 2. (a) (F (b) (F g) 0 N, (w) 1.02 103 N [down] g) 2.22 102 N [down], (w) 8.88 102 N [down] g) 2.22 102 N [down], (w) 6.65 102 N [down] Apply Newton’s second law. F N FN ma ma m (100.0 kg)19.6 s2 1.96 103 N Use the equation w F N to find the astronaut’s apparent weight. w F N (1.96 103 N) 1.96 103 N Paraphrase In deep space, the astronaut’s true weight is zero and the magnitude of her apparent weight is 1.96 103 N. Free Fall Let’s revisit the elevator scenario on pages 222 and 223. Suppose the elevator cable breaks (Figure 4.43). Assuming that frictional forces are negligible, the elevator, student, and scale all fall toward Earth with an acceleration of g. The student is now in free fall, the condition in which the only force acting on an object is F g. free fall: situation in which the only force acting on an object that has mass is the gravitational force a g 0 0 6 0 0 7 0 Fg Fg 600 N 300 200 100 400 0 500 600 700 Figure 4.43 The elevator, student, and scale are in free fall. Figure 4.44 The free-body diagram for the student in Figure 4.43 226 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 227 To understand free fall, draw the free-body diagram for the student (Figure 4.44) and write the equation for the net force: F net F ma F g g ma mg a g So in free fall, a g and both the student and the scale are accelerating at g downward. In Figure 4.43 on page 226, the scale reads zero because 0. Since it no longer exerts a normal force on the student, so F 0, the student’s apparent weight is also zero. Sometimes an F object in free fall is described as being “weightless.” However, this description is incorrect. In free fall, F w 0 but F 0. N N N g Observe the motion of water in a cup while in free fall by doing 4-4 QuickLab. 4-4 QuickLab 4-4 QuickLab Water in Free Fall Problem What is the motion of water in a cup when the cup is dropped from several metres above Earth’s surface? CAUTION: Do this activity outside. Have someone steady the ladder and be careful when climbing it. Materials paper cup pointed pen or pencil water dishpan stepladder Procedure 1 Make two holes on opposite sides of the cup near the bottom using the pen or pencil. Cover the holes with your thumb and forefinger. Then fill the cup with water. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. 2 Hold the cup at shoulder height above a dishpan, and uncover the holes. Observe what happens to the water (Figure 4.45). Have a partner sketch the path the water takes. 3 Repeat step 1 but climb the ladder and drop the cup toward the dishpan from a height of several metres. Observe the motion of the water during the fall. Questions 1. Describe the path and motion of the water (a) when the cup was held stationary, and cup with two holes and filled with water Figure 4.45 (b) when the cup was dropped from the ladder. Give a reason for your observations. Chapte |
r 4 Gravity extends throughout the universe. 227 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 228 Weightlessness Videos transmitted from a space shuttle or the space station often show astronauts floating in their cabin (Figure 4.46). Are the astronauts weightless in space? The answer is no. Then why do they appear to be weightless? Since the shuttle is some distance above Earth, g is less than its 1 r 2 value at Earth’s surface, because g . While the shuttle orbits Earth at high speed in an almost circular path, Earth exerts a gravitational force on the shuttle and everything in it. So the shuttle is able to remain in orbit. If an astronaut were standing on a scale in the shuttle, the scale would read zero, because the shuttle and everything in it are in free fall. The astronaut would feel “weightless” because the gravitational force exerted by Earth pulls the shuttle and the astronaut toward Earth. Suppose an astronaut is in a rocket in deep space and the acceleration of the rocket is zero. The astronaut would experience no measurable gravitational forces from any celestial bodies, and the astronaut’s acceleration would be zero. In this situation, the astronaut would have a true weight of zero and an apparent weight of zero, a condition called true weightlessness. true weightlessness: situation in which w 0 for an object and F 0 on the object g Figure 4.46 At the altitude of the shuttle, the value of g is about 90% of its value at Earth’s surface. 228 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 229 4.3 Check and Reflect 4.3 Check and Reflect Knowledge 1. (a) What is the difference between true weight and apparent weight? (b) Describe a situation in which the true weight of an object is zero but its apparent weight is not zero. 2. A person orbiting Earth in a spacecraft has an apparent weight of zero. Explain if the person still experiences a gravitational force. 8. An astronaut in a rocket has an apparent weight of 1.35 103 N [down]. If the acceleration of the rocket is 14.7 m/s2 [up] near Earth’s surface, what is the astronaut’s true weight? The acceleration due to gravity on Earth’s surface is about 9.81 m/s2 [down]. 9. A 50-kg astronaut experiences an acceleration of 5.0g [up] during liftoff. (a) Draw a free-body diagram for the astronaut during liftoff. 3. List two factors that affect the magnitude (b) What is the astronaut’s true weight of the gravitational field strength at Earth’s surface. Applications 4. Is there a place in the universe where true weightlessness actually exists? 5. Calculate the gravitational field strength at the location of a 70-kg astronaut 2.0rEarth from Earth’s centre. Use the data from Table 4.1 on page 218. 6. Graph the equation g GmMoon r 2 using technology. Refer to Student References 5: Graphing Data on pp. 872–874. Plot g on the y-axis (range of 02.0 N/kg) and r on the x-axis (range of 1–5rMoon). Toggle through to read values of g corresponding to specific values of rMoon to answer these questions: (a) Describe the graph of g vs. rMoon. How is it similar to Figure 4.10 on page 201? (b) What is the value of g (i) on the surface? (ii) at 1 2 rMoon above the surface? (iii) at rMoon above the surface? the (c) At what distance is g 1 100 and apparent weight? 10. Calculate the acceleration of the elevator in Figures 4.38 and 4.40 on page 223. Extensions 11. Draw a flowchart to summarize the steps needed to find the apparent weight of an object. Refer to Student References 4: Using Graphic Organizers on page 869. 12. Research how geophysicists and geologists use gravitational field strength to locate minerals, oil, and natural gas in Canada. Prepare a half-page report on your findings. Begin your search at www.pearsoned.ca/school/physicssource. 13. Suppose you are wearing a spacesuit. Where could you walk faster, on Earth or on the Moon? Explain your answer. 14. Draw a concept map to identify and link the concepts needed to understand gravitational acceleration and gravitational field strength near a celestial body other than Earth. Refer to Student References 4: Using Graphic Organizers on page 869. Create and solve a problem to demonstrate your understanding of these concepts. gravitational field strength on the surface of the Moon? e TEST 7. At the top of Mount Robson in British Columbia, a 7.5-kg turkey weighs 73.6 N [down]. Calculate the magnitude of the gravitational field strength at this location. To check your understanding of gravitational field strength, true weight, apparent weight, and free fall, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 4 Gravity extends throughout the universe. 229 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 230 CHAPTER 4 SUMMARY Key Terms and Concepts gravitational force weight gravitational mass action-at-a-distance force field gravitational field gravitational field strength Key Equations True weight: F mg g Newton’s law of gravitation: Fg GmAmB r 2 torsion balance true weight apparent weight free fall true weightlessness Gravitational field strength (or gravitational acceleration): g GMsource r 2 Apparent weight: w F N Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. gravitational force mass is one of two types Gravitation involves weight is gravitational field strength equation equation other examples inertial strong nuclear using exerted by a using celestial body where direction acceleration standard masses and a and a G is value 6.67 1011 N•m2/kg2 Figure 4.47 230 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 231 CHAPTER 4 REVIEW Knowledge 1. (4.2) What is the significance of the term “universal” in Newton’s law of universal gravitation? 2. (4.1, 4.3) A brick placed on an equal arm balance requires 5.0 kg to just balance it. When the brick is hung from a spring scale, the scale reads 48 N. The balance, standard masses, spring scale, and brick are moved to a planet where the gravitational field strength is 2.0 times that on Earth. What will be the reading on the balance and on the spring scale in this new location? 3. (4.1, 4.3) How does gravitational field strength vary with the mass of a celestial body? Assume that the radius is constant. 4. (4.3) The gravitational field strength at Earth’s surface is about 9.81 N/kg [down]. What is the gravitational field strength exactly 1.6rEarth from Earth’s centre? Applications 5. A 1.0-kg object, initially at rest, is dropped toward Earth’s surface. It takes 2.26 s for the object to fall 25 m. Determine how long it takes a 2.0-kg object to fall this distance from rest on Jupiter. Use the data from Table 4.1 on page 218. 6. Describe the steps you would use to determine the distance from Earth’s centre where the gravitational force exerted by Earth on a spacecraft is balanced by the gravitational force exerted by the Moon. Assume that you know the distance from Earth’s centre to the centre of the Moon. Do not do the calculations. 7. Use the data from Table 4.1 on page 218. Calculate the true weight of a 60.0-kg astronaut on (a) the surface of Mars, and (b) the surface of Saturn. 8. Objects A and B experience a gravitational force of magnitude 2.5 108 N. Determine the magnitude of the gravitational force if the separation distance is halved, mA increases by 8 times, and mB is reduced to 1 4 of its original value. 9. Suppose a 65-kg astronaut on Mars is standing on a scale calibrated in newtons in an elevator. What will be the reading on the scale when the acceleration of the elevator is (a) zero? (b) 7.2 m/s2 [up]? (c) 3.6 m/s2 [down]? 10. A 50-kg rock in a Nahanni River canyon breaks loose from the edge of a cliff and falls 500 m into the water below. The average air resistance is 125 N. (a) What is the average acceleration of the rock? (b) How long does the rock take to reach the water? (c) What is the true weight of the rock? (d) Does the rock have an apparent weight? Explain. Extensions 11. Research why astronauts do exercises in space and why they have difficulty walking when they return to Earth. Write a short paragraph of your findings. Begin your search at www.pearsoned.ca/ school/physicssource. 12. Pilots in high-speed planes are subject to g forces. Healthy people can withstand up to 3–4 gs. Beyond that limit, the blood will pool in the lower half of the body and not reach the brain, causing the pilot to lose consciousness. Research how Dr. Wilbur Franks from Toronto found a solution to this problem. What connection does this problem have to human survival during a space flight? Begin your search at www.pearsoned.ca/school/physicssource. Consolidate Your Understanding 13. Write a paragraph summarizing Newton’s law of gravitation. Include a numerical example that illustrates the law. Show a detailed solution. Think About It Review your answers to the Think About It questions on page 195. How would you answer each question now? e TEST To check your understanding of gravitation concepts, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 4 Gravity extends throughout the universe. 231 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 232 UNIT II PROJECT Tire Design, Stopping Distance, and Vehicle Mass Scenario Imagine that you work for Alberta Infrastructure and Transportation. A level section of highway in the mountains has an abnormally large number of accidents. The surface of one lane of the highway is concrete and the other is asphalt. You are a member of a research team formed to determine the stopping distances in the summer for both wet and dry days for different types of tires and for different masses of vehicles. Your team is to prepare a written report on your findings for the Traffic Branch of the Ministry. Assume that the vehicles are travelling at the posted speed limit of 90 km/h when the brakes are applied and that the vehicles are not equipped with anti-lock braking systems (ABS). Assume tha |
t the reaction time of drivers before they apply the brakes is 1.8 s. Planning Form a team of three to five members. Summarize the question your group is researching. Make hypotheses about how tire tread, surface condition, and vehicle mass might affect stopping distance. Assign roles to different team members. Some examples are team leader, materials manager, liaison officer, record keeper, and safety officer. Brainstorm strategies for researching and reporting on the question and create a timeline. Research tire designs that are designed to work well on both wet and dry road surfaces. Use the Internet and consult local tire suppliers. Materials • a digital camera and a computer • force-measuring equipment • mass-measuring equipment • new and used tires having different treads and designs • vehicle brochures Procedure 1 Research the Internet and interview different tire suppliers to identify tires designed to work well at above-freezing temperatures on both wet and dry pavement. Assessing Results After completing the project, assess its success based on a rubric designed in class* that considers research strategies experiment techniques clarity and thoroughness of the written report effectiveness of the team’s presentation quality and fairness of the teamwork 2 Visit a local tire supplier and borrow new and used tires of different designs. Photograph the tires to record the tread designs. 3 Design and conduct an experiment to determine the coefficients of static and kinetic friction for the tires on wet and dry asphalt, and wet and dry concrete. Recall that you will need the mass of the tires and the local value of gravitational field strength. CAUTION: Take all your measurements on parking lots or sidewalks, and beware of traffic. Consult the local police department and ask for supervision while doing your experiments. 4 Research the masses of at least four small, medium, and large vehicles travelling highways that would use these tires. Determine the average mass of each class of vehicle. Sales brochures from vehicle dealers have mass information. 5 Determine the average deceleration on both wet and dry roadways for the vehicles of different mass equipped with these tires. Remember that some drivers may lock their brakes while braking. 6 Determine the stopping distance for the different vehicles under different conditions. Remember to consider driver reaction time. 7 Write a report of your findings. Use graphs and tables where appropriate. Thinking Further Write a three-paragraph addition to your team’s report hypothesizing how driver education, changing the posted speed limit, and requiring that vehicles be equipped with ABS brakes might affect the results. *Note: Your instructor will assess the project using a similar assessment rubric. 232 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 233 UNIT II SUMMARY Unit Concepts and Skills: Quick Reference Concepts CHAPTER 3 Force Net force Newton’s first law Newton’s second law Summary Forces can change velocity. Resources and Skill Building 3.1 The Nature of Force Force is a push or a pull on an object. Force is a vector quantity measured in newtons (1 N 1 kg•m/s2). Net force is the vector sum of two or more forces acting simultaneously on an object. A free-body diagram helps you write the net force acting on an object. 3-1 QuickLab 3-2 QuickLab Examples 3.1–3.4 Examples 3.2–3.4 3.2 Newton’s First Law of Motion Newton’s first law states that an object will continue being at rest or moving at constant speed in a straight line unless acted upon by a non-zero net force. 3-3 QuickLab 3.3 Newton’s Second Law of Motion Newton’s second law states that when a non-zero net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object. Newton’s third law 3.4 Newton’s Third Law of Motion Newton’s third law states that if object A exerts a force on object B, then B exerts a force on A that is equal in magnitude and opposite in direction. Types of friction Factors affecting friction Coefficients of static and kinetic friction 3.5 Friction Affects Motion Friction is a force that opposes either the motion of an object or the direction the object would be moving in if there were no friction. Static friction is present when an object is stationary but experiences an applied force. Kinetic friction is present when an object is moving. The magnitude of the force of friction acting on an object is directly proportional to the normal force on the object. The coefficients of friction are proportionality constants that relate the magnitude of the force of friction to the magnitude of the normal force. Temperature, moisture, and the smoothness or roughness of the contact surfaces, and the materials forming the contact surface are some factors that affect the value of the coefficients of friction. 3-5 Inquiry Lab Figures 3.33–3.35 3-6 Design a Lab Examples 3.5–3.11 3-7 QuickLab Figures 3.53–3.57, 3.63, 3.65 3-8 QuickLab Examples 3.12, 3.13 3-9 Design a Lab 3-10 QuickLab Figures 3.70, 3.74, 3.78–3.80 Examples 3.14–3.16 3-11 Inquiry Lab Table 3.4 Examples 3.17–3.20 CHAPTER 4 Gravity extends throughout the universe. Gravitational force 4.1 Gravitational Forces due to Earth Gravitational force is a fundamental force, and can be described as an action-at-a-distance force or as a field. Gravitational field strength Gravitational field strength is the ratio of gravitational force to mass at a specific location. The units of gravitational field strength are N/kg. 4-1 QuickLab 4-2 QuickLab Figure 4.10 Newton’s law of universal gravitation Calculating g anywhere in the universe Variations of g True weight, apparent weight, free fall, and weightlessness 4.2 Newton’s Law of Universal Gravitation Newton’s law of universal gravitation states that the gravitational force of attraction between any two masses is directly proportional to the product of the masses and inversely proportional to the square of the separation distance between the centres of both masses. 4.3 Relating Gravitational Field Strength to Gravitational Force Newton’s law of gravitation can be used to determine the magnitude of gravitational field strength anywhere in the universe. The magnitude of gravitational field strength at a location is numerically equal to the magnitude of gravitational acceleration. The value of g at Earth’s surface depends on latitude, altitude, the composition of Earth’s crust, and Earth’s rotation about its axis. The true weight of an object is equal to the gravitational force acting on the mass, and depends on location. Apparent weight is the negative of the normal force acting on an object. Free fall is the condition where the only force acting on an object is the gravitational force. True weightlessness is the condition in which w 0 for an object and F 0 on the object. g Figures 4.11–4.13, 4.16, 4.24 Examples 4.1–4.4 4-3 Design a Lab Examples 4.5–4.6 Figures 4.34, 4.35 Figures 4.36–4.40, 4.43, 4.44, 4.46 4-4 QuickLab Examples 4.7, 4.8 Unit II Dynamics 233 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 234 UNIT II REVIEW Vocabulary 1. Using your own words, define these terms, concepts, principles, or laws. action-at-a-distance force action force apparent weight coefficient of friction field free-body diagram free fall gravitational field strength gravitational force gravitational mass inertia inertial mass kinetic friction net force Newton’s first law Newton’s law of gravitation Newton’s second law Newton’s third law normal force reaction force static friction tension true weight Knowledge CHAPTER 3 2. An object experiences three forces: F 1 is 60 N [22.0], F 3 is 83 N [300]. 2 is 36 N [110], and F Explain, using words and diagrams, how to calculate the net force on the object. What is the net force? 3. An object experiences zero net force. Work with a partner to describe the possibilities for its motion. 4. A person with a plaster cast on an arm or leg experiences extra fatigue. Use Newton’s laws to explain to a classmate the reason for this fatigue. 5. Use inertia and Newton’s first law to explain how the spin cycle in a washing machine removes water from wet clothes. 234 Unit II Dynamics 6. A load is placed on a 1.5-kg cart. A force of 6.0 N [left] causes the cart and its load to have an acceleration of 3.0 m/s2 [left]. What is the inertial mass of the load? 7. What happens to the acceleration of an object if the mass is constant and the net force (a) quadruples? (b) is divided by 4? (c) becomes zero? 8. Two people, A and B, are pulling a wagon on a horizontal, frictionless surface with two ropes. Person A applies a force [50] on one rope. Person B applies a force of 25 N [345] on the other rope. If the net force on the wagon is 55.4 N [26], calculate the magnitude of person A’s applied force. A θ 1 50º x θ 2 345º B 9. A book is at rest on a table. The table is exerting an upward force on the book that is equal in magnitude to the downward force exerted by the book on the table. What law does this example illustrate? 10. A pencil exerts a force of 15 N [down] on a notebook. What is the reaction force? What object is exerting the reaction force? 11. Explain why the coefficients of static and kinetic friction are numerals without units. 12. Draw a free-body diagram for a stationary 5.0-kg block resting on a rough incline forming an angle of 30.0 with the horizontal. (a) Explain why the block is stationary. (b) Explain why a free-body diagram is helpful to describe the situation. 13. How does the ability of a car slowing down on wet asphalt compare to it slowing down on wet concrete? Use the data from Table 3.4 on page 183. 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 235 CHAPTER 4 26. A car is stopped at a stoplight facing due east. 14. Compare the acceleration due to gravity and the gravitational field strength at the top of a tal |
l skyscraper on Earth. 15. Consider the quantities gravitational force, mass, and gravitational field strength. Which of these quantities affects the inertia of an object? 16. Suppose an athlete were competing in the 2010 Winter Olympics in Vancouver and Whistler, British Columbia. Whistler has an elevation of 2182 m at the top and 652 m at the base. If the ski jumping and bobsled events are held near the top of the mountain rather than at the base, how might the results of these events be affected? 17. Two bags, each containing 10 oranges of equal mass, are hung 4 m apart. In a small group, describe two situations, one involving mass and the other involving separation distance, that would double the gravitational force exerted by one bag on the other. Explain your answer. 18. A student working on a satellite problem got an Ns2 m answer of 57.3 . What physical quantity was the student solving for? 19. Use an example to explain the meaning of the statement: “The gravitational force exerted by Mars on a space probe varies inversely as the square of the separation distance between the centre of Mars and the centre of the probe.” 20. Is an object in free fall weightless? Explain your reasoning. 21. Compare the gravitational force exerted by Earth (mass M) on two satellites (masses m and 2m) in orbit the same distance from Earth. 22. Compare the magnitude of Earth’s gravitational field strength at the equator and at the North Pole. Explain your answer to a classmate. 23. On Earth, how does the mass of an object affect the values of the quantities below? Explain your answers. (a) acceleration due to gravity (b) gravitational field strength Applications 24. Two horizontal forces act on a soccer player: 150 N [40.0] and 220 N [330]. Calculate the net force on the player. 25. Calculate the acceleration of a 1478-kg car if it experiences a net force of 3100 N [W]. When the light turns green, the car gradually speeds up from rest to the city speed limit, and cruises at the speed limit for a while. It then enters a highway on-ramp and gradually speeds up to the highway speed limit all the while heading due east. Sketch a free-body diagram for the car during each stage of its motion (five diagrams in total). 27. A net force of magnitude 8.0 N acts on a 4.0-kg object, causing the velocity of the object to change from 10 m/s [right] to 18 m/s [right]. For how long was the force applied? 28. Two people, on opposite banks, are towing a boat down a narrow river. Each person exerts a force of 65.0 N at an angle of 30.0 to the bank. A force of friction of magnitude 104 N acts on the boat. (a) Draw a free-body diagram showing the horizontal forces acting on the boat. (b) Calculate the net force on the boat. 29. A force acting on train A causes it to have an acceleration of magnitude 0.40 m/s2. Train A has six cars with a total mass of 3.0 105 kg, and a locomotive of mass 5.0 104 kg. Train B has a locomotive of the same mass as train A, and four cars with a total mass of 2.0 105 kg. If the same force acts on train B, what will be its acceleration? Ignore friction. 30. A submarine rescue chamber has a mass of 8.2 t and safely descends at a constant velocity of 10 cm/s [down]. If g 9.81 m/s2, what is the upward force exerted by the cable and water on the chamber? 31. A 240-kg motorcycle and 70-kg rider are travelling on a horizontal road. The air resistance acting on the rider-bike system is 1280 N [backward]. The road exerts a force of static friction on the bike of 1950 N [forward]. What is the acceleration of the system? 32. The velocity of a 0.25-kg model rocket changes from 15 m/s [up] to 40 m/s [up] in 0.60 s. Calculate the force that the escaping gas exerts on the rocket. 33. Two boxes, A and B, are in contact and initially stationary on a horizontal, frictionless surface. Box A has a mass of 60 kg and box B a mass of 90 kg. A force of 800 N [right] acts on box A causing it to push on box B. (a) What is the acceleration of both boxes? (b) What is the magnitude of the action-reaction forces between the boxes? Unit II Dynamics 235 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 236 34. A person exerts a force of 1.5 N [right] to pull a 2.0-kg block of glass at constant velocity along a horizontal surface on the Moon (gMoon kinetic friction for the glass on the surface? 1.62 m/s2). What is the coefficient of 35. Three oak blocks, mA 4.0 kg, mB 3.0 kg, are positioned next to each other on mC a dry, horizontal oak surface. Use the data from Table 3.4 on page 183. 6.0 kg, and (a) What horizontal force must be applied to accelerate the blocks at 1.4 m/s2 [forward], assuming the blocks are moving at a constant velocity? (b) Calculate the force exerted by mB on mC. (c) Calculate the force exerted by mB on mA. 38. Three objects, A, B, and C, are connected together by light strings that pass over light, frictionless pulleys. The coefficient of kinetic friction for object B on the surface is 0.200. (a) What is the acceleration of object B? (b) What is the tension in each string? Explain why the tensions are different. (c) Draw a free-body diagram for object B. (d) Identify four action-reaction pairs associated with object B. mB 2.0 kg B Fapp B A C A mA 6.0 kg C 4.0 kg mC 36. A 10.0-kg block is placed on an incline forming an angle of 30.0 with the horizontal. Calculate the acceleration of the block if the coefficient of kinetic friction for the block on the incline is 0.20. m 30.0° 37. A rehabilitation clinic has a device consisting of a light pulley attached to a wall support. A patient pulls with a force of magnitude 416 N. The rope exerts a force of friction of magnitude 20 N on the pulley. Calculate the acceleration of mA and mB. wall support shaft mA 15 kg mB 20 kg A B 236 Unit II Dynamics 39. The gravitational force on an object located 2rEarth from Earth’s centre is 200 N [toward Earth’s centre]. What is the gravitational force if the object is 10rEarth from Earth’s centre? 40. A 50-kg diver steps off a 9.0-m diving tower at the same time as a 100-kg diver. Work with a partner to compare the times taken for the two divers to reach the water. Ignore air resistance. 41. Skylab 1, the first American space station, had a mass of about 68 t. It was launched into orbit 435 km above Earth’s surface. Calculate the gravitational field strength at the location of Skylab 1 at this altitude. Use the data from Table 4.1 on page 218. 42. A spring scale is used to measure the gravitational force acting on a 4.00-kg silver block on Earth’s surface. If the block and spring scale are taken to the surface of Mars, by how much does the reading on the spring scale change? Use the data from Table 4.1 on page 218. 43. A 60-kg student is standing on a scale in an elevator on Earth. What will be the reading on the scale when the elevator is (a) (i) moving down at constant speed? (ii) at rest at a floor? (iii) accelerating at 4.9 m/s2 [up]? (iv) accelerating at 3.3 m/s2 [down]? (b) What is the student’s apparent weight and true weight in all the situations in part (a)? 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 237 44. A 60-kg skydiver falls toward Earth with an unopened parachute. The air resistance acting on the skydiver is 200 N [up]. What is the true weight and acceleration of the skydiver? 45. A group of tourists on a ledge overlooking Pulpit Rock, Northwest Territories, dislodge a 25-kg boulder. The rock takes 8.0 s to fall 300 m into the water below. At this location, the gravitational field strength is 9.81 N/kg [down]. (a) Calculate the average acceleration of the boulder. (b) Calculate the average air resistance acting on the boulder. (c) What was the average apparent weight of the boulder during its fall? Extensions 46. The value of g on the Moon is less than that on Earth. So a pendulum on the Moon swings slower than it would on Earth. Suppose a pendulum is 36 cm long. Use the equation T 2 l g to calculate the period, T, at the equator on Earth and on the Moon. Use the data from Table 4.1 on page 218. 47. During the last seconds of a hockey game, the losing team replaces their goalie with a good shooter. The other team shoots the 150-g puck with an initial speed of 7.0 m/s directly toward the unguarded net from a distance of 32 m. The coefficient of kinetic friction for the puck on the ice is 0.08. (a) What is the force of kinetic friction acting on the puck? (b) What is the acceleration of the puck? (c) How long does it take the puck to stop? (d) Will the puck reach the net if no other player touches it? 48. Construct a gathering grid to distinguish among Newton’s three laws. In the left column, identify the criteria you will use to compare and contrast the laws. Add three additional columns, one for each law. Then place checkmarks in the appropriate columns to compare the laws. 49. During the 2000 Sydney Olympics, some swimmers wore special swimsuits designed to reduce water resistance. Compare the arguments that people might make to defend or oppose the standardization of athletic equipment in the interests of fair play. 50. List two different stakeholders in the airbag debate and describe how their positions on the issue compare. 51. The G rocket of the former Soviet Union has a mass of about 3.8 106 kg and its first-stage engines exert a thrust of about 5.0 107 N [up]. (a) What is the true weight of the rocket on Earth’s surface? (b) Calculate the net force acting on the rocket at liftoff. (c) Calculate the initial acceleration of the rocket. (d) What should happen to the acceleration if the force exerted by the engines remains constant as the fuel burns? (e) Why is the first stage jettisoned after the fuel is consumed? 52. Suppose the mass of the person sitting next to you is 70 kg and the separation distance between you and that person is 1.0 m. The mass of Mars is 6.42 1023 kg and the separation distance between Mars and Earth is 2.3 1011 m. Compare the gravitational force exerted by Mars on you and the gravitational force exerted by the per |
son sitting next to you on you. 53. In a small group, research the materials being used to make artificial joints such as hips and knees. Find out how they are designed to provide enough friction for stability but not so much friction that the joints cannot move. Begin your search at www.pearsoned.ca/school/physicssource. 54. Manufacturers of skis recommend different waxes for different snow temperatures. Design and carry out an experiment to test the recommendations for three different waxes. 55. Research gait analysis is the study of how humans walk and run. This topic is central to physiotherapy, orthopedics, the design of artificial joints and sports footwear, and the manufacture of orthotics. How do Newton’s three laws apply to gait analysis? Interview people associated with rehabilitation and sports. Write a brief summary of your findings. Begin your search at www.pearsoned.ca/school/physicssource. e TEST To check your understanding of dynamics, follow the eTest links at www.pearsoned.ca/school/physicssource. Unit II Dynamics 237 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 238 U N I T III Circular Motion, Circular Motion, Work, and Work, and Energy Energy The International Space Station is a silent companion to Earth, placed into an orbit that is a precise balance of kinetic and gravitational potential energies. The International Space Station stays in orbit because physicists and engineers applied the laws of physics for circular motion and conservation of energy to determine the satellite’s speed and height above Earth. e WEB To learn more about the types of satellites placed in orbit, and the paths that they take, follow the links at www.pearsoned.ca/school/physicssource. 238 Unit III 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 239 Unit at a Glance C H A P T E R 5 Newton’s laws can explain circular motion. 5.1 Defining Circular Motion 5.2 Circular Motion and Newton’s Laws 5.3 Satellites and Celestial Bodies in Circular Motion C H A P T E R 6 In an isolated system, energy is transferred from one object to another whenever work is done. 6.1 Work and Energy 6.2 Mechanical Energy 6.3 Mechanical Energy in Isolated and Non-isolated Systems 6.4 Work and Power Unit Themes and Emphases • Energy, Equilibrium, and Systems • Nature of Science • Scientific Inquiry Focussing Questions This unit focusses on circular motion, work, and energy. You will investigate the conditions necessary to produce circular motion and examine some natural and human examples. You will consider energy, its transfer, and how it interacts with objects. As you study this unit, consider these questions: • What is necessary to maintain circular motion? • How does an understanding of conservation laws contribute to an understanding of the Universe? • How can mechanical energy be transferred and transformed? Unit Project Building a Persuader Apparatus • When you have finished this unit, you will understand how energy is transferred when objects interact. You will be able to use this understanding in the design and construction of a persuader apparatus that is able to protect its passenger from injury in different types of collisions. Unit III Circular Motion, Work, and Energy 239 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 240 Newton’s laws can explain circular motion. If humans hadn’t invented the wheel — the first circular motion machine — it’s hard to imagine what life would be like. The number of devices that make use of the properties of circular motion is almost too large to count. Bicycles, gears, drills, transmissions, clutches, cranes, watches, and electric motors are just a few examples. The wheel and the many technologies derived from it are a uniquely human creation (Figure 5.1). But the principles of circular motion have always existed in nature. In fact, as you read this, you are spinning around in a large circle as Earth rotates on its axis. In Alberta, your speed is approximately 1000 km/h because of this rotation. At the same time, you are flying through space at an amazing speed of approximately 107 000 km/h as Earth revolves around the Sun. And you thought your car was fast! How is circular motion unique? What are the properties of objects moving with circular motion? In this chapter, you will explore the physics of circular motion that define and control these many technologies, as well as the motion of the planets in the solar system. C H A P T E R 5 Key Concepts In this chapter, you will learn about: uniform circular motion planetary and satellite motion Kepler’s laws Learning Outcomes When you have completed this chapter, you will be able to: Knowledge describe uniform circular motion as a special case of two-dimensional motion explain centripetal acceleration and force explain, quantitatively, the relationships among speed, frequency, period, and radius for circular motion explain, qualitatively, uniform circular motion using Newton’s laws of motion explain, quantitatively, the motion of orbiting bodies by using circular motion to approximate elliptical orbits predict the mass of a celestial body from orbital data explain, qualitatively, how Kepler’s laws were used to develop Newton’s universal law of gravitation Science, Technology, and Society explain the process of scientific inquiry illustrate how science and technology are developed to meet society’s needs and expand human capabilities analyze circular motion in daily situations Figure 5.1 240 Unit III 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 241 5-1 QuickLab 5-1 QuickLab Characteristics of Circular Motion Problem In what direction will an object moving in a circular path go when released? Materials a marble a circular barrier (e.g., rubber tubing, embroidery hoop, flexible toy car tracks) Procedure 1 Place the circular barrier on an unobstructed section of a table or the floor, then place the marble against the inside rim of the barrier (Figure 5.2). 2 You will be rolling a marble around inside the barrier. Before you do, predict where you think the marble will go when you remove the barrier. Now roll the marble around the inside rim of the barrier. 3 As the marble is rolling around the rim, lift the barrier and make a note of the direction the marble rolls. Also pay attention to the position of the marble when you lift the barrier. 4 Sketch the circular path that the marble took inside the barrier, the position of the marble when you lifted the barrier, and the path that it rolled after the barrier was lifted. 5 Repeat steps 3 and 4 several times. Each time release the marble when it is in a different position. Figure 5.2 Questions 1. Was your prediction correct? Why or why not? 2. What similarities, if any, exist between your sketches? 3. What conclusions can you draw about the motion of the marble when it was released? 4. In each sketch that you made, draw a line from the centre of the circular motion to the position of the marble when it was released. What is the angle between this line and the path of the marble when it was released? Think About It 1. Can you predict the position the marble would have to be in for it to move away from your body when it was released? 2. What can you say about the direction of the velocity of the marble at any moment in its circular path? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 5 Newton’s laws can explain circular motion. 241 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 242 info BIT A boomerang has two axes of rotation: one for the spin given to the boomerang as it leaves the hand, and the other for the large circle in which it moves. axle axis of rotation Figure 5.3 The axle of a wheel is part of the axis of rotation. axle: shaft on which a wheel rotates axis of rotation: imaginary line that passes through the centre of rotation perpendicular to the circular motion uniform circular motion: motion in a circular path at a constant speed 5.1 Defining Circular Motion The bicycle is not the most sophisticated machine ever created, but it does have a lot going for it. It is easy to maintain; it does not require any fuel; it is a very efficient form of transportation; and it is full of parts moving in circular motion. Perhaps most importantly for this lesson, it is easy for us to examine. When you pedal a bike, the force is transferred through the chain and the wheel turns (Figure 5.3). The wheel transmits the force to the road, which, according to Newton’s third law, pushes back, and the bicycle moves forward. The wheel’s axle is referred to as the axis of rotation because the entire wheel rotates around this shaft. If the wheels of the bike are moving at a constant speed, they are moving with uniform circular motion. That means their rotational speed is uniform. Because the wheel is circular, it can also provide a constant and uniform force to the road. This is important if you want the bike to move forward at a uniform speed. In the sections that follow, we will restrict our study of circular motion to two dimensions. In other words, the circular motion described in this chapter has only one axis of rotation and the motion is in a plane. Speed and Velocity in Circular Motion If you ride through puddles after a rainfall, you’ll come home with muddy water splashed on your back. Why is this? It has to do with the properties of an object moving in circular motion. As the wheel passes through the puddle, some of the water adheres to it. As the wheel rotates upward out of the puddle, excess water flies off. When it flies off, it moves along a path that is tangential to the wheel. Recall from Unit I that a tangent is a line that touches the circle at only one point and is perpendicular to the radius of the circle. In the case of the bicycle, that is the point where the water drops break free of the wheel. Unless the splashguard is big enough, it will not protect you from all the w |
ater that is flying off the wheel (Figure 5.4). path of the water drops when they leave the wheel at point A radial line A A tangential line Figure 5.4 Water that flies off the wheel at point A hits the rider in the back. The direction that the water drops fly is determined by the place where they leave the wheel. Figure 5.5 Water leaves the wheel at point A and flies off at a tangent. 242 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 243 PHYSICS INSIGHT A line tangent to the circle represents the velocity vector. It is perpendicular to the radial line at that point. v A r Figure 5.6 The velocity vector v shows the velocity of a particle at any instant as it moves on a circular path. A line drawn from point A in Figure 5.5 to the centre of the circle is called a radial line. The radial line and the tangential line are perpendicular to each other when they intersect at the same point on the circle. The tangential line represents the direction that the water drops are moving at any instant. The speed of the water drops is determined by their speed at the wheel. It is worth taking some time to review the difference between speed and velocity. Speed is a magnitude only (a scalar quantity) that does not have a direction. If a wheel is rotating with uniform circular motion, then the speed is constant, even though the wheel is continually turning and the water on the wheel is continually changing direction. Velocity, on the other hand, has a magnitude and a direction: it is a vector quantity. Because of this, even though the wheel is spinning at constant speed, the velocity of the water on the wheel is continually changing as the wheel’s direction of motion changes. That is, the water on the wheel is continually accelerating. It is correct to say that the speed of the water drops on the wheel represents the magnitude of the velocity. Knowing that the object moves in a circular path is often sufficient. However, we must specify the object’s velocity if we need to know its speed and direction at any instant in its circular path. Since we can assume the speed and direction of the water drop are known at any instant, we know its velocity. A velocity vector can be drawn at position A, and the drawing can be simplified as shown in Figure 5.6. Concept Check 1. 2. Imagine a Frisbee in level flight. Where is its axis of rotation, and what is its orientation? Identify all the axes of rotation in a typical bicycle. Figure 5.7 A mountain bike has many axes of rotation. Centripetal Acceleration and Force Now imagine that you drive over a small pebble that gets lodged in the treads of your bicycle wheel. As you ride, the pebble circles around with the wheel, as shown in Figure 5.8. At one moment it is in position A, and a fraction of a second later, it is in position B. A short time has passed as the pebble moved from point A to point B. This small change in time is written as t. Chapter 5 Newton’s laws can explain circular motion. 243 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 244 The pebble has experienced a very small displacement from point A to point B, which is written as d . The velocities of the pebble at point A and point B are v B, respectively (Figure 5.9). C represents the centre of the circle (the axis of rotation). A and v vA B vB A d r r C vB –vA v C t B vB vA A Figure 5.8 A pebble caught in the wheel of a bike moves from position A to B in a small time t, and experiences a change in velocity. Figure 5.9 As the pebble moves from point A to point B, it moves through angle and experiences a change in velocity. Figure 5.10 The change in velocity, v, points inward. As t decreases, the angle between B will become smaller and vwill point A and v v toward the centre of the circle. The math to show this is beyond the scope of this book. The speed of the pebble does not change, but its direction does. Therefore, its velocity also changes. The change in velocity (vv) can best be shown by subtracting the velocity at A from the velocity at B (using the rules of graphical vector addition) as shown in Figure 5.10. Angle is the same in both Figures 5.9 and 5.10, and the two triangles are similar. Something subtle but significant happens when we subtract the two velocity vectors. The change in velocity, vv, points inward. As the interval of time, t, becomes smaller, vv begins to point toward the centre of the circle! This can be shown using calculus, but this is beyond the scope of this book. Velocity and Acceleration Toward the Centre The changing velocity (v ) represents the change in direction of the object, not its speed. If an object has a changing velocity, then it must be accelerating. Since the changing velocity is pointing inward toward the centre of the circle, the acceleration must also be in that direction (Figure 5.11). It is called centripetal acceleration (a c ). For an object to move with circular motion, it must experience centripetal acceleration. If the circular motion is uniform, then so is the centripetal acceleration. According to Newton’s second law, if a mass is accelerating, it must also experience a non-zero net force. This non-zero net force is c ). In our example, the pebble stuck in the called the centripetal force (F wheel is experiencing a force, exerted by the rubber treads, that attempts to pull it toward the centre of the circle. This is the centripetal force. Why doesn’t the pebble actually move toward the centre of the wheel? It does, in a way. Remember: if the pebble were to break free of the tire’s grip, it would fly off at a tangent to its circular motion. While it remains stuck in the tire, it is forced to follow the circular path because the centripetal force attempts to pull it toward the centre. In other words, centripetal force is pulling the pebble toward the centre of the circle while at the same time, the pebble is moving off in a direction at a tangent to the circle. The result is the circular path in which it actually moves. v ac Figure 5.11 Although the velocity of an object moving with uniform circular motion is tangential, the centripetal acceleration (and centripetal force) is acting toward the centre of the circle. centripetal acceleration: acceleration acting toward the centre of a circle; centre-seeking acceleration centripetal force: force acting toward the centre of a circle causing an object to move in a circular path 244 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 245 5-2 Inquiry Lab 5-2 Inquiry Lab Speed and Radius Question For an object moving with uniform circular motion, what relationship exists between the radius of its path and its speed? (Assume the object is experiencing a constant centripetal force.) Hypothesis State a hypothesis relating the radius and speed. Remember to use an “if/then” statement. Variables The variables in this lab are the radius of the circular path, mass of the rubber stopper, mass of the hanging weight, number of revolutions, elapsed time, period, and speed. Read the procedure and identify the controlled, manipulated, and responding variables. Materials and Equipment 1-hole rubber stopper (mass 25 g) 1.5 m of string or fishing line small-diameter plastic tube 100-g mass metre-stick felt marker safety goggles stopwatch CAUTION: Remember to swing the rubber stopper over your head and in a place clear of obstructions. Table 5.1 Data for 5-2 Inquiry Lab Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Procedure 1 Copy Table 5.1, shown at the bottom of this page, into your notebook. 2 Secure the rubber stopper to one end of the string. 3 Run the other end of the string through the plastic tube and attach the 100-g mass to it. 4 Hold the end of the string attached to the stopper at the zero mark of a metre-stick laid on a table. The zero mark of the ruler should line up with the centre of the stopper. While holding the stopper in position, pull the string taut along the ruler. 5 With the felt marker, mark the string at 20, 30, 40, 50, and 60 cm. 6 Adjust the string’s position so that the 20-cm mark is positioned on the lip of the plastic tube. Record 20 cm in the “Radius” column of the table. 7 Grasp the plastic tube in one hand and pinch the string to the lip of the tube using your thumb or forefinger. 8 Put on your safety goggles. Begin spinning the rubber stopper in a horizontal circle above your head as you release the string. Make sure the 100-g mass is hanging freely (Figure 5.12). At first, you may have to pull the string up or down using your other hand to position the mark as the stopper is spinning. Radius (m) Time for 20 Revolutions (s) Time 1 Time 2 Average Time (s) Period (s) Speed (m/s) Figure 5.12 Chapter 5 Newton’s laws can explain circular motion. 245 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 246 9 Adjust the rate at which you spin the rubber stopper so that the string does not slip up or down and the mark stays in position at the lip of the tube. Once you have reached a steady rate, your partner can begin timing. Do not pull the string up or down with the other hand. 10 Your partner should time 20 complete revolutions of the rubber stopper using a stopwatch. While your partner does this, be sure to monitor the speed of the stopper so that the mark does not move off the lip. Record the time in the “Time 1” column of the table. 2. For each trial, divide the average time by the number of revolutions the stopper made. Record the values for the time it takes to make one revolution in the “Period” column. 3. For each trial, determine the speed of the stopper. The distance travelled is the circumference of the circle and the time is the period. 2 r Use the equation v , and record the value in T the “Speed” column. 4. Identify the controlled, responding, and manipulated 11 Repeat step 10 and record the time in the “Time 2” variables. column of the table. 12 Increase the radius by 10 cm, |
and record this radius in the “Radius” column of the table. Repeat steps 7 to 12 until all radii are used. Analysis 1. For each trial, average the two times and place the result in the “Average Time” column. 5. Identify the force that acted as the centripetal force, and determine its value. 6. Plot a graph of velocity versus radius. Remember to plot the manipulated variable on the horizontal axis and the responding variable on the vertical axis. 7. Complete the statement, “The speed varies with …” 8. Was your hypothesis accurate? If not, how can you modify it to reflect your observations from this lab? motion of ball if released force exerted by hand on rope (action force) Figure 5.13 The hand exerts a centripetal force on the rope and ball, but feels the reaction force exerted by the rope. This leads to a false impression that the centripetal force is acting outward. force exerted by rope on hand (reaction force) Misconceptions About Centripetal Force A common misconception is that centripetal force acts radially outward from the centre of a circle. This is not what happens. The change in the velocity of the object is inward, and therefore, so is the centripetal acceleration and force. What causes confusion is that when you spin an object in a circle at the end of a rope, you feel the force pulling outward on your hand (Figure 5.13). This outward pull is mistakenly thought to be the centripetal force. Newton’s third law states that for every action there is an equal and opposite reaction. If we apply this law to our example, then the action force is the hand pulling on the rope to make the object move in a circle. This is the centripetal force. The reaction force is the force the rope exerts on your hand. It is outward. This is the force that people often believe is the centripetal force. The rope would not exert a force on your hand unless your hand exerted a force on the rope first. Another misconception is that centripetal force is a unique and separate force responsible for circular motion. This is not true. It is best to think of centripetal force as a generic term given to any force that acts toward the centre of the circular path. In fact, the translation from Latin of the word centripetal is “centre seeking.” 246 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 247 Many different forces could actually be the centripetal force. For example, when a car turns a corner, the frictional force of the tires on the road acts as the centripetal force. If you spin an object around in a horizontal circle on a rope, the tension of the rope is the centripetal force. The force of gravity the Sun exerts on the planets is another example of a centripetal force. Sometimes several forces working together act as a centripetal force. For an object spinning in a vertical circle on a rope, two forces, gravity and tension, work together to act as the centripetal force at the top of the circle. This is because the centripetal force is a net force. It is often convenient to use it in place of the actual force or forces acting toward the centre. Table 5.2 summarizes circular motion quantities and their directions. Concept Check A pebble caught in the tread of a tire experiences a centripetal force as the tire turns. What force is responsible for it? Table 5.2 Circular Motion Quantities and Their Direction Quantity Velocity (v) Centripetal acceleration (a) Centripetal force (F net) c or F Change in velocity (v) Direction tangential to the circle toward the centre toward the centre toward the centre 5.1 Check and Reflect 5.1 Check and Reflect PHYSICS INSIGHT Some texts refer to centrifugal force. This refers to the reaction force that exists as a result of centripetal force. If the centripetal force is removed, there is no centrifugal force. e SIM For an interactive demonstration that explores the relationship among centripetal acceleration, force, and velocity, visit www.pearsoned.ca/school/ physicssource. Knowledge 1. Give an example of an object that moves with uniform circular motion and one that moves with non-uniform circular motion. 2. Identify the force acting as the centripetal force in each of the following situations: (a) A car makes a turn without skidding. (b) A ball is tied to the end of a rope and spun in a horizontal circle. (c) The Moon moves in a circular orbit around Earth. Applications 3. What is the relationship between the speed and radius of an object moving with uniform circular motion if the centripetal force is constant? 4. What is the relationship between the speed and velocity of an object moving with uniform circular motion? Extensions 5. Imagine pedalling a bike at a constant rate. Describe the motion of the bicycle if the wheels were not circular but oval. 6. Consider the relative speed between a pebble stuck in the tread of a bicycle tire and the ground. Explain why the pebble is not dislodged when it comes in contact with the ground. Suggest a method to dislodge the pebble while still riding the bike. e TEST To check your understanding of the definition of circular motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 5 Newton’s laws can explain circular motion. 247 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 248 5.2 Circular Motion and Newton’s Laws info BIT A spinning ball moving relative to the air causes an unbalanced force and the curving of the ball. A backspin on a ball makes it stay aloft longer. A topspin drives the ball downward. Soccer players put a spin on the ball when they kick it that makes it curve around opposing players. Professional golfers routinely strike the golf ball so that it has a spin that curves it into the wind or prevents it from rolling when it hits the ground. A baseball pitcher can throw a curving fastball at 144 km/h, making the batter’s job of hitting the ball much more difficult (Figure 5.14). Figure 5.14 Pitchers can put a spin on a ball that causes it to curve. This curve is predictable. e WEB To learn more about the effect of putting a spin on a ball, follow the links at www.pearsoned.ca/school/ physicssource. In these sports and many others, putting a spin on the ball is an essential skill. Even though the pitch from a pitcher may be extremely difficult to predict, the behaviour of the ball isn’t. If players perform the same motion reliably when they kick, throw, or hit the ball, it will always behave the same way. That is why good players, when faced with the curving soccer ball or fastball, can anticipate where the ball will be and adjust their positions. It is accurate to say that the physical properties of a spinning ball or anything moving with circular motion can be predicted. In fact, the rotational velocity, frequency, centripetal force, and radius of a spinning object can be related mathematically. M I N D S O N Spinning Objects in Sports In groups of two or three, think of as many sports as you can that involve a spinning motion. It may be the player or an object that has the spin. Indicate what advantages the spinning motion has for the player and what type of motion is used to cause the spin. Discuss your answers with the class. 248 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 249 Period and Frequency of Circular Motion A baseball pitcher can throw a baseball at speeds of about 145 km/h. By flicking his wrist, he can give the ball a spin so that, in effect, it has two velocities: a velocity as it approaches the batter, and a rotational velocity because of its spin (Figure 5.15). The rotational velocity can be measured indirectly by measuring the time it takes for one complete rotation. One complete rotation is called a cycle or revolution, and the time for one cycle is the period (T), measured in s/cycle. (a) P (b) v (c) P v v P (d) v P cycle: one complete back-andforth motion or oscillation revolution: one complete cycle for an object moving in a circular path period: the time required for an object to make one complete oscillation (cycle) (e) P v Figure 5.15 Point P on the spinning baseball makes one complete rotation from (a) to (e). The time for this is called the period and is measured in s/cycle. This is frequently abbreviated to s for convenience. If an object is spinning quickly, the period may be a fraction of a second. For example, a hard drive in a computer makes one complete revolution in about 0.00833 s. This value is hard to grasp and is inconvenient to use. It is often easier to measure the number of rotations in a certain amount of time instead of the period. Frequency (f) is a measurement that indicates the number of cycles an object makes in a certain amount of time, usually one second. The SI units for frequency are cycles/s or hertz (Hz). You might have noticed that the units for frequency are cycles/s while the units for period are s/cycle. Each is the inverse of the other, so the relationship can be expressed mathematically as: T 1 f or f 1 T You may also have seen rotational frequency expressed in rpm. Even though this unit for measuring frequency is not an SI unit, it is commonly used commercially in products such as power tools. An rpm is a revolution per minute and is different from a hertz. It represents the number of revolutions in one minute instead of one second, so it is always 60 times bigger than the value in Hz. A simple method can be used to convert Hz to rpm and vice versa: Hz 60 s/min 60 s/min rpm frequency: the number of cycles per second measured in hertz (Hz) rpm: revolutions per minute Chapter 5 Newton’s laws can explain circular motion. 249 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 250 Example 5.1 The hard drive in Figure 5.16 stores data on a thin magnetic platter that spins at high speed. The platter makes one complete revolution in 0.00833 s. Determine its frequency in Hz and rpm. Practice Problems 1. The propeller of a toy airplane rotates at 300 rpm. What is its frequency in hertz? 2. An electri |
c motor rotates at a frequency of 40 Hz. What is its rotational frequency in rpm? 3. A medical centrifuge is a device that separates blood into its parts. The centrifuge can spin at up to 6.0 104 rpm. What is its frequency in hertz and what is its period? Answers 1. 5.00 Hz 2. 2.4 103 rpm 3. 1.0 103 Hz; 1.0 103 s Given T 0.00833 s Required frequency in Hz and rpm T 1 f f 1 0.00833 s 120 Hz hard drive case Now convert the SI units of frequency to rpm: Hz 60 s/min 60 s/min rpm Analysis and Solution The frequency is the inverse of the period. Solve the frequency of the hard drive in the SI unit for frequency (Hz) and then convert the Hz to rpm. hard drive platter direction of rotation read/write head Figure 5.16 120 Hz 60 s min 7.20 103 rpm Paraphrase The frequency of the hard drive is 120 Hz or 7.20 103 rpm. Speed and Circular Motion At the beginning of this chapter you learned that, at this moment, you are moving at approximately 107 000 km/h as Earth moves in its orbit around the Sun. It’s hard to imagine that Earth is moving that fast through our solar system. How was the speed determined? The answer is the simple application of an equation you learned in section 1.2 of chapter 1: v d t where d is the distance travelled and t is the time that it takes the object to travel that distance. 250 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 251 In the case of circular motion, the distance around a circle is the circumference (C), given by C 2r. The time it takes for one revolution is the period (T). Therefore, the speed of anything moving with uniform circular motion can be described by the equation: v 2r T (1) where r is the radius in metres, T is the period in seconds, and v is the speed in metres per second. Let’s look at Earth as it follows a circular orbit around the Sun. Earth has an orbital radius of approximately 1.49 108 km, and makes one complete revolution in 365.24 days. By substituting these values into the equation for speed, we can do the following calculation: PHYSICS INSIGHT Remember: The SI units for distance and time are metres and seconds, respectively. v 2r T 2(1.49 1011 m) 365.24 24 60 60 s 2.97 104 m/s Then convert this to kilometres per hour: m 36 m k 2.97 104 m s 1 0 0 10 1 s 00 1.07 105 km/h h Earth’s speed as it orbits the Sun is approximately 107 000 km/h. The speed of a planet as it rotates on its axis can be determined in the same way but varies depending on the latitude. We will explore the reasons for this later in “Centripetal Force, Acceleration, and Frequency” in section 5.2. Example 5.2 A pebble is stuck in the treads of a tire at a distance of 36.0 cm from the axle (Figure 5.17). It takes just 0.40 s for the wheel to make one revolution. What is the speed of the pebble at any instant? Given r 36.0 cm 0.360 m T 0.40 s Required speed (v ) of the pebble Analysis and Solution Determine the speed by using equation 1: r 2 v T 2(0.360 m) 0.40 s 5.7 m/s v 36.0 cm 36.0 cm 36.0 cm up left down right Figure 5.17 The speed of the pebble is determined by its distance from the axis of rotation, and the wheel’s period (T). Practice Problems 1. How much time does it take for the tires of a racecar to make one revolution if the car is travelling at 261.0 km/h and the wheels have a radius of 0.350 m? 2. In 2006, an Alberta astronomer discovered the fastest spinning collapsed star (called a pulsar) ever found. It has a radius of only 16.1 km and is spinning at a rate of 716 Hz (faster than a kitchen blender). What is its speed at its equator? Paraphrase The speed of the pebble caught in the tire tread is 5.7 m/s. Answers 1. 0.0303 s 2. 7.24 107 m/s Chapter 5 Newton’s laws can explain circular motion. 251 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 252 A Closer Look at Centripetal Acceleration and Force In Unit II, you learned that, just before a race, dragster drivers spin their tires to make them hot and sticky so that the coefficient of friction increases between the tire and the road. When the race starts, the tires will have a better grip and the dragster will be able to accelerate at a greater rate. While the dragster performs the tire-spin, the tires change shape (Figure 5.18). The rear tires start off being fat and thick. During the spin they become thin and their diameter increases. Clearly, this must have something to do with the spinning motion of the wheel, and therefore centripetal force, but what? Figure 5.18 (a) At first, the dragster’s wheels have a low rotational speed and don’t stretch noticeably. The centripetal acceleration and force are small. (b) The dragster’s wheels are spinning very fast and stretch away from the rim. The centripetal acceleration and force are large. Before the tires start spinning, they are in their natural shape. When they are spinning, they experience a strong centripetal acceleration and force. Both act toward the centre of the wheel. Each tire is fastened to a rim, so the rim is pulling the tire inward. However, the tire moves the way Newton’s first law predicts it will — it attempts to keep moving in a straight line in the direction of the velocity. Thus the tire, being made of rubber, stretches. Dragster tires and the tires of trucks and passenger cars are designed to endure high speeds without being torn apart. The faster a wheel spins, the greater the centripetal acceleration and force. Car tires are thoroughly tested to ensure they can handle a centripetal force much greater than the centripetal force at the speed that you would normally drive in the city or on the highway. However, if you drove at a speed that exceeded the tires’ capabilities, the tires could be torn apart. Is speed the only factor that affects centripetal acceleration and force, or are there other factors? Are the factors that affect centripetal acceleration the same as those that affect the centripetal force? To answer these questions, let’s start by taking a closer look at centripetal acceleration. 252 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 253 Factors Affecting the Magnitude of Centripetal Acceleration The rotational speed is one factor that determines the magnitude of the centripetal acceleration, but what other factors play a role? To answer this question, look at Figure 5.19. It shows the two diagrams you saw earlier as Figures 5.9 and 5.10 in section 5.1. Using these figures, we can derive an equation for centripetal acceleration. vA B vB A d r r C vB –vA v C Figure 5.19 An object following in a circular path moves through a displacement as there is a change in velocity v. The triangles formed by these two vectors will d help us solve for centripetal acceleration. As already stated, the two triangles are similar. Therefore, we can compare them. For convenience, the triangles have been redrawn below without the circles (Figure 5.20). Since vA and vB have the same value, we have dropped the designations A and B in Figure 5.20. We have omitted the vector arrows because we are solving only for the magnitude of the centripetal acceleration Figure 5.20 Triangles ABC and DEF are similar, so we can use a ratio of similar sides. Triangle ABC is similar to triangle DEF in Figure 5.20. Therefore, a ratio of similar sides can be created: d v r v or v d v r (2) Chapter 5 Newton’s laws can explain circular motion. 253 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 254 Remember, v is directed toward the centre of the circle. The time it took for the velocity vectors to move the small distance d can be designated t. In this time, the v vector was created. d , we can manipulate the equation so that: Since v t d t v (3) To find acceleration toward the centre of the circle (centripetal acceleration), divide the v by t. v t ac (4) Now we can substitute equations 2 and 3 into equation 4: v d r d v ac By taking the reciprocal of the denominator and multiplying the two fractions, we have: d v v d r ac Simplified, this becomes: ac 2 v r (5) where ac is the centripetal acceleration in metres per second squared toward the centre of the circle, v is the rotational speed of the object moving with uniform circular motion in metres per second, and r is the radius of the circular motion in metres. The centripetal acceleration depends only on the speed and radius of the circular motion. Mass does not affect it, just as the mass of a falling object does not affect the acceleration of gravity caused by Earth. A truck or a marble will both experience a gravitational acceleration of 9.81 m/s2 [down]. Two objects of different masses moving in a circular path will experience the same centripetal acceleration if they have the same radius and speed. Mass does not affect centripetal acceleration, but companies that manufacture racing tires, jet engines, and other equipment know they cannot ignore it. In fact, the mass of a tire or fan blade is important to them. These companies are continually looking for ways to reduce mass without decreasing the strength of their products. Why? The answer has to do with the centripetal force that these devices experience. If a fan blade or tire has a large mass, it will experience a large centripetal force and might break apart at high speeds. Reducing the mass decreases the centripetal force these parts experience, but often with a trade-off in strength. Next, we will examine the factors that influence centripetal force. 254 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 255 Example 5.3 A DVD disc has a diameter of 12.0 cm and a rotational period of 0.100 s (Figure 5.21). Determine the centripetal acceleration at the outer edge of the disc. Given D 12.0 cm 0.120 m T 0.100 s Required centripetal acceleration (ac) Analysis and Solution The magnitude of the centripetal acceleration depends on speed and radius. Convert the diameter to a radius by dividing it by 2. D 12.0 cm T 0.100 s Figure 5.21 D r 2 0 m 0.12 2 0.0600 m Determine the speed of the outer edge |
of the disc: r 2 v T . 0 2( m) 00 6 0 0 s 0 1 0 . 3.77 m/s Now use equation 5 to determine the centripetal acceleration: ac 2 v r m 2 3.77 s 0.0600 m 2.37 102 m/s2 Note that no vector arrows appear on ac or v because we are solving for their magnitude only. Paraphrase The centripetal acceleration at the edge of the DVD disc is 2.37 102 m/s2. Practice Problems 1. You throw a Frisbee to your friend. The Frisbee has a diameter of 28.0 cm and makes one turn in 0.110 s. What is the centripetal acceleration at its outer edge? 2. A child playing with a top spins it so that it has a centripetal acceleration of 125.0 m/s2 at the edge, a distance of 3.00 cm from the axis of rotation. What is the speed at the edge of the top? 3. A helicopter blade has a diameter of 14.0 m and a centripetal acceleration at the tip of 2527.0 m/s2. What is the period of the helicopter blade? Answers 1. 4.57 102 m/s2 2. 1.94 m/s 3. 0.331 s Chapter 5 Newton’s laws can explain circular motion. 255 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 256 e TECH For an interesting interactive simulation of the relationship between velocity and centripetal force, follow the links at www.pearsoned.ca/physicssource. e MATH To graph the relationship between centripetal force and speed and determine the mass of the object from the graph, visit www.pearsoned.ca/ physicssource. Factors Affecting the Magnitude of Centripetal Force Mass does not affect centripetal acceleration, but it does influence the force needed to move an object in a circular path. From Newton’s second law, we know that the net force is the product of mass and the net acceleration. Therefore, centripetal force must simply be the product of the mass and centripetal acceleration: Fnet ma mac Fc or Fc v 2 m r (6) where m is the mass in kilograms, v is the rotational speed in metres per second of the object moving with uniform circular motion, and r is the radius of the circular motion in metres. Notice that v and r are the same as in equation 5. Therefore, all the factors that affect centripetal acceleration also affect the centripetal force; namely, speed and the radius of rotation. However, the mass affects only the centripetal force. Example 5.4 Determine the magnitude of the centripetal force exerted by the rim of a dragster’s wheel on a 45.0-kg tire. The tire has a 0.480-m radius and is rotating at a speed of 30.0 m/s (Figure 5.22). Practice Problems 1. An intake fan blade on a jet engine has a mass of 7.50 kg. As it spins, the middle of the blade has a speed of 365.9 m/s and is a distance of 73.7 cm from the axis of rotation. What is the centripetal force on the blade? 2. A 0.0021-kg pebble is stuck in the treads of a dirt bike’s wheel. The radius of the wheel is 23.0 cm and the pebble experiences a centripetal force with a magnitude of 0.660 N. What is the speed of the wheel? Answers 1. 1.36 106 N 2. 8.5 m/s Given m 45.0 kg r 0.480 m v 30.0 m/s Required centripetal force exerted on the tire by the rim (Fc) Analysis and Solution Use equation 6 to solve for the centripetal force. Fc v 2 m r m 2 (45.0 kg)30.0 s 0.480 m 8.44 104 N v 30.0 m/s Fc r 0.480 m Figure 5.22 The dragster wheel experiences a centripetal force pulling it inward. Paraphrase The magnitude of the centripetal force exerted on the tire by the rim is 8.44 104 N. 256 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 257 5-3 Inquiry Lab 5-3 Inquiry Lab Speed and Centripetal Force Question What is the relationship between centripetal force and the speed of a mass moving in a horizontal circle? Hypothesis State a hypothesis relating centripetal force and speed. Remember to use an “if/then” statement. Variables The variables in this lab are the radius of the circular path, mass of the rubber stopper, mass of the hanging mass, number of revolutions, elapsed time, period, and speed. Read the procedure and identify the controlled, manipulated, and responding variables. Materials and Equipment 1-hole rubber stopper (mass 25 g) 1.5 m of string or fishing line 0.5-cm-diameter plastic tube 5 masses: 50 g, 100 g, 150 g, 200 g, 250 g metre-stick felt marker safety goggles stopwatch CAUTION: Remember to swing the rubber stopper over your head and in a place clear of obstructions. Table 5.3 Data for 5-3 Inquiry Lab Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Procedure 1 Copy Table 5.3, at the bottom of this page, into your notebook. 2 Secure the rubber stopper to one end of the string. 3 Run the other end of the string through the plastic tube and attach the 50-g mass to it. Record this mass in the “Mass” column of the table. The force of gravity exerted on this mass is the centripetal force. 4 Hold the end of the string attached to the stopper at the zero mark of a metre-stick laid on a table. The zero mark of the ruler should line up with the centre of the stopper. While holding the stopper in position, pull the string taut along the ruler. 5 With the felt marker, mark the string at 40 cm. 6 Hold the rubber stopper in one hand and the plastic tube in the other. Adjust the string’s position so that the 40-cm mark is positioned on the lip of the plastic tube. With the hand holding the plastic tube, pinch the string to the lip of the tube using your thumb or forefinger. 7 Put on your safety goggles. Begin spinning the rubber stopper in a horizontal circle above your head as you release the string (Figure 5.23). Make sure the mass is hanging freely. At first, you may have to pull the string up or down using your other hand to position the mark as the stopper is spinning. 8 Adjust the rate at which you spin the rubber stopper so that the string does not slip up or down and the mark stays in position at the lip at the top of the tube. Once you have reached a steady rate, your partner can begin timing. Do not pull the string up or down with the other hand. Mass (kg) Time for 20 Revolutions (s) Time 1 Time 2 Average Time (s) Period (s) Speed (m/s) Centripetal Force (101N) Figure 5.23 Chapter 5 Newton’s laws can explain circular motion. 257 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 258 9 Your partner should time 20 complete revolutions of the rubber stopper using a stopwatch. While your partner does this, be sure to monitor the speed of the stopper so that the mark does not move off the lip. Record the time in the “Time 1” column of Table 5.3. 10 Repeat step 9 and record the time in the “Time 2” column of Table 5.3. 11 Increase the hanging mass by 50 g, and record this mass in the “Mass” column of the table. Repeat steps 6 to 10 until all the masses are used. Analysis 1. For each trial, average the two times, and place the result in the “Average Time” column of the table. 2. For each trial, divide the average time by the number of revolutions the stopper made. Record the value in the “Period” column of the table. 3. For each trial, determine the speed of the stopper 2 r using the equation v . Record the value in the T “Speed” column of the table. 4. For each trial determine the force of gravity acting on the mass hanging from the string using the equation Fg mg. Record these values in the “Centripetal Force” column of the table. 5. Identify the controlled, responding, and manipulated variables. 6. Plot one graph of speed versus centripetal force and a second graph of the square of the speed versus centripetal force. Remember to plot the manipulated variable on the horizontal axis and the responding variable on the vertical axis. 7. Complete the statement, “The speed varies with ….” 8. Was your hypothesis accurate? If not, how can it be modified to reflect your observations in this lab? inward outward Ff centre of curvature Figure 5.24 A car turning left. The force of friction of the tires on the road is the centripetal force. PHYSICS INSIGHT On a horizontal surface the force of gravity (Fg) is equal and opposite to the normal force (FN). A Horizontal System in Circular Motion Imagine that a car is following a curve to the left on a flat road (Figure 5.24). As the car makes the turn its speed remains constant, and it experiences a centripetal force. The centripetal force is caused by the wheels turning to the left, continually changing the direction of the car. If it weren’t for the frictional force between the tire and the road, you would not be able to make a turn. Hence, the frictional force between the tires and the road is the centripetal force. For simplicity, this is written as: Fc Ff Recall that the magnitude of the force of friction is repre FN, where FN is the normal force, sented by the equation Ff or perpendicular force exerted by the surface on the object. For an object on a horizontal surface, the normal force is equal and opposite to the force of gravity. Recall from Unit II that the coefficient of friction, , is the magnitude of the force of friction divided by the magnitude of the normal force. Think of it as a measure of how well two surfaces slide over each other. The lower the value, the easier the surfaces move over one another. Assume that the driver increases the speed as the car turns the corner. As a result, the centripetal force also increases (Figure 5.25). Suppose the force of friction cannot hold the tires to the road. In other words, the force of friction cannot exert the needed centripetal force because of the increase in speed. In that case, the car skids off the road tangentially. Recall that kinetic friction is present when a car slides with its wheels locked. If a car turns a corner without skidding, static friction is present. 258 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 259 outward inward up down Ff FN Fg F f The maximum Figure 5.25 F c frictional force that can be exerted between the road and the tires determines the maximum speed the car can go around the turn without skidding. For horizontal surfaces, the normal force is equal and opposite to the force of |
gravity (F N F g). Example 5.5 Determine the maximum speed at which a 1500.0-kg car can round a curve that has the radius of 40.0 m, if the coefficient of static friction between the tires and the road is 0.60. Given m 1500.0 kg 0.60 s r 40.0 m g 9.81 m/s2 Required maximum speed (v ) Analysis and Solution First draw a free-body diagram to show the direction of the forces (Figure 5.26). The normal force is equal to the force of gravity on a horizontal surface. Fc Ff v 2 m FN r FN mg v 2 m (mg) r up outward inward FN down Ff m 1500.0 kg Fg Figure 5.26 v 2 gr v gr (0.60)9.81(40.0 m) m s2 15 m/s Paraphrase The fastest that the car can round the curve is 15 m/s or 55 km/h. If it attempts to go faster, the force of static friction will be insufficient to prevent skidding. Practice Problems 1. An Edmonton Oiler (m 100 kg) carves a turn with a radius of 7.17 m while skating and feels his skates begin to slip on the ice. What is his speed if the coefficient of static friction between the skates and the ice is 0.80? 2. Automotive manufacturers test the handling ability of a new car design by driving a prototype on a test track in a large circle (r 100 m) at ever-increasing speeds until the car begins to skid. A prototype car (m 1200 kg) is tested and found to skid at a speed of 95.0 km/h. What is the coefficient of static friction between the car tires and the track? 3. A 600.0-g toy radio-controlled car can make a turn at a speed of 3.0 m/s on the kitchen floor where the coefficient of static friction is 0.90. What is the radius of its turn? Answers 1. 7.5 m/s 2. 0.710 3. 1.0 m Chapter 5 Newton’s laws can explain circular motion. 259 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 260 Centripetal Force and Gravity The designers of amusement park rides know their physics. Rides will toss you around, but leave you unharmed. Many of these rides spin you in circles — the Ferris wheel and the roller coaster, for example. The roller coaster often has a vertical loop somewhere along its track (Figure 5.27). Why is it that when you reach the top of the loop, where the car is inverted, you don’t fall out? It isn’t because of the harness that they put over you before you start the ride. That just keeps you strapped into the car so you don’t do something silly like stand up while the roller coaster is moving. No, the answer lies in the physics of the roller coaster’s design. A Vertical System in Circular Motion All roller coasters, regardless of their appearance, are designed so that each car has sufficient velocity to remain in contact with the track at the top of the loop. At the top, the centripetal force is exerted by two forces working in the same direction: the track on the car, which is the normal force (F g). Both forces push the car toward the centre of the loop. N) and the force of gravity (F The speed of the car determines the amount of centripetal force needed to maintain a certain radius. As you saw from equation 6, the centripetal force is directly related to the square of the speed. Here is equation 6 again: Fc mv 2 r (6) But the force of gravity is independent of speed, and will always pull the car downward with the same force. To demonstrate the role that gravity plays as a portion of the centripetal force, we can look at a hypothetical situation of a roller coaster going around a loop as shown in Figures 5.28(a), (b), and (c). Assume a roller coaster car like the one in Figure 5.28 on the opposite page experiences a force of gravity that is 1000 N. This value won’t change regardless of the car’s position on the track or its speed. Remember, the centripetal force is the net force. In this case, it is equal to the sum of the gravitational force (F g) and the track’s force (F Figure 5.28 illustrates how speed affects the roller coaster car’s motion when it is sent through the loop three times. Each time, it is sent with less speed. N) on the car. Figure 5.27 Why doesn’t this car fall off the track at the top of the loop? info BIT Most roller coaster loops use a shape called a clothoid, where the curvature increases at the top. This reduces the speed needed to move safely through the loop. It also adds thrills by having relatively longer vertical parts in the loop. PHYSICS INSIGHT centripetal force F c force due to gravity F g force that the track F exerts on the car N 260 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 261 FN Fg Fc Fg FN 1500 N 1000 N 500 N Fg Fc Fg FN 1000 N 1000 N 0 N Fc Fg FN 800 N 1000 N 200 N Since the normal force cannot pull upward, it cannot generate 200 N. 200 N more is needed to keep the car on a track of this radius, so the car falls off. The first time through the loop, the speed is such that the roller Figure 5.28(a) coaster requires a centripetal force of 1500 N to keep it moving in a circular path. At the top of the loop, the roller coaster car will experience a centripetal force that is the sum of the force of gravity and the force exerted by the track, pushing the car inward to the centre of the circle. The centripetal force acts down, so it is 1500 N. The force of gravity is constant at 1000 N so the track pushes inward with 500 N to produce the required centripetal force. The car goes around the loop with no problem. Figure 5.28(b) Suppose the next time the car goes around the track, it is moving more slowly, so that the centripetal force required is only 1000 N. In this case, the force of gravity alone can provide the required centripetal force. Therefore, the track does not need to exert any force on the car to keep it moving on the track. There is no normal force, so the force of gravity alone is the centripetal force. The car goes around the loop again with no problem. Figure 5.28(c) Now suppose the last time the car goes around the track, it is moving very slowly. The required centripetal force is just 800 N, but the force of gravity is constant, so it is still 1000 N; that is, 200 N more than the centripetal force required to keep the car moving in a circular path with this radius. If the track could somehow pull upward by 200 N to balance the force of gravity, the car would stay on the track. This is something it can’t do in our hypothetical case. Since the gravitational force cannot be balanced by the track’s force, it pulls the car downward off the track. The slowest that any car can go around the track would be at a speed that requires a centripetal force that has a magnitude equal to gravity. Gravity would make up all of the centripetal force. This can be expressed as: Fc Fg Remember that this equality doesn’t mean that the centripetal force is a different force than the force of gravity. It means that it is the force of gravity. All roller coasters are designed so that the cars’ speed is enough to create a centripetal force greater than the force of gravity to minimize the chance of the car leaving the track. The wheels of roller coaster cars also wrap around both sides of the track so the track can indeed pull upwards. Concept Check The centripetal force exerted on the Moon as it orbits Earth is caused by Earth’s gravity. What would happen to the Moon’s orbit if the Moon’s velocity increased or slowed down? Chapter 5 Newton’s laws can explain circular motion. 261 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 262 Example 5.6 A 700.0-kg roller coaster car full of people goes around a vertical loop that has a diameter of 50.0 m (Figure 5.29). What minimum speed must the roller coaster car have at the top of the vertical loop to stay on the track? up right left down Fg D 50.0 m Figure 5.29 When the roller coaster is moving with the minimum speed to maintain its circular path, the force of gravity alone is the centripetal force. The track exerts no force on the car. Analysis and Solution For the roller coaster to stay on the track at the top of the loop with the minimum speed, the centripetal force is the force of gravity. Practice Problems 1. Neglecting friction, what is the minimum speed a toy car must have to go around a vertical loop of radius 15.0 cm without falling off? 2. What is the maximum radius a roller coaster loop can be if a car with a speed of 20.0 m/s is to go around safely? Answers 1. 1.21 m/s 2. 40.8 m e LAB For a probeware activity that investigates circular motion in a vertical plane, follow the links at www.pearsoned.ca/physicssource. PHYSICS INSIGHT All objects fall at a rate of 9.81 m/s2 regardless of their mass. To determine the radius of the loop, divide the diameter by 2. 0 m 50. r 2 25.0 m Use the equality of the centripetal force and gravity to solve for the speed: Fnet Fc Fg Fg m v 2 mg r 2 v g r v rg 25.0 m9.81 15.7 m/s m s2 The roller coaster car must have a minimum speed of 15.7 m/s to stay on the track. You can swing a full pail of water around in a circle over your head without getting wet for the same reason that a roller coaster can go around a loop without falling off the track. Let’s examine the case of a mass on the end of a rope, moving in a vertical circle, and see how it compares to the roller coaster. A bucket of water is tied to the end of a rope and spun in a vertical circle. It has sufficient velocity to keep it moving in a circular path. Figures 5.30(a), (b), and (c) show the bucket in three different positions as it moves around in a vertical circle. 262 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 263 (a) (b) Fg (c) F T Fg Fg F c F g F T e SIM To learn more about centripetal force for vertical circular motion, visit www.pearsoned.ca/physicssource. Figure 5.30(a) The bucket is at the top of the circle. In this position, two forces are acting on the bucket: the force of gravity and the tension of the rope. Both are producing the centripetal force and are acting downward. The equation to represent this situation is: F F F c g T Figure 5.30(b) When the bucket has moved to the position where the rope is parallel to the ground, the force of gravity is perpen |
dicular to the tension. It does not contribute to the centripetal force. The tension alone is the centripetal force. We can write this mathematically as: F F c T Figure 5.30(c) As the bucket moves through the bottom of the circle, it must have a centripetal force that overcomes gravity. The tension is the greatest here because gravity is acting opposite to the centripetal force. The equation is the same as in (a) above, but tension is acting upward, so when the values are placed into the equation this time, F g is negative. The effect is demonstrated in Example 5.7. T is positive and F Concept Check 1. A bucket filled with sand swings in a vertical circle at the end of a rope with increasing speed. At some moment, the tension on the rope will exceed the rope’s strength, and the rope will break. In what position in the bucket’s circular path is this most likely to happen? Explain. Is it necessary to know the position of an object moving in a vertical circle with uniform speed if you are determining centripetal force? Explain. 2. Forces Affecting an Object Moving in a Vertical Circle In summary, an object moving in a vertical circle is affected by the following forces: • The centripetal force is the net force on the object in any position. • The centripetal force is determined by the object’s mass, speed, and radius. In the case of the roller coaster and bucket of water, their mass and radius of curvature are constant so only their speed affects the centripetal force. • The force of gravity is one of the forces that may contribute to the centripetal force. • The force of gravity remains constant regardless of the position of the object. Chapter 5 Newton’s laws can explain circular motion. 263 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 264 Example 5.7 A bucket of water with a mass of 1.5 kg is spun in a vertical circle on a rope. The radius of the circle is 0.75 m and the speed of the bucket is 3.00 m/s. What is the tension on the rope in position C, as shown in Figure 5.31? Given r 0.75 m m 1.5 kg v 3.00 m/s g 9.81 m/s2 Required T) tension (F up left down A right r 0.75 m B c C m 1.5 kg FT Fg v 3.00 m/s Figure 5.31 Analysis and Solution The centripetal force acting on the bucket will not change as the bucket moves in a vertical circle. The tension will change as the bucket moves in its circular path because gravity will work with it at the top and against it at the bottom. In position C, the force of gravity works downward (negative), but the centripetal force and tension act upward (positive). Tension must overcome gravity to provide the centripetal force (Figure 5.32). Practice Problems 1. Using the information in Example 5.7, determine the tension in the rope at position A in Figure 5.31. 2. Using the information in Example 5.7, determine the tension in the rope in position B in Figure 5.31. 3. A 0.98-kg rock is attached to a 0.40-m rope and spun in a vertical circle. The tension on the rope when the rock is at the top of the swing is 79.0 N [down]. What is the speed of the rock? Answers 1. 3.3 N [down] 2. 18 N [left] 3. 6.0 m/s Remember that the centripetal force is the net force and is the vector sum of all the forces acting on the bucket. Therefore, the equation is: F net Fc Fg FT FT Fc Fg up left down right FT Fg Figure 5.32 Fnet Fg FT (mg) mv 2 r m 2 (1.5 kg )3.00 s 0.75 m m (1.5 kg ) 9.81 s2 18 N [14.715 N] 33 N Paraphrase The tension on the rope at position C is 33 N [up]. This is the maximum tension the system will experience because gravity acts in the opposite direction of the centripetal force. 264 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 265 Centripetal Force, Acceleration, and Frequency At your local hardware store, you will find a variety of rotary power tools that operate by circular motion. The table saw, circular saw, impact wrench, reciprocating saw, and rotary hammer are a few (Figure 5.33). One of the selling features listed on the box of most of these tools is the rotational frequency (in rpm). At the beginning of this chapter, you learned that rpm refers to the frequency of rotation measured in revolutions per minute. This is an Imperial measurement that has been around for hundreds of years. It has probably persisted because people have a “feel” for what it means. Even though revolutions per minute (rpm) is not considered an SI unit for frequency, it is a very useful measurement nevertheless. Why is the frequency of rotation often a more useful measure than the speed? The answer has to do with the nature of a rotating object. Figure 5.33 Most tools, motors, and devices that rotate with high speed report the frequency of rotation using rpm instead of the speed. The Effect of Radius on Speed, Period, and Frequency Imagine a disc spinning about its axis with uniform circular motion (Figure 5.34). Positions A, B, and C are at different radii from the axis of rotation, but all the positions make one complete revolution in exactly the same time, so they have the same period. Of course, if the periods for points A, B, and C are the same, so are their frequencies, and we can make the following generalization: For any solid rotating object, regardless of its shape, the frequency of rotation for all points on the object will be the same. Compare the speeds of points A, B, and C. Point A is the closest to the axis of rotation and has the smallest radius, followed by B, and then C with the largest radius. As already discussed, all three points make one complete revolution in the same amount of time, so point C, which has the farthest distance to cover, moves the fastest. Point B moves more slowly than C because it has less distance to travel. Point A has the slowest speed because it has the least distance to cover in the same amount of time. In essence, the speed of the spinning disc changes depending on which point you are referring to. In other words, the speed of a point on a disc changes with respect to its radius. C B A Figure 5.34 The speeds at A, B, and C are all different, whereas the rotational frequency of this disc is the same at any point. Chapter 5 Newton’s laws can explain circular motion. 265 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 266 r 3.83 106 m 60º N P Q 30º N 0º 90º N 53º N equator r 6.38 106 m S Figure 5.35 The speed of any point on Earth depends on its latitude (which determines its rotational radius). Point P moves more slowly than point Q, but they both have the same period and frequency. At the beginning of this chapter, you learned that the speed of an Albertan is roughly 1000 km/h as Earth rotates on its axis. However, not every point on Earth’s surface moves with the same speed. Remember: Speed changes with radius, and on Earth, the distance from the axis of rotation changes with latitude (Figure 5.35). The fastest motion is at the equator and the slowest is at the poles, but every point on Earth has the same period of rotation — one day. Determining Centripetal Force Using Period and Frequency In earlier example problems, such as Example 5.3, a given rotational frequency or period had to be converted to speed before the centripetal acceleration or force could be determined. It would be simpler to derive the equations for centripetal acceleration and force using rotational frequency or period to save a step in our calculations. The equations for centripetal acceleration and force are: ac v 2 r and Fc mv 2 r Recall that: 2r T v By substituting the velocity equation into the centripetal acceleration equation, the result is: ac (2r)2 rT 2 42r 2 rT 2 ac 42r T 2 The centripetal force is Fc 42mr T 2 Fc ma, so the centripetal acceleration is: (7) (8) Period is just the inverse of frequency, so it is relatively simple to express equations 7 and 8 in terms of frequency: ac 42rf 2 and Fc 42mrf 2 To convert rpm to hertz, simply divide by 60. (9) (10) 266 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 267 Example 5.8 The compressor blades in a jet engine have a diameter of 42.0 cm and turn at 15 960 rpm (Figure 5.36). Determine the magnitude of the centripetal acceleration at the tip of each compressor blade. ac v D 42.0 cm Practice Problems 1. A space station shaped like a wheel could be used to create artificial gravity for astronauts living in space. The astronauts would work on the rim of the station as it spins. If the radius of the space station is 30.0 m, what would its frequency have to be to simulate the gravity of Earth (g 9.81 m/s2)? 2. A 454.0-g mass, attached to the end of a 1.50-m rope, is swung in a horizontal circle with a frequency of 150.0 rpm. Determine the centripetal force acting on the mass. Answers 1. 9.10 102 Hz 2. 1.68 102 N Figure 5.36 The centripetal acceleration at the tip of a blade can be determined from the frequency of the blade’s rotation. Analysis and Solution First convert the frequency to SI units (Hz) and determine the radius. Then use equation 9. m n i 9 15 60 f s 0 6 m in rev 1 1 266.00 Hz D r 2 0 m 0.42 2 0.210 m ac 42rf 2 42(0.210 m)(266.00 Hz)2 5.87 105 m/s2 The magnitude of the centripetal acceleration at the tip of each compressor blade is 5.87 105 m/s2. Chapter 5 Newton’s laws can explain circular motion. 267 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 268 5.2 Check and Reflect 5.2 Check and Reflect Knowledge 1. A car heading north begins to make a right turn. One-quarter of a whole turn later, in what direction is the centripetal force acting? 2. Does the Moon experience centripetal acceleration? Why? 3. What two things could you do to increase the centripetal force acting on an object moving in a horizontal circle? 4. An object moves in a vertical circle at the end of a rope. As the object moves, explain what happens to: (a) the force of gravity, and (b) the tension. 5. What force acts as the centripetal force for a plane that is making a horizontal turn? Applications 6. A car’s wheels have a radius of 0.5 m. If the car travels at a speed of 15.0 m/s, |
what is the period of rotation of the wheels? 7. A propeller blade has a period of rotation of 0.0400 s. What is the speed of the outer tip of the propeller blade if the tip is 1.20 m from the hub? 8. A 1500-kg car is making a turn with a 100.0-m radius on a road where the coefficient of static friction is 0.70. What is the maximum speed the car can go without skidding? 9. A car rounds a curve of radius 90.0 m at a speed of 100.0 km/h. Determine the centripetal acceleration of the car. 10. NASA uses a centrifuge that spins astronauts around in a capsule at the end of an 8.9-m metallic arm. The centrifuge spins at 35 rpm. Determine the magnitude of the centripetal acceleration that the astronauts experience. How many times greater is this acceleration than the acceleration of gravity? 268 Unit III Circular Motion, Work, and Energy 11. What minimum speed must a toy racecar have to move successfully through a vertical loop that has a diameter of 30.0 cm? 12. Determine the centripetal acceleration acting on a person standing at the equator (r Earth 6.38 106 m). 13. An ant climbs onto the side of a bicycle tire a distance of 0.40 m from the hub. If the 0.010-g ant can hold onto the tire with a force of 4.34 10–4 N, at what frequency would the tire fling the ant off assuming the wheel is spun on a horizontal plane? Extensions 14. Two pulleys are connected together by a belt as shown in the diagram below. If the pulley connected to the motor spins at 200.0 rpm, what is the frequency of the larger pulley? (Hint: Both pulleys have the same velocity at their outer edge.) r1 10 cm pulley belt r2 25 cm 15. Two NASCAR racecars go into a turn beside each other. If they remain side by side in the turn, which car has the advantage coming out of the turn? e TEST To check your understanding of circular motion and Newton’s laws, follow the eTest links at www.pearsoned.ca/school/physicssource. 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 269 5.3 Satellites and Celestial Bodies in Circular Motion Johannes Kepler (1571–1630) was a German mathematician with a strong interest in astronomy. When he started working for renowned Danish astronomer Tycho Brahe (1546–1601) in 1600, he was given the problem of determining why the orbit of Mars didn’t completely agree with mathematical predictions. Brahe probably gave Kepler this job to keep him busy since he didn’t share Kepler’s ideas that the planets revolved around the Sun. At this time, most people, including scientists, believed that the planets and the Sun revolved around Earth. info BIT Comets are objects in space that have very elliptical orbits. They are often difficult to detect because they are relatively small and their orbits take them a great distance from the Sun. It is likely there are many comets orbiting the Sun that we have yet to discover. Kepler’s Laws Brahe had made very meticulous observations of the orbital position of Mars. Kepler used this data to determine that the planet must revolve around the Sun. Kepler also hypothesized that Mars had an elliptical orbit, not a circular one. Until this time, all mathematical predictions of a planet’s position in space were based on the assumption that it moved in a circular orbit. That is why Brahe’s observations disagreed with mathematical predictions at the time. By recognizing that planets move in elliptical orbits, Kepler could account for the discrepancy. Kepler’s First Law semi-minor axis semi-major axis 1 Sun focus focus 2 minor axis major axis An ellipse is an elongated circle. Figure 5.37 is an example of an ellipse. There are two foci, as well as major and minor axes. In Kepler’s model, the Sun is at one focus and the planet’s orbit is the path described by the shape of the ellipse. It is clear from Figure 5.37 that the planet will be closer to the Sun in position 1 than in position 2. This means that the planet’s orbital radius must be changing as time goes by. Why then is a planet’s orbital radius often written as a fixed number? There are two reasons: • The radius used is an average value. Mathematically, this average radius can be shown to be the same length as the semi-major axis. • The orbit of the planet, although an ellipse, is very close to being a circle, so the orbital radius really doesn’t change much from position 1 to position 2. Figure 5.37 is an exaggeration of a planet’s orbit for the purposes of clarity. The degree to which an ellipse is elongated is called the eccentricity. It is a number between 0 and 1, with 0 being a perfect circle and anything above 0 being a parabola, which flattens out to a line when it reaches 1. Table 5.4 shows the eccentricities of the orbits of the planets and other celestial bodies in our solar system. Figure 5.37 Kepler described the motion of planets as an ellipse with a semi-major axis and a semiminor axis. The average orbital radius is the semi-major axis. PHYSICS INSIGHT In an ellipse, the separation of the foci determines the shape of the ellipse. Chapter 5 Newton’s laws can explain circular motion. 269 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 270 Table 5.4 Eccentricities of Orbits of Celestial Bodies Kepler’s Second Law Celestial Body Mercury Venus Earth Mars Ceres Jupiter Saturn Uranus Neptune Pluto Eris Sedna Eccentricity 0.205 0.007 0.017 0.093 0.080 0.048 0.054 0.047 0.009 0.249 0.437 0.857 e WEB Sedna was discovered in 2004 and is the largest object yet found in a region of space known as the Oort cloud. On July 29, 2005, an object now named Eris was confirmed. It is larger than Pluto but also in the Kuiper belt. To learn more about these celestial bodies, the Kuiper belt, and the Oort cloud, follow the links at:www.pearsoned.ca/school/ physicssource. Kepler went on to state two more revolutionary ideas relating to the orbits of planets in the solar system. His second law stated that planets move through their elliptical orbit in such a manner as to sweep out equal areas in equal times. Imagine a line that extends from the Sun to the planet. As the planet moves in its orbit, the line moves with it and sweeps out an area. In the same amount of time, at any other position in the planet’s orbit, the planet will again sweep out the same area (Figure 5.38). area 1 Sun area 2 Figure 5.38 Kepler’s second law states that a line drawn from the Sun to the planet sweeps out equal areas in equal times. One consequence of this rule is that a planet’s speed must change throughout its orbit. Consider Figure 5.39 where area 1 is equal to area 2. As the planet approaches the Sun, the orbital radius decreases. If it is to sweep out an area equal to area 2, the planet must speed up and cover a larger distance to compensate for the smaller radius. As the planet gets farther from the Sun, the orbital radius gets larger. The planet slows down and sweeps out the same area again in the same amount of time (Figure 5.39). v 109 289 km/h v 109 289 km/h Sun v 105 635 km/h v 105 635 km/h Figure 5.39 Earth’s speed varies from 105 635 km/h at its farthest distance from the Sun to 109 289 km/h when it is closest. Its speed changes to compensate for the change in its orbital radius. 270 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 271 Kepler’s Third Law Kepler’s third law states that the ratio of a planet’s orbital period squared divided by its orbital radius cubed is a constant that is the same for all the planets orbiting the Sun. Written mathematically, it is: 2 T a K 3 ra (11) e SIM where Ta is the orbital period of planet A, ra is the orbital radius of planet A, and K is Kepler’s constant. Since the constant K applies to all planets orbiting the Sun, we can equate the ratio T 2/r 3 between any two planets and write the equation: 2 2 T T b a 3 3 rb ra (12) Until Kepler noticed this relationship between the planets, there was really no indication that the third law would be true. If it hadn’t been for Tycho Brahe’s accurate measurements of the planets’ orbital positions throughout their year, it is unlikely that Kepler would have made this discovery. Once the relationship was known, it was easy to verify, and it further bolstered the credibility of the heliocentric (Suncentred) model of the solar system. It is important to note that when Kepler derived his third law, he applied it only to planets orbiting the Sun. However, this law can be extended to moons that orbit a planet. In fact, in the most general sense, Kepler’s third law is applicable to all celestial bodies that orbit the same focus. For bodies orbiting a different focus, Kepler’s constant will be different. Kepler’s three laws can be summarized this way: 1. All planets in the solar system have elliptical orbits with the Sun at one focus. 2. A line drawn from the Sun to a planet sweeps out equal areas in equal times. 3. The ratio of a planet’s orbital period squared to its orbital radius cubed is a constant. All objects orbiting the same focus (e.g., planets, 2 2 T T b a the Sun) have the same constant. 3 3 rb ra Determining Kepler’s Constant To learn more about Kepler’s second law regarding eccentricity, orbital period, and speed of a planet, visit www.pearsoned.ca/ school/physicssource. e WEB The Titius-Bode law is another mathematical description that predicts the orbital radius of the planets. To learn more about the Titius-Bode law, follow the links at www.pearsoned.ca/ school/physicssource. Kepler’s third law states that the constant K is the same for all planets in a solar system. The period and orbital radius of Earth are well known, so they are used to compute the constant. The mean (average) orbital distance for Earth from the Sun is 1.50 1011 m, and Earth’s orbital period is 31 556 736 s. Kepler’s constant can be calculated as shown below: PHYSICS INSIGHT The orbital radius is always measured from the centre of the orbiting body to the centre of the body being orbited. 2 T E K 3 rE (3.15567 107 s)2 (1.50 1011 m)3 2.95 1019 s2/m3 Chapter 5 Newton’s laws can explain circular m |
otion. 271 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 272 info BIT Some scientists speculate that Pluto and Charon might have been objects from the Kuiper belt that were attracted into an orbit of the Sun by Neptune because of their small size and large orbital radius. The Kuiper belt is a large band of rocky debris that lies 30 to 50 AU from the Sun. e MATH Kepler’s constant applies only to bodies orbiting the same focus. To explore this concept in more depth and to determine Kepler’s constant for Jupiter’s moons, visit www.pearsoned.ca/physicssource. info BIT The SI unit symbol for year is a. However, you will not often find Kepler’s constant written with this value. The reason is twofold. If different units for distance and time are used (something other than metres and seconds), then the constant can be made to equal 1. This has obvious mathematical benefits. The other reason is that using metres and seconds as the units of measurement is impractical when dealing with the scale of our solar system. To represent astronomical distances, it becomes necessary to use units bigger than a metre or even a kilometre. For example, measuring the distance from Earth to the Sun in kilometres (150 000 000 km) is roughly like measuring the distance from Edmonton to Red Deer in millimetres. Astronomical Units A more suitable measurement for astronomical distances has been adopted. It is called the astronomical unit (AU). One astronomical unit is the mean orbital distance from Earth to the Sun (the length of the semi-major axis of Earth’s orbit). This is a more manageable unit to use. For example, Neptune is only 30.1 AU away from the Sun on average. Kepler’s constant, using units of years (a) and AU, can be deter- mined as follows: 2 T E K 3 rE )2 a (1 )3 U A (1 1 a2/AU3 The advantage of using the units of an Earth year and astronomical units becomes clear as Kepler’s constant works out to 1. Any other planet in our solar system must also have the same constant because it orbits the same focus (the Sun). Be careful not to use this value of Kepler’s constant for all systems. For example, if Kepler’s third law is used for moons orbiting a planet, or planets orbiting a different sun, the constant will be different. Example 5.9 Practice Problems 1. Use Kepler’s third law to determine the orbital period of Jupiter. Its orbital radius is 5.203 AU. 2. Pluto takes 90 553 Earth days to orbit the Sun. Use this value to determine its mean orbital radius. 3. A piece of rocky debris in space has a mean orbital distance of 45.0 AU. What is its orbital period? Mars has an orbital radius of 1.52 AU (Figure 5.40). What is its orbital period? Mars Sun r 1.52 AU Figure 5.40 Mars has a mean orbital radius of 1.52 AU (not drawn to scale). Kepler’s third law can be used to determine its period. 272 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 273 Analysis and Solution Start with the equation for determining Kepler’s constant where TMars is the the orbital period of Mars, and r Mars is the mean orbital radius of Mars. Answers 1. 11.87 Earth years 2. 39.465 AU 3. 302 Earth years T 2 Mars TMars Mars Kr 3 1(1.52 AU)3 a2 AU3 1.87 a The orbital period of Mars is 1.87 Earth years. In other words, it takes 1.87 Earth years for Mars to go around the Sun once. A close examination of Table 5.5 shows that as the orbital radius of a planet increases, so does its orbital period. Planets nearest the Sun have the highest orbital speeds. The years of these planets are the shortest. Planets farthest from the Sun have the longest years. Kepler’s laws don’t apply just to planets orbiting the Sun. They apply to all bodies that orbit the same focus in an ellipse. This means that moons orbiting a planet are also subject to Kepler’s laws and have their own Kepler’s constant. Earth has only one Moon (natural satellite) but Jupiter and Saturn have many, and more are being found all the time. Table 5.6 shows the planets and some of their known moons. Table 5.5 Solar Celestial Bodies Celestial Body Sun Mercury Venus Earth Mars Ceres Jupiter Saturn Uranus Neptune Pluto Eris Sedna Mass (kg) 1.99 1030 3.30 1023 4.87 1024 5.97 1024 6.42 1023 9.5 1020 1.90 1027 5.69 1026 8.68 1025 1.02 1026 1.20 1022 ? 4.21 1021 Equatorial Radius (m) 6.96 108 2.44 106 6.05 106 6.38 106 3.40 106 4.88 105 7.15 107 6.03 107 2.56 107 2.48 107 1.70 106 ~1.43 106 8.5 105 Orbital Period (days) — 87.97 224.7 365.24 686.93 1679.8 4330.6 10 755.7 30 687.2 60 190 90 553 204 540 3 835 020 Mean Orbital Radius (m) — 5.79 1010 1.08 1011 1.49 1011 2.28 1011 4.14 1011 7.78 1011 1.43 1012 2.87 1012 4.50 1012 5.91 1012 ~1.01 1013 7.14 1013 Distance (AU) — 0.387 0.723 1 1.524 2.766 5.203 9.537 19.191 30.069 39.482 67.940 479.5 PHYSICS INSIGHT Kepler’s third law can only be applied to bodies orbiting the same object. info BIT As of August 2006, the International Astronomical Union (IAU) finally developed a definition of a planet. To be a planet, a celestial body must orbit a star, be large enough that its own gravity forms it into a spherical shape, and have cleared the neighbourhood around its orbit. Pluto fails to satisfy the last criterion as its orbit is near many other Kuiper belt objects. There are now officially only eight planets in our solar system. Chapter 5 Newton’s laws can explain circular motion. 273 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 274 Concept Check 1. What type of orbit would a planet have if the semi-minor axis of its orbit equalled the semi-major axis? 2. Why can’t Kepler’s third law be applied to Earth’s Moon and Jupiter’s moon Callisto using the same value for Kepler’s constant? 3. Assume astronomers discover a planetary system in a nearby galaxy that has 15 planets orbiting a single star. One of the planets has twice the mass, but the same orbital period and radius as Earth. Could Kepler’s constant for our solar system be used in the newly discovered planetary system? Explain. 4. Assume astronomers discover yet another planetary system in a nearby galaxy that has six planets orbiting a single star, but all of the planets’ orbits are different from Earth’s. Explain how Kepler’s third law could apply to this planetary system. 5. Compare and contrast planets orbiting a star with points on a rotating solid disc (Figure 5.34). Table 5.6 The Planets and Their Large Moons Planet Moons Mass (kg) Equatorial Radius (m) Orbital Period (Earth days) Mean Orbital Radius (m) Eccentricity Discovered (Year) Earth Mars Jupiter (4 most massive) Saturn (7 most massive) Uranus (5 most massive) Neptune (3 most massive) Moon 7.35 1022 1.737 106 Phobos Deimos Io Europa Ganymede Callisto Mimas Enceladus Tethys Dione Rhea Titan Iapetus Miranda Ariel Umbriel Titania Oberon Proteus Triton Nereid 1.063 1016 2.38 1015 1.340 104 7.500 103 8.9316 1022 4.79982 1022 1.48186 1023 1.07593 1023 3.75 1019 7 1019 6.27 1020 1.10 1021 2.31 1021 1.3455 1023 1.6 1021 6.6 1019 1.35 1021 1.17 1021 3.53 1021 3.01 1021 5.00 1019 2.14 1022 2.00 1019 1.830 106 1.565 106 2.634 106 2.403 106 2.090 105 2.560 105 5.356 105 5.600 105 7.640 105 2.575 106 7.180 105 2.400 105 5.811 105 5.847 105 7.889 105 7.614 105 2.080 105 1.352 106 1.700 105 274 Unit III Circular Motion, Work, and Energy 27.322 0.3189 1.262 1.769 3.551 7.154 16.689 0.942 1.37 1.887 2.74 4.52 15.945 79.33 1.41 2.52 4.14 8.71 13.46 1.12 5.8766 360.14 3.844 108 9.378 106 2.346 107 4.220 108 6.710 108 1.070 109 1.883 109 1.855 108 2.380 108 2.947 108 3.774 108 5.270 108 1.222 109 3.561 109 1.299 108 1.909 108 2.660 108 4.363 108 5.835 108 1.176 108 3.548 108 5.513 109 0.0549 0.015 0.0005 0.004 0.009 0.002 0.007 0.0202 0.00452 0.00 0.002 0.001 0.0292 0.0283 0.0027 0.0034 0.005 0.0022 0.0008 0.0004 0.000016 0.7512 — 1877 1877 1610 1610 1610 1610 1789 1789 1684 1684 1672 1655 1671 1948 1851 1851 1787 1787 1989 1846 1949 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 275 Example 5.10 Mars has two moons, Deimos and Phobos (Figure 5.41). Phobos has an orbital radius of 9378 km and an orbital period of 0.3189 Earth days. Deimos has an orbital period of 1.262 Earth days. What is the orbital radius of Deimos? Mars Phobos Deimos TP 0.3189 d TD 1.262 d rP 9378 km rD Figure 5.41 The orbital period and radius of Phobos can be used with the orbital period of Deimos to determine its orbital radius (not drawn to scale). Given TP rP TD 0.3189 d 9378 km 1.262 d Required orbital radius of Deimos (rD) Analysis and Solution Both Phobos and Deimos orbit the same object, Mars, so Kepler’s third law can be used to solve for the orbital radius of Deimos. The units of days and kilometres do not need to be changed to years and astronomical units because Kepler’s third law is simply a ratio. It is important to be consistent with the units. If we use the units of kilometres for the measure of orbital radius, then the answer will be in kilometres as well. 2 2 T T P D 3 3 rP rD rD 1.262 d)2(9378 km)3 (0.3189 d)2 rD 1.2916 1013 km3 31.2916 1013 km3 2.346 104 km Practice Problems 1. Titan is one of the largest moons in our solar system, orbiting Saturn at an average distance of 1.22 109 m. Using the data for Dione, another moon of Saturn, determine Titan’s orbital period. 2. The Cassini-Huygens probe began orbiting Saturn in December 2004. It takes 147 days for the probe to orbit Saturn. Use Tethys, one of Saturn’s moons, to determine the average orbital radius of the probe. 3. Astronomers are continually finding new moons in our solar system. Suppose a new moon X is discovered orbiting Jupiter at an orbital distance of 9.38 109 m. Use the data for Callisto to determine the new moon’s orbital period. Answers 1. 16.0 d 2. 5.38 106 km 3. 186 d Paraphrase The orbital radius of Deimos is 2.346 104 km. This answer is reasonable since a moon with a larger orbital period than Phobos will also have a larger orbital radius. Chapter 5 Newton’s laws can explain circular motion. 275 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 276 5-4 Design a Lab 5-4 Design a Lab Orbital Period an |
d Radius Kepler determined that planets have slightly elliptical orbits. He also determined empirically that the period and radius of the planet’s orbit were related. The purpose of this lab is to design a method to highlight the relationship between the period and radius of planets’ orbits around the Sun. The Question How can the relationships between the orbital period and orbital radius of planets orbiting the Sun be shown by graphical means? Design and Conduct Your Investigation This lab should be designed to investigate the relationship between period and radius. State a hypothesis relating the orbital period to the radius for planets orbiting the Sun. Remember to use an “if/then” statement. The lab should show how a planet’s orbital radius affects its orbital period. To do this, use Table 5.5 to select data for at least five planets’ periods and orbital radii. Organize the relevant data from Table 5.5 into a chart in a suitable order that can be used to plot a graph using a graphing calculator or other acceptable means. Choose appropriate units for the manipulated and responding variables. Look at the type of graph that results to draw a conclusion relating period and radius. Test the relationship you discovered from this lab with the relationship that exists between the orbital radius and period of a planet’s moons. Do this by picking a planet from Table 5.6 that has several moons and perform the same procedure and analysis. Compare the relationships of orbital radius and period of the planets with that of the moons. Comment on any similarities or differences. e WEB In August 2006, the IAU decided to create three classifications of solar system objects: planets, dwarf planets, and small solar system objects. Pluto is now classified as a dwarf planet, along with Eris and Ceres (formerly an asteroid). To be a dwarf planet, the celestial body must orbit the Sun and be large enough that its own gravity forms it into a spherical shape, but not large enough that it has cleared the neighbourhood around its orbit. To learn about dwarf planets, follow the links at: www.pearsoned.ca/school/ physicssource. Moon v path of moon g g Earth Figure 5.42 The Moon is falling to Earth with the acceleration of gravity just like the apple does. But the Moon also has a tangential velocity (v) keeping it at the same distance from Earth at all times. Newton’s Version of Kepler’s Third Law Some stories suggest that Newton was sitting under a tree when an apple fell to the ground and inspired him to discover gravity. This is definitely not what happened, as gravity was already known to exist. But the falling apple did lead him to wonder if the force of gravity that caused the apple to fall could also be acting on the Moon pulling it toward Earth. This revelation might seem obvious but that’s only because we have been taught that it’s true. In his day, Kepler theorized that magnetism made the Moon orbit Earth, and planets orbit the Sun! For most of the 1600s, scientists had been trying to predict where planets would be in their orbit at specific times. They failed to grasp the underlying mechanism responsible for the elliptical orbits that Kepler had shown to exist. In 1665, Newton finally recognized what no one else did: the centripetal force acting on the Moon was the force of gravity (Figure 5.42). The Moon was being pulled toward Earth like a falling apple. But the Moon was also moving off tangentially so that the rate at which it was falling matched the rate at which Earth curved away from it. In fact, the same mechanism was responsible for the planets orbiting the Sun. 276 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 277 By recognizing that the centripetal force and force of gravity were the same, Newton solved the mystery of planetary motion. His derivations mathematically proved Kepler’s third law. The implications of Newton’s work were huge because scientists now had the mathematical tools necessary to explore the solar system in more depth. Determining the Speed of a Satellite Earth Fg Moon r 3.844 108 m Figure 5.43 The Moon experiences a centripetal force that is the force of gravity of Earth on the Moon (not drawn to scale). Two of Newton’s derivations deserve close examination. The first derivation uses the orbital radius of a body orbiting a planet or the Sun to determine the body’s velocity. Recall from Chapter 4 that Newton had determined the equation for the force of gravity that one object exerts on another: Fg Gm1m2 r 2 He correctly reasoned that the gravitational force exerted by the Moon and Earth must be the centripetal force acting on the Moon (Figure 5.43). So: Fg Fc If we substitute the equations for centripetal force and gravity into this equation, we obtain: onv 2 GmMo mMo on 2 r r mEarth where G is the universal gravitational constant, mMoon is the mass of the Moon in kilograms, mEarth is the mass of Earth in kilograms, v is the speed of the Moon in metres per second, and r is the orbital radius of the Moon in metres. PHYSICS INSIGHT Since Fc Fg mv 2 r GMm r 2 and 2 r 2 m4 T 2 r GMm r 2 4 2r m 2 T GMm r 2 T 2 r 3 42 GM which is a constant containing the mass of the object causing the orbit. info BIT Pluto and its moon Charon are close enough in mass that they have a common centre of gravity between them in space (see Extrasolar Planets on page 283). Charon does not orbit Pluto — both bodies orbit their common centre of gravity. Since their centre of gravity is in space and not below the surface of Pluto, they form a binary system. Astronomers have been aware of binary stars in our universe for many years (i.e., two stars that have a common centre of gravity in space around which they revolve). Pluto and Charon are unique in our solar system. Chapter 5 Newton’s laws can explain circular motion. 277 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 278 PHYSICS INSIGHT Equation 13 uses the mass of Earth, but is not restricted to it. In a more general sense, the mass refers to the object being orbited. The mass of the Moon cancels, leaving: v 2 r Gm arth E 2 r Solving for v gives: v GmEarth r (13) In its current form, this equation determines the speed of any object orbiting Earth. The speed of an object orbiting any planet, the Sun, or another star for that matter, can be determined by using the mass of the object being orbited in place of the mass of Earth. Example 5.11 Earth’s Moon is 3.844 105 km from Earth (Figure 5.44). Determine the orbital speed of the Moon. Earth m 5.97 1024 kg Practice Problems 1. Neptune’s average orbital radius is 4.50 1012 m from the Sun. The mass of the Sun is 1.99 1030 kg. What is Neptune’s orbital speed? 2. The moon Miranda orbits Uranus at a speed of 6.68 103 m/s. Use this speed and the mass of Uranus to determine the radius of Miranda’s orbit. The mass of Uranus is 8.68 1025 kg. Answers 1. 5.43 103 m/s 2. 1.30 108 m Moon r 3.844 105 km Figure 5.44 The orbital speed of the Moon can be determined from its orbital radius and the mass of Earth (not drawn to scale). Analysis and Solution Convert the radius of the Moon’s orbit to SI units. Then use the mass of Earth in equation 13. 0 10 r (3.844 105 km) k 0 m m 3.844 108 m v GmEarth r 6.67 1011 N 2 (5.97 1024 kg) m 2 g k 3.844 108 m 1.02 103 m/s The orbital speed of the Moon is 1.02 103 m/s or 3.66 103 km/h. 278 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 279 Measuring the Orbital Height of a Satellite Recall from Chapter 4 that the radius is always measured from the centre of one object to the centre of the other. This means that the orbital radius refers to the distance from centre to centre when measuring the distance from Earth to the Moon or from the Sun to the planets. Figure 5.45 shows the Earth-Moon system drawn to scale. For an artificial (human-made) satellite, the height of its orbit is usually measured from Earth’s surface. To determine the velocity of an artificial satellite, you must first find its proper orbital height (from the centre of Earth). This is done by adding Earth’s radius to the height of the satellite above Earth’s surface. 384 400 km Earth farthest communication satellites from Earth Moon Figure 5.45 The Earth-Moon system drawn to scale. The radius of Earth doesn’t need to be considered when comparing the distance between Earth and the Moon because it is insignificant compared to the great distance separating the two bodies. The distance between Earth and a communication satellite (35 880 km away) is small, so Earth’s radius must be included in calculations involving orbital radius and period. Example 5.12 LandSat is an Earth-imaging satellite that takes pictures of Earth’s ozone layer and geological features. It orbits Earth at the height of 912 km (Figure 5.46). What are its orbital speed and its period? Practice Problems 1. The International Space Station orbits Earth at a height of 359.2 km. What is its orbital speed? 2. The Chandra X-ray satellite takes X-ray pictures of high-energy objects in the universe. It is orbiting Earth at an altitude of 114 593 km. What is its orbital period? Answers 1. 7.69 103 m/s 2. 4.19 105 s 912 km rE 6.38 106 m Figure 5.46 LandSat follows a polar orbit so that it can examine the entire Earth as the planet rotates below it. The radius of Earth must be added to LandSat’s height above the surface to determine the satellite’s orbital radius (not drawn to scale). Given height of the satellite above Earth’s surface 912 km Required LandSat’s orbital speed (v ) and period (T ) Chapter 5 Newton’s laws can explain circular motion. 279 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 280 Analysis and Solution The radius of LandSat’s orbit must be measured from the centre of Earth. To do this, add Earth’s radius to the satellite’s height above the planet’s surface. Then determine the speed and period of the satellite. r rEarth 912 000 m (6.38 106 m) (9.12 105 m) 7.292 106 m Fg Fc v 2 msatellite r Gmsatelli |
te r 2 mEarth v GmEarth r 6.67 1011 2 m N (5.97 1024 kg) 2 g k 7.292 106 m 7.390 103 m/s r 2 T v 2(7.292 106 m) m 7.390 103 s 6.20 103 s Paraphrase The speed of the satellite is 7.39 103 m/s (2.66 104 km/h). It orbits Earth once every 6.20 103 s (103 minutes). e TECH For an interactive simulation of the effect of a star’s mass on the planets orbiting it, follow the links at www.pearsoned.ca/ physicssource. Determining the Mass of a Celestial Body A second derivation from Newton’s version of Kepler’s third law has to do with determining the mass of a planet or the Sun from the period and radius of a satellite orbiting it. For any satellite in orbit around a planet, you can determine its speed if you know the mass of the planet. But just how do you determine the mass of the planet? For example, how can you “weigh” Earth? Newton realized that this was possible. Let’s look at the equality Fg again, but this time we will use equation 8 for centripetal force. Fc Recall that: Fc 42mMoonr T 2 Moon where mMoon is the mass of the Moon in kilograms, and TMoon is the orbital period of the Moon in seconds. 280 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 281 PHYSICS INSIGHT Equation 14 uses the mass of Earth and the period of its Moon. This equation can be used for any celestial body with a satellite orbiting it. In a more general sense, the mass refers to the object being orbited, and the period is the period of the orbiting object. Fg, then Since Fc 4 2m GmMo M on 2 2 r T Mo on oonr mEarth Solve for mEarth mEarth 4 2 (14) You can determine Earth’s mass by using the orbital radius and period of its satellite, the Moon. From Table 5.6 on page 274, the Moon’s period and radius are: • period of the Moon (TMoon) 27.3 days or 2.36 106 s • radius of the Moon’s orbit (r Moon) 384 400 km or 3.844 108 m mEarth 42(3.844 108 m)3 m N (2.36 106 s)26.67 1011 2 g k 2 6.04 1024 kg The mass of Earth is 6.04 1024 kg, which is close to the accepted value of 5.97 1024 kg. Of course, this equation is not restricted to the Earth-Moon system. It applies to any celestial body that has satellites. For example, the Sun has eight planets that are natural satellites. Any one of them can be used to determine the mass of the Sun. Concept Check 1. What insight did Newton have that helped him explain the 2. motion of the planets? If Newton were told that our solar system is orbiting the centre of our galaxy, how would he explain this? 3. What previously immeasurable quantity could be determined with the use of Newton’s version of Kepler’s third law? Orbital Perturbations At about the same time as Kepler was figuring out the mechanism of the solar system, Galileo Galilei (1564–1642) pointed a relatively new invention at the sky. He began using a telescope to closely examine Jupiter. Only a few planets are visible to the naked eye: Mercury, Venus, Mars, Saturn, and Jupiter. Until the early 1600s, any observations of these planets were done without the aid of a telescope. Within a few months of using only an 8-power telescope, Galileo had discovered four moons of Jupiter. It became apparent how useful a telescope would be in the field of astronomy. e WEB To learn more about Galileo’s discoveries with the telescope in 1609 and 1610, follow the links at www.pearsoned.ca/ school/physicssource. Chapter 5 Newton’s laws can explain circular motion. 281 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 282 orbital perturbation: irregularity or disturbance in the predicted orbit of a planet e WEB Telescopes and space technologies are continually improving, and new planetary objects are being found all the time. For the most up-to-date list of planets and celestial bodies in our solar system, follow the links at www.pearsoned.ca/school/ physicssource. Uranus Fg Planet X Figure 5.47 If a planet X existed and was behind Uranus in its orbit, it would pull Uranus back and outward. Within the space of 100 years, telescope technology improved dramatically, and the field of astronomy began its golden age. Astronomers plotted the positions of the planets more accurately than ever before and could peer deeper into the solar system. William Herschel (1738–1822) discovered the new planet Uranus in 1781, which created enormous interest. However, it wasn’t long before astronomers noticed something strange about the orbit of Uranus. The orbital path of Uranus deviated from its predicted path slightly, just enough to draw attention. Astronomers called this deviation, or disturbance, an orbital perturbation. The Discovery of Neptune It had been over 120 years since Kepler and Newton had developed the mathematical tools necessary to understand and predict the position of the planets and their moons. Confident in the reliability of these laws, astronomers looked for a reason for the perturbation in the orbit of Uranus. According to mathematical predictions, Uranus should have been farther along in its orbit and closer to the Sun than it actually was. Somehow its progress was being slowed, and it was being pulled away from the Sun. Recall that anything with mass creates a gravitational field. The strength of this field depends on the mass of the object and the separation distance from it. The orbit of Uranus was minutely perturbed. Could another as-yet-undiscovered planet be exerting a gravitational pull or tug on Uranus whenever the orbital path took these two planets close together? If there was a planet X farther out and behind Uranus, it would exert a gravitational pull that would slow Uranus down and pull its orbit outward (Figure 5.47). This could explain the perturbation in the orbit of Uranus and was precisely the assumption that two astronomers, Urbain Le Verrier (1811–1877) of France and John Adams (1819–1892) of Britain, made in 1845. By examining exactly how much Uranus was pulled from its predicted position, Le Verrier and Adams could use Newton’s law of gravitation to mathematically predict the size and position of this mysterious planet — if it existed. Working independently, both scientists gave very similar predictions of where to look for the planet. In September 1846, at the request of Le Verrier, German astronomer Johann Gottfried Galle (1812–1910) looked for the planet where Le Verrier predicted it would be and he found it! Le Verrier called the planet Neptune. It remained the solar system’s outermost planet for the next 84 years until astronomers discovered Pluto in 1930 by analyzing the orbital perturbations of Neptune. (In 2006, Pluto was reclassified as a dwarf planet.) The Search for Other Planets The search for more planets in our solar system continues. A large band of rocky debris called the Kuiper belt lies 30 to 50 AU from the Sun. Pluto is a resident in this belt. Many scientists believed that Pluto was unlikely to be the only large resident of the belt and that the belt very likely contained more planet-sized objects. 282 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 283 In July 2005, a Kuiper belt object about 1.5 times bigger than Pluto was found. The scientific community has now given it the status of a dwarf planet, but at the time of publication of this book, it did not yet have an official name. Do other larger-than-Pluto bodies exist in the Kuiper belt? Probably, but their low reflectivity and extreme distance make them difficult to detect. THEN, NOW, AND FUTURE Extrasolar Planets For many centuries, humans have wondered if they were the only intelligent life in the universe. It seems unlikely, given the multitude of stars in our galaxy alone. However, for any life to exist, it’s pretty safe to assume that it must have a planet to inhabit. Until 1995, there was no conclusive proof that any other star besides our Sun had planets. In October 1995, the first extrasolar planet was found orbiting a star similar in size to our Sun. It is called “extrasolar” because it is outside (“extra”) our solar system. This planet was named 51 Pegasi b after the star around which it revolves (Figure 5.48). It is huge but its orbital radius is extremely small. It is unlikely to support life as its gravitational field strength and temperature are extreme. Yet the discovery was a milestone since finding a planet orbiting a bright star 48 light years away is a very difficult task. A planet doesn’t produce any light of its own, and it would be relatively dark compared to the very bright star beside it. The light coming from a star is so bright it makes direct observation of a nearby planet difficult. Imagine staring into a car’s headlight on a dark night and trying to see an ant crawling on the bumper. Now imagine the car is 23 000 km away, and you get an idea of the magnitude of the problem. However, the increasing power of telescopes and innovative detection techniques are yielding new planet findings all the time. How is this accomplished? There are several different ways, but the most common and productive way is not to look for the planet directly but to look at the star that it is orbiting and watch for perturbations (also called wobble) in the star’s movement. That’s right, stars move. If they don’t have a planet or planets orbiting them, then they simply move in a linear fashion through space. If they have a planet or planets in orbit around them, not only do they move linearly, but they also wobble in a circular path. This is because a planet exerts a gravitational pull on its star just as the star exerts the same pull on the planet. They both revolve around each other, just like you and a very heavy object would if you spun the object around in a circle at the end of a rope. Since the star is much more massive than the planet, its orbital radius is very small. This perturbation of the star is detectable and is indirect evidence that a planet must be orbiting it. As you can imagine, for a star to have a noticeable wobble, the planet that orbits it must be relatively large and fairly cl |
ose to the star. This would make conditions on the planet inhospitable for life as we know it. But the search goes on and as our technology improves, who knows what we may find? Questions 1. Why is it so hard to detect an extrasolar planet? 2. What new technologies and techniques are making it possible to detect extrasolar planets? 3. Why is it unlikely that any life exists on the extrasolar planets discovered using these new techniques? ▲ Figure 5.48 An artist’s conception of 51 Pegasi b — the first extrasolar planet found orbiting a star similar to our own. Chapter 5 Newton’s laws can explain circular motion. 283 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 284 Figure 5.49 A satellite in low Earth orbit Artificial Satellites At present, there are well over 600 working artificial satellites orbiting Earth. About half of them are in a low or medium Earth orbit, ranging from 100 to 20 000 km above Earth’s surface (Figure 5.49). The other half are geostationary satellites that orbit Earth at a distance of precisely 35 880 km from Earth’s surface, directly above the equator. Depending on their design and orbit, satellites perform a variety of tasks. Weather, communication, observation, science, broadcast, navigation, and military satellites orbit Earth at this moment. These include a ring of GPS (global positioning system) satellites that are used to triangulate the position of a receiver wherever it may be. With the help of a GPS receiver, you could find your position on the planet to within a few metres. All satellites are designed to receive information from and transmit information to Earth. Each satellite has an antenna that is used to receive radio signals from satellite stations on the ground. At the same time, satellites send information back down to Earth for people to use. 5-5 Decision-Making Analysis 5-5 Decision-Making Analysis The Costs and Benefits of Putting a Satellite into Orbit Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork The Issue Satellites perform a variety of tasks that make them almost indispensable. They map Earth, find geological formations and minerals, help us communicate over great distances and in remote areas, and help predict the movement of weather systems such as hurricanes. Opponents of the continued unregulated use of satellites argue that the cost of deploying satellites is enormous; they don’t have a long lifespan; and their failure rate is high. Furthermore, once a satellite is in a medium or geostationary orbit, it will stay there even after it becomes inoperative. It turns into nothing more than very expensive and dangerous space junk. Background Information The first satellite in orbit was Sputnik in 1957. Since that time there have been over 4000 launches, and space has become progressively more crowded. The best estimates suggest that there are about 600 active satellites in orbit, and about 6000 pieces of space debris that are being tracked. The space debris can be very hazardous for missions carrying astronauts to low Earth orbit. If hit by even a small piece of orbiting debris, a spacecraft could be destroyed. To limit the overcrowding of space, satellite manufacturers are designing more sophisticated satellites that can handle a higher flow of information so that fewer satellites have to be deployed. This drives up the cost of manufacturing satellites because they are more technologically advanced than their predecessors. Unfortunately, as well as being more sophisticated, they are more prone to failure. Analyze and Evaluate 1. Identify two different types of satellites based on the type of job they perform. 2. For each type of satellite from question 1: (a) Identify the type of orbit that it occupies and explain the job that it performs. (b) Determine the approximate cost of deployment (getting it into space). (c) Determine its expected lifespan. 3. Suggest an alternative technology that could be used in place of each of these satellites. Analyze the effectiveness and cost of this technology compared with the satellite. 4. Propose possible changes that could be made to the way satellites are built and deployed that could lessen the overcrowding of space. 284 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 285 Geostationary Satellites Geostationary satellites may be the most interesting of all satellites because they appear to be stationary to an observer on Earth’s surface, even though they travel around Earth at high velocity. They are placed at a specific altitude so that they make one complete orbit in exactly 24 h, which is the same as Earth’s rotational period (Figure 5.50). These satellites are placed in the plane of the equator, so they will have exactly the same axis of rotation as Earth, and will stay fixed over the same spot on the planet (Figure 5.51). To an observer on the ground, geostationary satellites appear motionless. axis of rotation of Earth Earth plane of Earth’s equator equator equator equator P P P satellite satellite’s orbit in plane of equator Figure 5.50 Geostationary satellites orbit in the plane of the equator with the same axis of rotation as Earth. Communication satellites are geostationary. A communication signal, such as a telephone or TV signal, can be sent from the ground at any time of the day to the nearest geostationary satellite located over the equator. That satellite can then relay the signal to other geostationary satellites located over different spots on Earth, where the signal is then transmitted back down to the nearest receiving station. Weather satellites also make use of this orbit. They may be “parked” near a landmass such as North America, using cameras to map the weather. Weather forecasters receive a continuous stream of information from the satellites that allows them to predict the weather in their area. This type of orbit is in high demand and is filled with satellites from many different countries. Unfortunately, the orbit must be fixed at 35 880 km from Earth’s surface and be directly over the equator. This orbit risks being overcrowded. If satellites are placed too close together, their signals can interfere with each other. Satellites also tend to drift slightly in their orbit and therefore cannot be placed close to each other. Many derelict satellites that were placed in geostationary orbits are still there taking up room, since they continue to orbit even after they no longer function. The limited room available is filling up fast. At present, about half (at least 300) of the active satellites in orbit are geostationary. An artificial satellite obeys the same laws of physics as a natural satellite does. The orbital radius determines its orbital period and speed, and it is governed by the same laws that describe the motion of moons around a planet or planets around the Sun. That means low Earth orbit satellites make one complete orbit in about 90 min, while geostationary satellites take exactly one day. Figure 5.51 A geostationary satellite moves with a fixed point (P) on the Earth since both revolve around the same axis with the same period. To an observer on the ground, the satellite does not appear to be moving. info BIT Geostationary satellites are also referred to as geosynchronous satellites. e WEB To learn more about real-time tracking of satellites in orbit around Earth, follow the links at www.pearsoned.ca/school/ physicssource. Chapter 5 Newton’s laws can explain circular motion. 285 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 286 In the short space of time since the first satellite, Sputnik, was launched in October 1957, satellites have become indispensable (Figure 5.52). They have enabled us to see areas of our planet and atmosphere never seen before. They have enabled us to know our location anywhere on Earth within a few metres, and have allowed us to communicate with the remotest places on Earth. The future of satellites seems assured. Figure 5.52 Around Earth, space is crowded with the multitude of satellites currently in orbit. The large ring farthest from Earth is made up of geostationary satellites. 5.3 Check and Reflect 5.3 Check and Reflect Knowledge 11. Jupiter’s moon Io has an orbital period of 1. What is an astronomical unit? 2. Why is the orbital radius of a planet not constant? 3. An eccentricity of 0.9 indicates what kind of shape? 4. Where in a planet’s orbit is its velocity the greatest? The smallest? Why? 5. State the condition necessary for Kepler’s third law to be valid. 6. Is Kepler’s constant the same for moons orbiting a planet as it is for planets orbiting the Sun? Why? 7. Does our Sun experience orbital perturbation? Explain. Applications 8. Sketch a graph that shows the trend between the planets’ orbital radius and their period. 9. Venus has an average orbital period of 0.615 years. What is its orbital radius? 10. Another possible planet has been discovered, called Sedna. It has an average orbital radius of 479.5 AU. What is its average orbital speed? 1.769 d and an average orbital radius of 422 000 km. Another moon of Jupiter, Europa, has an average orbital radius of 671 000 km. What is Europa’s orbital period? 12. Determine the average orbital speed of Mimas using orbital data from Table 5.6 on page 274. 13. Using the orbital period and radius of Venus from Table 5.5 on page 273, determine the mass of the Sun. Extensions 14. As more satellites are put into space, the amount of orbital debris increases. What solutions can you suggest to decrease the amount of space junk? 15. Using a graphing calculator or other suitable software, plot a graph of velocity versus radius for an artificial satellite orbiting Earth. e TEST To check your understanding of satellites and celestial bodies in circular motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 286 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 287 CHAPT |
ER 5 SUMMARY Key Terms and Concepts axle axis of rotation uniform circular motion centripetal acceleration centripetal force cycle revolution period frequency rpm Key Equations v 2r T ac v 2 r Fc mv 2 r 42r T 2 ac 42mr T 2 Fc 2 T a K 3 ra Conceptual Overview satellite artificial satellite orbital perturbation extrasolar planet Kepler’s laws ellipse eccentricity orbital period orbital radius 2 2 T T b a 3 3 rb ra The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. where the Uniform Circular Motion applies to where the object experiences Centripetal Acceleration is tangent to the acts toward the of the Centre which is a Net Force which could be a Combination of Forces Tension Figure 5.53 Planetary/Satellite Motion where explained Newton’s Version of Kepler’s 3rd Law was realized by and he derived Newton that was used to determine the Chapter 5 Newton’s laws can explain circular motion. 287 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 288 CHAPTER 5 REVIEW Knowledge 1. (5.1) What is the direction of centripetal acceleration? 2. (5.1) Describe what produces the centripetal Applications 14. If the frequency of a spinning object is doubled, what effect does this have on the centripetal acceleration? force in the following cases: (a) A boat makes a turn. (b) A plane makes a horizontal turn. (c) A satellite orbits Earth. 3. (5.1) What force is responsible for the tug your hand feels as you spin an object around at the end of a rope? 4. (5.1) Sketch a diagram of a mass moving in a vertical circle. Draw the velocity, centripetal acceleration, and centripetal force vectors at a point on the circle’s circumference. 5. (5.2) Why do spinning tires tend to stretch? 6. (5.2) A heavy mass attached to the end of a cable is spinning in a vertical circle. In what position is the cable most likely to break and why? 7. (5.2) What force acts as the centripetal force for a motorcycle making a turn? 8. (5.2) Explain why a truck needs a larger turning radius than a car when they move at the same speed. 9. (5.3) In what position of its orbit does a planet move the fastest? 10. (5.3) In the orbit of a planet, what does the semi-major axis of an ellipse represent? 11. (5.3) Equation 14, page 281, can be confusing because it uses the mass of one object and the period of another. (a) Explain what mEarth and TMoon represent. (b) Explain how equation 14 is used in the most general case. 12. (5.3) Use Kepler’s second law to explain why a planet moves more quickly when it is nearer to the Sun than when it is farther away in its orbital path. 13. (5.3) Why can’t Kepler’s constant of 1 a2/AU3 be used for moons orbiting planets or planets in other solar systems? 15. A slingshot containing rocks is spun in a vertical circle. In what position must it be released so that the rocks fly vertically upward? 16. People of different masses are able to ride a roller coaster and go through a vertical loop without falling out. Show the mathematical proof of this. 17. An eagle circles above the ground looking for prey. If it makes one complete circle with the radius of 25.0 m in 8.0 s, what is its speed? 18. A ride at the fair spins passengers around in a horizontal circle inside a cage at the end of a 5.0-m arm. If the cage and passengers have a speed of 7.0 m/s, what is the centripetal force the cage exerts on an 80.0-kg passenger? 19. What is the minimum speed that a glider must fly to make a perfect vertical circle in the air if the circle has a radius of 200.0 m? 20. A child spins an 800.0-g pail of water in a vertical circle at the end of a 60.0-cm rope. What is the magnitude of the tension of the rope at the top of the swing if it is spinning with the frequency of 2.0 Hz? 21. A driver is negotiating a turn on a mountain road that has a radius of 40.0 m when the 1600.0-kg car hits a patch of wet road. The coefficient of friction between the wet road and the wheels is 0.500. If the car is moving at 30.0 km/h, determine if it skids off the road. 22. The blade of a table saw has a diameter of 25.4 cm and rotates with a frequency of 750 rpm. What is the magnitude of the centripetal acceleration at the edge of the blade? 23. The tip of a propeller on an airplane has a radius of 0.90 m and experiences a centripetal acceleration of 8.88 104 m/s2. What is the frequency of rotation of the propeller in rpm? 24. Using information and equations from this chapter, verify that the speed of Earth in orbit around the Sun is approximately 107 000 km/h. 288 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 289 25. A car rounds a corner of radius 25.0 m with the 34. A star in the Andromeda galaxy is found to have centripetal acceleration of 6.87 m/s2 and a hubcap flies off. What is the speed of the hubcap? 26. An electron (m 9.11 1031 kg) caught in a magnetic field travels in a circular path of radius 30.0 cm with a period of 3.14 108 s. (a) What is the electron’s speed? (b) What is the electron’s centripetal acceleration? 27. Halley’s comet has an orbital period of 76.5 a. What is its mean orbital radius? 28. An extrasolar planet is found orbiting a star in the Orion nebula. Determine the mass of the star if the planet has an orbital period of 400.0 Earth days and an orbital radius of 1.30 1011 m. 29. The Milky Way is a spiral galaxy with all the stars revolving around the centre where there is believed to be a super-massive black hole. The black hole is 2.27 1020 m from our Sun, which revolves around it at a speed of 1234 m/s. (a) What is the period of our solar system’s orbit around the black hole? (b) How massive is the black hole? (c) What centripetal acceleration does our solar system experience as a result of the black hole’s gravity? 30. Newton hypothesized that a cannonball fired parallel to the ground from a cannon on the surface of Earth would orbit Earth if it had sufficient speed and if air friction were ignored. (a) What speed would the cannonball have to have to do this? (Use Kepler’s third law.) (b) If the mass of the cannonball were doubled, what time would it take to orbit Earth once? 31. Use the Moon’s period (27.3 d), its mass, and its distance from Earth to determine its centripetal acceleration and force. 32. Neptune has a moon Galatea that orbits at an orbital radius of 6.20 107 m. Use the data for Nereid from Table 5.6 on page 274 to determine Galatea’s orbital period. 33. Determine the orbital speed of Ariel, a moon of Uranus, using Table 5.5 (page 273) and Table 5.6 (page 274). a planet orbiting it at an average radius of 2.38 1010 m and an orbital period of 4.46 104 s. (a) What is the star’s mass? (b) A second planet is found orbiting the same star with an orbital period of 6.19 106 s. What is its orbital radius? Extensions 35. Paraphrase two misconceptions about centripetal force mentioned in this book. How do these misconceptions compare with your preconceptions of centripetal force? 36. Assuming a perfectly spherical Earth with uniform density, explain why a person standing at the equator weighs less than the same person standing at the North or South Pole. Consolidate Your Understanding Create your own summary of uniform circular motion, Kepler’s laws, and planetary and satellite motion by answering the questions below. If you want to use a graphic organizer, refer to Student References 4: Using Graphic Organizers pp. 869–871. Use the Key Terms and Concepts listed above and the Learning Outcomes on page 240. 1. In a few sentences, summarize how frequency, period, and velocity affect centripetal acceleration. 2. Explain to a classmate in writing why the velocity at different radii on a spinning disc will vary while the rotational frequency remains constant. Think About It Review your answers to the Think About It questions on page 241. How would you answer each question now? e TEST To check your understanding of circular motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 5 Newton’s laws can explain circular motion. 289 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 290 C H A P T E R 6 Key Concepts In this chapter, you will learn about: ■ work, mechanical energy, and power ■ the work-energy theorem ■ isolated and non-isolated systems ■ the law of conservation of energy Learning Outcomes When you have finished this chapter, you will be able to: Knowledge ■ use the law of conservation of energy to explain the behaviours of objects within isolated systems ■ describe the energy transformations in isolated and non-isolated systems using the work–energy theorem ■ calculate power output Science, Technology, and Society ■ explain that models and theories are used to interpret and explain observations ■ explain that technology cannot solve all problems ■ express opinions on the support found in Canadian society for science and technology measures that work toward a sustainable society 290 Unit III In an isolated system, energy is transferred from one object to another whenever work is done. Figure 6.1 Tension mounts as the motor pulls you slowly to the top of the first hill. Slowly you glide over the top, then suddenly, you are plunging down the hill at breathtaking speed. Upon reaching the bottom of the hill, you glide to the top of the next hill and the excitement begins all over again. As you race around the roller coaster track, each hill gets a bit lower until, at last, you coast to a gentle stop back at the beginning. You probably realize that because of friction the trolley can never regain the height of the previous hill, unless it is given a boost. It seems obvious to us that as objects move, kinetic energy is always lost. Energy is the most fundamental concept in physics. Everything that occurs in nature can be traced back to energy. The complicating factor is that there are so many forms of energy it is often very difficult to keep track of what happens to the energy when it is transf |
erred. Energy is a scalar quantity. This chapter concentrates on gravitational potential energy, kinetic energy, and elastic potential energy. In this chapter you will take the first steps to understanding the role of energy in nature. Specifically, you will learn how energy is given to and taken from objects when they interact with each other. 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 291 6-1 QuickLab 6-1 QuickLab Energy Changes of a Roller Coaster e WEB 7 Start the simulation. This activity uses the roller coaster simulation found at www.pearsoned.ca/school/physicssource. Problem How does the energy of a roller coaster vary as it travels on its track? Materials computer connected to the Internet clear plastic ruler Procedure 1 Click on the start button for the simulation. 2 Observe the motion of the cart. 3 Click on “continue” and note what happens to the motion of the trolley as it moves along the track. 4 Repeat step 3 until the simulation is complete. 5 Reset the simulation. 6 Use a see-through plastic ruler to measure the lengths of the potential energy bar (blue) and the kinetic energy bar (green) before you start the simulation. Record your measurements. ▼ Table 6.1 8 Each time the trolley pauses, measure the length of the potential energy and the kinetic energy bars and record the results in a table similar to Table 6.1. Questions 1. What assumptions are you making when you measure the lengths of the energy bars? 2. What is the effect on the potential energy of the trolley as it moves upward and downward? 3. What is the effect on the kinetic energy of the trolley as it moves upward and downward? Is this true as the trolley moves upward to the top of the first hill? Explain. 4. From the table, what happens to the energy of the trolley as it moves from the start to position “a”? 5. For each of the positions at which the trolley pauses, how does the sum of lengths of the bars change? What does the sum of these lengths represent? 6. Is there an energy pattern as the trolley moves along the track? Describe the pattern. 7. Do you think that this pattern is representative of nature? Explain. Position Length of Potential Energy Bar (mm) Length of Kinetic Energy Bar (mm) Sum of Lengths (mm) start a b Think About It 1. If two cars are identical except for the size of their engines, how will that affect their performance on the highway? 2. What is the “law of conservation of energy”? When does this law apply? 3. When work is done on an object, where does the energy used to do the work go? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. e LAB For a probeware activity, go to www.pearsoned.ca/ school/physicssource. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 291 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 292 info BIT Inuit hunters devised unusual ways of storing potential energy in a bow. One way they accomplished this was to tie cords of sinew along the back of the bow (Figure 6.3). The sinew was more heavily braided where strength was needed and less heavily braided where flexibility was important. When the bow was bent, the cords would stretch like a spring to store energy. In the absence of a wood source, bows were often made of antler or bone segments. e WEB To see photographs and to learn more about the technology of Inuit bows, follow the links at www.pearsoned.ca/ school/physicssource. 6.1 Work and Energy Figure 6.2 When the string on a bow is pulled back, elastic potential energy is stored in the bow. energy: the ability to do work Figure 6.3 Sinew-backed bow of the Inuit Copper people of the Central Arctic. Maximum gravitational potential energy is stored in the skier at the top of the run. Gravitational potential energy changes to kinetic energy and heat during the run. Figure 6.4 During the downhill run, the skier’s gravitational potential energy is continually converted into kinetic energy and heat. 292 Unit III Circular Motion, Work, and Energy An archer is pulling back her bowstring (Figure 6.2). She does work on the bow transforming chemical energy in her muscles into elastic potential energy in the bow. When she releases the string, the bow does work on the arrow. The elastic potential energy of the bow is transformed into the energy of motion of the arrow, called kinetic energy. As skiers ride up a lift, the lift’s motor is transforming chemical energy of the fuel into gravitational potential energy of the individuals. As they go downhill, gravity does work on the skiers transforming their gravitational potential energy into kinetic energy and heat. In both these examples, work transfers energy. In the case of the archer, energy is transformed from chemical energy into elastic potential energy and then into kinetic energy (from the archer to the bow to the arrow). In the case of the skier, energy is transformed from the chemical energy of the motor’s fuel into the gravitational potential energy of the skier at the top of the run and then into a combination of changing kinetic energy, gravitational potential energy, and heat of the skier as she speeds downhill. All these processes involve work. 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 293 Work Is Done When Force Acts Over a Displacement ) acts on an object resulting in a displacement (d ), a When a force (F transfer of energy occurs. This energy transfer is defined as the work done by the force. In introductory courses the quantity of work is usually defined by the equation W Fd. and d Work is a scalar quantity. However, the relative directions of the vectors F ) does not act are important. If the applied force (F parallel to the displacement (Figure 6.5), you must resolve the force into components that are parallel (F) and perpendicular (F) to the displacement. Only the component of the force parallel to the displacement actually does work. The component of the force acting perpendicular to the displacement does no work. work: a measure of the amount of energy transferred when a force acts over a given displacement. It is calculated as the product of the magnitude of applied force and the displacement of the object in the direction of that force. PHYSICS INSIGHT The unit of work and energy is the joule (J). It is a derived unit. 1 J 1 Nm kgm2 s2 1 F d F F Figure 6.5 When a force acts on an object, resulting in a displacement, only the component of the force that acts parallel to the displacement does work. If the box moves , does work. horizontally, only the horizontal component, F Thus, the equation for work is often written as W Fd where F is the magnitude of the component of the force that acts parallel to the displacement. In Figure 6.5, where the angle between the direction of the force and the direction of the displacement is , the component of the force parallel to the displacement is given by F Fcos F d If you replace F by Fcos , the calculation for work becomes Figure 6.6 W (Fcos )d Let’s look at two special cases. First, when the force acts parallel to the displacement, the angle 0° so that cos 1, making F F. This results in the maximum value for the work that the force could do over that displacement (Figure 6.6). Second, if the force acts perpendicular to the displacement, there is no parallel component. Mathematically, since 90° then cos 0 making F 0. In this case, the applied force does no work on the object (Figure 6.7). Figure 6.7 d F Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 293 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 294 Concept Check When a centripetal force acts on an object, the object shows no increase in speed and therefore no increase in kinetic energy. In terms of the work done by the centripetal force, explain why this is true. Example 6.1 Figure 6.8 shows a force of 150 N [0°] acting on an object that moves over a displacement of 25.0 m [25.0°] while the force acts. What is the work done by this force? d F 25.0° Figure 6.8 Practice Problems 1. You pull a sled along a horizontal surface by applying a force of 620 N at an angle of 42.0° above the horizontal. How much work is done to pull the sled 160 m? 2. A force acts at an angle of 30.0° relative to the direction of the displacement. What force is required to do 9600 J of work over a displacement of 25.0 m? 3. A force of 640 N does 12 500 J of work over a displacement of 24.0 m. What is the angle between the force and the displacement? 4. A bungee jumper with a mass of 60.0 kg leaps off a bridge. He is in free fall for a distance of 20.0 m before the cord begins to stretch. How much work does the force of gravity do on the jumper before the cord begins to stretch? Answers 1. 7.37 104 J 2. 443 N 3. 35.5º 4. 1.18 104 J Given 1.50 102 N [0°] F 25.0 m [25.0°] d Required work done by the force (W) Analysis and Solution From Figure 6.8, the angle between the force and the displacement is 25.0°. Draw a component diagram (Figure 6.9). The component that does work is F (cos 25.0°). Solve using the equation for work. W (Fcos )d (1.50 102 N)(cos 25.0°)(25.0 m) 3.399 103 N·m 3.40 103 J F F 25.0° F Figure 6.9 Component diagram Paraphrase and Verify The work done by the force is 3.40 103 J. If the force had acted parallel to the displacement, the maximum amount of work done would have been 3.75 103 J. Since cos 25.0° is about 0.91, the answer of 3.40 103 J, or 0.91 times the maximum value for W, is reasonable. 294 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 295 Gravitational Potential Energy An object is said to have potential energy if it has the ability to do work by way of its position or state. There are several forms of potential energy. Imagine a ride at the fair where the passengers are lifted vertically before being allowed to drop in free fall, as i |
n Figure 6.10. Ignoring friction, the work done by the machinery to lift the passengers and car to a height, h, is equal to the change in gravitational potential energy, EP. To lift the object straight up at a constant speed, the force applied must be equal but opposite to the force of gravity on the object. The equation for calculating work can be used to develop the equation for change in gravitational potential energy. EP W Fd where F is the magnitude of the force acting parallel to the displacement, and d is the magnitude of the displacement. To lift an object of mass m upward at a constant speed, the force is equal in magnitude and parallel, but opposite in direction, to the gravitational force, F g. mg where g is the magRecall from Unit II that Fg nitude of the acceleration due to gravity, which has a constant value of 9.81 m/s2 near Earth’s surface. It follows that F mg If the object is moved through a change in height h, so that d h, the change in potential energy equation becomes EP mgh Figure 6.10 A motor works transferring energy to the ride car. The gravitational potential energy gained produces the exciting free fall. So, when an object is moved upward, h increases and h is positive, and the potential energy increases (positive change). When an object is moved downward, h decreases and h is negative, and the potential energy decreases (negative change). gravitational potential energy: the energy of an object due to its position relative to the surface of Earth Concept Check Does the above equation for change in gravitational potential energy apply to objects that move over very large changes in height (e.g., change as experienced by a rocket)? Explain. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 295 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 296 Example 6.2 Practice Problems 1. An elevator car has a mass of 750 kg. Three passengers of masses 65.0 kg, 30.0 kg, and 48.0 kg, ride from the 8th floor to the ground floor, 21.0 m below. Find the change in gravitational potential energy of the car and its passengers. 2. A book with a mass of 1.45 kg gains 25.0 J of potential energy when it is lifted from the floor to a shelf. How high is the shelf above the floor? 3. The Mars rover lifts a bucket of dirt from the surface of Mars into a compartment on the rover. The mass of the dirt is 0.148 kg and the compartment is 0.750 m above the surface of Mars. If this action requires 0.400 J of energy, what is the gravitational acceleration on Mars? Answers 1. 1.84 105 J 2. 1.76 m 3. 3.60 m/s2 If the car and its passengers in Figure 6.10 have a mass of 500 kg, what is their change in gravitational potential energy when they are lifted through a height of 48.0 m? Given m 500 kg g 9.81 m/s2 h 48.0 m up down Required change in gravitational potential energy (EP) 48.0 m Analysis and Solution Sketch the movement of the car as in Figure 6.11 and solve for EP. EP mgh (5.00 102 kg)9.81 2.35 105 kg 2.35 105 (N m) 2.35 105 J m s2 m s2 (48.0 m) m Figure 6.11 Paraphrase The change in gravitational potential energy of the car and its passengers is a gain of 2.35 105 J. The object moved upward, gaining gravitational potential energy. If Ep1 represents the potential energy of an object at height h1 and Ep2 its potential energy when it is lifted to a height h2, then the change in potential energy is, by definition, Ep Since Ep Ep2 Ep1 mgh Ep2 Ep1 mg(h2 h1) Consider an object at ground level as having zero potential energy. If the object is raised from the ground level, h1 0) 0. It follows that Ep2 Ep2 0 mg(h2 mgh2 In general, the potential energy of an object at height h, measured from the ground, is Ep mgh 296 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 297 Choosing a Reference Point You see from the equation for the change in potential energy on page 295 that E depends only on the change in height, h. The values of h may be measured from any convenient reference point, as long as the reference point is kept the same for all the measurements when solving a problem. The change in height, and therefore the change in gravitational potential energy, is the same regardless of your frame of reference. Look at the book resting on the shelf in Figure 6.12. The value of h for the shelf can be defined relative to the floor (hf), relative to the table (ht), or even relative to the ceiling above the shelf (hc), in which case hc will have a negative value. Usually, you choose the frame of reference that most simplifies your measurements and calculations for h. For example, if you were trying to determine how much gravitational potential energy the book would lose as it fell from the shelf to the tabletop, then it would be logical to use the tabletop as your reference point. If you used another position as a reference point, your calculations might be slightly more complex, but the final answer for the amount of gravitational potential energy the book loses would be the same. Change in gravitational potential energy depends only on change in vertical height. The change in gravitational potential energy of an object depends only on the change in height. For example, the change in gravitational potential energy of a cart rolling down a frictionless ramp as in Figure 6.13 depends only on the vertical measurement, h. The actual distance an object travels, while it moves through a given change in height, does not affect its change in gravitational potential energy. d h Figure 6.13 As the cart rolls down the ramp, only the change in height h affects its change in gravitational potential energy. hc ht hf Figure 6.12 The book has gravitational potential energy due to its position on the shelf. reference point: an arbitrarily chosen point from which distances are measured PHYSICS INSIGHT The calculation of h from Figure 6.13 involves the use of the trigonometric h . ratio sin d Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 297 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 298 Example 6.3 Figure 6.14 shows a toy car track set up on a tabletop. (a) What is the gravitational potential energy of the car, which has a mass of 0.0250 kg, relative to the floor? (b) Calculate the change in gravitational potential energy of the car when it arrives at the bottom of the hill. Given m 0.0250 kg g 9.81 m/s2 2.15 m h1 0.950 m h2 up down toy car glide track Figure 6.14 table Practice Problems 1. A pile driver drops a mass of 550 kg from a height of 12.5 m above the ground onto the top of a pile that is 2.30 m above the ground. Relative to ground level, what is the gravitational potential energy of the mass (a) at its highest point? (b) at its lowest point? (c) What is the change in the gravitational potential energy of the mass as it is lifted from the top of the pile to its highest point? 2. A roller coaster trolley begins its journey 5.25 m above the ground. As the motor tows it to the top of the first hill, it gains 4.20 105 J of gravitational potential energy. If the mass of the trolley and its passengers is 875 kg, how far is the top of the hill above the ground? 3. A winch pulls a 250-kg block up a 20.0-m-long inclined plane that is tilted at an angle of 35.0° to the horizontal. What change in gravitational potential energy does the block undergo? Answers 1. (a) 6.74 104 J (b) 1.24 104 J (c) 5.50 104 J 2. 54.2 m 3. 2.81 104 J h1 2.15 m floor h2 0.950 m Required (a) gravitational potential energy at the top of the hill relative to the floor (Ep1) (b) change in the gravitational potential energy as the car moves from the top to the bottom of the hill (Ep) Analysis and Solution (a) To find gravitational potential energy relative to the floor, use that surface to define h 0 and make all height measurements from there. Ep1 (2.15 m) m s2 mgh1 (0.0250 kg)9.81 m2 0.527 kg 2 s 0.527 J (b) To find the change in gravitational potential energy, use the data and Figure 6.14 to calculate the change in height (h h2 h h2 h1). h1 Ep 0.950 m 2.15 m 1.20 m mg(h) (0.0250 kg)9.81 0.294 J (1.20 m) m s2 Paraphrase and Verify (a) The gravitational potential energy relative to the floor is 0.527 J. (b) The change in gravitational potential energy is 0.294 J. As the car rolls down the hill it loses 0.294 J of gravitational potential energy. Note: You could calculate Ep2 first and then use Ep Ep2 Ep1 298 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 299 Hooke’s Law In 1676, Robert Hooke, an English physicist, showed that the stretch produced by a force applied to a spring was proportional to the magnitude of the force. This relationship is known as Hooke’s Law and applies to any elastic substance when a force is exerted upon it. Thus, if a mass is suspended from a spring (Figure 6.15) the position (x) of the mass changes in proportion to the force (the weight (Fg) of the mass) exerted on the spring. x 0 x d x 2d x 3d x 4d d d d d Stretch Produced by a Force Applied to a Spring Fg 0 Fg W Fg 2W Fg 3W Fg 4W In an experiment to test this prediction, students suspended a series of masses from a spring and measured the position for each mass. Their data are shown in Table 6.2. Figure 6.15 The stretch produced by a force applied on a spring is proportional to the magnitude of the force. ▼ Table 6.2 Students’ experimental data Mass m (g) 0 200 400 600 800 1000 Weight Fg (N) Position x (m) 0 1.96 3.92 5.87 7.85 9.81 0 0.050 0.099 0.146 0.197 0.245 10 ) Force vs. Position for a Mass Suspended on a Spring 0 0.05 0.15 0.10 Position x (m) 0.20 0.25 Figure 6.16 Graph of data from Table 6.2 The students then plotted a graph of the magnitude of the applied force (Fg) as a function of the position (x) of the spring. The resulting line is a straight line with a constant slope (Figure 6.16). The equation of this line is F kx where k is the slope of the line. The slope of the line is determined by th |
e properties of the spring and is defined as the elastic or spring constant (k). This constant tells us how hard it is to stretch/compress the spring from the equilibrium position at x 0. For the graph in Figure 6.16, the slope is found as shown below: k F x Fi Ff xi xf 10.0 N 3.0 N 0.250 m 0.075 m 40 N m This force-position graph is characteristic for all springs whether the force stretches or compresses the spring. When a heavier or lighter spring is used, the slope of the line changes but the line is still straight. You will deal with Hooke’s Law in greater depth when you study simple harmonic motion in Chapter 7. PHYSICS INSIGHT A spring becomes nonelastic at a certain critical stretch value called the elastic limit. If the force applied does not exceed the elastic limit, the material will return to its original shape. If a spring is stretched beyond its elastic limit, its shape will be permanently distorted or the spring may break. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 299 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 300 Elastic Potential Energy elastic potential energy: the energy resulting from an object being altered from its standard shape, without permanent deformation Force vs. Position for an Elastic System ) ( N F e c r o F F 0 Area Ep Position x (m) x Figure 6.17 The area under the force-position curve is equal to the work done by the force to stretch the spring to that position. PHYSICS INSIGHT Ep, the change in elastic potential energy depends on the square of the stretch in the spring. That is: Ep 1 k (x2 2 2 x1 2) k (x)2 1 2 When the archer in Figure 6.2 draws her bow she stores another form of potential energy, elastic potential energy, in the bow. Both gravitational potential energy and elastic potential energy form part of mechanical energy. The study of elastic potential energy requires the use of Hooke’s law. The amount of energy stored in a spring is equal to the work done to stretch (or compress) the spring, without causing any permanent deformation. The force is not constant, so the equation for work used earlier, (W Fd) does not apply, because that equation requires a constant force acting over the displacement. However, when force-position graphs are used, work is equivalent to the area under the curve. The units for this area are N·m, equivalent to joules, the unit for work. You can therefore determine the amount of work done to stretch the spring from its equilibrium position to the position x by calculating the area of the shaded portion of Figure 6.17. the elastic or spring constant k: k slope F x Calculation of Elastic Potential Energy The area under the curve in Figure 6.17 is the shaded triangle whose area is calculated by A hb. The base (b) is equal to the magnitude 1 2 of the position (x), and the height (h) is equal to the magnitude of the force (F) at that position. Thus, in terms of force and position, the equation for the area under the curve is 1 W Fx 2 From Hooke’s law, the magnitude of the force (F) is equal to F kx, so the work done to stretch the spring can be written as: 1 W (kx)(x) 2 1 kx2 2 The work done to stretch (or compress) a spring from its equilibrium position to any position (x) results in storing elastic potential energy (Ep) in the spring. Therefore, the equation for the elastic potential energy stored in the spring is given by Ep 1 kx2 2 Concept Check Explain why it is incorrect to try to find the change in elastic potential energy of a stretched spring from the measurement of the change in the stretch. 300 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/28/08 11:13 AM Page 301 Example 6.4 A spring is stretched to a position 35.0 cm from its equilibrium position. At that point the force exerted on the spring is 10.5 N. (a) What is the elastic potential energy stored in the spring? (b) If the stretch in the spring is allowed to reduce to 20.0 cm, what is the change in the elastic potential energy? Given x1 F1 x2 35.0 cm 0.350 m 10.5 N 20.0 cm 0.200 m Required (a) elastic potential energy in the spring stretched ) to 0.350 m (Ep1 (b) change in the elastic potential energy when the stretch is reduced from 0.350 m to 0.200 m (Ep) Analysis and Solution (a) Calculate the value for k, the elastic constant for the spring, using Hooke’s law. F1 k kx1 F1 x1 10.5 N 0.350 m N m 30.0 Graph Showing Change in Elastic Potential Energy 10. From the data given, plot the graph of change in elastic potential energy, Figure 6.18. Next, 0 0.200 0.350 Position x (m) Figure 6.18 use EP kx2, to find the elastic potential energy for 1 2 a stretch of 0.350 m. This is equivalent to finding the area of the large triangle in Figure 6.18. Ep1 1 kx1 2 2 N m 30.0 1 2 1.8375 N m 1.84 Nm 1.84 J (0.350 m)2 m2 (b) To find the change in the elastic potential energy, first find the elastic potential energy at a stretch of 0.200 m and then subtract from that value the answer to part (a). This is equivalent to finding the shaded area of the graph in Figure 6.18. Practice Problems 1. A force of 125 N causes a spring to stretch to a length of 0.250 m beyond its equilibrium position. (a) What is the elastic potential energy stored in the spring? (b) If the spring contracts to a stretch of 0.150 m, what is the change in elastic potential energy? 2. An engineer is designing the suspension system for a car. He decides that the coil spring used in this car should compress 4.00 cm when a force of 1000 N is applied to it. (a) What is the spring constant of the spring? (b) If the spring is compressed a distance of 14.0 cm, what force must have been exerted on it? 3. The elastic constant for a spring is 750 N/m. (a) How far must you stretch a spring from its equilibrium position in order to store 45.0 J of elastic potential energy in it? (b) If you wanted to double the elastic potential energy stored in the spring, how much farther would you need to stretch it? 4. A spring has an elastic constant of 4.40 104 N/m. What is the change in elastic potential energy stored in the spring when its stretch is increased from 12.5 cm to 15.0 cm? 5. When a spring is stretched by 0.400 m from its equilibrium position, its elastic potential energy is 5.00 102 J. (a) What is the magnitude of the force required to produce this amount of stretch? (b) If the force causing the stretch is changed to 1000 N, how much change in elastic potential energy results? Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 301 06-PearsonPhys20-Chap06 7/28/08 11:15 AM Page 302 Answers 1. (a) 15.6 J (b) 10.0 J 2. (a) 2.50 104 N/m (b) 3.50 103 N 3. (a) 0.346 m (b) 0.143 m 4. 1.51 102 J 5. (a) 2.50 103 N (b) 420 J kinetic energy: the energy due to the motion of an object PHYSICS INSIGHT There are two kinds of kinetic energy. The kinetic energy studied here is more correctly referred to as translational kinetic energy, since the objects are moving along a line. Earth has both translational kinetic energy (because it orbits the Sun) and rotational kinetic energy (because it spins on its axis). The elastic potential energy for a stretch of 0.200 m is: EP2 2 1 kx2 2 N 30.0 1 m 2 0.600 Nm 0.600 J (0.200 m)2 The change in the elastic potential energy is: EP EP2 EP1 0.600 J 1.84 J 1.24 J Paraphrase (a) The energy stored in the spring at the initial stretch is 1.84 J. (b) When the stretch is reduced from 0.350 m to 0.200 m, the elastic potential energy stored in the spring reduced by 1.24 J to 0.600 J. Kinetic Energy Examine Figure 6.19. When the archer releases the arrow, the bowstring exerts a non-zero force on the arrow, which accelerates the arrow toward its target. As the arrow gains speed, it gains kinetic energy (Ek). v Figure 6.19 Figure 6.20 When an object is in free fall, gravity is working to increase its kinetic energy. When the hiker in Figure 6.20 drops the rock off the cliff, the force of gravity accelerates the rock downward increasing its speed and thus its kinetic energy. Kinetic energy is a scalar quantity. The kinetic energy of an object is calculated using the equation Ek 1 mv2 2 Concept Check If the kinetic energy of an object doubles, by what factor does its speed increase? 302 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 303 Example 6.5 On a highway, a car of mass 1.2 103 kg, travelling at 20 m/s, has kinetic energy equal to a loaded van of mass 4.8 103 kg. What is the speed of the van? Given m1 v1 Ekcar 1.2 103 kg; m2 20 m/s Ekvan 4.8 103 kg; Required the speed of the van (v2) Analysis and Solution The two vehicles have equal kinetic energy mv 2. Find the kinetic energy of the car and then use that value to solve for the speed of the van. 1 2 Ekcar 1 mvcar 2 2 Ekvan 2.4 105 J m 2 (1.2 103 kg)20 s 1 2 mvvan 1 2 2 2.4 105 kgm2 s2 (2.4 105 J)(2) 4.8 103 kg vvan 2.4 105 J 10 m/s Paraphrase The van is travelling at 10 m/s. M I N D S O N Energy of Impact Practice Problems 1. A 45.0-kg girl pedals a 16.0-kg bicycle at a speed of 2.50 m/s. What is the kinetic energy of the system? 2. A car travelling at 80.0 km/h on a highway has kinetic energy of 4.2 105 J. What is the mass of the car? 3. A skateboarder with a mass of 65.0 kg increases his speed from 1.75 m/s to 4.20 m/s as he rolls down a ramp. What is the increase in his kinetic energy? Answers 1. 1.91 102 J 2. 1.7 103 kg 3. 474 J Project LINK How will the concept of the kinetic energy of a moving vehicle relate to the design of your persuader apparatus? 3. Investigate the incidence of meteor collisions in Canada. Where is the meteor impact crater that is closest to where you live? Approximately how many meteors have landed in Alberta? What was the greatest kinetic energy for a meteor that landed (a) in Alberta (b) in Canada? There is evidence that many meteors have hit Earth’s surface. The vast quantity of kinetic energy that these meteors have at the time of impact is revealed by the size of |
the craters that they create (Figure 6.21). 1. What types of measurements would scientists need to make in order to estimate the kinetic energy of the meteor at the instant of impact? 2. What types of experiments could be done to verify the scientists’ assumptions? Figure 6.21 Meteor impact craters are found in all regions of Earth. This one, called the Barringer crater, is in Arizona. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 303 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 304 Example 6.6 A man on a trampoline has a mass of 75.0 kg. At the instant he first touches the surface of the trampoline (at its rest position) he is descending with a speed of 8.00 m/s. At his lowest point, the man is 0.650 m below the trampoline’s rest position. (a) What is the kinetic energy of the man when he first contacts the trampoline? (b) If you assume that, at his lowest point, all of the man’s kinetic energy is transformed into elastic potential energy, what is the elastic constant for the trampoline? Practice Problems 1. A bow that has an elastic constant of 2500 N/m is stretched to a position of 0.540 m from its rest position. (a) What is the elastic potential energy stored in the bow? (b) If all of the elastic potential energy of the bow were to be transformed into kinetic energy of a 95.0-g arrow, what would be the speed of the arrow? 2. Cannon A fires a 1.5-kg ball with a muzzle velocity of 550 m/s, while cannon B fires cannon balls with one-third the mass but at twice the muzzle velocity. Which of these two cannons would be more effective in damaging a fortification? Explain why. 3. It is estimated that the meteor that created the crater shown in Figure 6.21 on the previous page had a radius of 40 m, a mass of approximately 2.6 108 kg, and struck Earth at a speed of nearly 7.20 104 km/h. (a) What was the kinetic energy of the meteor at the instant of impact? (b) When one tonne (t) of TNT explodes, it releases about 4.6 109 J of energy. In terms of tonnes of TNT, how much energy did the meteor have at impact? Answers 1. (a) 365 J 2. EkA 3. (a) 5.2 1016 J : EkB (b) 87.6 m/s 3 : 4. Ball B will do more damage. (b) 1.1 107 t v m 75.0 kg Given m 75.0 kg v 8.00 m/s x 0.650 m Required (a) kinetic energy of the man (Ek) (b) the elastic constant of the trampoline (k) 0.650 m Figure 6.22 Analysis and Solution (a) Find the initial kinetic energy, by using Ek Ek 1 mv2 2 1 mv2 2 1 2 m (75.0 kg)8.00 s m2 s2 2.40 103 kg 2 2.40 103 J (b) Assume that the elastic potential energy at 0.650 m is 2.40 103 J and solve for the elastic constant. Ep 1 kx2 2 Solve for k. k 2Ep x2 2(2.40 103 J) (0.650 m)2 1.136 104 N m 1.14 104 N m Paraphrase (a) The kinetic energy of the man is 2.40 103 J. (b) The elastic constant of a spring that stores 2.40 103 J of elastic potential energy when it is stretched 0.650 m is 1.14 104 N/m. 304 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 305 6.1 Check and Reflect 6.1 Check and Reflect 1. If a force does not act parallel to the resulting displacement, what is the effect on the work done by the force? 8. A spring has an elastic constant of 650 N/m. Initially, the spring is compressed to a length of 0.100 m from its equilibrium position. 2. Describe how a non-zero force can act on an object over a displacement and yet do no work. 3. Explain why the frame of reference affects the calculated value of an object’s gravitational potential energy but not the change in its gravitational potential energy. 4. What is meant by elastic potential energy? 5. A force of 1500 N [up] acts to lift an object of 50.0-kg mass to a height of 24.0 m above its original position. (a) How much work did the force do on the object? (b) What was the gain in the object’s gravitational potential energy? (c) What might account for the difference in the two answers? 6. A force of 850 N [30] acts on an object while it undergoes a displacement of 65.0 m [330]. What is the work the force does on the object? 7. You are working on the 5th floor of a building at a height of 18.0 m above the sidewalk. A construction crane lifts a mass of 350 kg from street level to the 12th floor of the building, 22.0 m above you. Relative to your position, what is the gravitational potential energy of the mass (a) at street level? (b) when it is on the 12th floor? (c) What is its change in gravitational potential energy as it is raised? (a) What is the elastic potential energy stored in the spring? (b) How much further must the spring be compressed if its potential energy is to be tripled? 9. Two cars (A and B) each have a mass of 1.20 103 kg. The initial velocity of car A is 12.0 m/s [180] while that of car B is 24.0 m/s [180]. Both cars increase their velocity by 10.0 m/s [180]. (a) Calculate the gain in kinetic energy of each car. (b) If both cars gain the same amount of velocity, why do they gain different amounts of kinetic energy? 10. A cart with a mass of 3.00 kg rolls from the top of an inclined plane that is 7.50 m long with its upper end at a height of 3.75 m above the ground. The force of friction acting on the cart as it rolls is 4.50 N in magnitude. (a) What is the change in gravitational potential energy when the cart moves from the top of the inclined plane to ground level? (b) What is the work done by friction? 11. An ideal spring with an elastic constant of 2000 N/m is compressed a distance of 0.400 m. (a) How much elastic potential energy does this compression store in the spring? (b) If this spring transfers all of its elastic potential energy into the kinetic energy of a 2.00-kg mass, what speed would that mass have? Assume the initial speed of the mass is zero. e TEST To check your understanding of potential and kinetic energy, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 305 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 306 6.2 Mechanical Energy mechanics: the study of kinematics, statics, and dynamics mechanical energy: the sum of potential and kinetic energies In physics, the study of mechanics includes kinematics (the study of motion), statics (the study of forces in equilibrium), and dynamics (the study of non-zero forces and the motion that results from them). Gravitational potential energy, elastic potential energy, and kinetic energy form what is called mechanical energy. When work is done on a system, there may be changes in the potential and kinetic energies of the system. This relationship is expressed as the work-energy theorem. info BIT Doubling the speed of a vehicle means quadrupling the necessary stopping distance. This relationship between stopping distance and speed is based on the physics of work and kinetic energy. The Work-Energy Theorem A 1.50 105-kg jet plane waits at the end of the runway. When the airtraffic controller tells the pilot to take off, the powerful engines can each produce more than 2.5 105 N of thrust to accelerate the plane along the runway. In order to produce the speed of about 250 km/h required for takeoff, the engines would need to convert more than 3 108 J of chemical energy from the fuel supply into kinetic energy. Figure 6.23 The work done by the jet’s engines must convert enough chemical energy into kinetic energy to produce a velocity sufficient for takeoff. e LAB For a probeware activity, go to www.pearsoned.ca/ school/physicssource. PHYSICS INSIGHT In the example of the airplane takeoff, is the angle between the direction of Fnet or a, and d. As explained by Newton’s Laws of Motion, the non-zero net force causes the jet plane to accelerate along the runway. Since a change in kinetic energy must involve a change in speed, kinetic energy changes are always the result of the acceleration, which in turn is caused by a non-zero net force. In terms of work and energy, this means that changes in kinetic energy (Ek) are always the result of the work done by a non-zero net force (Wnet). Ek Wnet (Fnet)(cos )(d) (ma)(cos )(d) In all cases, work done by a non-zero net force results in a change in kinetic energy but the applied forces on an object may cause changes in its potential energy, its kinetic energy, or both. For example, once the jet plane is in the air, the thrust produced by its engines must increase its speed (Ek) as well as cause it to gain altitude (Ep). 306 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 307 Zero and Non-zero Net Forces and Effects on Energy A motor that is pulling a block up a frictionless inclined plane at a constant speed (Figure 6.24) is exerting a force that causes a change in gravitational potential energy but not kinetic energy. The constant speed indicates that the applied force (Fapp) is exactly balanced by the Fg component of Fg. There is zero net force; Fapp Fg. p u p u h ill FN a 0 v constant Fapp h ill o w n d o w n d Fg Fg Fg Since Fnet 0 Fapp Fg FN Fg Fg Fapp Fg Figure 6.24 If all the forces acting on a block combine to produce a net force of zero, the block moves up the incline at a constant speed. It increases its gravitational potential energy but not its kinetic energy. If, however, the force applied is now increased so that there is a Fg (Figure 6.25), the forces are no longer non-zero net force and Fapp balanced, the block accelerates up the incline, and both kinetic energy and potential energy change. Now the work done on the block is transferred to both its kinetic energy and its gravitational potential energy. This is expressed mathematically as W E or, in more detail, as W Ek Ep This is known as the work-energy theorem. p u FN p u h ill a 0 v Fapp h ill o w n d o w n d Fg Fg Fg Since Fnet 0 Fapp Fg a 0 Hence v increases and Ek increases. FN Fg Fg Fapp Fg Figure 6.25 If the forces acting on a block are such that there is a non-zero net force Fg. Both the kinetic energy and the gravitational |
potential energy up the plane, then Fapp will increase as the block moves up the incline. The work-energy theorem states that the work done on a system is equal to the sum of the changes in the potential and kinetic energies of the system. PHYSICS INSIGHT The symbol EP could refer to gravitational potential energy, elastic potential energy, or the sum of the gravitational and elastic potential energies. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 307 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 308 Concept Check A block is sliding down an inclined plane. If there is no friction, describe a situation where the net work might still be negative. Example 6.7 A block is moved up a frictionless inclined plane by a force parallel to the plane. At the foot of the incline, the block is moving at 1.00 m/s. At the top of the incline, 0.850 m above the lower end, the block is moving at 4.00 m/s. The block has a mass of 1.20 kg. What is the work done on the block as it moves up the incline? Practice Problems 1. A mountain climber rappels down the face of a cliff that is 25.0 m high. When the climber, whose mass is 72.0 kg, reaches the bottom of the cliff he has a speed of 5.00 m/s. What is the work done on the climber by the rope? 2. A force of 150 N [up] acts on a 9.00-kg mass lifting it to a height of 5.00 m. (a) What is the work done on the mass by this force? (b) What is the change in gravitational potential energy? (c) What change in kinetic energy did the mass experience? 3. Draw a free-body diagram for the forces on the mass in question 2. (a) Calculate the net force acting on the mass. (b) Calculate the work done on the mass by the net force. (c) How does this relate to the answer to question 2(c)? Answers 1. 1.68 104 J 2. (a) 750 J (b) 441 J (c) 309 J 3. (a) 61.7 N [up] (b) 309 J (c) Ek 309 J Given m 1.20 kg 1.00 m/s v1 4.00 m/s v2 h 0.850 m g 9.81 m/s2 1.00 m/s v1 p u h ill n w o d w o d h ill p u n Fapp 4.00 m/s v2 h 0.850 m Required work done on the block as it moves up the incline (W) Figure 6.26 Analysis and Solution The work-energy theorem states that the work will be equal to the sum of the changes in the kinetic and potential energies. For the change in kinetic energy find the difference in the final and initial kinetic energy using the final and initial speeds. Change in gravitational potential energy can be found from the change in height (Figure 6.26). W Ek (Ek2 mv2 1 2 Ep Ek1) (mgh) 1 2 mv1 2 2 (mgh) 1 2 (1.20 kg)(4.00 m/s)2 (1.20 kg)(1.00 m/s)2 1 2 (1.20 kg)(9.81 m/s2)(0.850 m) (9.60 J 0.60 J) (10.01 J) 19.0 J Paraphrase and Verify The work caused the block to gain a total of 19.0 J, the sum of 9.00 J of kinetic and 10.0 J of potential energy as it moved up the ramp. 308 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 309 Calculations of Mechanical Energy To calculate the mechanical energy of an object is simply to find the total of the kinetic energy and all forms of potential energy. Em Ek Ep Because gravitational potential energy is defined relative to a reference point, mechanical energy will also depend on that reference point. Example 6.8 A cannon ball is fired from Earth’s surface. At the peak of its trajectory, it has a horizontal speed of 160 m/s and is 1.20 103 m above the ground. With reference to the ground, what is the mechanical energy of the cannon ball at the highest point on its trajectory, if the mass of the cannon ball is 5.20 kg? v2 h up v1 down Figure 6.27 Given m 5.20 kg v 160 m/s h 1.20 103 m g 9.81 m/s2 Required mechanical (total) energy of the cannon ball (Em) Analysis and Solution At the top of its trajectory, the cannon ball has kinetic energy due to its horizontal motion, and gravitational potential energy because of its height above the ground. Em Ep Ek 1 2 mv2 mgh Practice Problems 1. A rocket is accelerating upward. When the rocket has reached an altitude of 5.00 103 m, it has reached a speed of 5.40 103 km/h. Relative to its launch site, what is its mechanical energy, if the mass of the rocket is 6.50 104 kg? 2. What is the speed of a 4.50-kg cannon ball if, at a height of 275 m above the ground, its mechanical energy relative to the ground is 6.27 104 J? 3. As a roller coaster trolley with a mass of 600 kg coasts down the first hill, it drops a vertical distance of 45.0 m from an initial height of 51.0 m above the ground. If, at the bottom of the hill, its speed is 30.0 m/s: (a) what is the trolley’s mechanical energy relative to the top of the hill, and (b) what is the trolley’s mechanical energy relative to the ground? (5.20 kg)9.81 2 m s 1 2 (5.20 kg)160 6.656 104 J 6.121 104 J 1.28 105 J m s2 (1.20 103 m) Answers 1. 7.63 1010 J 2. 150 m/s 3. (a) 5.13 103 J (b) 3.05 105 J Paraphrase and Verify The total energy of the cannon ball at the top of the trajectory is 1.28 105 J. The gravitational potential energy is positive because the cannonball is higher than the reference point. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 309 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 310 6.2 Check and Reflect 6.2 Check and Reflect Knowledge Extensions 1. What are the forms of energy that make up 8. The figure below shows the force versus mechanical energy? 2. Why does your choice of a frame of reference affect the calculated value of the mechanical energy? 3. What is the relationship between the net force and kinetic energy? 4. State the work-energy theorem. Applications 5. A net force of 5.75 103 N [180] acts on a mass of 23.0 kg. If, while the force acts, the mass travels through a displacement of 360 m [210], what work did the net force do on the object? Into what form of energy was this work transferred? 6. At a height of 75.0 m above the ground, a cannon ball is moving with a velocity of 240 m/s [up]. If the cannon ball has a mass of 12.0 kg, what is its total mechanical energy relative to the ground? What effect would there be on the answer, if the velocity of the cannon ball were downward instead of upward? Explain. 7. A mass of 8.50 kg is travelling 7.50 m/s [up]. It is acted on by a force of 340 N [up] over a displacement of 15.0 m [up]. (a) What work does the applied force do on the object? (b) What is its gain in potential energy? (c) What is its change in kinetic energy? (d) What is its speed at the end of the 15.0-m displacement? 310 Unit III Circular Motion, Work, and Energy displacement graph for an elastic spring as it is compressed a distance of 0.240 m from its equilibrium position by a force of magnitude 2.40 103 N. A 7.00-kg mass is placed at the end of the spring and released. As the spring expands, it accelerates the mass so that when the spring’s compression is still 0.180 m from its equilibrium position, the mass has a speed of 6.00 m/s. (a) What is the mechanical energy in this system when the spring is compressed to 0.240 m? (b) What is the mechanical energy in the system when the spring is compressed to 0.180 m? (c) How much work has been done on the mass by this system as the spring expanded from a compression of 0.240 m to 0.180 m? (d) How does the work done on the mass by the spring compare to the kinetic energy of the mass? equilibrium position of spring Force vs. Displacement for an Elastic Spring 2400 1800 1200 600 ) ( N F e c r o F 0 0.06 0.12 Displacement d (m) 0.18 0.24 e TEST To check your understanding of the work–energy theorem and mechanical energy, follow the eTest links at www.pearsoned.ca/school/physicssource. 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 311 6.3 Mechanical Energy in Isolated and Non-isolated Systems Isolated Systems Imagine two people are in an isolated (sealed) room. They may complete as many money transfers as they like but the total amount of money in the room before and after each transfer will be the same. We can say that the total amount of money in this system is conserved, in that it does not change during transactions. PHYSICS INSIGHT An Open System can exchange both energy and matter with its surroundings. A Closed System can exchange energy but not matter with its surroundings. An Isolated System cannot exchange energy or matter with its surroundings. Figure 6.28 In an isolated room, the total amount of money in the room, before and after a transaction, is constant. Figure 6.29 In a non-isolated room, the amount of money in the room may change. Now imagine that the room is not isolated. In this case, money may be taken out of (or put into) the room so that the total amount of money in the room is not necessarily constant. In this system, it cannot be said that money is conserved. It would be much more complex to keep track of the money transfers that occur in this non-isolated room compared with those occurring in the isolated room. In physics, when the energy interactions of a group of objects need to be analyzed, we often assume that these objects are isolated from all other objects in the universe. Such a group is called an isolated system. Isolated Systems and Conservation of Energy While objects within an isolated system are free to interact with each other, they cannot be subjected to unbalanced forces from outside that system. In terms of mechanical energy, that means that no force from outside the system may work to transfer energy to or from any object inside the system. The quantity of energy in the system must be constant. Even though friction may seem like an internal force, its effect is to allow energy to escape from a system as heat. Thus, an isolated system must also be frictionless. These ideas will be further explored later in the chapter. info BIT In everyday terms, energy conservation means to use as little energy as possible to accomplish a task. In physics, energy conservation refers to systems, such as an ideal pendulum, in which the total amount of energy is constant. isolated system: a group of objects assumed to be isolate |
d from all other objects in the universe Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 311 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 312 e TECH Consider the transformation of energy from potential energy to kinetic energy in a falling object, and in a ball bouncing on a trampoline. Follow the eTech links at www.pearsoned.ca/school/ physicssource. e SIM Find out more about the mechanical energy, gravitational potential energy, and kinetic energy of a satellite-Earth system or a projectile-Earth system. Go to www.pearsoned.ca/school/ physicssource. Conservation of Mechanical Energy Because the mechanical energy (the sum of potential and kinetic energies) for an isolated system must be a constant, it follows that if you calculate the mechanical energy (Em) at any two randomly chosen times, the answers must be equal. Hence, Em2 Em1 (1) Within an isolated system, energy may be transferred from one object to another or transformed from one form to another, but it cannot be increased or decreased. This is the law of conservation of energy. Relationship between kinetic and potential energy in an isolated system The law of conservation of energy is one of the fundamental principles of science and is a powerful mathematical model for analysis and prediction of the behaviour of objects within systems. Viewed from a slightly different perspective, conservation of energy states that, in terms of mechanical energy, any gain in kinetic energy must be accompanied by an equal loss in potential energy. Ek Ep (2) Statements (1) and (2) are equivalent. This can be verified as follows. If total energy remains constant regardless of time or position, then Em2 Em1. But mechanical energy is the sum of the kinetic and potential energies. Therefore, Ek2 Ep2 Ek1 Ep1 Hence, Ek1 Ek2 Ep2 Ep1 Ek (Ep2 Ep1) Thus, Ek Ep is true. Energy vs. Position for a Block Sliding Down an Inclined Plane Em Ek Em Ep Ek Ep Ek Em Ep ) J ( E y g r e n E Ep Ek Position d (m) Figure 6.30 In an isolated system the loss in gravitational potential energy is equal to the gain in kinetic energy. As a block slides down a frictionless inclined plane, its gravitational potential energy will decrease and its kinetic energy will increase. Figure 6.30 shows the energy-position graph for this isolated system. As the block moves down the plane, the sum of the heights of the potential and kinetic energy curves (value of blue line plus value of red line) at any point is equal to the block’s mechanical energy (E m). The mechanical energy (shown by the purple line) is constant; therefore, energy is conserved. This graph is typical of an isolated system. 312 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 313 Example 6.9 A frictionless roller coaster car has a mass (m) of 8.00 102 kg. At one point on its journey, the car has a speed of 4.00 m/s and is 35.0 m above the ground. Later, its speed is measured to be 20.0 m/s. (a) Calculate its total initial mechanical energy relative to the ground. (b) What is its gravitational potential energy in the second instance? v1 Given m 8.00 102 kg 4.00 m/s 35.0 m 20.0 m/s v2 g 9.81 m/s2 h1 Required (a) mechanical energy (Em1) (b) gravitational potential energy when the speed is 20.0 m/s (Ep2) Analysis and Solution A frictionless roller coaster can be treated as an isolated system. (a) The mechanical energy at any point is the sum of its kinetic and potential energies. Em1 Em1 Ep1 Ek1 1 mv1 2 2 mgh1 (8.00 102 kg)4.00 1 2 m s 2 (8.00 102 kg)9.81 m s2 (35.0 m) 2.810 105 J 2.81 105 J (b) The system is defined as isolated, meaning that energy is conserved. By the law of conservation of energy, the mechanical energy at any two points must be equal. The gravitational potential energy at the second point must be equal to the mechanical energy less the kinetic energy at the second point. Ek2 Em2 Ep2 Ep2 Em1 Em1 Em1 Em1 Ek2 1 mv2 2 2 m 1 2 (8.00 102 kg)20.0 2.810 105 J s 2 2.810 105 J 1.600 105 J 1.21 105 J Practice Problems 1. In an isolated system, a crate with an initial kinetic energy of 250 J and gravitational potential energy of 960 J is sliding down a frictionless ramp. If the crate loses 650 J of gravitational potential energy, what will be its final kinetic energy? 2. A mass of 55.0 kg is 225 m above the ground with a velocity of 36.0 m/s [down]. Use conservation of energy to calculate its velocity when it reaches a height of 115 m above the ground. Ignore the effects of air resistance. 3. A human “cannon ball” in the circus is shot at a speed of 21.0 m/s at an angle of 20 above the horizontal from a platform that is 15.0 m above the ground. See Figure 6.31. (a) If the acrobat has a mass of 56.0 kg, what is his gravitational potential energy relative to the ground when he is at the highest point of his flight? Ignore the effects of air resistance. (b) If the net in which he lands is 2.00 m above the ground, how fast is he travelling when he hits it? human cannon ball cannon 15.0 m net 2.00 m Figure 6.31 Answers 1. 900 J 2. 58.8 m/s [down] 3. (a) 9.69 103 J (b) 26.4 m/s Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 313 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 314 e SIM Learn about the relationships among the mechanical, kinetic, and gravitational potential energies of a pendulum. Go to www.pearsoned.ca/school/ physicssource. Paraphrase and Verify (a) The total initial mechanical energy relative to the ground is 2.81 105 J. (b) The gravitational potential energy at a speed of 20.0 m/s is 1.21 105 J. The kinetic energy increased from 6.40 103 J to 1.60 105 J, while the gravitational potential energy decreased from 2.74 105 J to 1.21 105 J. As kinetic energy increases, potential energy decreases. When the speed is 20.0 m/s, the car must be below its starting point. M I N D S O N Energy and Earth’s Orbit At its closest point to the Sun (perihelion), around January 4th, Earth is about 147 million kilometres from the Sun. At its farthest point from the Sun (aphelion), around July 5th, Earth is about 152 million kilometres from the Sun. • In terms of the conservation of energy, what conclusions can be made about Earth’s speed as it moves around the Sun? • What assumptions must you make to support your conclusions? A Simple Pendulum A simple pendulum is an excellent approximation of an isolated system. During its downswing, Earth’s gravity does work on the pendulum to transfer gravitational potential energy into kinetic energy. On the upswing, gravity transfers kinetic energy back into gravitational potential energy. The mechanical energy of the pendulum is constant (Figure 6.32). Newton’s third law of motion states that for every action force there is an equal but opposite reaction force. This means that as Earth’s gravity acts on the pendulum converting gravitational potential energy into kinetic energy, the pendulum must also act to convert gravitational potential energy to kinetic energy for Earth; Earth must be part of the isolated system that contains the pendulum. Earth’s mass compared to that of the pendulum is enormous, so its reaction to the pendulum is immeasurably small. This explains why we can ignore the effects of the pendulum on Earth and analyze the pendulum as if it were an isolated system. Treating the pendulum as an isolated system greatly simplifies the calculations of the system’s mechanical energy. It means that the force of gravity works on the pendulum without changing the energy in the system. In fact, while work done by the force of gravity may transfer energy from one form to another it never causes a change in mechanical energy (Figure 6.33). Forces that act within systems but do not change their mechanical energy are defined as conservative forces. This type of force will be discussed in more detail later in this chapter. Em Ep 0 at Ep max max Ep min Ek Em 0 Ek at Ep min where h 0 min Ep max Ek Figure 6.32 As a pendulum swings, gravity acts to convert energy back and forth between gravitational potential energy and kinetic energy. PHYSICS INSIGHT If h 0 had been defined to occur at the lowest point of the pendulum’s swing, the gravitational potential energy at the lowest point would be zero and the mechanical energy would be equal to the kinetic energy. Then, at the highest point on the swing, where movement stops and kinetic energy is zero, the mechanical energy would be equal to the potential energy. 314 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 315 Because the pendulum acts as an isolated system, energy is conserved. To calculate the mechanical energy of a pendulum it is necessary to know its mass, its height above the reference point, and its speed. If all of those values are known at any one point on its swing, then the mechanical energy of the pendulum is known at all points on its swing. Once the mechanical energy is known, it can be used to predict the pendulum’s motion at any instant along its path, and to correlate kinetic and potential energy with the amplitude of the swing. Energy Conservation in a Simple Pendulum ) Em Ep Ek Em Ep Ek 0.5a 0 0.5a a Position x (m) Figure 6.33 The force of gravity acts to change the gravitational potential energy and kinetic energy of the pendulum so that the mechanical energy remains constant. hmax h 0 a 0 hmax a Example 6.10 An ideal pendulum, as shown in Figure 6.32, is suspended by a string that is 2.00 m long. It is pulled sideways and released. At the highest point of its swing the pendulum bob is 25.0 cm above the floor. At the lowest point of its swing the pendulum bob is 5.00 cm above the floor. The mass of the pendulum bob is 250 g. (a) What is the mechanical energy of the pendulum, relative to the floor, when the bob is at its highest point? (b) What is the mechanical energy of the pendulum, relative to the floor, when the bob is at its lowest point? (c) Wha |
t is the kinetic energy of the bob when it is at its lowest point? (d) What is the speed of the pendulum bob when the bob is at its lowest point? Given m 250 g 0.250 kg g 9.81 m/s2 h1 h2 v1 25.0 cm 0.250 m 5.00 cm 0.0500 m 0 Required (a) sum of gravitational potential and kinetic energies of the pendulum at the highest point (Em1) (b) mechanical energy of the pendulum at the lowest point (Em2) (c) kinetic energy of the bob at the lowest point (Ek2) (d) speed of the bob at the lowest point (v2) Practice Problems 1. When the pendulum bob in Example 6.10 is 15.0 cm above the floor, calculate: (a) its mechanical energy relative to the floor (b) its kinetic energy (c) its speed 2. A model rocket has a mass of 3.00 kg. It is fired so that when it is 220 m above the ground it is travelling vertically upward at 165 m/s. At that point its fuel runs out so that the rest of its flight is without power. Assume that the effect of air friction is negligible and that all potential energies are measured from the ground. (a) What is the mechanical energy of the rocket, relative to the ground, when it is 220 m above the ground? (b) When it reaches the highest point on its trajectory, what will its gravitational potential energy be? (c) How far above the ground is the rocket at its highest point? (d) When it hits the ground, what is its speed? Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 315 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 316 3. A roller coaster trolley and its passengers have a mass of 840 kg. The trolley comes over the top of the first hill with a speed of 0.200 m/s. The hill is 85.0 m above the ground. The trolley goes down the first hill and up to the crest of the second hill 64.0 m above the ground. Ignore the effect of frictional forces. What is the kinetic energy of the trolley at the top of the second hill? 4. A pole-vaulter with a mass of 56.0 kg tries to convert the kinetic energy of her approach into height. (a) What is the maximum height she can expect to attain if her approach speed is 8.00 m/s? Assume that the centre of mass of the vaulter is initially 0.850 m above the ground. (b) Describe the energy changes that occur from the time the vaulter starts to run until she reaches the highest point of her jump. Answers 1. (a) 0.613 J (b) 0.245 J (c) 1.40 m/s 2. (a) 4.73 104 J (b) 4.73 104 J (c) 1.61 103 m (d) 178 m/s 3. 1.73 105 J 4. (a) 4.11 m Analysis and Solution (a) At its highest point, the speed of the pendulum is zero. Thus, the mechanical energy at that point is equal to its gravitational potential energy. Em1 Ep1 Ek1 Em1 mgh1 1 mv1 2 2 (0.250 kg)9.81 kgm2 s 2 0.6131 0.613 J m s2 1 (0.250 m) (0.250 kg)(0)2 2 (b) In an isolated system, the mechanical energy is constant. Thus, by the law of conservation of energy the mechanical energy at its lowest point is equal to the mechanical energy at its highest point. Em2 Em2 0.613 J Em1 (c) The kinetic energy at the lowest point is the difference between the mechanical and gravitational potential energies at that point. Ek2 Ep2 (mgh2) Em2 Ek2 Ep2 Em2 Em2 0.6131 J (0.250 kg)9.81 0.6131 J (0.1226 J) 0.491 J m s2 (0.0500 m) (d) The speed at the lowest point can be found from 2 Ek2 kinetic energy. 1 mv2 2 2(0.491 J) v2 2Ek2 m 0.250 kg 1.98 m s Paraphrase and Verify (a) At the highest point, the total energy of the pendulum is 0.613 J. (b) As the bob swings lower, gravitational potential energy is lost and kinetic energy is gained. The total energy remains 0.613 J. (c) At the lowest point, the kinetic energy is 0.491 J, the difference between its total energy and its gravitational potential energy. (d) At the lowest point of its swing, the bob has a speed of 1.98 m/s. 316 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 317 6-2 Inquiry Lab 6-2 Inquiry Lab Conservation of Mechanical Energy Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question Is energy conserved during the motion of a pendulum? Hypothesis State a hypothesis concerning the energy status of a pendulum. Remember to write this in the form of an if/then statement. Variables The variables in this lab are the values used to calculate the gravitational potential energy (mass, gravitational acceleration, and height) and the kinetic energy (mass and speed) at various points on the swing of the pendulum. Consider and identify which are controlled variable(s), which manipulated variable(s), and which responding variables. (NOTE: The voltage to the timer must not vary from trial to trial. If you are using a variable-voltage power supply, adjust the voltage so that the timer runs smoothly, and leave the power supply untouched for the remainder of the experiment. Stop and start the timer by disconnecting and reconnecting the lead attached to the black post of the power supply rather than turning the power supply off and on.) 3 Record your results in a table of data (Table 6.3). Table 6.3 Calibration Data Test Number Total Time t (s) Number of intervals N Time/Interval t (s) 1 2 Materials and Equipment string (at least 2.0 m long) pendulum bob (a 1-kg mass or greater) metre-stick ticker tape timer masking tape stopwatch (for timer calibration) Procedure For some interval timers, the period of the timer varies with the operating voltage. If your timer is of that type, begin by calibrating the timer. This is done by pulling a strip of ticker tape through the timer for a measured time, as set out below. Calibrate the timer: 1 Start the tape moving steadily through the timer, then connect the timer to the power supply for an exact measure of time (3 to 5 s works well). Be sure the tape does not stop moving while the timer is running. 2 Count the number of intervals between the dots (N) and divide that number into the measured time (t) to determine the time lapse per interval (t). Do at least one more calibration trial (without changing the voltage) to check if the time per interval (t) remains constant. Set up the apparatus: 4 Suspend the pendulum from a suitable solid point and allow the pendulum bob to come to rest at its lowest point. Place a marker (e.g., a piece of masking tape) on the floor below the centre of mass of the bob to indicate the pendulum’s rest position. Measure and record the length (l) of the pendulum, and mass (m) of the pendulum bob. (NOTE: The length of a pendulum is measured from the point at which it pivots to the centre of mass of the pendulum bob. If the shape of the bob is a symmetrical solid, such as a sphere or a cylinder, the centre of mass is at its geometric centre.) 5 Pull the pendulum sideways so that its horizontal displacement (x) is about one-half its length. Place a marker, such as an iron stand, at this point, which will be xmax for the experiment. Ensure the path of the pendulum is clear, then release it to check the path of its swing. One team member should be positioned to catch the pendulum so that it does not swing back. CAUTION: Make sure that the path of the pendulum is clear before you allow it to begin its swing. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 317 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 318 6 Position the ticker tape timer at a distance approximately equal to the length of the pendulum from the pendulum’s rest position. Locate the timer so that its height is just above the lowest point of the pendulum bob’s path. Align the timer so that the tape does not bind as the bob pulls it through the timer. See Figure 6.34. Anchor the timer firmly so that it does not shift during trials. power supply pendulum bob timer timer tape Figure 6.34 7 With ticker tape attached to the pendulum but without the timer running, do a trial run of the system. Attach the tape to the pendulum bob at its centre of mass, so that the pull of the tape does not cause the bob to twist. Use a length of ticker tape that will reach from the timer to a point slightly beyond the bob’s rest position. Move the pendulum bob sideways to its starting point. Hold the bob in place while you pull the tape tight, then gently allow the tape to take up the weight of the bob. Be sure that the tape is not twisted so that it does not rip as it passes through the timer. Release the tape and allow the pendulum to pull it through the timer. Collect data: 8 Once the timer is positioned so that the tape moves smoothly through it, you are ready to do a trial with the timer running. First, with the bob at its rest position, have one team member hold the bob stationary and pull the tape through the timer until it is just taut. Place a mark on the tape at the location where the timer records its dots. This mark on the tape records the position of the bob when its horizontal displacement is equal to zero (x 0). 9 Move the pendulum bob sideways to the starting point of its swing (xmax). Again hold the bob steady while a team member pulls the tape tight. Gently allow the tape to take up the weight of the pendulum bob. Start the timer, then release the pendulum. 10 Lay the tape out on a table and place a line across the tape through the first dot the timer put on the tape. At that position, the speed is zero (v 0) and the position is the maximum displacement (xmax). Measure the length of the tape from x 0 to xmax. 11 Locate the two dots that define the interval containing the mark that indicates the position of the bob at its rest position (x 0). Label this interval as interval 1. Measure the length (x1) of this interval (the space between the two dots on either side of the mark) and record it in a data table (Table 6.4). Calculate the speed v of the pendulum for interval 1 by dividing x1 by the interval time t. 12 Along the length of the tape, between x 0 and xmax, choose at least four more time intervals and draw a line across the tape at the midpoint of each chosen interval. S |
tarting from interval 1, number the selected intervals as 2, 3, etc. For each of the chosen intervals, measure (a) the length of the interval (x), and (b) the distance (x) to the midpoint of the interval from the line indicating x 0. Record your measurements in a data table (Table 6.4). Analysis 1. Use a table similar to Table 6.4 to organize your data. 2. Calculate the height (h) of the pendulum above its rest position by using the relationship h l l 2 x2. (See the diagram in Figure 6.35.) 3. Calculate the values for the gravitational potential (Ep), the kinetic (Ek) and the mechanical (Em) energies for each of the intervals you marked on your tape. 4. On the same set of axes, plot graphs for Ep and Ek against the horizontal displacement (x) of the pendulum. Describe the relationship between Ek and the position of the pendulum that is indicated by the graph. Does Ek change uniformly as the pendulum swings? Does Ep change uniformly as the pendulum swings? What relationship does the graph suggest exists between Ep and Ek for the pendulum? 5. On the same set of axes, plot a graph of the total mechanical energy (Em) of the system against horizontal position (x). What does the graph suggest is the nature of the total mechanical energy for the pendulum? Suggest a reason for this relationship. 318 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 319 6. Within experimental error, can the mechanical energy 7. How is your hypothesis affected by your data? Explain. of the system be considered constant? If the mechanical energy is assumed constant, what value would you choose to be the most representative of this energy? Explain why. For each of the intervals that you chose for analysis, what is the percent error in the mechanical energy at that interval? Does your analysis indicate a systemic error change for the pendulum as it swings? What would be the cause of this error? Table 6.4 Pendulum Data Horizontal Displacement x (m) 0 Height h (m) 0 Ep (J) 0 Interval Number Figure 6.35 Interval Length x (cm) Interval Speed v (m/s) Ek (J) Em (J) (Ep + Ek) Conservative and Non-conservative Forces To understand the law of conservation of energy you must understand that some forces, such as gravity and elastic forces, act within systems without affecting the mechanical energy of the system. When such forces operate, energy is conserved. These are called conservative forces. Other forces, such as friction, and forces applied from outside a system, cause the energy of the system to change so that energy is not conserved. These are known as non-conservative forces. Figure 6.36 shows a system of two ramps joining point P to point Q. The drop h is the same for the two ramps, but ramp A is shorter than ramp B, because of the hills in ramp B. If a frictionless car is released from P and moves down one ramp to Q, the amount of kinetic energy the car gains in moving from P to Q does not depend on which ramp (A or B) the car coasts down. If energy is conserved, the kinetic energy at point Q is equal to the potential energy at point P, so the speed of the car at point Q will be the same whether it comes down ramp A or ramp B. Since a conservative force does not affect the mechanical energy of a system, the work done by a conservative force to move an object from one point to another within the system is independent of the path the object follows. car point P h ramp A ramp B point Q Because Ep at P Ek at Q, the speed of the car at Q is the same no matter which ramp it coasts down. Figure 6.36 If a conservative force acts on an object, then the work it does is independent of the path the object follows between two points. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 319 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 320 PHYSICS INSIGHT No matter how closely conditions approximate an isolated system, the force of friction is never truly zero. Friction is continually converting kinetic energy to thermal energy. This thermal energy, which cannot be converted back into mechanical energy, radiates out of the system as heat. non-isolated system: a system in which there is an energy exchange with the surroundings Friction Is a Non-conservative Force In the absence of friction, the car in Figure 6.36 would return to point P with no change in its mechanical energy. However, if there is friction, any motion of the car will be subject to it. When you analyze the work done by friction, you can see that path length does affect the work done on the car. The term d, the distance through which the force of friction acts, is not the displacement, but is always the actual distance the object travels. Wf is the work done by the force of friction, Ff, on the system. Therefore: Ff Wf but dB d dA, so WfB WfA and the car on Ramp B would lose more energy. Since the potential energy at the bottom of the ramp is the same regardless of the route, the loss in mechanical energy must be a loss in kinetic energy. Therefore, friction is not a conservative force. Because thermal energy is being radiated out of the system, the system is, by definition, a non-isolated system. The amount of work done by friction will cause the mechanical energy of the system to change so that Em Wf Therefore, Em1 Em1 Em2 Em2 Wf or Wf The direction of the force of friction is always exactly opposite to the direction of the motion; therefore, the calculated value of Wf is always negative. Friction always reduces the mechanical energy of a system. Concept Check What assumptions must be made if you wish to use the law of conservation of energy to solve a problem in physics? M I N D S O N That’s the Way the Ball Bounces With each successive bounce, the height attained by a bouncing ball becomes less. Assuming the elastic forces that cause the ball to bounce are conservative in nature, and no outside forces act on the ball, you might expect it to behave as an isolated system. If so, the energy of the ball should be conserved. Use the concept of systems and conservation of energy to explain why the height decreases with each bounce. 320 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 321 PHYSICS INSIGHT In a non-isolated system, Wf has a negative value and decreases the total mechanical energy. The effect of friction is opposite to forces that are adding mechanical energy. Energy Changes in Non-isolated Systems Not all external forces remove mechanical energy from a system. Motors, in general, are used to add mechanical energy to a system. A ski-lift motor, for example, increases the gravitational potential energy of the skiers. More generally, if several external forces (A, B, C, . . .), as well as friction, act on a system, then the total work done by all of these forces produces the change in mechanical energy. Em2 Em1 Em1 W (WA WB WC . . . Wf) This is simply another version of the work–energy theorem. Comparison of Energy-Position Graphs for Isolated and Non-isolated Systems Figure 6.37 shows the energy-position graphs for a block sliding down an inclined plane in a non-isolated system. In an energy-position graph, the mechanical energy Em is the sum of the potential energy Ep and kinetic energy Ek. So, the height of the mechanical energy line above the axis is the sum of the heights of the potential and kinetic energy lines. The purple line is the sum of the values of the red line and the blue line. Energy vs. Position for a Non-isolated System ) Position d (m) Em Ep Ek Em Ek Ep Figure 6.37 Friction acts on a non-isolated system to remove mechanical energy. By rearranging the equation for the definition of work, E Fd E d F it can be easily seen that force is equal to the slope of an energy-position Nm m graph. The units of the slope are , or units of force. In particular: • The component of the force of gravity parallel to the motion can be determined by calculating the slope of the gravitational potential energy-position graph. • The net force can be determined by calculating the slope of the kinetic energy-position graph. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 321 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 322 e MATH To determine the forces along an incline in an isolated system by using an energy-position graph, visit www.pearsoned.ca/physicssource. Project LINK How will the design of your persuader apparatus allow for the energy changes in a system during a collision? For example, in the isolated system in Figure 6.30 on page 312, the slope of the potential energy curve gives the component of gravity parallel to the inclined plane. The slope of the kinetic energy curve gives the net force. The slope of the mechanical energy curve is zero indicating that no outside forces act on this system. In the non-isolated system shown in Figure 6.37, Em is not constant so friction is present. The slope of the total energy curve gives the force of friction. As an example of a non-isolated system, imagine a cart accelerating down an inclined plane. The force of friction removes energy from the system, but it is not sufficient to stop the cart from speeding up. The magnitude of the change in kinetic energy is less than the magnitude of the change in gravitational potential energy. In this case, the mechanical energy decreases by the amount of energy that friction removes from the system. The graph would be similar to Figure 6.37. 6-3 Design a Lab 6-3 Design a Lab The Energy Involved in a Collision The Question What happens to the energy of the system when two carts collide? Design and Conduct Your Investigation Design an experiment to investigate the energy of a system in which two carts collide. In one case, compare the energy before and after the collision for simulated “elastic” collisions in which the carts interact via a spring bumper. In a second case, compare the energy before and after a colli |
sion when the carts stick together in what is called an “inelastic” collision. You will need to develop a list of materials and a detailed procedure. Use the work-energy theorem to explain your results and form conclusions. e LAB If probeware is available, perform 6-3 Design a Lab using a motion sensor. For a probeware activity, go to www.pearsoned.ca/school/ physicssource. Concept Check A block slides down an inclined plane, radiating energy out of the system as heat due to friction. Yet when you measure the mechanical energy at the bottom of the ramp, the total energy in the system is unchanged. Explain how this might occur. 322 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 323 6.3 Check and Reflect 6.3 Check and Reflect Knowledge 1. What is meant by an isolated system? 2. If energy is conserved in a system, how can work be done in the system? 3. Describe the changes in the forms of energy as an acrobat bounces on a trampoline so that she goes higher after each bounce. 4. Can a system be classified as isolated if a non-conservative force acts on it? Explain. 5. A golfer drives a golf ball from the top of a cliff. The ball’s initial velocity is at an angle above the horizontal. If there were no air friction, describe the energy transformations from the time the golfer starts her swing until the golf ball lands on the ground at a distance from the bottom of the cliff. Include the energy transformation at the point of impact. 9. The figure below shows the energy-position graphs for two different systems. For each graph, describe what is happening to the object(s) in the system in terms of their energies. Describe for each the nature of the forces acting on the object(s). Energy vs. Position ) Position d (m) Energy vs. Position Position d (m) Em Ep Ek Em Ep Ek Em Ek Ep 6. The pendulum of a clock is given a tiny Extensions push at the beginning of each swing. Why? 10. A 3.60-m-long pendulum with a 1.25-kg Applications 7. Two masses are suspended by a light string over a frictionless pulley. Mass A is 2.40 kg, mass B is 1.50 kg. Can this be considered an isolated system? Explain. On release, mass A falls to the tabletop, 1.40 m below. What is the kinetic energy of this system the instant before mass A hits the tabletop? bob is pulled sideways until it is displaced 1.80 m horizontally from its rest position. (a) Use the Pythagorean theorem to calculate the bob’s gain in height. If the bob is released, calculate the speed of the bob when it (b) passes through its rest position (c) is 0.250 m above its rest position 8. Draw graphs showing the gravitational potential, kinetic, and mechanical energy of the system in question 7 against the change in position of mass A if there is (a) no friction, (b) a force of friction, but mass A still accelerates. A 1.40 m 3.60 m m 1.25 kg h 1.80 m B e TEST Diagram for questions 7 and 8. To check your understanding of mechanical energy in isolated and non-isolated systems, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 323 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 324 info BIT In metric terms, 1.00 hp 746 W or about 0.75 kW. 6.4 Work and Power Power, Work, and Time Toward the end of the 18th century, horses were the main source of energy used to drive the pumps that removed water from mines. Thus, when James Watt (1736–1819) wanted to know how his newly improved steam engine compared with existing methods of pumping water out of mines, he compared its effectiveness to that of horses. Today, even though it is a rather awkward unit, we still use his concept of horsepower (hp) to identify the power output of motors, especially in the automotive industry. Figure 6.38 This drag racer’s 7000 hp engine burns a special fuel mixture called nitromethane. Each of its eight cylinders generates approximately three times the power of a normal car engine. The distortion of the tires, as seen above, is evidence of the magnitude of the forces exerted during acceleration. The high-performance race car in Figure 6.38 accelerates to speeds over 530 km/h in about 4.4 s. A family sedan with a 250-hp engine can accelerate to 100 km/h, from rest, in about 8.0 s. Aside from acceleration, what aspect of a car’s performance is affected by the horsepower rating of its motor? On the highway, cars with 100-hp motors and cars with 300-hp motors both easily travel at the speed limit. What factors decide how much power is required to move a car along the highway? power: the rate of doing work In physics, power (P) is defined as the rate of doing work. Thus, the equation for power is P W t or P E t The unit of power, the watt (W), is named in recognition of James Watt’s contributions to physics. Using the equation for power we see that a power output of one watt results when one joule of work is done per second Efficiency efficiency: ratio of the energy output to the energy input of any system Efficiency may be defined in terms of energy or in terms of power. In both cases, the ratio of the output (useful work) to the input (energy expended) defines the efficiency of the system. Thus, efficiency can be calculated as either, Efficiency Energy output (Eout) Energy input (Ein) , or Power output (Pout) Power input (Pin) 324 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 325 Concept Check In terms of kg, m, and s, what is the unit for power? Example 6.11 An elevator and its occupants have a mass of 1300 kg. The elevator motor lifts the elevator to the 12th floor, a distance of 40.0 m, in 75.0 s. (a) What is the power output of the elevator? (b) What is the efficiency of the system if the motor must generate 9.40 kW of power to do the specified work? Given m 1.300 103 kg g 9.81 m/s2 h 40.0 m t 75.0 s Pin 9.40 103 W up down 40.0 m Required (a) power output of the elevator (b) efficiency of the system 1300 kg Figure 6.39 Analysis and Solution (a) The work done by the elevator is equal to its gain in gravitational potential energy. The power output of the elevator is the change in potential energy divided by the time40.0 m) (1.300 103 kg)9.81 s2 75.0 s 6.802 103 J s 6.80 103 W Practice Problems 1. The engine of a crane lifts a mass of 1.50 t to a height of 65.0 m in 3.50 min. What is the power output of the crane? Convert the SI unit answer to hp. 2. If a motor is rated at 5.60 kW, how much work can it do in 20.0 min? 3. A tractor, pulling a plough, exerts a pulling force of 7.50 103 N over a distance of 3.20 km. If the tractor’s power output is 25.0 kW, how long does it take to do the work? Answers 1. 4.55 kW (6.11 hp) 2. 6.72 106 J 3. 960 s (16.0 min) (b) Efficiency is the ratio of the power output to the power input. P t o u Efficiency .724 Paraphrase and Verify (a) The power output of the elevator is 6.80 103 W. The answer has the right order of magnitude for the given data. The power output is equivalent to about sixty-eight 100-W light bulbs. (b) The efficiency of the system is 0.724 (72.4%). Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 325 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 326 6-4 Inquiry Lab 6-4 Inquiry Lab Measuring the Power Output of a Motor The Question How much power can a small electric motor generate? The Problem The problem in the lab is to measure the power output of a motor by timing how long it takes for the motor to do a fixed amount of work. Variables The variables for measuring power are the work done against gravity (EP) and the time (t) it takes to do the work. Calculating EP requires mass (m), gravitational acceleration (g), and change in height (h). Materials small dc electric motor alligator clip leads iron stand 1-kg mass test-tube clamps low-voltage power supply dowel (about 3 cm long and 1 cm in diameter) thread (about 2.5 m long) tape paper clip washers scale (sensitive to at least 0.1 g) stopwatch metre-stick Procedure Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork CAUTION: Close the test-tube clamp just tight enough to hold the motor in place. If you tighten it too much, it could warp the body of the motor. cardboard disks electric motor test-tube clamp mounted on an iron stand dowel light string upper mark washers paper clip lower mark leads to variable voltage power supply Figure 6.40 CAUTION: Check with your instructor to be sure the connections are correct before you plug in the power supply. If the motor is incorrectly connected, it could be damaged when the current is turned on. 4 Connect the power supply to the electric motor. Once your instructor has approved your connection, disconnect the lead connected to the red post of the power supply and turn on the power supply. 1 Use the balance to determine the mass of the paper 5 Place five washers on the string, as shown in clip. Record your measurement. 2 Place 10 washers on the balance scale and determine their mass. Calculate the average mass of the washers. Record your measurement. 3 Assemble the apparatus as shown in Figure 6.40. Set up a measuring scale behind the string holding the washers. The distance between the upper and lower timing marks on the scale may be adjusted if your apparatus permits, but should be 1.5 m or greater. Figure 6.40. Complete the circuit by holding the insulated alligator clip lead on the red post of the power supply and observe the speed with which the motor lifts the washers. Adjust the number of washers until the motor moves the load upward at a uniform speed. (If the speed is too great, it will be difficult to time the motion of the washers. If the speed is too slow, then the motor may not run smoothly.) 326 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 327 6 Pull the thread to unwind thread from the dowel until |
2. Make a graph of the power output versus the mass the washers rest on the floor. Start the motor. Measure the time the washers take to travel between the lower and upper timing marks. 7 Record your measurements in a table such as Table 6.5. 8 Vary the number of washers on the paper clip and repeat the trial. Do trials with at least three different masses. 9 Calculate the work that the motor did in lifting the washers the measured distance. 10 Calculate the power output for each trial. Analysis 1. Does the power output of the motor vary with the force it is exerting? being lifted. 3. For what mass does the motor produce the most power? 4. What is the advantage of lifting the weights over a long distance? 5. Suggest reasons why the motor might generate more power when different masses are used. 6. Does the motor feel warm after it has done some work? What does that tell you about this system? 7. Would it make sense to use a very large motor to lift very tiny masses? Explain. 8. In terms of car engines, what are the implications for engine size? Table 6.5 Power Output of a Motor Trial Number Number of Washers Mass m (kg) Time t (s) Change in Potential Energy Ep mgh (J) Power P Ep/t (W) Power and Speed When a motor, such as the electric motor in 6-4 Inquiry Lab, applies a constant force to move an object at a constant speed, the power output of the motor can be shown to be the product of the force and the speed. When the force is constant, the work done can be found by the equation W Fd Inserting the equation for work into the equation for power gives: d F P t d F t But the expression d/t is just average speed vave; therefore, P (F)(vave) e WEB Why is it that if you double the speed of a car, the rate at which it consumes fuel more than doubles? Is lowering the speed limit the most effective way to conserve energy or would design changes (e.g., hybrid fuel systems or fuel cells) be more effective? With a group of classmates, investigate how best to improve the energy efficiency of automobiles. Use your library and the Internet. Present the results of your investigation in a report using presentation software such as PowerPoint ™. Begin your search at www.pearsoned.ca/school/ physicssource. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 327 06-PearsonPhys20-Chap06 7/25/08 8:22 AM Page 328 Example 6.12 A car, of mass 2000 kg, is travelling up a hill at a constant speed of 90.0 km/h (25.0 m/s). The force of air resistance, which opposes this motion, is 450 N. The slope of the hill is 6.0 (Figure 6.41). (a) Draw a free-body diagram to show the external forces acting on the car as it moves up the hill. (b) Determine the forward force needed to maintain the car’s speed. (c) Assuming that all the power output of the car engine goes into maintaining the car’s forward motion, calculate the power output of the engine. Given m 2.000 103 kg F 4.50 102 N [downhill] air g 9.81 m/s2 v 25.0 m/s 6. Fair 6.0º Figure 6.41 Required (a) a free-body diagram for the car (b) forward force (Ff ) (c) power (p) Analysis and Solution (a) Figure 6.42(a) shows the free-body diagram. (b) Since the car is not FN v Ff Fg FN Ff Fair Fg accelerating, the net force on the car must be zero, 0, both parallel and Fnet perpendicular to the incline of the hill. The forward force (Ff) must be equal to the sum of the magnitudes of the force of air resistance (Fair) and the component of the gravitational force that acts parallel to the incline (Fg). Figure 6.42(a) Fg Fg In the parallel direction g F F net F F Fg Fair Fnet Ff air f Now, Fg mg sin (2.000 103 kg)9.81 2.051 103 N Forces Ff Fair Fg m s2 (sin 6.0) Figure 6.42(b) Therefore, 0 Ff Ff (2.051 103 N) (4.50 102 N) 2.051 103 N 4.50 102 N 2.50 103 N Practice Problems 1. What is the power output of an electric motor that lifts an elevator with a mass of 1500 kg at a speed of 0.750 m/s? 2. An engine’s power is rated at 150 kW. Assume there is no loss of force due to air resistance. What is the greatest average speed at which this engine could lift a mass of 2.00 t? 3. A 1250-kg race car accelerates uniformly from rest to 30.0 m/s in 4.00 s on a horizontal surface with no friction. What must be the average power output of its motor? 4. Each car in a freight train experiences a drag force of 6.00 102 N due to air resistance. (a) If the engine of the train is to pull a train of 75 cars at a constant speed of 72.0 km/h, what power is required to move the cars? (b) If the engine operates at 15.0% efficiency, what must be the power generated by the engine to move these cars? Answers: 1. 11.0 kW 2. 7.65 m/s 3. 141 kW 4. (a) 900 kW (b) 6.00 103 kW 328 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 329 (c) Calculate the power output of the engine using the forward force and the speed with which the car moves along the ramp. P Ffvave (2.50 103 N)(25.0 m/s) 6.25 104 W 62.5 kW Paraphrase (a) The free-body diagram shows the external forces acting on the car. (b) If the car moves at a constant speed, then the forward force must be a force of 2.50 103 N. (c) The power output of the car is 62.5 kW. M I N D S O N Power and Dance At the moment of takeoff, Cossack dancers must generate considerable power to perform their spectacular leaps (Figure 6.43). Discuss techniques you could use to measure the power the dancers must generate to make such a jump. University Faculties of Kinesiology study this and other aspects of how humans move. 1. What factors involved in the jump will you need to determine? 2. What equipment might you require to measure those factors? 3. How would you measure the dancer’s maximum power output compared with the power he can generate over a sustained period of time? Figure 6.43 e WEB To learn more about power generated in human activities, follow the links at www.pearsoned.ca/school/ physicssource. THEN, NOW, AND FUTURE Fuel for the Future? While the automobile in Figure 6.44 may look like a normal car, nothing could be further from the truth. When this vehicle travels along one of Vancouver’s streets, its motor is barely audible. Perhaps even more surprising is the fact that the exhaust this car produces is pure water. While the motor that drives the car is actually an electric motor, it is the source of the electricity that is getting all the attention. The “battery” in this car is called a fuel cell. At present, fuel cells are not an economically viable replacement for the internal combustion engine although successful trials of fuel-cell buses have been made in several cities around the world. For further information, go to the Internet. Start your research at www.pearsoned.ca/school/physicssource. Questions 1. What are the advantages and disadvantages of a fuel-celldriven motor over an internal combustion engine? 2. Which of the advantages and disadvantages identified above can be further improved on or overcome by scientific and technological research? Explain. 3. What are the limitations of science and technology to finding answers to the problems associated with energy production and use? Figure 6.44 This car’s exhaust is pure water. The impact of fossil fuel (oil and coal)-burning systems on the environment has made the search for alternative energy sources much more attractive. This search is further enhanced by the realization that the supply of fossil fuels is finite. Even though Canada has, at present, an abundant supply of fossil fuels, it is still a world leader in fuel- cell research. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 329 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 330 6-5 Problem-Solving Lab 6-5 Problem-Solving Lab Power and Gears Recognize a Need Modern bicycles have many gears to enable riders to make best use of their efforts. In the automotive industry, manufacturers use a device called a dynamometer (Prony brake) to measure the power output of the motors they install in their vehicles. A Prony brake for bicycles would be a useful thing. The Problem How does the gear used by a cyclist affect the power output at the drive wheel of the bicycle? In which gear do cyclists generate the greatest power? Criteria for Success A successful experiment will determine if there is a relationship between the power at the drive wheel of the bicycle and the gear in which the bicycle is being ridden. Brainstorm Ideas Investigate the design of a Prony brake, then brainstorm how that design might be adapted to measure the power output of a bicycle. Remember, for your results to be useful 6.4 Check and Reflect 6.4 Check and Reflect Knowledge 1. What is the relationship between the amount of work that is done and the power output of the machine that does the work? 2. A farmer says, “My tractor with its 60-hp engine easily pulls a plough while my car with a 280-hp engine cannot even budge it.” How can you explain this fact? 3. What is the relationship between the speed of an object and the power required to move it? Applications 4. You lift a 25.0-kg mass to your waist (0.800 m) in 1.20 s. What is your power output? 5. An airplane’s engine exerts a thrust of 1.20 104 N to maintain a speed of 450 km/h. What power is the engine generating? 330 Unit III Circular Motion, Work, and Energy Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork the design must allow a rider to “ride” the bicycle in a normal manner. If a computer and probeware are available, consider using probeware in your experimental design, to measure the speed of the Prony brake. e WEB Use the Internet to investigate Prony brake design. Begin your search at www.pearsoned.ca/school/physicssource. Build a Prototype Build a Prony brake that can measure the power output of a student riding a bicycle. Test and Evaluate Make measurements of the power output of a student riding a bicycle using various gear sett |
ings. Communicate Prepare a report of your research using a computer spreadsheet program to organize your data and to generate a graph of the power output versus the gear level. Print your graph, in colour if possible, as part of your report. 6. An electric motor has a power rating of 1.50 kW. If it operates at 75% efficiency, what work can it do in an hour? 7. A motor of a car must generate 9.50 kW to move the car at a constant speed of 25.0 m/s. What is the force of friction on the car? Extension 8. A cannon fires a ball with a muzzle velocity of 240 m/s. The cannon ball has a mass of 3.60 kg. The barrel of the cannon is 1.20 m long and exerts a force of friction on the cannon ball of 650 N. What is the average power provided to fire the cannon ball? e TEST To check your understanding of power, work, and efficiency, follow the eTest links at www.pearsoned.ca/school/physicssource. 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 331 CHAPTER 6 SUMMARY Key Terms energy work gravitational potential energy Key Equations W (F cos )d Ek 1 mv2 2 reference point elastic potential energy kinetic energy mechanics mechanical energy work-energy theorem isolated system non-isolated system conservation of energy power efficiency Ep mgh EP mgh W Ek Ep Em Ek Ep Ep 1 kx2 2 P W t E t Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to produce a full summary of the chapter. Mechanics involves energy changes of position of motion rate of change calculated using can be either W Fd Ep mgh or can be either or non-isolated in which energy is conserved or Em2 Em1 W or or Ek Ep Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 331 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 332 CHAPTER 6 REVIEW Knowledge 1. (6.1) (a) When a force acts on an object to do work, why do you need to know the angle between the direction of the force and the direction of the displacement? (b) If you know how the magnitude of a force changes while it acts over a displacement, how can you find the amount of work it does? (c) Describe the nature of the energy transfers for the work done on a bungee jumper from the time he leaps off the platform until his velocity is zero at the lowest point of his jump. (d) Two students calculate the gravitational potential energy of a mass resting on a shelf. One student calculates that it has 12.0 J of energy while the other calculates the gravitational potential energy to be 35.0 J. Is it possible that they are both right? Explain. (e) Two masses, A and B, are at rest on a horizontal frictionless surface. Mass A is twice as great as mass B. The same force acts on these masses over the same displacement. Which mass will have the greater (i) speed, and (ii) kinetic energy at the end of the displacement? Explain. (f) Many fitness facilities have treadmills as exercise machines. Is running on a treadmill work since the runner is not really moving? (g) Explain why it takes less force to push a cart up an inclined plane onto a platform than it does to lift the cart straight up from the floor. Assume you are able to move the cart by either method. Does it also take less work to lift it or to roll it up the plane? (h) An object sits on a tabletop. In terms of describing the object’s gravitational potential energy, which reference point is better: the ground outside the room, the floor of the room, or the tabletop? Explain. 2. (6.2) (a) What is the effect of the work done by a net force? (b) If a force acts upward on an object, does all the work done by this force become potential energy? 3. (6.3) (a) Explain why the force of gravity but not the force of friction is called a conservative force. (b) An ideal spring is one where no energy is lost to internal friction. Is the force exerted by the spring considered a conservative force? Explain. Is a mass that is oscillating up and down on the end of an ideal spring a good approximation of an isolated system? (c) Since no system on Earth is truly an isolated system, why is it advantageous to assume that a system is isolated? (d) A truly isolated system does not exist on Earth. How does that affect the fundamental principle of conservation of energy? (e) If a system is not isolated, how can one calculate the change in mechanical energy in the system? (f) The mechanical energy of a system is measured at two different times and is the same each time. Is this an isolated system? Explain. 4. (6.4) (a) An elevator takes 2.50 min to travel from the ground floor to the 10th floor of an apartment block. The tenants want the landlord to increase the speed of the elevator but the landlord argues that speeding up the elevator means that it will need to work harder and that would take more energy. Is he correct? Explain. (b) The transmission of an automobile allows the work done in the engine to be transmitted to the wheels. For a given power output by the engine, the wheels are not rotated as fast by a low gear as they are by a high gear. What advantage does having gears give the driver of a car? Applications 5. What is the change in kinetic energy if a net force of 3.80 103 N [0] acts on a mass while it undergoes a displacement of 95.0 m [335]? 6. For gravitational potential energy, when the height doubles so does the potential energy. However, for elastic potential energy, if the stretch of the spring doubles, the energy does not. (a) How does the stored elastic potential energy (c) What forms of energy are considered to be change if the stretch doubles? part of mechanical energy? (d) Can two people calculate the mechanical energy of an object and get two different correct answers? Explain. (b) Explain in terms of force-position graphs for gravitational and elastic potential energies why this happens. 332 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 333 7. The figure below shows the graph of the force as a function of displacement for an elastic spring stretched horizontally 25.0 cm from its equilibrium position. A mass of 0.400 kg is attached to the spring and released. If the mass is sliding on a horizontal frictionless surface, what is the speed of the mass when the spring has contracted to (a) 10.0 cm from its equilibrium position, and (b) its equilibrium position? Force vs. Displacement 50 40 30 20 10 ) ( N F e c r o F 0 0.05 0.10 0.15 0.20 0.25 Displacement d (m) Graph for question 7 8. A bungee jumper with a mass of 65.0 kg leaps from a bridge. At the lowest point of the jump he is 30.0 m below the point from which the jump began. If, at equilibrium, the bungee cord is 15.0 m long, what is the elastic constant for the cord? HINT: Assume an isolated system. At the lowest point the bungee cord must convert all of the jumper’s lost gravitational potential energy into elastic potential energy in the cord. 9. A motorcycle stuntman wants to jump over a line of city buses. He builds a takeoff ramp that has a slope of 20, with its end 3.20 m above the ground. The combined mass of the motorcycle and rider is 185 kg. To clear the buses, the cyclist needs to be travelling 144 km/h when he leaves the end of the ramp. How high above the ground is the motorcycle at its highest point? 10. A skydiver reaches a maximum speed (or terminal velocity) of 36.0 m/s due to the force of air resistance. If the diver has a mass of 65.0 kg, what is the power output of the air resistance acting on him? Extensions 11. Even when making short flights, jets climb to altitudes of about 10 000 m. Gaining altitude requires a great expenditure of fuel. Prepare a short research report to explain the advantages and disadvantages of travelling at these altitudes. Consolidate Your Understanding You have been hired to tutor a student on the topics in this chapter. Describe how you would answer the questions below. In each instance, include an example. 1. What is the difference between work and energy? 2. When an object moves up a hill, how does the length of the hill affect the increase in the gravitational potential energy? 3. If a cart, at rest, is allowed to coast from the top of a hill to the bottom, is the kinetic energy at the bottom of the hill always equal to the loss in gravitational potential energy? 4. A given force is to act on a block to accelerate it from rest as it slides up the length of an inclined plane. Using accurate spring balances and rulers, how could you gather data to enable you to calculate, with reasonable accuracy, the kinetic energy of the block when it reaches the top of the incline? You may not use timing devices. 5. What is the difference between an isolated and a non-isolated system in terms of work and energy? 6. What factor of a car’s motion is most directly affected by the power of its engine? Think About It Review the answers you gave to the questions in Think About It on page 291. How would you change these answers? e TEST To check your understanding of energy, work, and power, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. 333 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 334 UNIT III PROJECT Building a Persuader Apparatus Scenario It is 1965. Seatbelts are oddities used only by airline passengers at takeoff and landings. The term “airbag” hasn’t even been invented. Speed limits and traffic deaths are on the rise. Imagine that you are part of a team of engineers who design and build automobiles. Your company challenges you to design and build a model that can be used to convince its shareholders and the public that it is possible to build much safer automobiles. The automobile company has challenged your design team to produce safety features for its vehicles that will allow its passengers to survive crashes under severe conditions. Your team must determine how best to protect the passenger whil |
e, at the same time, keeping the size and mass of the car itself to reasonable proportions. Your presentation to the company will be used to persuade them of the benefits of your design. Finally, your report should persuade the public of the advantages of using the safety equipment you recommend for automobiles. Planning Your team should consist of three to five members. Your first task is to formulate your research question. This done, you will need to identify the assumptions about the nature of the collisions in which your vehicle may be involved. Since not all collisions may be head-on, your passenger (a fresh raw egg) should survive unscathed from a wide variety of crash scenarios. While all team members should be active participants in all aspects of the project, you should identify and draw on any special talents of your team. Create a team structure that assigns responsibilities such as team manager, data analyst, and record keeper. Begin by brainstorming possible design features and research strategies. Where might you find information on existing safety features? Which features are the most effective? How will you compare results for the various types of crashes? You may wish to draw on information from all topics in this course to improve the safety features of your vehicle. Create timelines for all phases of the project. Create a report that incorporates written, graphic, and photographic analysis of your project. Materials • material for construction of the vehicle • mass-measuring equipment • equipment to provide known energy crash conditions • egg passengers • digital camera • computer Assessing Results Assess the success of your project based on a rubric* designed in class that considers: research strategies thoroughness of the experimental design effectiveness of the experimental technique effectiveness of the team’s public presentation Procedure 1 Research existing safety features used in automotive production. Identify which features are the most effective in reducing crash injuries. Keep a record of the sources of your information. If information comes from the Internet, be sure to identify the site, and who sponsors the information. Be alert to Internet sites that may contain biased information. Identify the most common types of injuries resulting from automobile accidents. 2 Design the persuader vehicle and gather the materials required for its construction. 3 Design the experiment that your team will use to test the effectiveness of your vehicle’s design. Make sure that your experimental design makes it possible to accurately compare the energy of the vehicle when it is involved in different crash scenarios. CAUTION: Your vehicle will need to gain considerable energy, which may or may not result in unexpected behaviour when it crashes. Take proper precautions to ensure that the vehicle path is clear during trials. 4 Test your vehicle’s safety features under a variety of crash conditions. Assess how effective your system is when it is involved in crashes happening from different directions. 5 Prepare a report using an audiovisual format that will dramatically emphasize for your audience the value of the safety features that you recommend. Thinking Further Write a short appendix to your report (two or three paragraphs) to identify possible directions that future research might take to make automobile travel even safer. Suggest steps that government, technology, and industry should take in making automobile travel safer. *Note: Your instructor will assess the project using a similar assessment rubric. 334 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 335 UNIT III SUMMARY Unit Concepts and Skills: Quick Reference Concepts Chapter 5 Summary Newton’s laws can explain circular motion. Resources and Skill Building Speed and velocity Centripetal acceleration and force Velocity and circular motion Centripetal acceleration Centripetal force — a horizontal system in circular motion Centripetal force — a vertical system in circular motion Centripetal force — acceleration and frequency 5.1 Defining Circular Motion The velocity of an object moving with circular motion is tangent to the circle and 90 to the radial line. Figures 5.3–5.6, 5.8–5.11, 5.13; QuickLab 5-1; Inquiry Lab 5-2 Centripetal acceleration and centripetal force are both directed toward the centre of the circle. Figures 5.8–5.11; Table 5.2; 5.2 Circular Motion and Newton’s Laws The velocity of circular motion can be determined by dividing the circumference by the period. The centripetal acceleration of an object is determined by the velocity squared divided by the radius. Newton’s second law states F ma and can be applied to centripetal acceleration. A car making a turn experiences a centripetal acceleration and force that is created by the force of friction between the tires and the road. The minimum speed necessary to move an object through a vertical loop equates centripetal force with gravitational force. Centripetal force can be equated to the gravitational force for planetary objects. eSIM Example 5.2; Minds On Figures 5.18–5.20; Example 5.3 Inquiry Lab 5-3; eTECH; Example 5.4; Figures 5.24, 5.25 Figures 5.27–5.30; eTECH; Example 5.5; eSIM Centripetal acceleration and force can be determined using period and frequency instead of speed. Figures 5.32–5.34; Example 5.7 Kepler’s laws 5.3 Satellites and Celestial Bodies in Circular Motion Kepler formulated three laws that explained the motion of planets in the solar system. Figures 5.36–5.38; Tables 5.4–5.6; Examples 5.8, 5.9; eSIM; Design a Lab 5-4 Newton’s version of Kepler’s third law Fc for Earth–Moon system. He also Newton recognized the reason that Kepler’s laws were correct: Fg found a way to determine the mass of an object from the period of a celestial body orbiting it. Figures 5.41, 5.42; Examples 5.10, 5.11; eTECH Orbital perturbations The discovery of Uranus and Pluto occurred because of the apparent disturbances in the orbit of the planets. Extrasolar planets have been discovered by examining perturbations in stars’ movements. Then, Now, and Future; Figures 5.46, 5.47 Artificial satellites Humans have placed a variety of artificial satellites into orbit to meet society’s needs. 5-5 Decision-Making Analysis; Figures 5.48–5.51 Chapter 6 In an isolated system, energy is transferred from one object to another whenever work is done. Work Potential energy Kinetic energy 6.1 Work and Energy Work is the transfer of energy that occurs when a force acts over a displacement. It is a scalar quantity measured in joules. (1 J 1 N· m) Potential energy is the energy a body has because of its position or configuration. It is a scalar quantity measured in joules. Kinetic energy is the energy a body has because of its motion. It is a scalar quantity measured in joules. Example 6.1 QuickLab 6-1; Example 6.2; Example 6.3; Example 6.4 QuickLab 6-1; Example 6.5; Example 6.6 Work-energy theorem 6.2 Mechanical Energy Work done by a net force causes a change in kinetic energy. The work–energy theorem states that the work done on a system is equal to the sum of the changes in the potential and kinetic energies. Mechanical energy Mechanical energy is the sum of the potential and kinetic energies. Example 6.7 Example 6.7 Example 6.8 Isolated systems 6.3 Mechanical Energy in Isolated and Non-isolated Systems The law of conservation of energy states that in an isolated system, the mechanical energy is constant. Example 6.9; Example 6.10 Conservation of energy A simple pendulum is a good approximation of an isolated system in which energy is conserved. Inquiry Lab 6-2 Conservative forces Non-isolated systems Power A conservative force does not affect the mechanical energy of a system. Example 6.10; Inquiry Lab 6-2 In non-isolated systems, the mechanical energy may change due to the action of nonconservative forces. Design a Lab 6-3 6.4 Work and Power Power is defined as the rate of doing work. Power is calculated by finding the ratio of the work done to the time required to do the work. It is measured in watts. (1 W 1 J/s) Power may be calculated by taking the product of the force doing the work and the average speed. Example 6.11; Inquiry Lab 6-4; Example 6.12 Problem-Solving Lab 6-5 Unit III Circular Motion, Work, and Energy 335 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 336 UNIT III REVIEW Vocabulary 1. Using your own words, define these terms: artificial satellite axis of rotation axle centripetal acceleration centripetal force conservation of energy conservative force cycle eccentricity efficiency elastic potential energy ellipse energy frequency gravitational potential energy isolated system Kepler’s constant Kepler’s laws kinetic energy mean orbital radius mechanical energy non-isolated system orbital period orbital perturbations period potential energy power reference point revolution rpm satellite uniform circular motion work work-energy theorem Knowledge CHAPTER 5 2. An object is moving in a circular path with a uniform centripetal acceleration that doesn’t change. What will happen to the velocity if the radius is reduced? 3. The centripetal acceleration of a car as it goes around a turn is inward, but the car will not skid in that direction if it is moving too quickly. Explain. 336 Unit III Circular Motion, Work, and Energy 4. A bucket of water is spun in a vertical circle on the end of a rope. (a) Explain what force or forces act as the centripetal force when the bucket is in the highest position. (b) In which position is the rope most likely to break? Why? 5. Explain why centripetal force is inward when the force acting on your hand as you spin an object in a circular path is outward. 6. Using what you have learned about the force of gravity and circular motion, provide a thorough explanation why the magnitude of centripetal force changes for planets orbiting the Sun. 7. A roller coaster goes around a vertical loop with just enough velocity to ke |
ep it on the track. (a) (b) In which position or positions is the force of gravity the centripetal force? Explain. In which position or positions is there a force exerted on the track by the roller coaster? Explain. (c) Using the equation Fg mg and equation 6 from Chapter 5 on page 256, show why mass does not affect the speed required for the roller coaster to successfully enter and exit the loop as shown in the diagram below. D C A B 8. What is the relationship between frequency and radius for a rotating disc? 9. The motor of a table saw is rated for its horsepower and rotational frequency. Explain why rotational frequency is used instead of rotational speed. 10. What physical quantities must be known for the mass of Earth to be determined? 11. Kepler showed that planets follow elliptical orbits. (a) Which planet has the least elliptical orbit? (b) Which planet’s semi-major axis is the closest in length to its semi-minor axis? 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 337 12. Three identical coins 24. A slingshot is used to propel a stone vertically C B A are placed on a rotating platter as shown. As the frequency of rotation increases, identify which coin will begin to slide off first. Explain your answer. 13. Briefly explain how Neptune was discovered. Use the terms orbital perturbation, force of gravity, and orbital velocity in your explanation. 14. Your friend argues that Neptune would not have been discovered as soon as it was if Neptune were a much smaller planet and Uranus much bigger. Is she right? Defend your answer. 15. What difficulties do astronomers face when searching for extrasolar planets that might have life as we know it? CHAPTER 6 16. Express a joule in terms of kilograms, metres, and seconds. 17. If work is a scalar quantity, why is it affected by the directions of the force and displacement? 18. What happens to an object’s gravitational potential energy when it is in free fall? 19. Explain why doubling the speed of an object does not result in a doubling of its kinetic energy. 20. A large mass and a small mass with the same kinetic energy are sliding on a horizontal frictionless surface. If forces of equal magnitude act on each of these bodies to bring them to rest, which one will stop in the shorter distance? Explain. 21. Describe the energy changes of a roller coaster car from the time when it is just coming over the crest of one hill until it arrives at the crest of the next hill. 22. A cart is pulled up an inclined plane by a force that is just large enough to keep the cart moving without a change in its speed. Is this an isolated system? Explain why or why not. Is the force used to move the cart up the incline a conservative force? 23. A cart at the top of an inclined plane is allowed to roll down the plane. Under what conditions can this system be considered isolated? If the conditions that make this an isolated system do not exist, is the force that moves the cart down the plane still considered a conservative force? Explain. upward. Describe the energy changes that are involved from the time the stone is placed in the slingshot until the stone reaches its maximum height. 25. If a force that acts on an object results in a change in the object’s kinetic energy, what can be said about the nature of this force? 26. How do you calculate work from a force- displacement graph? 27. According to the work-energy theorem, how much work is done on an isolated system? 28. Does power affect the amount of work you are able to do? Explain why or why not. Applications 29. Electrons in an electric (AC) circuit vibrate at 60 Hz. What is their period? 30. A cell phone is set to vibrate when it rings. It vibrates with a period of 0.0160 s. What is the frequency of the ring? 31. A toy top spins at 300.0 rpm. What is the frequency (in Hz) and the period of the top? 32. The Moon orbits Earth once every 27.3 days at a mean orbital radius of 3.844 105 km. What is its speed? 33. A child sits in a pretend airplane on a ride at an amusement park. The airplane is at the end of a long arm that moves in a circular path with a radius of 4.0 m at a speed of 1.57 m/s. What is the period of the ride? 34. A person sliding down a water slide at a speed of 5.56 m/s encounters a turn with a radius of 10.0 m. Determine the acceleration that he experiences in the turn. 35. A pilot of a jet airplane makes a sharp turn to create an acceleration of 4.00 times the acceleration of gravity. If the turn has a radius of 500.0 m, what is the speed of the plane? 36. A cork (m 2.88 g) is caught in a small whirlpool created in the basin of a sink. What is the centripetal force acting on the cork, if its speed is 0.314 m/s at a radius of 4.00 cm? 37. When braking to a stop, the maximum force that friction exerts on a 1250-kg auto is 3200 N. (a) If the original speed of the auto is 12.0 m/s, what is the minimum stopping distance? If the speed of the car were twice as great, how would that affect the minimum stopping distance? (b) Unit III Circular Motion, Work, and Energy 337 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 338 38. A force of 250 N [up] is applied to a mass of 15.0 kg over a displacement of 9.60 m [up]. (a) How much work does the force do on the mass? (b) What is the change in gravitational (c) potential energy? If it is an isolated system, explain the difference between the answers for (a) and (b). 39. A car with a mass of 2.00 103 kg is travelling at a velocity of 15.0 m/s [0] on a horizontal stretch of highway. The driver presses on the accelerator so that the force propelling the car forward is increased to 3.30 103 N [0]. The force acts over a displacement of 55.0 m [0] during which force of friction on the car is 5.00 102 N in magnitude. (a) Draw a free-body diagram to analyze the forces acting on the car. What is the net force on the car? (b) What is the work done by net force over the displacement? (c) What is the final kinetic energy of the car? (d) What is the final speed of the car? 40. A block with a mass of 0.800 kg is initially at rest on a frictionless inclined plane. A force of 5.00 N, applied parallel to the inclined plane, moves the block a distance of 4.50 m up the plane. If, at the end of the effort, the block has a speed of 6.00 m/s up the incline, what is the change in height through which it moved? 41. A varying force acts on a 25.0-kg mass over a displacement as shown in the graph below. The mass has an initial velocity of 12.0 m/s [0]. Recall that the area of a force-displacement graph is equivalent to the work done by the force. (a) What is the ) ] ˚ 0 [ 60 45 N (a) What is the change in gravitational potential energy for block A? (b) What was the change in height through which block A moved? B m 0 5 1 . 1.50 m A 43. For question 42, draw the graph that shows the potential, kinetic, and mechanical energies for the system as a function of the displacement assuming (a) there is no friction, (b) there is friction, but block A still accelerates up the hill. 44. A pendulum bob with a mass of 0.750 kg is initially at rest at its equilibrium position. You give the bob a push so that when it is at a height of 0.150 m above its equilibrium position it is moving at a speed of 2.00 m/s. (a) How much work did you do on the bob? If you pushed on the bob with a force of (b) 40.0 N parallel to the displacement, how far did you push it? 45. A billiard ball with a speed of 2.00 m/s strikes a second ball initially at rest. After the collision, the first ball is moving at a speed of 1.50 m/s. If this is an elastic collision (i.e., energy is conserved), what is the speed of the second ball after the collision? Assume that the balls have identical masses of 0.200 kg. 46. A cannon ball (m 3.00 kg) is fired at a velocity of 280 m/s at an angle of 20 above the horizontal. The cannon is on a cliff that is 450 m above the ocean. (a) What is the mechanical energy of the work that the force did on the mass? ( F e c r o F 30 15 (b) What is the final speed of the mass? 0 10 20 Displacement d (m [0˚]) 30 cannon ball relative to the base of the cliff? (b) What is the greatest height above the ocean 40 that the cannon ball reaches? (c) What is the speed of the cannon ball when it lands in the ocean? 42. In the following diagram, block A is at rest on a frictionless inclined plane. It is attached to block B by a light cord over a frictionless pulley. Block A has a mass of 4.50 kg and B has a mass of 5.50 kg. When they are released, block A moves up the incline so that after it has moved a distance of 1.50 m along the incline it has a speed of 3.00 m/s. 47. A Styrofoam™ ball is dropped from a height of 5.00 m. The mass of the ball is 0.200 kg. When the ball hits the ground it has a speed of 3.00 m/s. (a) What change in mechanical energy does the ball undergo while it falls? (b) What is the average force that air friction exerted on the falling ball? 338 Unit III Circular Motion, Work, and Energy 06-PearsonPhys20-Chap06 7/24/08 12:56 PM Page 339 48. What is the average power output if an engine lifts a 250-kg mass a distance of 30.0 m in 20.0 s? 49. What is the effective power required to maintain a constant speed of 108 km/h if the force opposing the motion is 540 N in magnitude? 50. An airplane engine has an effective power output of 150 kW. What will be the speed of the plane if the drag (air friction opposing the motion of the plane) exerts a force of 2.50 103 N? Extensions 51. A 90.9-kg gymnast swings around a horizontal bar in a vertical circle. When he is directly over top of the bar, his arms experience a tug of 108.18 N. What is the speed of his body in this position? (Assume that the gymnast’s mass is centred 1.20 m from the bar.) 52. Suppose a solar system, with three planets circling a star, exists in a galaxy far far away. One of the planets orbits the star with the period of 6.31 107 s and a radius of 3.00 1011 m. It has a moon that orbits it with a period of 1.73 106 s at a radius of 6.00 108 m. (a) What is the |
mass of the planet’s star? (b) What is the mass of the planet? (c) What is the speed of the planet’s moon? 53. A soil-moving machine called a bucket wheel loader has a large metallic wheel with a radius of 3.05 m that has many scoops attached to it. The scoops are designed to dig into the ground and lift soil out as the wheel turns around. If the wheel turns with a frequency of 0.270 Hz, will the soil fall out of a scoop when it gets to the top of the wheel? 54. Three blocks (A, B, and C), with masses 6.00 kg, 4.00 kg, and 2.00 kg, respectively, are initially at rest on a horizontal frictionless surface as shown in diagram (a) on the right. A force of 48.0 N [90] acts on block A over a displacement of 7.50 m [90]. Block A is connected to block B by a string that is 1.00 m long and block B is connected to block C by a string that is 1.50 m long. Initially, the three blocks are touching each other. As the blocks move and the strings become taut, they end up as shown in diagram (b). Is this an isolated or a non-isolated system? Explain. (a) What is the speed of the blocks after the force has acted for the full 7.50 m? (b) What is the speed of block A when the force has acted over a displacement of 2.00 m? Hint: Find the work done by the force. (a) 1.50 m string connecting B to C 1.00 m string connecting A to B force meter (b) Mass A Mass B Mass C 0 m 1. 0 force meter 0 m 1. 5 F F Skills Practice 55. Your cousin doesn’t understand how a satellite can stay in orbit without falling toward Earth. Using the knowledge you have gained from this unit, provide a short explanation. 56. Figure 6.21 on page 303 shows the impact crater for a meteor that landed in Arizona. How widespread is the evidence of meteors striking Earth? Where is the impact crater closest to where you live? Do Internet research to identify locations of impact craters in Alberta and Canada. On a map of Alberta, show the location of meteor impacts. What clues on maps show meteor landings? For each crater, identify how much kinetic energy the meteor would have had when it struck Earth. 57. Explain how you would experiment to determine the quantity of external work done on a system for a cart accelerating down an inclined plane. Could you confirm this quantity by measuring forces? Self-assessment 58. Describe to a classmate one misconception you had about circular motion before studying this unit. Explain what you know about this concept now. 59. Describe to a classmate the relationship between the roles of science and technology in the development of new energy resources. 60. Can science provide solutions to all of the problems associated with the impact of energy consumption on the environment? Give reasons for your answer. e TEST To check your understanding of circular motion, work, and energy, follow the eTest links at www.pearsoned.ca/school/physicssource. Unit III Circular Motion, Work, and Energy 339 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 340 U N I T IV Oscillatory Oscillatory Motion and Motion and Mechanical Mechanical Waves Waves An earthquake more than two thousand kilometres away sent this tsunami speeding across the Indian Ocean. Waves, a form of simple harmonic oscillations, can efficiently transport incredible amounts of energy over great distances. How does a wave move through its medium? How does understanding simple harmonic motion help us understand how waves transport energy? 340 Unit IV 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 341 Unit at a Glance C H A P T E R 7 Oscillatory motion requires a set of conditions. 7.1 Period and Frequency 7.2 Simple Harmonic Motion 7.3 Position, Velocity, Acceleration, and Time Relationships 7.4 Applications of Simple Harmonic Motion C H A P T E R 8 Mechanical waves transmit energy in a variety of ways. 8.1 The Properties of Waves 8.2 Transverse and Longitudinal Waves 8.3 Superposition and Interference 8.4 The Doppler Effect Unit Themes and Emphases • Change, Energy, and Matter • Scientific Inquiry • Nature of Science Focussing Questions As you study this unit, consider these questions: • Where do we observe oscillatory motion? • How do mechanical waves transmit energy? • How can an understanding of the natural world improve how society, technology, and the environment interact? Unit Project • By the time you complete this unit, you will have the knowledge and skill to research earthquakes, the nature of earthquake shock waves, and the use of the Richter scale to rate earthquake intensity. On completion of your research, you will demonstrate the operation of a seismograph. e WEB To learn more about earthquakes and their environmental effects, follow the links at www.pearsoned.ca/school/physicssource. Unit IV Oscillatory Motion and Mechanical Waves 341 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 342 Oscillatory motion requires a set of conditions. On October 15, 1997, NASA launched the Cassini-Huygens space probe toward Saturn — a distance of 1 500 000 000 km from Earth. The probe’s flight path took it by Venus twice, then Earth, and finally past Jupiter on its way to Saturn. This route was planned so that, as the probe approached each planet, it would be accelerated by the planet’s gravitational force. Each time, it picked up more speed, allowing it to get to Saturn more quickly (Figure 7.1). Recall from Chapter 4 that increasing the probe’s speed this way is referred to as gravity assist. The entire journey of 3 500 000 000 km took seven years. For this incredible feat to succeed, scientists had to know where the planets would be seven years in the future. How could they do this? They relied on the fact that planets follow predictable paths around the Sun. Nature is full of examples of repetitive, predictable motion. Water waves, a plucked guitar string, the orbits of planets, and even a bumblebee flapping its wings are just a few. In this chapter, you will examine oscillatory motion. Oscillatory motion is a slightly different form of motion from the circular motion you studied in Chapter 5. You will better understand why bungee jumpers experience the greatest pull of the bungee cord when they are at the bottom of a fall and why trees sway in the wind. You may notice the physics of many objects that move with oscillatory motion and gain a new insight into the wonders of the natural world. C H A P T E R 7 Key Concepts In this chapter, you will learn about: ■ oscillatory motion ■ simple harmonic motion ■ restoring force ■ oscillating spring and pendulum ■ mechanical resonance Learning Outcomes When you have completed this chapter, you will be able to: Knowledge ■ describe oscillatory motion in terms of period and frequency ■ define simple harmonic motion as being due to a restoring force that is directly proportional and opposite to the displacement of an object from an equilibrium position ■ explain the quantitative relationships among displacement, acceleration, velocity, and time for simple harmonic motion ■ define mechanical resonance Science, Technology, and Society ■ explain that the goal of science is knowledge about the natural world Figure 7.1 The Cassini-Huygens probe successfully orbits Saturn after a seven-year journey through the solar system. 342 Unit IV 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 343 7-1 QuickLab 7-1 QuickLab Oscillatory Motion of Toys Problem What is the time taken by one complete back-and-forth motion of a toy? Procedure 1 Fully wind the spring mechanism of one of the toys. 2 Release the winding knob and start your stopwatch. Materials several different wind-up toys, such as: • swimming frog • hopping toy • monkey with cymbals • walking toy yo-yo stopwatch (or wristwatch with a second hand) Figure 7.2 Think About It 3 Count the number of complete back-and-forth movements the toy makes in 10 s. These back-andforth movements are called oscillations. 4 Record the number of oscillations made by the toy. 5 Repeat steps 1 to 4 for each toy. For the yo-yo, first achieve a steady up-and-down rhythm. Then do steps 3 and 4, counting the up-and-down motions. This type of repetitive movement is also an oscillation. Questions 1. In what ways are the motions of all the toys similar? 2. Divide the number of oscillations of each toy by the time. Be sure to retain the units in your answer. What does this number represent? (Hint: Look at the units of the answer.) 3. Which toy had the most oscillations per second? Which had the least? 4. Divide the time (10 s) by the number of oscillations for each toy. Be sure to retain the units in your answer. How long is the interval of oscillation of each toy? 5. Which toy had the longest time for one oscillation? Which had the shortest time? 1. How are the oscillations per second and the time for one oscillation related? 2. What do you think influences the number of oscillations per second of the toys you tested? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 7 Oscillatory motion requires a set of conditions. 343 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 344 7.1 Period and Frequency On a warm summer day in your backyard, you can probably hear bees buzzing around, even if they are a few metres away. That distinctive sound is caused by the very fast, repetitive up-and-down motion of the bees’ wings (Figure 7.3). Take a closer look at the bumblebee. The motion of a bee’s wings repeats at regular intervals. Imagine that you can examine the bee flying through the air. If you start your observation when its wings are at the highest point (Figure 7.4(a)), you see them move down to their lowest point (Figure 7.4(c)), then back up again. When the wings are in the same position as when they started (Figure 7.4(e)), one complete oscillation has occurred. An oscillation is a repetitive backand-forth motion. One complete oscillation is called a cycle. Figure 7.3 The wings of a bee in flight make a droning sound b |
ecause of their motion. oscillatory motion: motion in which the period of each cycle is constant info BIT Earthquake waves can have periods of up to several hundred seconds. (a) (b) (c) (d) (e) Figure 7.4 The bee’s wings make one full cycle from (a) to (e). The time for this motion is called the period. The time required for the wings to make one complete oscillation is the period (T). If the period of each cycle remains constant, then the wings are moving up and down with oscillatory motion. Recall from Chapter 5 that the number of oscillations per second is the frequency (f), measured in hertz (Hz). The equation that relates frequency and period is: f 1 T (1) Table 7.1 shows the period of a bee’s wings as it hovers, along with other examples of periods. ▼ Table 7.1 Periods of Common Items Object Bumblebee wings Hummingbird wings Medical ultrasound technology Middle C on a piano Electrical current in a house Period 0.00500 s 0.0128 s 1 106 to 5 108 s 0.0040 s 0.0167 s 344 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 345 A piston in the engine of a car also undergoes oscillatory motion if it is moving up and down in equal intervals of time. The piston shown in Figure 7.5 moves from position (a) (its highest point) through (b) to position (c), where it is at its lowest point. It begins moving back up again through (d) to (e), where it returns to its highest position. The range of movement from (a) to (e) is one cycle. A single piston in a Formula 1 racecar can achieve a frequency of 300 cycles/second or 300 Hz (18 000 rpm). The piston makes 300 complete cycles in only 1 s. Conversely, the period of the piston, which is the time for one complete cycle, is the inverse of the frequency. It is a mere 0.003 s or about 100 times faster than the blink of an eye! (a) (b) (c) (d) (e) e MATH Using a graphing calculator, plot period as a function of frequency. Use the following values of frequency: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. If possible, print this graph, label it, and add it to your notes. What is the shape of this graph? info BIT A Formula 1 racecar has 10 cylinders but the engine size is limited to 3.0 L, no bigger than many engines in medium-sized cars. The fuel efficiency of F1 cars is approximately 1.3 km/L. Figure 7.5 The piston makes one complete cycle from positions (a) to (e). The time it takes to do this is its period. The number of times it does this in 1 s is its frequency. Example 7.1 What is the frequency of an automobile engine in which the pistons oscillate with a period of 0.0625 s? Analysis and Solution f 1 T 1 0.0625 s 16.0 Hz The frequency of the engine is 16.0 Hz. Practice Problems 1. Earthquake waves that travel along Earth’s surface can have periods of up to 5.00 minutes. What is their frequency? 2. A hummingbird can hover when it flaps its wings with a frequency of 78 Hz. What is the period of the wing’s motion? Answers 1. 3.33 103 Hz 2. 0.013 s M I N D S O N Examples of Oscillatory Motion Working with a partner or group, make a list of three or four natural or humanmade objects that move with the fastest oscillatory motion that you can think of. Make a similar list of objects that have very long periods of oscillatory motion. Beside each object estimate its period. The lists must not include the examples already mentioned. Be prepared to share your lists with the class. Chapter 7 Oscillatory motion requires a set of conditions. 345 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 346 7-2 Inquiry Lab 7-2 Inquiry Lab Relating Period and Frequency Your teacher may want to do this Inquiry Lab in the gym instead of the science lab. Question What is the relationship between the period and the frequency of a bouncing ball? Hypothesis Write a hypothesis that relates the period of the ball’s bounce to its frequency. Remember to use an “if/then” statement. Variables The variables in this lab are the height of the bounce, period, and frequency. Read the procedure, then determine and label the controlled, manipulated, and responding variables. Materials and Equipment stopwatch chair basketball metre-stick tape Procedure 1 Copy Table 7.2 into your notes. ▼ Table 7.2 Bounce, Period, and Frequency Bounce Height (cm) Time for 20 Bounces (s) Period (s/bounce) Frequency (bounces/s) Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 2 Find a convenient place to bounce the basketball near the wall. Using the metre-stick, place tape at heights of 50, 75, 100, 125, and 150 cm. Mark the heights on the tape. 3 Using just a flick of the wrist, begin bouncing the ball at the 50-cm mark. The top of the ball should just make it to this height at the top of its bounce. The ball should bounce with a steady rhythm. 4 While one person is bouncing the ball, another person uses the stopwatch to record the time taken for 20 complete bounces. Record this time in Table 7.2. 5 Reset the stopwatch and begin bouncing the ball to the next height up. Record the time for 20 complete bounces as you did in step 4. To achieve the proper height you may have to stand on a chair. Analysis 1. Using the data for 20 bounces, determine the period and frequency for each height. Record the values in the table. 2. Draw a graph of frequency versus period. What type of relationship is this? 3. What effect did increasing the bounce height have on the period of a bounce? 50 75 100 125 150 346 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 347 In 7-2 Inquiry Lab, the ball will make many more bounces in a certain length of time if it travels a shorter distance than a longer one. Its frequency will be high. By necessity, the amount of time it takes to make one bounce (its period) will be small. The next section explores oscillatory motion by going one step further. You will examine a type of oscillatory motion in which the range of motion is related to the applied force. 7.1 Check and Reflect 7.1 Check and Reflect Knowledge 1. What conditions describe oscillatory motion? 2. Which unit is equivalent to cycles/s? 3. Define period and frequency. 4. How are period and frequency related? 5. Is it possible to increase the period of an oscillatory motion without increasing the frequency? Explain. 6. Give three examples of oscillatory motion that you have observed. Applications 7. What is the frequency of a swimming water toy that makes 20.0 kicking motions per second? 8. Do the oars on a rowboat move with 12. A dog, happy to see its owner, wags its tail 2.50 times a second. (a) What is the period of the dog’s wagging tail? (b) How many wags of its tail will the dog make in 1 min? Extensions 13. Use your library or the Internet to research the frequency of four to six different types of insect wings. Rank these insect wings from highest to lowest frequency. 14. Which of these motions is oscillatory? Explain. (a) a figure skater moving with a constant speed, performing a figure eight (b) a racecar racing on an oval track (c) your heartbeat oscillatory motion? Explain. 15. Many objects exhibit oscillatory motion. 9. Determine the frequency of a guitar string that oscillates with a period of 0.004 00 s. 10. A dragonfly beats its wings with the frequency of 38 Hz. What is the period of the wings? 11. A red-capped manakin is a bird that can flap its wings faster than a hummingbird, at 4800 beats per minute. What is the period of its flapping wings? Use your library or the Internet to find the frequency or range of frequencies of the objects below. (a) fluorescent light bulbs (b) overhead power lines (c) human voice range (d) FM radio range (e) lowest note on a bass guitar e TEST To check your understanding of period and frequency, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 7 Oscillatory motion requires a set of conditions. 347 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 348 info BIT A human eardrum can oscillate back and forth up to 20 000 times a second. 7.2 Simple Harmonic Motion Children on swings can rise to heights that make their parents nervous. But to the children, the sensation of flying is thrilling. At what positions on a swing does a child move fastest? When does the child’s motion stop? Many objects that move with oscillatory motion exhibit the same properties that a child on a swing does. A piston moves up and down in the cylinder of an engine. At the extreme of its motion, it stops for a brief instant as it reverses direction and begins to accelerate downward until it reaches the bottom of its stroke. There it stops again and accelerates back, just as the swing does. In order for the piston or a child on a swing to experience acceleration, it must experience a non-zero net force. This section explores how the net force affects an object’s motion in a special type of oscillatory motion called simple harmonic motion. 7-3 QuickLab 7-3 QuickLab Determining the Stiffness of a Spring Problem How does the force applied to a spring affect its displacement? Materials spring (with loops at each end) spring scale metre-stick tape Procedure 1 Make a two-column table in your notebook. Write “Displacement (cm)” as the heading of the left column, and “Force (N)” as the heading of the right column. 2 Determine the maximum length the spring can be pulled without permanently deforming it. If you are unsure, ask your instructor what the maximum displacement of the spring is. Divide this length by 5. This will give you an idea of the even increments through which to pull your spring. 3 Lie the spring flat on the surface of the table so that it lies in a straight line. Leave room for it to be stretched. Do not pull on the spring. 4 Fix one end of the spring to the desk by holding the loop at the end of the spring with your hand. Do not let this end of the spring move. 5 Attach the spring scale to the free end of the spring but do not pull on it yet. 6 Align the 0-cm mark of the metre-stick wit |
h the other end of the spring at exactly where the spring scale is attached. Tape the stick to the desk (Figure 7.6). spring scale Figure 7.6 7 Pull the spring, using the spring scale, by the incremental distance determined in step 2. Record the values of the displacement and force in your table. 8 Repeat step 7, until you have five values each for displacement and force in your table. 9 Gently release the tension from the spring. Clean up and put away the equipment at your lab station. 348 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 349 Questions 1. Determine the controlled, manipulated, and responding 4. Determine the slope of the line. What are the units of the slope? variables. 2. Plot a graph of force versus displacement. Be sure to use a scale that allows you to use the full graph paper. Draw a line of best fit. 3. What kind of relationship does the line of best fit represent? 5. Extrapolate where the line intercepts the horizontal axis. Why does it intercept there? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Hooke’s Law Robert Hooke (1635–1703) was a British scientist best remembered for using a microscope to discover plant cells, but his talents extended into many areas (Figure 7.7). He is credited with inventing the universal joint, which is used today on many mechanical devices (including cars); the iris diaphragm used to adjust the amount of light that enters a camera lens; the respirator to help people breathe; and the compound microscope, just to name a few of his inventions. In the field of oscillatory motion, he is acknowledged for his work with elastic materials and the laws that apply to them (Figure 7.8). In 1676, Hooke recognized that the more stress (force) is applied to an object, the more strain (deformation) it undergoes. The stress can be applied in many ways. For example, an object can be squeezed, pulled, or twisted. Elastic materials will usually return to their original state after the stress has been removed. This will not occur if too much force is applied or if the force is applied for too long a time. In those cases, the object will become permanently deformed because it was strained beyond the material’s ability to withstand the deformation. The point at which the material cannot be stressed farther, without permanent deformation, is called the elastic limit. A spring is designed to be a very elastic device, and the deformation of a spring (whether it is stretched or compressed) is directly related to the force applied. The deformation of a spring is referred to as its displacement. From his observations, Hooke determined that the deformation (displacement) is proportional to the applied force. This can also be stated as “force varies directly with displacement.” It can be written mathematically as: F x This relationship is known as Hooke’s law, which states: The deformation of an object is proportional to the force causing the deformation. Figure 7.7 Robert Hooke lived at the same time as Sir Isaac Newton. Figure 7.8 Hooke’s notes show the simple devices he used to derive his law. Chapter 7 Oscillatory motion requires a set of conditions. 349 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 350 Hooke’s law: the deformation of an object is proportional to the force causing it Figure 7.9(a) shows a spring that conforms to Hooke’s law. It experiences a displacement that is proportional to the applied force. When no mass is applied to the spring, it is not compressed, and it is in its equilibrium position. As mass is added in increments of 10 g, the displacement (deformation) of the spring increases proportionally in increments of 5 cm, as shown in Figures 7.9(b), (c), and (d). spring constant: amount of stiffness of a spring (a) x 5 cm (b) 10 g (c) (d) x 0 x 10 cm 20 g x 15 cm 30 g Figure 7.9 The spring pictured above conforms to Hooke’s law. If the mass is doubled, the displacement will also double, as seen in (b) and (c). If the force (mass) is tripled, the displacement will triple, as seen in (b) and (d). Each spring is different, so the force required to deform it will change from spring to spring. The stiffness of the spring, or spring constant, is represented by the letter k. Using k, you can write the equation for Hooke’s law as: kx F where F is the applied force that extends or compresses the spring, k is the spring constant, and x is the displacement from the equilibrium position. Graphing Hooke’s Law Figure 7.10 shows how you can use a force meter to measure the applied force required to pull the spring from its equilibrium position to successively farther displacements. For simplicity, we’ll assume that all the springs used in this text are “ideal” springs, meaning that they have no mass. FA FA x 0 Figure 7.10 A force meter is attached to a spring that has not been stretched. The spring is then pulled through several displacements. Each time, the force required for the displacement is recorded. 350 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 351 Figure 7.11 Graph of data from Table 7.3 Table 7.3 shows the data collected for this spring and the results plotted on the graph shown in Figure 7.11. ▼ Table 7.3 Data for Figure 7.11 Force vs. Displacement of a Spring Displacement (m) 0.00 0.10 0.20 0.30 0.40 Force (N) 0.0 5.0 10.0 15.0 20.0 20 ) N ( e c r o F 10 0.0 0.1 0.2 0.3 0.4 Displacement (m) Notice that the relationship is linear (a straight line), which means force is proportional to the displacement. The slope of the line can be determined by the following calculations: slope F x (20.0 N 0.0 N) (0.40 m 0.00 m) 50 N/m This slope represents the spring constant k. The variables F and x are vectors but here we are calculating their scalar quantities so no vector arrows are used. In this example, an applied force of 50 N is needed to stretch (or compress) this spring 1 m. Therefore, the units for the spring constant are newtons per metre (N/m). By plotting the force-displacement graph of a spring and finding its slope, you can determine the spring constant of any ideal spring or spring that obeys Hooke’s law. Of the many objects that display elastic properties, springs are arguably the best to examine because they obey Hooke’s law over large displacements. Steel cables are also elastic when stretched through relatively small displacements. Even concrete displays elastic properties and obeys Hooke’s law through very small displacements. In simpler terms, the property of elasticity gives a material the ability to absorb stress without breaking. This property is vital to consider when structural engineers and designers build load-bearing structures such as bridges and buildings (Figure 7.12). You will learn more about the factors that must be considered in bridge and building design later in this chapter. Figure 7.12 The Jin Mao Tower in Shanghai, China, is 88 storeys high. Skyscrapers are built with elastic materials so they can sway in high winds and withstand the shaking of an earthquake. The main building materials for the Jin Mao Tower are concrete and steel. Chapter 7 Oscillatory motion requires a set of conditions. 351 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 352 Example 7.2 Practice Problems 1. A spring is stretched through several displacements and the force required is recorded. The data are shown below. Determine the spring constant of this spring by plotting a graph and finding the slope. Displacement Force (m) 0.00 0.10 0.20 0.30 0.40 0.50 0.60 (N) 0.0 20.0 50.0 80.0 95.0 130.0 150.0 2. Determine the spring constant of a spring that has the forcedisplacement graph shown in Figure 7.15. Force vs. Displacement 40 30 20 10 ) Displacement (m) Figure 7.15 Answers 1. 2.5 102 N/m 2. 15 N/m To determine the spring constant of a spring, a student attaches a force meter to one end of the spring, and the other end to a wall as shown in Figure 7.13. She pulls the spring incrementally to successive displacements, and records the values of displacement and force in Table 7.4. Plot the values on a graph of force as a function of displacement. Using a line of best fit, determine the spring constant of the spring. x 0 Figure 7.13 Given ▼ Table 7.4 Data for Figure 7.14 Displacement (m) 0.00 0.10 0.20 0.30 0.40 Force (N) 0.00 0.25 0.35 0.55 0.85 Required spring constant (k) Analysis and Solution Using the values from Table 7.4, plot the graph and draw a line of best fit. Force vs. Displacement of a Spring 1..5 (x2, y2) (0.30, 0.60) (x1, y1) (0.10, 0.20) 0.0 0.1 0.2 0.3 Displacement (m) 0.4 Figure 7.14 352 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 353 The slope of the line gives the spring constant (k). Pick two points from the line and solve for the slope using the equation below. Note the points used in the equation are not data points. slope k F x point 1 (0.10, 0.20) point 2 (0.30, 0.60) (0.60 N 0.20 N) (0.30 m 0.10 m) k 2.0 N/m Paraphrase The spring constant of the spring is 2.0 N/m. The Restoring Force Imagine that you have applied a force to pull a spring to a positive dis- placement (x ) as shown in Figure 7.16. While you hold it there, the spring exerts an equal and opposite force in your hand, as described by Newton’s third law in Chapter 3. However, this force is to the left, in the negative direction, and attempts to restore the spring to its equilibrium position. This force is called the restoring force. left right FA x 0 FR x Figure 7.16 The spring system is pulled from its equilibrium position to displacement xx. The displacement is positive, but the restoring force is negative. The restoring force always acts in a direction opposite to the displacement. Therefore, Hooke’s law is properly written with a negative sign when representing the restoring force. kx F (2) In this case, while the spring is held in this position, the applied force and the restoring force have equal magnitudes but opposite direct |
ions, so the net force on the system is zero. In the next section, a mass will be attached to the spring and it will slide on a frictionless horizontal surface. The restoring force will be the only force in the system and will give rise to a repetitive back-and-forth motion called simple harmonic motion. restoring force: a force acting opposite to the displacement to move the object back to its equilibrium position Chapter 7 Oscillatory motion requires a set of conditions. 353 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 354 Example 7.3 A spring has a spring constant of 30.0 N/m. This spring is pulled to a distance of 1.50 m from equilibrium as shown in Figure 7.17. What is the restoring force? Practice Problems 1. Determine the restoring force of a spring displaced 55.0 cm. The spring constant is 48.0 N/m. 2. A spring requires a force of 100.0 N to compress it a displacement of 4.0 cm. What is its spring constant? Answers 1. 26.4 N 2. 2.5 103 N/m Analysis and Solution Draw a diagram to represent the stretched spring. Displacement to the right is positive, so the restoring force is negative because it is to the left, according to Hooke’s law. kx F 30.0 45.0 N (1.50 m) N m The restoring force is 45.0 N [left]. left right FR FA x 1.50 m equilibrium Figure 7.17 Simple Harmonic Motion of Horizontal Mass-spring Systems Figure 7.18 shows a mass attached to an ideal spring on a horizontal frictionless surface. This simple apparatus can help you understand the relationship between the oscillating motion of an object and the effect the restoring force has on it. x 0 left right Fnet 0 m Figure 7.18 The mass is in its equilibrium position (x 0) and is at rest. There is no net force acting on it. Any displacement of the mass to the right is positive, and to the left, negative. e SIM Observe a simulation of simple harmonic motion in a horizontal mass-spring system. Follow the eSIM links at www.pearsoned.ca/school/ physicssource. The position of the mass is represented by the variable x and is measured in metres. In Figure 7.18, there is no tension on the spring nor restoring force acting on the mass, because the mass is in its equilibrium position. Figure 7.19 shows how the restoring force affects the acceleration, displacement, and velocity of the mass when the mass is pulled to a positive displacement and released. 354 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 355 The mass has been pulled to its maximum displacement, Figure 7.19(a) called its amplitude (symbol A). When the mass is released, it begins oscillating with a displacement that never exceeds this distance. The greater the amplitude, the more energy a system has. In this diagram, x A. At maximum displacement, the restoring force is at a maximum value, and therefore, so is the acceleration of the mass, as explained by Newton’s second law (F ma). When the mass is released, it accelerates from rest (v 0) toward its equilibrium position. As the mass approaches this position, its velocity is increasing. But the restoring force is decreasing because the spring is not stretched as much. Remember that force varies directly with displacement. Figure 7.19(b) As the mass returns to its equilibrium position (x 0), it achieves its maximum velocity. It is moving toward the left (the negative direction), but the restoring force acting on it is zero because its displacement is zero. The mass continues to move through the equilibrium position and begins to compress the spring. As it compresses the spring, the restoring force acts on the mass toward the right (the positive direction) to return it to its equilibrium position. This causes the mass to slow down, and its velocity approaches zero. Figure 7.19(c) After passing through the equilibrium position, the mass experiences a restoring force that opposes its motion and brings it to a stop at the point of maximum compression. Its amplitude here is equal, but opposite to its amplitude when it started. At maximum displacement, the velocity is zero. The restoring force has reached its maximum value again. The restoring force is positive, and the displacement is negative. The restoring force again accelerates the mass toward equilibrium. Figure 7.19(d) The mass has accelerated on its way to the equilibrium position where it is now. The restoring force and acceleration are again zero, and the velocity has achieved the maximum value toward the right. At equilibrium, the mass is moving to the right. It has attained the same velocity as in diagram (b), but in the opposite direction. Figure 7.19(e) The mass has returned to the exact position where it was released. Again the restoring force and acceleration are negative and the velocity is zero. The oscillation will repeat again as it did in diagram (a). (a) (b) (c) (d) (e) F max a max v 0 m x is positive F 0 a 0 v max m x 0 F max a max v 0 m x is negative F 0 a 0 v max m x 0 F max a max v 0 m x is positive In Figure 7.19(e), the mass has returned to the position where it started, and one full oscillation has occurred. Throughout its entire motion, the mass-spring system obeys Hooke’s law. In other words, at any instant, the restoring force is proportional to the displacement of the mass. Any object that obeys Hooke’s law undergoes simple harmonic motion (SHM). SHM is oscillatory motion where the restoring force is proportional to the displacement of the mass. An object that moves with SHM is called a simple harmonic oscillator. simple harmonic motion: oscillatory motion where the restoring force is proportional to the displacement of the mass simple harmonic oscillator: an object that moves with simple harmonic motion Chapter 7 Oscillatory motion requires a set of conditions. 355 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 356 (a) (b) Fspring equilibrium Fnet 0 Fg Figure 7.20 The spring in (a) has no mass attached. In (b), the spring stretches until the force exerted by the mass is equal and opposite to the force of gravity, and equilibrium is reached. The net force (or restoring force) is the vector sum of the force of gravity and the tension of the spring. In this case, it is zero. Figure 7.21 The net force (the restoring force) is the vector sum of the upward force exerted by the spring and the downward force of gravity. The force of gravity is always negative and constant, but the force exerted by the spring varies according to the displacement, so the net force changes as the position of the mass changes from (a) to (e). The values of F and v are identical to the horizontal mass-spring system. net, a, PHYSICS INSIGHT For any frictionless simple harmonic motion, the restoring force is equal to the net force. Simple Harmonic Motion of Vertical Mass-spring Systems Figure 7.20(a) shows a spring without a mass attached, anchored to a ceiling. Assume that the spring itself is massless, so it will not experience any displacement. When a mass is attached, the spring is pulled down and deforms as predicted by Hooke’s law. The mass will come to rest when the downward force of gravity is equal to the upward pull (tension) of the spring (Figure 7.20(b)). The displacement of the spring depends on its spring constant. A weak spring has a small spring constant. It will stretch farther than a spring with a large spring constant. In Figure 7.20(b), the net force (or restoring force) acting on the mass is zero. It is the result of the upward tension exerted by the spring balancing the downward force of gravity. This position is considered the equilibrium position and the displacement is zero. If the mass is lifted to the position shown in Figure 7.21(a) and released, it will begin oscillating with simple harmonic motion. Its amplitude will equal its initial displacement. Regardless of the position of the mass, the force of gravity remains constant but the tension of the spring varies. In the position shown in Figures 7.21(b) and (d), the net (restoring) force is zero. This is where the spring’s tension is equal and opposite to the force of gravity. In the position shown in Figure 7.21(c), the displacement of the spring is equal to the amplitude, and the tension exerted by the spring is at its maximum. The mass experiences the greatest restoring force, which acts upward. up down x 0 (a) (b) (c) (d) (e) Fs Fg Fnet max Fnet a max v 0 Fg Fs Fnet Fs Fg Fnet 0 0 Fnet a 0 v max Fs Fg Fs Fg Fnet max Fnet a max v 0 Fs Fnet Fg Fs Fg Fnet 0 0 Fnet a 0 v max Fs Fg Fs Fg Fnet max Fnet a max v 0 Fs Fg Fnet When the mass is below the equilibrium position, the upward force exerted by the tension of the spring is greater than the gravitational force. So the net force — and therefore the restoring force — is upward. Above the equilibrium position, the downward force of gravity exceeds the upward tension of the spring, and the restoring force is downward. The values of velocity, acceleration, and restoring force change in exactly the same way that they do in a horizontal massspring system. 356 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 357 Example 7.4 A spring is hung from a hook on a ceiling. When a mass of 510.0 g is attached to the spring, the spring stretches a distance of 0.500 m. What is the spring constant? Practice Problems 1. Five people with a combined mass of 275.0 kg get into a car. The car’s four springs are each compressed a distance of 5.00 cm. Determine the spring constant of the springs. Assume the mass is distributed evenly to each spring. 2. Two springs are hooked together and one end is attached to a ceiling. Spring A has a spring constant (k) of 25 N/m, and spring B has a spring constant (k) of 60 N/m. A mass weighing 40.0 N is attached to the free end of the spring system to pull it downward from the ceiling. What is the total displacement of the mass? Answers 1. 1.35 104 N/m 2. 2.3 m s and F g Given x 0.500 m m 510.0 g 0.5100 kg g 9.81 m/s2 Required spring constant (k) Analysis and Solution Draw a diagram to |
show the mass-spring system and the forces acting on the mass. up down x 0.500 m equilibrium Fs Fg Figure 7.22 The mass is not moving so the net force on the mass is zero. F are therefore equal in magnitude. kx mg mg x m (0.5100 kg )9.81 s2 0.500 m k 10.0 N/m Paraphrase The spring constant is 10.0 N/m. Chapter 7 Oscillatory motion requires a set of conditions. 357 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 358 Examples of Simple Harmonic Motion Provided the cord doesn’t go slack, a person making a bungee jump will bob up and down with SHM, as shown in Figures 7.23 and 7.24. The cord acts as the spring and the person is the mass. up down x 0 Fcord Fg Fcord Fg Fnet Fcord Fg Fnet Fcord Fnet Fg Fcord Fg Fnet 0 Fnet Fcord Fg Figure 7.24 The bungee jumper bouncing up and down on the cord after a jump in (a) is a vertical mass-spring system. The cord acts as a spring and the jumper is the mass. The restoring (net) force acting on the bungee jumper is the same as it was for the vertical mass-spring system. When the oscillating finally stops, the jumper will come to a stop in the equilibrium position. The reeds of woodwind instruments, such as the clarinet, behave as simple harmonic oscillators. As the musician blows through the mouthpiece, the reed vibrates as predicted by the laws of SHM. Once a simple harmonic oscillator is set in motion, it will slowly come to rest because of friction unless a force is continually applied. We will examine these conditions in section 7.4 on resonance. SHM is repetitive and predictable, so we can state the following: • The restoring force acts in the opposite direction to the displacement. • At the extremes of SHM, the displacement is at its maximum and is referred to as the amplitude. At this point, force and acceleration are also at their maximum, and the velocity of the object is zero. • At the equilibrium position, the force and acceleration are zero, and the velocity of the object is at its maximum. Figure 7.23 A bungee jumper experiences SHM as long as the cord does not go slack. e WEB After the first few oscillations following the jump, the bungee jumper oscillates with simple harmonic motion. To learn more about simple harmonic motion in the vertical direction, follow the links at www.pearsoned.ca/school/ physicssource. 358 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 359 Concept Check 1. Must the line on a graph of force versus displacement for 2. a spring always intercept the origin? Explain. In what situation might the line on a graph of force as a function of displacement for a spring become non-linear? 3. A student wants to take a picture of a vertical mass-spring system as it oscillates up and down. At what point in the mass’s motion would you suggest that she press the button to take the clearest picture? Instead of plotting a force-displacement graph for a spring, a student plots a restoring force-displacement graph. Sketch what this graph might look like. 4. 5. How would you write the equation for Hooke’s law to reflect the shape of the graph above? Simple Harmonic Motion of a Pendulum The Cassini-Huygens space probe featured at the beginning of this chapter is named in honour of two distinguished scientists. Among many other notable accomplishments, the Italian astronomer Giovanni Cassini (1625–1712) observed the planets Mars and Jupiter and measured their periods of rotation. Christiaan Huygens (1629–1695), a Dutch mathematician and astronomer, invented the first accurate clock. It used a swinging pendulum and was a revolution in clock making (Figure 7.25). For small displacements, a swinging pendulum exhibits SHM. Since SHM is oscillatory, a clock mechanism that uses a pendulum to keep time could be very accurate. Up until Huygens’s time, clocks were very inaccurate. Even the best clocks could be out by as much as 15 minutes a day. They used a series of special gears and weights that didn’t always produce a uniform rate of rotation — a necessity for an accurate mechanical clock. Huygens recognized that if he could take advantage of the uniform oscillations of a pendulum, he could produce a much better clock. When completed, his pendulum clock was accurate to within one minute a day. This may not be very accurate by today’s standards, but was easily the best of its time. Pendulum clocks became the standard in time keeping for the next 300 years. Let’s examine cases where an ideal pendulum swings through a small angle, as explained in Figure 7.26 (a) to (e). In this book, all pendulums are considered ideal. That is, we assume that the system is frictionless, and the entire mass of the pendulum is concentrated in the weight. While this is not possible in reality, it is reasonable to make these assumptions here because they provide reasonably accurate results. Figure 7.25 A replica of Huygens’s pendulum clock Chapter 7 Oscillatory motion requires a set of conditions. 359 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 360 (a) left right v 0 max FR a max Fg Fg FR (b) left right v max 0 FR a 0 (c) left right v 0 max FR a max Fg Fg FR (d) left right v max 0 FR a 0 (e) left right v 0 max FR a max Figure 7.26(a) The mass (called a “bob”) is attached to the string and has been pulled from its equilibrium (rest) position through a displacement angle of . It has a mass m. When the bob’s displacement is farthest to the right, the restoring force is a maximum negative value and velocity is zero. When the pendulum is released, gravity becomes the restoring force. Given the direction of the force of gravity, the acceleration due to gravity is straight down. However, the motion of the pendulum is an arc. A component of gravity acts along this arc to pull the bob back toward equilibrium and is, by definition, the restoring force (F R). We can express FR in terms of Fg with the following equation: FR Fg(sin ) Figure 7.26(b) As the bob accelerates downward, its velocity begins to increase and the restoring force (F R) becomes less and less. When it reaches the equilibrium position, no component of gravity is acting parallel to the motion of the bob, so the restoring force is zero, but the velocity has reached its maximum value. Figure 7.26(c) restoring force has also reached a maximum value but it acts toward the right. The bob’s velocity is zero again. The bob has reached its maximum displacement to the left. The The bob passes through the equilibrium position and begins to move upward. As it does so, the restoring force becomes larger as the displacement of the bob increases. But just like the mass-spring system, the restoring force is acting in a direction opposite to the bob’s displacement. At the other extreme of the bob’s displacement, the restoring force has slowed the bob to an instantaneous stop. In this position, the displacement and restoring force are a maximum, and the bob’s velocity is zero. Figure 7.26(d) The bob has achieved its maximum velocity, but this time it is to the right. The bob’s displacement is again zero, and so is the restoring force. On its back swing, the bob moves through the equilibrium position again, as shown here. The velocity is a maximum value, just as it was in Figure 7.26(b), but now it’s in the opposite direction. The restoring force once more brings the bob’s motion to a stop for an Figure 7.26(e) instant at the position farthest to the right. The restoring force is a maximum negative value, and the bob’s velocity is zero. The pendulum has made one complete oscillation as shown in diagrams (a) to (e). Note that the motion of the pendulum bob and the mass-spring systems are similar. Figure 7.19(a–e) on page 355 and Figure 7.26(a–e) on this page are comparable because both systems undergo the same changes to force, velocity, and acceleration at the same displacements. 360 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 361 Motion with Large Amplitudes Earlier in this chapter, you read that a pendulum acts as a simple harmonic oscillator for small angles. Why is that? How is its motion different from SHM at larger angles? The best way to answer these questions is to plot a graph of force versus displacement like those done for springs earlier in this chapter (e.g., Figure 7.11 on page 351). The displacement of the pendulum can be measured by its angle from the vertical. If the graph is linear, then the restoring force is proportional to the displacement, and the pendulum has moved in SHM, as described by Hooke’s law. To create this graph, use the equation for restoring force that you saw in the explanation of Figure 7.26(a) on the previous page: FR Fg(sin ) (3) From this equation, we can plot the values for angles up to 90 for a bob with a mass of 1.0 kg. As the graph in Figure 7.27 shows, the line is not linear, so the restoring force does not vary proportionally with the displacement. Strictly speaking, a pendulum is not a true simple harmonic oscillator. However, the line is almost linear up to about 20. At angles of less than 15, the deviation from a straight line is so small that, for all practical purposes, it is linear. ▼ Table 7.5 Data for Figure 7.27 Angle (°) Restoring Force (N) 0 10 20 30 40 50 60 70 80 90 0 1.70 3.36 4.91 6.31 7.51 8.50 9.22 9.66 9.81 Force vs. Displacement of a Pendulum 10 10 20 40 50 60 30 Displacement (°) 70 80 90 Figure 7.27 For the pendulum to be a true simple harmonic oscillator, its graph of restoring force versus displacement should be linear, as the dotted line suggests. After 15, its line departs from the straight line, and its motion can no longer be considered SHM. Chapter 7 Oscillatory motion requires a set of conditions. 361 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 362 Example 7.5 Practice Problems 1. Determine the restoring force of a pendulum that is pulled to an angle of 12.0 left of the vertical. The mass of the bob is 300.0 g. 2. At what angle must a pendulum be displaced to create a restoring force of 4.00 N [left] o |
n a bob with a mass of 500.0 g? Answers 1. 0.612 N [right] 2. 54.6 Determine the magnitude of the restoring force for a pendulum bob of mass 100.0 g that has been pulled to an angle of 10.0 from the vertical. Given g 9.81 m/s2 m 100.0 g 0.1000 kg Required restoring force (FR) Analysis and Solution Draw a diagram of the pendulum in its displaced position to show the forces acting on the bob. left right 10.0° FT Fg Fg FR Figure 7.28 The restoring force F arc path of the pendulum. R is the component of F g that is tangential to the FR Fg(sin ) mg(sin ) (0.1000 kg)9.81 0.170 N m s2 (sin 10.0) Paraphrase The magnitude of the restoring force acting on the pendulum is 0.170 N. When Christiaan Huygens designed the first pendulum clock, his primary concern was to have the clock operate with a very consistent period. For a uniform period, he could use gear ratios in the mechanism to translate the motion of the pendulum to meaningful units of time, such as minutes and hours. Which factors influence the period of a pendulum, and which do not? To discover how a pendulum’s mass, amplitude, and length influence its period, do 7-4 Inquiry Lab. 362 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 363 Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 2 Attach the thermometer clamp to the top of the retort stand. Attach the thread to the thermometer clamp. Make sure the thread is a little shorter than the height of the clamp from the table. 3 Squeeze one end of the thread in the clamp and use a slip knot on the other end to attach the first mass. The mass should hang freely above the table. 4 Measure the length of the thread from the clamp to the middle of the mass. Record this as the length of the pendulum at the top of Table 7.6. 5 Pull the mass on the thread back until it makes an angle of 15 with the vertical, as measured with the protractor. 6 Remove the protractor, and release the mass as you start to time it. Count the number of complete oscillations it makes in 20 s. Record this number in your table. 7 Remove the mass and replace it with the next mass. Loosen the clamp and adjust the length of the thread so that it is the same length as for the previous mass. (Remember to measure length to the middle of the mass.) Repeat steps 5 to 7 until all the masses are used. Analysis 1. Determine the frequency and period of each mass. Record the numbers in your table. 2. Plot a graph of period versus mass. Remember to place the manipulated variable on the horizontal axis. 3. What conclusion can you draw about the relationship between the mass and the period of a pendulum? Explain your answer and show any relevant calculations. 7-4 Inquiry Lab 7-4 Inquiry Lab A Pendulum and Simple Harmonic Motion Question What is the relationship between the period of a pendulum and its mass, amplitude, and length? Materials and Equipment thermometer clamp retort stand 1.00-m thread (e.g., dental floss) 4 masses: 50 g, 100 g, 150 g, 200 g ruler (30 cm) or metre-stick protractor stopwatch or watch with a second hand Hypothesis Before you begin parts A, B, and C, state a suitable hypothesis for each part of the lab. Remember to write your hypotheses as “if/then” statements. Variables The variables are the length of the pendulum, the mass of the pendulum, elapsed time, and the amplitude of the pendulum. Read the procedure for each part and identify the controlled, manipulated, and responding variables each time. Part A: Mass and Period Procedure 1 Copy Table 7.6 into your notebook. ▼ Table 7.6 Mass and Period Length of Pendulum Mass (g) No. of Cycles/20 s Frequency (Hz) Period (s) 50 100 150 200 Chapter 7 Oscillatory motion requires a set of conditions. 363 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 364 Part B: Amplitude and Period Part C: Length and Period Procedure Procedure 1 Copy Table 7.7 into your notebook. 1 Copy Table 7.8 into your notebook. ▼ Table 7.7 Amplitude and Period ▼ Table 7.8 Length and Period Length of Pendulum Amplitude () No. of Cycles/20 s Frequency (Hz) Period (s) Length (m) No. of Cycles/20 s Frequency (Hz) Period (s) 5 10 15 20 2 Use the same apparatus as in part A. 3 Attach a 200-g mass to the free end of the thread. 4 Measure the length of the thread from the clamp to the middle of the mass. Record this length at the top of Table 7.7. 5 Pull the mass on the thread back until it makes an angle of 5 with the vertical, as measured with the protractor. 6 Remove the protractor, and release the mass as you start to time it. Count the number of complete oscillations it makes in 20 s. Record this number in Table 7.7. 7 Repeat steps 5 and 6, each time increasing the amplitude by 5. Analysis 1. Determine the frequency and period for each amplitude and record the numbers in the appropriate columns in the table. 2. Plot a graph of period versus amplitude. Remember to place the manipulated variable on the horizontal axis. 3. What conclusion can you draw about the relationship between the amplitude and the period of a pendulum? Show any relevant calculations. 2 Use the same apparatus as in part A. Start with a pendulum length of 1.00 m. 3 Attach a 200-g mass to the free end of the thread. 4 Measure the length of the thread from the clamp to the middle of the mass. Record this length in Table 7.8. 5 Pull the mass on the thread back until it makes an angle of 15 with the vertical, as measured with the protractor. 6 Remove the protractor, and release the mass as you start to time it. Count the number of complete oscillations it makes in 20 s. Record this number in Table 7.8. 7 Repeat steps 4 to 6, but each time decrease the length of the pendulum by half. Analysis 1. Determine the frequency and period for each length and record the values in the appropriate column in the table. 2. Plot a graph of period versus length. Remember to place the manipulated variable on the horizontal axis. 3. What conclusion can you draw about the relationship between the length and the period of a pendulum? Show any relevant calculations. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Pendulums and mass-spring systems are not the only simple harmonic oscillators. There are many other examples: a plucked guitar string, molecules vibrating within a solid, and water waves are just a few. In section 7.4, you will explore some human-made examples of SHM and learn about an interesting property called resonance. 364 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 365 7.2 Check and Reflect 7.2 Check and Reflect Knowledge 7. Two students are given the task of 1. The restoring force of a vertical mass- spring system is determined by the mass attached to the spring and the spring constant k. What two factors determine the restoring force of a pendulum? 2. Copy the following tables into your notes. Then fill in the blanks by using the words, “zero” or “maximum.” determining the spring constant of a spring as accurately as possible. To do this, they attach a force meter to a spring that is lying on a desk and is anchored at the other end. One student pulls the spring through several displacements, while the other records the force applied, as shown in the table below. Using this table, plot a graph of force versus displacement. Find the spring constant by determining the slope of the line of best fit. Pendulum Displace- Accelerment ation System max x max a max v min F Velocity Restoring Force Displacement (m) 0.00 0.10 0.20 0.30 0.40 Force (N) 0.00 0.15 0.33 0.42 0.60 Displace- Acceler- ment ation Velocity Restoring Force Massspring System max x max a max v min F 8. Determine the restoring force for a pendulum bob with a mass of 0.400 kg that is pulled to an angle of 5.0 from the vertical. 9. A toy car, with a wind-up spring motor, on a horizontal table is pulled back to a displacement of 20.0 cm to the left and released. If the 10.0-g car initially accelerates at 0.55 m/s2 to the right, what is the spring constant of the car’s spring? (Hint: The restoring force is F ma.) 3. Explain why a pendulum is not a true simple harmonic oscillator. Applications Extension 4. A mass of 2.0 kg is attached to a spring with a spring constant of 40.0 N/m on a horizontal frictionless surface. Determine the restoring force acting on the mass when the spring is compressed to a displacement of 0.15 m. 5. A spring hangs vertically from a ceiling and has a spring constant of 25.0 N/m. How far will the spring be stretched when a 4.0-kg mass is attached to its free end? 6. An applied force of 25.0 N is required to compress a spring 0.20 m. What force will pull it to a displacement of 0.15 m? 10. Obtain three different types of rulers: plastic, metal, and wooden. Fix one end of each ruler to the side of a desk so the ruler juts out horizontally a distance of 25 cm from the edge. Hang enough weight on the end that sticks out to make the ruler bend downward by 2 to 3 cm. Record the deflection of the ruler and the mass used in each case. (Note: The deflection does not have to be the same for each ruler.) Use these data to determine the spring constant for each ruler. Rank the rulers from highest spring constant to lowest. e TEST To check your understanding of simple harmonic motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 7 Oscillatory motion requires a set of conditions. 365 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 366 info BIT An accelerometer is a device used to measure acceleration. It is designed like a mass-spring system. The force exerted by the accelerating object causes the mass to compress the spring. The displacement of the mass is used to determine the positive or negative acceleration of the object. Accelerometers are commonly used in airbag systems in cars (see Chapter 3). If the car slows down too quickly, the displacement of the mass is large and it triggers the airbag to depl |
oy. PHYSICS INSIGHT An object does not have to be moving to experience acceleration. 7.3 Position, Velocity, Acceleration, and Time Relationships One way for ball players to practise their timing is by attempting to throw a ball through a tire swinging on a rope. Someone just beginning this kind of practice might throw too early or too late, missing the tire altogether. Part of the difficulty has to do with the continually changing velocity of the tire. Choosing the best time to throw the ball is an exercise in physics. With practice, the human brain can learn to calculate the proper time to throw without even being aware that it is doing so. Throwing the ball through the tire is much more difficult than it sounds because the tire is a simple harmonic oscillator for small amplitudes. Not only is the velocity continually changing, but so is the restoring force and the acceleration. The only constant for a swinging tire is its period. In this section, you will mathematically analyze acceleration, velocity, and period for SHM in a mass-spring system, and then determine the period of a pendulum. Both mass-spring systems and pendulums are simple harmonic oscillators, as described in section 7.2, but they are different from each other. The mass-spring system has a spring constant k, but the pendulum does not. For this reason, we will look at each separately, starting with the mass-spring system. Acceleration of a Mass-spring System In section 7.2, you learned that two equations can be used to describe force in the mass-spring system: Newton’s second law and Hooke’s law. They can be written mathematically as: • Newton’s second law: F • Hooke’s law: F kx ma (2) net Since both equations refer to the restoring force, you can equate them: F F net ma kx x a k m (4) where aa is the acceleration in metres per second squared; k is the spring constant in newtons per metre; x is the displacement in metres; and m is the mass of the oscillator in kilograms. 366 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 367 e MATH To see how the spring constant, mass, position and acceleration are related graphically, visit www.pearsoned.ca/school/ physicssource. Figure 7.29 The acceleration of a simple harmonic oscillator depends on its position. In position (a), the oscillator moves from its maximum displacement and maximum positive acceleration through to position (b), where the displacement and acceleration are zero. It then moves to position (c), where the oscillator again experiences a maximum acceleration and displacement in the other direction. maximum height v 0 a 9.81 m/s2 The acceleration of a horizontal mass-spring simple harmonic oscillator can be determined by its spring constant, displacement, and mass. It’s logical that the acceleration of the mass depends on how stiff the spring is and how far it is stretched from its equilibrium position. It is also reasonable to assume that, if the mass is large, then the acceleration will be small. This assumption is based on Newton’s second law. The acceleration depends on the displacement of the mass, so the acceleration changes throughout the entire motion as shown in Figure 7.29. Since acceleration of a simple harmonic oscillator is not uniform, only the instantaneous acceleration of the mass can be determined by equation 4. () () (b) (c) () Acceleration (a) (a) () (b) () Displacement () (c) x 0 The Relationship Between Acceleration and Velocity of a Mass-spring System The acceleration of a simple harmonic oscillator is continually changing, so it should come as no surprise that the velocity changes too. As we have just seen, the maximum acceleration occurs when the oscillator is at its maximum displacement. At this position, it is tempting to think that the velocity will be at its maximum as well, but we know that this is not the case. Remember, the acceleration is at its greatest magnitude at the extremes of the motion, yet the oscillator has actually stopped in these positions! In some ways, a ball thrown vertically into the air is similar (Figure 7.30). The acceleration of gravity acts on the ball to return it to the ground. When the ball reaches its maximum height, it comes to a stop for an instant, just like the mass-spring system you studied earlier in this chapter. Figure 7.30 At its maximum height, the ball stops for a brief instant, yet the acceleration of gravity acting on it is not zero. This is similar to the mass-spring system. Chapter 7 Oscillatory motion requires a set of conditions. 367 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 368 () () (a) (b) (c) () Velocity (b) (a) (c) Displacement () x 0 Figure 7.31 The velocity of a simple harmonic oscillator is not uniform. The mass experiences its greatest acceleration at the extremes of its motion where the velocity is zero. Only after the mass accelerates from position (a) to (b) does its velocity reach its maximum value. The mass then decelerates from (b) to (c) where it comes to a stop again. Since the acceleration of the oscillator decreases as it approaches the equilibrium position (Figure 7.31(a)), the velocity does not increase at a uniform rate. The velocity-displacement graph looks like Figure 7.31(b). Figure 7.32 shows a diagram of a simple harmonic oscillator (a mass-spring system) as it moves through one-half of a complete oscillation from (a) to (b) to (c). Below the diagram are the acceleration-displacement and velocity-displacement graphs. The diagram of the oscillator and the graphs are vertically aligned so the graphs show you what is happening as the massspring system moves. In the diagram at the top of Figure 7.32, you can see the oscillator in position (a). It is at its farthest displacement to the left and the spring is compressed. The velocity-displacement graph shows that the oscillator’s velocity in this position is zero (graph 2). You can also see from the accelerationdisplacement graph that the acceleration at that moment is positive and a maximum, but the displacement is negative (graph 1). () () Acceleration and Displacement — Always Opposite Graph 1: Acceleration vs. Displacement (a) (a) (b) (c) () Acceleration () (b) () Displacement () (c) () Velocity (b) Graph 2: Velocity vs. Displacement (a) (c) Displacement () x 0 Figure 7.32 The mass-spring system experiences a changing acceleration and velocity as it makes one-half of a full oscillation from position (a) to position (c). In fact, if you look closely at graph 1, you might notice how the acceleration and displacement are always opposite to one another, regardless of the position of the mass. The acceleration is positive while the displacement is negative, and vice versa. This isn’t surprising, however, because it is what the negative sign in the equation for Hooke’s law illustrates: kx. F Look again at Figure 7.32 and follow the mass as it moves from position (a) to positions (b) and (c). As the oscillator accelerates from position (a) to the right, it picks up speed. The velocity-displacement graph (graph 2) shows that the velocity is positive and increasing as the oscillator approaches position (b), yet the acceleration is decreasing, as shown in graph 1. The oscillator goes through the equilibrium position with a maximum positive velocity, but now the acceleration becomes negative as the spring tries to pull the oscillator back (graph 1). This is why the oscillator slows down and the velocity-displacement graph returns to zero in position (c). 368 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 369 Consider a vertical mass-spring system. A bungee jumper will experience a positive acceleration when she is below the equilibrium position and a negative acceleration when above it (Figures 7.33 and 7.34). (a) (b) (c) equilibrium x max x 0 x max a v a v a v Displacement Figure 7.33 Displacement Figure 7.34 After a jump, the bungee jumper is shown in three positions: At the lowest point (a), in the equilibrium position (b), and at her maximum displacement (c). In each case the circled region on the graphs indicates her acceleration and velocity. Maximum Speed of a Mass-spring System Now you know that a simple harmonic oscillator will experience its greatest speed at the equilibrium position. What factors influence this speed, and how can we calculate it? In our examples, the mass-spring system is frictionless, and no external forces act on it. This is referred to as an isolated system and the law of conservation of energy applies. We will use this concept to derive the equation for the maximum speed. Recall from Chapter 6 that the total mechanical energy in an isolated system remains constant. That means that the kinetic and potential energy of the system may vary, but their sum is always the same. Chapter 7 Oscillatory motion requires a set of conditions. 369 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 370 In other words, at any position in the motion of a mass-spring system, the sum of kinetic and potential energies must be equal to the total energy of the system. Recall that kinetic energy is expressed as: Ek 1 mv2 2 Recall that elastic potential energy is expressed as: Ep 1 kx2 2 Let’s begin by looking at the energy of the system in two positions: • When the mass is at the maximum displacement (Figure 7.35(a)). • When the mass is at the minimum displacement (Figure 7.35(b)). v 0 Ek 0 Ep max m x max The oscillator at its maximum displacement. Figure 7.35(a) Potential energy has reached a maximum value (Ep oscillator’s displacement is a maximum (x A). The kinetic energy is zero (v 0). ) because the max v max Ek max Ep 0 m x 0 Figure 7.35(b) The oscillator at its minimum displacement (x 0). Kinetic energy has reached a maximum value (Ekmax has a maximum velocity. The potential energy is zero (x 0). ) because the oscillator Remember that the total energy of the system remains constant regardless of the oscillator’s position. The equation for the total energy is: ET |
Ep Ek The kinetic energy of the oscillator at its maximum displacement is zero so the total energy of the oscillator at that position must be: ET Epmax The potential energy of the oscillator at its minimum displacement is zero so the total energy of the oscillator at that position must be: ET Ekmax Because the total energy is always the same, we can write: Ekmax Epmax or 1 2 mv 2 max 1 kx 2 2 max 370 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 371 max mv2 1 kA2 2 If we use A to represent xmax, we can write: 1 2 We can then simplify this equation to solve for vmax: 1 2 1 kA2 2 mv 2 max mv 2 max kA2 v 2 max kA2 m Then we take the square root of each side: vmax kA2 m or vmax A k m (5) e WEB To find out how these factors are taken into account in bungee jumping, follow the links at www.pearsoned.ca/school/ physicssource. Factors That Influence the Maximum Speed of a Mass-spring System Three factors influence the maximum speed of a mass-spring system: • the amplitude of the oscillations: If the oscillator moves through a large amplitude, the restoring force increases in proportion to the amplitude. As the restoring force increases, so does the acceleration, and the oscillator will achieve a greater velocity by the time it reaches the equilibrium position. • the stiffness of the spring: A stiffer spring with a higher spring constant exerts a stronger restoring force and creates a greater maximum velocity for the same reasons that increasing the amplitude does. • the mass of the oscillator: Changing the mass of an oscillator has a different effect. If the mass increases, the velocity of the oscillations decreases. This is because the oscillator has more inertia. A larger mass is harder to accelerate so it won’t achieve as great a speed as a similar mass-spring system with less mass. Concept Check 1. When acceleration is negative, displacement is positive and vice versa. Why? 2. Why is the velocity-time graph of a simple harmonic oscillator a curved line? 3. The acceleration-displacement graph and velocity-displacement graph are shown in Figure 7.32 on page 368 for half of an oscillation only. Sketch three more acceleration-displacement and velocitydisplacement graphs for the second half of the oscillation. 4. Suppose the amplitude of an object’s oscillation is doubled. How would this affect the object’s maximum velocity? Chapter 7 Oscillatory motion requires a set of conditions. 371 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 372 PHYSICS INSIGHT When a mass is hanging from a vertical spring at rest in the equilibrium position, the downward g is force of gravity F equal and opposite to the upward force exerted s, so the by the spring F restoring force is zero. The force of gravity acting on the mass doesn’t change. If the spring is displaced from the equilibrium position, the restoring force will just be the force of the spring due to its displacement x. This is kx. expressed as F R Example 7.6 A 100.0-g mass hangs motionless from a spring attached to the ceiling. The spring constant (k) is 1.014 N/m. The instructor pulls the mass through a displacement of 40.0 cm [down] and releases it. Determine: (a) the acceleration when the mass is at a displacement of 15.0 cm [up], and (b) the maximum speed of the mass. up v down a x 15.0 cm equilibrium x 40.0 cm FR Figure 7.36 Given m 100.0 g 0.1000 kg k 1.014 N/m x 40.0 cm [down] 0.400 m [down] Required (a) acceleration (a) when x 15.0 cm [up] 0.150 m [up] (b) maximum speed (vmax) Practice Problems 1. A 0.724-kg mass is oscillating on a horizontal frictionless surface attached to a spring (k 8.21 N/m). What is the mass’s displacement when its instantaneous acceleration is 4.11 m/s2 [left]? 2. A 50.0-g mass is attached to a spring with a spring constant (k) of 4.00 N/m. The mass oscillates with an amplitude of 1.12 m. What is its maximum speed? 3. An instructor sets up an oscillating vertical mass-spring system (k 6.05 N/m). The maximum displacement is 81.7 cm and the maximum speed is 2.05 m/s. What is the mass of the oscillator? Answers 1. 0.362 m [right] 2. 10.0 m/s 3. 0.961 kg Analysis and Solution (a) The mass will begin to oscillate when released. Acceleration R net is a vector quantity so direction is important. F F ma kx kx a m N (0.150 m) 1.014 m 0.1000 kg 1.52 m/s2 (b) The maximum speed occurs when the mass is in the equilibrium position, whether it is moving up or down. The displacement of the mass before it is released is the amplitude (A) of the mass’s oscillation. vmax A k m N 1.014 m 0.1000 kg 0.400 m 1.27 m/s Paraphrase (a) The mass has an acceleration of 1.52 m/s2 [down] when it is 15.0 cm above the equilibrium position. (b) The maximum speed of the mass is 1.27 m/s. 372 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:09 PM Page 373 Period of a Mass-spring System The next time you are travelling in a vehicle at night, watch for bicycles moving in the same direction as your vehicle. Notice the peculiar motion of the pedals as they reflect the light from your headlights (Figure 7.37). From a distance, these reflectors don’t appear to be moving in a circular path, but seem to be moving up and down. The apparent up-and-down motion of the pedals is the same kind of motion as a mass-spring system oscillating back and forth, so it is simple harmonic motion. This observation proves useful because it is an example of how circular motion can be used to describe simple harmonic motion. The next few pages show how to derive equations for the period and maximum speed of a simple harmonic oscillator. Two conditions are necessary if circular motion is to be used to replicate simple harmonic motion: 1. The period of both the circular motion and the simple harmonic motion must be the same. 2. The radius of the circular motion must match the amplitude of the oscillator. For example, look at Figure 7.38, where a mass moving in a circular path with a radius r is synchronized with a mass-spring simple harmonic oscillator. This illustration demonstrates how circular motion can be used to describe SHM. Figure 7.37 From a distance, the reflectors on the bicycle pedals would appear to be moving up and down instead of in a circle. (a) (b) (c) (d) (e Figure 7.38 A mass moving in a circle is a simple harmonic oscillator that corresponds to the mass-spring oscillator shown below it. One-half of a complete cycle is shown here. For our purposes, the following conditions are true: • The radius of the circular motion is equal to the amplitude of the oscillator (r A, as shown in Figure 7.38(a)). • The mass in circular motion moves at a constant speed. • The periods of the mass in circular motion and the oscillator in the mass-spring system are the same. Chapter 7 Oscillatory motion requires a set of conditions. 373 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 374 Deriving the Equation for the Period of a Mass-spring System Recall that the maximum velocity of a simple harmonic oscillator occurs when it is in its equilibrium position, which is position (c) in Figure 7.38. At the exact moment that the mass in circular motion is in position (c), its velocity is in the same direction as the velocity of the mass-spring system, and they are both moving at the same speed. But if the mass moving in a circular path is moving at a constant speed, then it must always be moving at the maximum speed of the mass-spring oscillator! Therefore, the maximum speed (vmax) of the mass-spring system is equal to the speed (v) of the circular mass system. The speed of an object moving in a cir- cular path was derived in Chapter 5. It is: v 2r T (6) Figure 7.39 The strings of a piano all have different masses. Even if they vibrate with the same amplitude they will have a different period of vibration because each string has a different mass. A heavy string will vibrate with a longer period (and lower frequency) than a lighter string. The speed of the circular motion (v) matches the maximum speed on the mass-spring oscillator (vmax), and the radius of the circle matches its amplitude. Therefore, we can customize the equation for the mass-spring oscillator: 2A T vmax (7) If we equate equation 5 and equation 7, we get: A k m 2A T We can then solve for T: A k m 2A T 2π k T T 2 m m k (8) This equation describes the period of a simple harmonic oscillator, where T is the period of the oscillator in seconds; k is the spring constant in newtons per metre; and m is the mass of the oscillator in kilograms. Figure 7.39 is an example of an application of this equation. PHYSICS INSIGHT The period of a simple harmonic oscillator does not depend on displacement. 374 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 375 Factors Affecting the Period of an Oscillating Mass The larger the oscillating mass is, the longer its period of oscillation is. This seems reasonable since a large mass takes longer to speed up or slow down. It would also seem reasonable that the period should be inversely related to the spring constant, as the equation suggests. After all, the stiffer the spring, the more force it exerts over smaller displacements. Therefore, you could expect the mass to oscillate more quickly and have a smaller period. What is interesting about this equation is not what influences the period but what does not. It may seem odd that the displacement of the mass has no influence on the period of oscillation. This means that if you were to pull a mass-spring system to a displacement x and then let go, it would have the same period of oscillation as it would if you pulled it to a displacement of 2x and released it! The two identical mass-spring systems in Figure 7.40 have different amplitudes but the same period. k1 10.0 N/m k2 10.0 N/m amplitude 20 cm m1 x 0 amplitude 10 cm m2 x 0 1 10° θ 2 5° θ Figure 7.40 Two identical mass-spring systems have the same spring constant and mass, but different amplitudes. Which has the longest p |
eriod? They have the same period because displacement doesn’t affect period. Figure 7.41 Two identical pendulums have the same mass and length, but different amplitudes. Which one has the longest period? They have the same period because displacement doesn’t affect the period of a simple harmonic oscillator. This relationship is true for any simple harmonic oscillator, including a pendulum with a small amplitude. It is easy enough to test. Take two pendulums with the same mass and length (Figure 7.41). Pull both bobs back to different displacements. Remember to keep the displacements small so the pendulums oscillate with SHM. Release them at the same time. You will discover that both make one full oscillation in unison. This means they return to the point of release at exactly the same time. The pendulum that begins with the larger displacement has farther to travel but experiences a larger restoring force that compensates for this. Chapter 7 Oscillatory motion requires a set of conditions. 375 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 376 Example 7.7 What is the period of oscillation of a mass-spring system that is oscillating with an amplitude of 12.25 cm and has a maximum speed of 5.13 m/s? The spring constant (k) is 5.03 N/m. Practice Problems 1. A mass of 2.50 kg is attached to a horizontal spring and oscillates with an amplitude of 0.800 m. The spring constant is 40.0 N/m. Determine: a) the acceleration of the mass when it is at a displacement of 0.300 m the maximum speed the period b) c) 2. A 2.60-g mass experiences an acceleration of 20.0 m/s2 at a displacement of 0.700 m on a spring. What is k for this spring? 3. What is the mass of a vertical mass-spring system if it oscillates with a period of 2.0 s and has a spring constant of 20.0 N/m? 4. What is the period of a vertical mass-spring system that has an amplitude of 71.3 cm and maximum speed of 7.02 m/s? The spring constant is 12.07 N/m. Answers 1. a) 4.80 m/s2 b) 3.20 m/s c) 1.57 s 2. 0.0743 N/m 3. 2.0 kg 4. 0.638 s A 12.25 cm left right Given A 12.25 cm 0.1225 m k 5.03 N/m vmax 5.13 m/s Required period of the oscillations (T ) equilibrium Analysis and Solution To determine the period of the oscillator, you need to know the oscillator’s mass. Use the maximum speed equation (equation 5) to find the mass: Figure 7.42 EPmax Ekmax A2 2 k mv max 2 2 mv 2 max kA2 0.1225 m)2 5.03 m m 2 5.13 s 2.868 103 kg Then use equation 8 to determine the period: T 2 m k 2 2.868 103 kg N 5.03 m 0.150 s Paraphrase The period of the mass-spring oscillator is 0.150 s. 376 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 377 Concept Check 2. 1. What effect does doubling the displacement have on the period of oscillation of a simple harmonic oscillator? Explain your answer. In order to compare circular motion to the motion of a simple harmonic oscillator, what two conditions must be satisfied? If the mass and spring constant of a mass-spring oscillator were doubled, what effect would this have on the period of the oscillations? 3. 4. Two mass-spring systems with identical masses are set oscillating side by side. Compare the spring constants of the two systems if the period of one system is twice the other. The Period of a Pendulum Christiaan Huygens recognized that a pendulum was ideally suited for measuring time because its period isn’t affected by as many of the factors that influence a mass-spring system. A pendulum doesn’t have a spring constant, k, like the mass-spring system does, and unlike the mass-spring system, the mass of the pendulum does not affect its period. Because of these factors, a new equation for a pendulum’s period must be derived. In doing so, you will discover why its mass is irrelevant and what factors play a role in its period of oscillation. Take a closer look at the pendulum when it is at a small displacement of 15 or less, as shown in Figure 7.43. For a small angle (), the displacement of the bob can be taken as x. The sine of angle is expressed as: x l sin Recall that the restoring force for a pendulum is FR above expression for sin in this equation: Fg sin . Use the Fg FR x l θ l x Recall also that in a mass-spring system, the restoring force is F kx. We want to solve for the period (T ), which is a scalar quantity, so the negative sign in Hooke’s law can be omitted. The two equations for restoring force can then be equated: Figure 7.43 For a pendulum with a small displacement of 15 or less, the displacement is x. kx Fg x l Fg mg kx (mg) x l k mg l Chapter 7 Oscillatory motion requires a set of conditions. 377 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 378 None of the values of mass, gravitational field strength, or length change for a pendulum. They are constants, which are represented by k. Substitute them into equation 8 (page 374): T 2 m k g) m (m 2 T 2 l g l (9) PHYSICS INSIGHT The period of a pendulum does not depend on its mass or amplitude. e SIM Learn more about the motion of a pendulum and the factors that affect it. Follow the eSIM links at www.pearsoned.ca/school/ physicssource. where l is the length of the pendulum string in metres; and g is the gravitational field strength in newtons per kilogram. Recall that the length of the pendulum is always measured from the point where it is attached at the top, to the centre of mass of the bob, not the point at which the string or wire is attached to the bob. Also recall that the period of the pendulum’s swing does not depend on the mass of the pendulum bob. This may not seem logical but it is indeed the case — just as the acceleration of an object in free fall doesn’t depend on the mass of the object. The Pendulum and Gravitational Field Strength Equation 9 is useful when it is manipulated to solve for g, the gravitational field strength. As you learned in section 4.3 of Chapter 4, the gravitational field strength varies with altitude and latitude. The magnitude of the gravitational field is 9.81 N/kg at any place on Earth’s surface that corresponds to the average radius of Earth. However, very few places on the surface of Earth are at exactly the average radius. To determine the exact value of g at any point, you can use a pendulum. If you manipulate equation 9 and solve for g, you get: g 42l T 2 (10) Due to the changing nature of Earth’s gravity, Christiaan Huygens’s pendulum clock (introduced in section 7.2) was only accurate if it was manufactured for a specific place. For example, pendulum clocks designed to operate in London could not be sold in Paris because the accuracy could not be maintained. The difference in gravitational field strength between London and Paris meant that the period of oscillation would be slightly different. The difference in g between two locations could be quite small, but the cumulative effect on a pendulum clock would be significant. An extreme example of the varying value of g at different geographic locations can be illustrated by using a pendulum to determine the gravitational field strength at the top of Mount Everest. 378 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 379 Example 7.8 What is the gravitational field strength at the top of Mount Everest at an altitude of 8954.0 m, if a pendulum with a length of 1.00 m has a period of 2.01 s? Analysis and Solution Use equation 10 to determine g. Note that no vector arrow is required with the symbol g because you are calculating a scalar quantity. g 42l T 2 (42)(1.00 m) (2.01 s)2 9.77 m/s2 9.77 N/kg T 2.01 s 8954.0 m Figure 7.44 Mount Everest g 9.81 N/kg The gravitational field strength at the top of Mount Everest is 9.77 N/kg, which is very close to the accepted value of 9.81 N/kg. The extra height of Mount Everest adds very little to the radius of Earth. Practice Problems 1. What is the gravitational field strength on Mercury, if a 0.500-m pendulum swings with a period of 2.30 s? 2. A pendulum swings with a period of 5.00 s on the Moon, where the gravitational field strength is 1.62 N/kg [down]. What is the pendulum’s length? 3. What period would a 30.0-cm pendulum have on Mars, where the gravitational field strength is 3.71 N/kg [down]? Answers 1. 3.73 N/kg [down] 2. 1.03 m 3. 1.79 s At the top of Mount Everest, a pendulum will swing with a slightly different period than at sea level. So a pendulum clock on Mount Everest, oscillating with a longer period than one at sea level, will report a different time. Huygens’s clocks also suffered from another problem: the pendulum arm would expand or contract in hot or cold weather. Since the length of the arm also determines the period of oscillation, these clocks would speed up or slow down depending on the ambient temperature. Given their limitations, pendulum clocks were not considered the final solution to accurate timekeeping. Further innovations followed that you may want to research on your own. e WEB To learn more about pendulum clocks and the evolution of timekeeping, create a timeline of the evolution of clock design. In your timeline include what the innovation was, who invented it, and the year it was introduced. Begin your search at www.pearsoned.ca/school/ physicssource. Concept Check 1. An archer is doing target practice with his bow and arrow. He ties an apple to a string and sets it oscillating left to right, down range. In what position of the apple should he aim so that he increases his chances of hitting it? Explain your answer. 2. What factors affect the accuracy of pendulum clocks? Why? Chapter 7 Oscillatory motion requires a set of conditions. 379 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 380 7.3 Check and Reflect 7.3 Check and Reflect Knowledge 1. Explain the effect that changing each of the following factors has on the period of a mass-spring system: (a) amplitude (b) spring constant (c) mass 2. Explain what effect changing each of the following factors has on the period of a pendulum: (a) amplitude (b) gravitational field strength (c) mass 9. A pendulu |
m bob (m 250.0 g) experiences a restoring force of 0.468 N. Through what angle is it displaced? 10. A 50.0-cm pendulum is placed on the Moon, where g is 1.62 N/kg. What is the period of the pendulum? Extensions 11. A horizontal mass-spring system oscillates with an amplitude of 1.50 m. The spring constant is 10.00 N/m. Another mass moving in a circular path with a radius of 1.50 m at a constant speed of 5.00 m/s is synchronized with the mass-spring system. Determine the mass-spring system’s: 3. Describe the positions that a mass-spring system and pendulum are in when: (a) period (b) mass (a) acceleration is a maximum (b) velocity is a maximum (c) restoring force is maximum 4. Why is the acceleration of a simple harmonic oscillator not uniform? 5. A mass-spring system has a negative displacement and a positive restoring force. What is the direction of acceleration? Applications 6. What length of pendulum would oscillate with a period of 4.0 s on the surface of Mars (g 3.71 N/kg)? 7. A mass of 3.08 kg oscillates on the end of a horizontal spring with a period of 0.323 s. What acceleration does the mass experience when its displacement is 2.85 m to the right? 8. A 50.0-kg girl bounces up and down on a pogo stick. The girl has an instantaneous acceleration of 2.0 m/s2 when the displacement is 8.0 cm. What is the spring constant of the pogo stick’s spring? (c) maximum acceleration 12. A quartz crystal (m 0.200 g) oscillates with simple harmonic motion at a frequency of 10.0 kHz and has an amplitude of 0.0500 mm. What is its maximum speed? 13. A horizontal mass-spring system has a mass of 0.200 kg, a maximum speed of 0.803 m/s, and an amplitude of 0.120 m. What is the mass’s position when its acceleration is 3.58 m/s2 to the west? 14. Suppose an inquisitive student brings a pendulum aboard a jet plane. The plane is in level flight at an altitude of 12.31 km. What period do you expect for a 20.0-cm pendulum? (Hint: First determine the gravitational field strength as shown in Chapter 4.) e TEST To check your understanding of position, velocity, acceleration, and time relationships in mass-spring systems and pendulums, follow the eTest links at www.pearsoned.ca/school/physicssource. 380 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 381 info BIT When you walk with a drink in your hand at the right speed, your motion creates resonance in the liquid. This makes waves that splash over the edge of the cup. To prevent this, people walk slowly so resonance doesn’t occur, often without knowing why this works. resonant frequency: the natural frequency of vibration of an object 7.4 Applications of Simple Harmonic Motion People’s arms swing as they walk. An annoying rattle can develop in a car when it reaches a certain speed. A child can make large waves in the bathtub by sliding back and forth. Many things can be made to vibrate, and when they do, they seem to do it with a period of motion that is unique to them. After all, how often do you think about your arms swinging as you walk? You don’t — they seem to swing of their own accord and at their own frequency. The water in the bathtub will form very large waves when the child makes a back-and-forth motion at just the right rate. Any other rate won’t create the waves that splash over the edge and soak the floor, which, of course, is the goal. In all these cases, the object vibrates at a natural frequency. Resonant frequency is the natural frequency of vibration of an object. In other words, objects that are caused to vibrate do so at a natural frequency that depends on the physical properties of the object. All objects that can vibrate have a resonant frequency, including a pendulum. Maintaining a Pendulum’s Resonant Frequency A pendulum swings back and forth at its resonant frequency. Since the acceleration of gravity does not change if we stay in the same place, the only factor that affects the resonant frequency is the pendulum’s length. All pendulums of the same length oscillate with the same natural (resonant) frequency. Huygens made use of this fact when he designed his pendulum clock (Figure 7.45). He knew that all pendulum clocks would keep the same time as long as the length of the pendulum arms was the same. Their resonant frequencies would be identical. However, Huygens faced some challenges in making a pendulum clock. The arm of the pendulum would expand or contract with temperature, affecting its period. But this was a relatively minor issue compared to another difficulty that had to be overcome — friction. Unless something was done, friction would very quickly stop the pendulum from swinging. To compensate for the effects of friction, he designed his clocks so that the pendulum was given a small push at just the right moment in its swing. The timing of these pushes coincided with the resonant frequency of the pendulum. By doing this, Huygens could make the pendulum swing for as long as the periodic force was applied. Figure 7.45 The interior of Huygens’s clock Chapter 7 Oscillatory motion requires a set of conditions. 381 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 382 forced frequency: the frequency at which an external force is applied to an oscillating object PHYSICS INSIGHT The forced frequency that is the same as the resonant frequency will increase the amplitude of the SHM, but will not change the resonant frequency. mechanical resonance: the increase in amplitude of oscillation of a system as a result of a periodic force whose frequency is equal or very close to the resonant frequency of the system Forced Frequency To visualize how this works, imagine a child on a swing. A swing is an example of a pendulum, with the child as the bob. The swing moves back and forth at its natural frequency, which depends only on its length. To keep the swing going with the same amplitude, all the parent has to do is push at just the right moment. The timing of the pushes must match the frequency of the swing. As anyone who has pushed a swing can attest, it takes very little energy to keep a swing swinging to the same height. The frequency at which the force is applied to keep the swing moving is called the forced frequency. If the forced frequency matches or is close to the resonant frequency of the object, then very little force is required to keep the object moving. The resonant frequency won’t change though, because it depends only on the length of the pendulum. If the parent decides to push a little harder each time the swing returns, then the swing’s amplitude will increase, but not its frequency. A larger force than is needed to overcome friction will create a larger amplitude of motion. If the forced frequency isn’t close to the resonant frequency, then the object will not vibrate very much and will have a small amplitude. Imagine trying to increase the frequency of a pendulum by increasing the forced frequency. Much of the force won’t be transferred to the pendulum because the pendulum won’t be in the right position when the force is applied. The pendulum will bounce around but there will be no increase in its amplitude of vibration, and its motion will become harder to predict. The flowchart in Figure 7.46 on the next page summarizes the relationship between forced frequency and resonant frequency. Mechanical Resonance A forced frequency that matches the resonant frequency is capable of creating very large amplitudes of oscillation. This is referred to as mechanical resonance. This can be a good or bad thing. The larger the amplitude, the more energy the system has. Huygens’s pendulum clock didn’t need to have large oscillations, so a very small force could keep the pendulum swinging. A small weight-driven mechanism was used to provide the force needed. The force simply had to be applied with the same frequency as the pendulum. Huygens managed to do this without much difficulty. His pendulum clocks were a great success but weren’t completely practical since they had to be placed on solid ground. A pendulum clock would not work aboard a ship because sailing ships of the time were buffeted by the waves more than today’s large ocean-going vessels are. The motion of the ship on the waves would disturb a pendulum’s SHM, so sailors could not take advantage of the increased accuracy these clocks provided. 382 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 383 The key to successfully navigating across an ocean (where there are no landmarks) was to use an accurate clock on the ship. This clock could be synchronized to a clock in Greenwich, England, which is situated on the prime meridian of 0° longitude. As the ship travelled east or west, the sailors could compare their local time, using the Sun and a sundial, to the ship’s clock, which was still synchronized to the time on the prime meridian. The difference in time between the two clocks could be used to compute their longitudinal position. However, it wasn’t until the 1700s that a brilliant clockmaker, John Harrison, successfully made a marine chronometer (ship’s clock) that was immune to the buffeting of waves and temperature. It contained several ingenious innovations and, for better or worse, made the pendulum obsolete in navigation. NO The amplitude of vibration is not increased or may be decreased. Is the forced frequency close to the resonant frequency? YES A small force is required to keep the object vibrating with the same amplitude. The force is increased. The amplitude of vibration increases. Figure 7.46 Flowchart of the effect of forced frequency on resonant amplitude Chapter 7 Oscillatory motion requires a set of conditions. 383 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 384 7-5 QuickLab 7-5 QuickLab Investigating Mechanical Resonance Problem How can we cause a pendulum to begin oscillating at its resonant frequency using a forced frequency? 8 Repeat step 7 three more times. Each time, lengthen the thread of mass 2 by 10 cm. Fo |
r your last trial, the thread of mass 2 should be 40.0 cm long. Materials retort stands string thread 2 identical masses (200 g each) string thread mass 2 retort stand mass 1 two equal masses Figure 7.47 Procedure Part A 1 Read the questions in the next column before doing the lab. 2 Set up the two retort stands about 75 cm apart. 3 Tie the string to both retort stands at the same height (50.0 cm) on each stand, as shown in Figure 7.47. Clamp the ends of the string to the retort stands so they don’t slip. The string should be taut. 4 Pull the two retort stands farther apart if you need to remove slack from the string. 5 Attach the thread to one mass and tie the other end to the string so that the distance from the mass to the string is 30 cm. This is mass 1. 6 Repeat step 5 for the second mass (mass 2) so that it has a length of 10 cm and is attached to the string about 15 cm from the first mass. 7 Make sure neither mass is moving, then pull mass 2 back a small distance and release it. Observe the motion of mass 1 as mass 2 oscillates. Make a note of the maximum amplitude that mass 1 achieves. Part B 9 Adjust the thread length of mass 2 so that it is as close to the thread length of mass 1 as possible. 10 Make sure both masses are motionless. Pull back and release mass 2. Note the amplitude of vibration that mass 1 achieves. 11 Pull the retort stands farther apart and hold them there so the tension in the string is increased, and the string is almost horizontal. 12 Make sure both masses are motionless, then pull back mass 2 and release it. Note the amplitude of vibration that mass 1 achieves. Questions Part A 1. At what thread length did mass 2 create the maximum oscillation of mass 1? Explain why this happened, in terms of frequency. 2. At what thread length did mass 2 create the minimum oscillation of mass 1? Explain why this happened, in terms of frequency. 3. Why did mass 1 have a large amplitude of vibration in only one case? Part B 4. What effect did increasing the tension on the string have on the amplitude achieved by mass 1? 5. Why did increasing the tension alter the maximum oscillation of mass 1? 6. Write a sentence describing the effect that increasing the tension had on the resonant amplitude of mass 1. Use the terms forced frequency and resonant amplitude in your answer. e LAB For a probeware lab, go to www.pearsoned.ca/school/physicssource. 384 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 385 Resonance Effects on Buildings and Bridges A forced frequency that matches the resonant frequency can create problems for designers of bridges and skyscrapers. A bridge has a resonant frequency that can be amplified by the effect of wind. Air flows around the top and bottom of a bridge and can cause it to vibrate. The bridge will vibrate at its resonant frequency, with a large amplitude, even though the force applied by the wind may be relatively small. As the bridge vibrates, it may flex more than it is designed to and could conceivably vibrate to pieces. A skyscraper is also susceptible to forced vibrations caused by the wind. Most skyscrapers have a huge surface area and catch a lot of wind. Even though a building is a rigid structure, the force of the wind can make it sway back and forth. The wind causes a phenomenon called “vortex shedding.” It can create a forced vibration that matches the natural frequency of the building’s back-and-forth vibration. The unfortunate result is to increase the sway (amplitude) of the building. The occupants on the top floors of the skyscraper will feel the effects the most. Over time, the continual large sway could weaken the building’s structural supports and reduce its lifespan. Reducing Resonance Effects To counter resonance effects on bridges and buildings, engineers build them in such a way as to reduce the amplitude of resonance. Bridge designers make bridges more streamlined so that the wind passes over without imparting much energy. They also make bridges stiff, so a larger force is needed to create a large amplitude. The second-largest bridge in the world, the Great Belt East Bridge of Denmark is built with a smooth underside, like an airplane wing, that greatly increases its streamlined shape (Figure 7.48). It is not likely that a forced vibration would cause it to resonate. Skyscraper designers employ many strategies to lessen resonant vibration. One very effective approach is to use a large mass at the top of the building, called a “tuned mass damper,” which is free to oscillate back and forth (Figure 7.49). Controlled by computers, it can be made to vibrate at the resonant frequency of the building. When the building sways left, the mass moves right, and when the building sways right, the mass moves left. This has the effect of cancelling the vibration of the building. Any process that lessens the amplitude of an object’s oscillations is referred to as “damping.” Figure 7.48 The Great Belt East Bridge of Denmark is 6.8 km long and is constructed with a smooth underside. This allows air to flow by without inducing a resonant frequency. Chapter 7 Oscillatory motion requires a set of conditions. 385 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 386 building movement cables damper movement Figure 7.49 The Taipei 101 building in Taiwan was completed in 2004 and stands 101 stories high. The inset shows a tuned mass damper in the building designed by Motioneering Inc. of Guelph, Ontario. It has a huge mass and vibrates opposite to the direction of the building, cancelling much of the amplitude of the resonant vibration. THEN, NOW, AND FUTURE Stressed-out Airplanes Ask any mechanical engineers, and they will tell you the importance of designing equipment to minimize vibration. Vibration causes excess wear on parts and stress on materials. Nowhere is this more evident than on an airplane. Yancey Corden knows this better than most people. Yancey is an aircraft maintenance engineer, and one of his jobs is to inspect aircraft for excess metal fatigue. Yancey was born on the Pavilion Reserve in south central British Columbia but grew up north of Williams Lake. His father maintained their car, boat, and other equipment around the home. Yancey watched and helped his father, and during this time, his interest in mechanics grew. Not long after finishing high school, Yancey enrolled in the aircraft maintenance engineer program at the British Columbia Institute of Technology located in Burnaby. He is now qualified with an M1 certification, which allows him to work on planes under 12 500 kg, and an M2 certification, which allows him to work on larger planes. He received specialized training in structural maintenance. This shift of force. This happens because the airplane springs upward due to the lighter load, and as a result, the wings tend to flutter up and down. This vibration causes stress fractures on the wing, and it is Yancey’s job to find them. If a problem is found, Yancey designs the solution. This could involve fabricating a new part or simply fixing the existing one. He enjoys his job because each day is different and brings new challenges. He is thankful that he had the foresight to maintain good marks when he went to high school because the physics and science courses he took directly applied to his training. He is very proud of his heritage but he also believes it is important to focus on who you are and where you are going. Questions 1. What factors contribute to metal fatigue on a firefighting airplane? 2. What steps must be taken to gain a licence as an aircraft maintenance engineer? 3. To what factors does Yancey attribute his success? Figure 7.50 Yancey Corden involves an in-depth knowledge of the skin and frame of the plane. In 2003 Yancey moved to Alberta where he works in Red Deer for a company called Air Spray. Air Spray maintains and repairs Lockheed L-188 airplanes. These planes were originally manufactured as passenger craft in the 1950s, but because of their rugged design, they have been converted to firefighting aircraft today. They carry over 10 000 kg of water and fire retardant and are capable of dumping the entire amount in three seconds. When a dump of water and fire retardant occurs, the wings and airframe of the plane undergo a huge 386 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 387 Quartz Clocks The technology of clock design and manufacture has taken huge leaps since the 1600s when Huygens built his first pendulum clock. Today, quartz clocks are the most accurate timepieces commercially available. They have an accuracy of about 1/2000 of a second a day. A quartz clock works on the principle of resonance. Inside each quartz clock is a tiny crystal of quartz. Quartz is a mineral that naturally forms into crystals. It also has a property unique to just a handful of materials: it will bend when a voltage is applied to it. If a pulse of voltage is applied to it, the crystal will begin to vibrate at its resonant frequency, just as a cymbal vibrates when hit by a drumstick. Once the quartz crystal is set vibrating, the circuitry of the clock times successive voltage pulses to synchronize with the frequency of the crystal. The synchronized voltage provides the forced frequency to keep the crystal oscillating just as the pendulums of Huygens’s clocks needed a synchronized forced frequency to keep them from running down. The difference is that the pendulum clock receives the forced frequency through mechanical means, while the quartz crystal clocks get the forced vibration from electrical means. Resonant Frequency of a Quartz Crystal The crystal’s resonant frequency depends on its size and shape and is not affected significantly by temperature. This makes it ideal for keeping time. As the crystal gets larger, more voltage is required to make it oscillate, and its resonant frequency decreases. A piece of quartz could be cut to oscillate once every second, but it |
would be far too large for a wristwatch and would require a large voltage to operate. If the crystal size is decreased, less voltage is required to make it oscillate. info BIT A substance that deforms with an applied voltage is called a piezoelectric material. A piezoelectric material will also create a voltage if stressed. Some butane lighters use a piezoelectric material to create a flame. As the lighter trigger is pressed, butane gas is released and the piezoelectric material undergoes stress. The piezoelectric material creates a voltage that causes a spark to jump a very small gap at the end of the lighter, igniting the butane. e WEB Atomic clocks keep time extremely precisely. Do they use a principle of resonance to keep such accurate time? Begin your search at www.pearsoned.ca/school/ physicssource. Quartz crystals are cut to a size and shape small enough to fit into a watch and use a small voltage (Figure 7.51). In most of today’s quartz watches, the crystal vibrates with a resonant frequency of about 30 kHz and operates at 1.5 V. A small microprocessor in the watch combines these oscillations to make one oscillation per second so the watch can display time in a meaningful way. The topic of resonant frequencies is large and can’t possibly be fully covered in this unit. You will learn more about resonance in musical instruments in Chapter 8. Figure 7.51 The quartz crystal in a wristwatch is enclosed in the small metal cylinder (lower right). Chapter 7 Oscillatory motion requires a set of conditions. 387 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 388 12. What factors affect the resonance of a quartz crystal? 13. (a) What are two advantages of a quartz clock over a pendulum clock? (b) Are there any disadvantages of a quartz clock compared with a pendulum clock? Extensions 14. Use the knowledge you have gained about the design of a pendulum clock and the equation for its period in section 7.3 to answer the following question. What would the length of the pendulum’s arm have to be so that it would oscillate with a resonant frequency of 1.00 Hz in Alberta (g 9.81 N/kg)? Under what conditions would it be most accurate? 15. Use your local library or the Internet to find out what automobile manufacturers do to reduce resonant frequencies in cars. 16. Investigate other methods not mentioned in the text that bridge designers use to lessen resonant vibrations. 17. Tuned mass dampers are not just used on buildings; cruise ships also have them. Explain why a cruise ship might have them and how they would be used. 18. Use your local library or the Internet to explore orbital resonance. In one or two paragraphs, explain how it applies to Saturn’s rings. e TEST To check your understanding of applications of simple harmonic motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 7.4 Check and Reflect 7.4 Check and Reflect Knowledge 1. What provides the force necessary to start a building or bridge oscillating? 2. What is forced frequency? 3. Explain what engineers use to reduce resonant vibrations of buildings and how these devices or structures work. 4. Explain the effect of applying a force to a vibrating object with the same frequency. 5. Identify two limitations of Huygens’s pendulum clock. 6. Can a pendulum clock built to operate at the equator have the same accuracy at the North Pole? Explain. 7. What is damping? Use an example in your explanation. Applications 8. How could a person walking across a rope bridge prevent resonant vibration from building up in the bridge? 9. An opera singer can shatter a champagne glass by sustaining the right musical note. Explain how this happens. 10. Tuning forks are Y-shaped metal bars not much bigger than a regular fork. They can be made to vibrate at a specific frequency when struck with a rubber hammer. A piano tuner uses tuning forks to tune a piano. Explain, in terms of resonance, how this might be done. 11. Students are asked to find ways to dampen or change the resonant frequency of a pendulum. Here is a list of their suggestions. Identify the ones that would work and those that would not. In each case, justify your answer. (a) Apply a forced frequency that is different from the resonant frequency. (b) Place the pendulum in water. (c) Increase the mass of the pendulum bob. (d) Move the pendulum to a higher altitude. 388 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 389 CHAPTER 7 SUMMARY Key Terms and Concepts period frequency oscillation cycle oscillatory motion Hooke’s law spring constant restoring force simple harmonic motion simple harmonic oscillator resonant frequency amplitude forced frequency mechanical resonance Key Equations kx F vmax A k m T 2 m k T 2l g Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. where the time for each interval is the Oscillatory Motion can produce which is the inverse of frequency measured in can take the form of simple harmonic motion mechanical resonance that obeys whereas written mathematically as forced frequency can cause an object to cycles/s F kx where F is the where x is the with a large displacement amplitude Figure 7.52 Chapter 7 Oscillatory motion requires a set of conditions. 389 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 390 CHAPTER 7 REVIEW Knowledge 1. (7.1) What is oscillatory motion? Use an example 16. Determine the spring constant from the following graph: in your answer. 2. (7.1) Under what conditions must a ball be bounced so it has oscillatory motion? 3. (7.2) What is the defining property of an elastic material? 4. (7.2) What force, or forces, act on an isolated, frictionless simple harmonic oscillator? 5. (7.2) State the directional relationship that exists between the restoring force and displacement of a simple harmonic oscillator. 6. (7.2) What quantity does the slope of a force- displacement graph represent? 7. (7.2) What can be said about a pendulum’s position if the restoring force is a non-zero value? 8. (7.3) Why isn’t acceleration uniform for a simple harmonic oscillator? 9. (7.3) Why is it acceptable to consider a pendulum a simple harmonic oscillator for small displacements, but not for large displacements? 10. (7.4) If the forced frequency and the resonant frequency are similar, what effect does this have on an oscillator? Applications 11. Determine the restoring force acting on a 1.0-kg pendulum bob when it is displaced: (a) 15 (b) 5 12. Determine the frequency of a guitar string that oscillates with a period of 0.0040 s. 13. What is the period of a ball with a frequency of 0.67 Hz? 14. After a diver jumps off, a diving board vibrates with a period of 0.100 s. What is its frequency? 15. What is the restoring force on a 2.0-kg pendulum bob displaced 15.0? Force vs. Displacement 160 140 120 100 80 60 40 20 ) N ( e c r o F 0 0.2 0.4 0.6 0.8 Displacement (m) 17. A spring hangs from the ceiling in a physics lab. The bottom of the spring is 1.80 m from the floor. When the teacher hangs a mass of 100 g from the bottom of the spring, the spring stretches 50.0 cm. (a) What is its spring constant? (b) What force must a person apply to pull the 100.0-g mass on the bottom of the spring down through a displacement of 20.0 cm? (c) The 100.0-g mass is removed and a 300.0-g mass is attached. What is the distance of the mass above the floor? 18. Two different springs, A and B, are attached together at one end. Spring A is fixed to the wall as shown. The spring constant of A is 100.0 N/m and B is 50.0 N/m. What is the combined stretch of the two springs when a force of 25.0 N [right] is applied to the free end of spring B? A B wall 390 Unit IV Oscillatory Motion and Mechanical Waves 07-Phys20-Chap07.qxd 7/24/08 1:10 PM Page 391 19. Students stretch an elastic band attached to a force meter through several displacements and gather the following data. Use a graphing calculator or another acceptable method to plot the graph of this data and determine if the elastic band moves as predicted by Hooke’s law. Displacement (cm) 0.00 10.0 20.0 30.0 40.0 50.0 Force (N) 0.00 3.80 15.2 34.2 60.8 95.0 20. How long must the arm of a pendulum clock be to swing with a period of 1.00 s, where the gravitational field strength is 9.81 N/kg? 21. What is the period of a 10.0-kg mass attached to a spring with a spring constant of 44.0 N/m? 22. Determine the maximum velocity of a 2.00-t crate suspended from a steel cable (k 2000.0 N/m) that is oscillating up and down with an amplitude of 12.0 cm. 23. A 0.480-g mass is oscillating vertically on the end of a thread with a maximum displacement of 0.040 m and a maximum speed of 0.100 m/s. What acceleration does the mass have if it is displaced 0.0200 m upwards from the equilibrium position? 24. Determine the period of oscillation of a pendulum that has a length of 25.85 cm. 25. An astronaut who has just landed on Pluto wants to determine the gravitational field strength. She uses a pendulum that is 0.50 m long and discovers it has a frequency of vibration of 0.182 Hz. What value will she determine for Pluto’s gravity? 26. A student is given the relationship for a pendulum: T 2X (a) What does X represent? (b) The student records the period of the pendulum and finds it is 1.79 s. What is the pendulum’s length? Extensions 27. A spring (k 10.0 N/m) is suspended from the ceiling and a mass of 250.0 g is hanging from the end at rest. The mass is pulled to a displacement of 20.0 cm and released. (a) What is the maximum velocity of the mass? (b) What is the period of oscillation of the mass if it is displaced 15.0 cm and released? 28. A horizontal mass-spring system has a mass M attached to a spring that oscillates back and forth at a frequency of 0.800 Hz. Determine the frequency in the following cases. (a) The mass is doubled. (b) The amplitude is tripled. 29. Identify which of the following examples is SHM and which is not. Ex |
plain. (a) a bouncing ball (b) a hockey player moving a puck back and forth with his stick (c) a plucked guitar string Consolidate Your Understanding Create your own summary of oscillatory motion, simple harmonic motion, restoring force, and mechanical resonance by answering the questions below. If you want to use a graphic organizer, refer to Student References 4: Using Graphic Organizers on pp. 869–871. Use the Key Terms and Concepts listed above and the Learning Outcomes on page 342. 1. Prepare a quick lesson that you could use to explain Hooke’s law to a peer using the following terms: restoring force, displacement, linear relationship. 2. Construct a two-column table with the title “Mass-spring System.” The first column has the heading, “Factors Affecting Period” and the second column has the heading, “Factors Not Affecting Period.” Categorize the following factors into the two columns: mass, spring constant, amplitude, restoring force, velocity. Think About It Review your answers to the Think About It questions on page 343. How would you answer each question now? e TEST To check your understanding of oscillatory motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 7 Oscillatory motion requires a set of conditions. 391 08-PearsonPhys20-Chap08 7/24/08 2:22 PM Page 392 C H A P T E R 8 Key Concepts In this chapter you will learn about: mechanical waves — longitudinal and transverse universal wave equation reflection interference acoustical resonance Doppler effect Learning Outcomes When you have finished this chapter, you will be able to: Knowledge describe how transverse and longitudinal waves move through a medium explain how the speed, wavelength, frequency, and amplitude of a wave are related describe how interference patterns can be used to determine the properties of the waves explain the Doppler effect describe the difference between transverse and longitudinal waves describe how waves are reflected explain the relationship between rays and waves apply the universal wave equation to explain how frequency, wavelength, and wave velocity are related explain the effects of constructive and destructive interference Science, Technology, and Society explain that the goal of technology is to provide solutions to practical problems 392 Unit IV Mechanical waves transmit energy in a variety of ways. Figure 8.1 What do bats and dolphins have in common? The phrase “blind as a bat” states a common fallacy. Bats have some vision using light, but when placed in pitch-black rooms crisscrossed with fine wires, they can easily fly around and unerringly locate tiny flying insects for food. Dolphins have shown that they can quickly locate and retrieve objects even when they are blindfolded. We usually assume that vision requires light but both bats and dolphins have evolved the ability to “see” using sound waves. Research in science and technology has developed “eyes” that enable humans also to see using sound waves, that is, navigate with senses other than sight. Medicine uses ultrasound (frequencies above the audible range) to look at objects such as a fetus or a tumour inside the body. Submarines can circumnavigate the globe without surfacing by using sound waves to explore their underwater environment. In Chapter 6, you studied how mass transfers energy when it moves through space. Waves, on the other hand, are able to transmit vast quantities of energy between two places without moving any mass from one location to another. Radio waves carry information, sound waves carry conversations, and light waves provide the stimulus for the cells that enable vision. This chapter introduces you to the nature and properties of waves. By experimenting with various forms of wave motion, you will learn about this common, but often misunderstood method of energy transmission. 08-PearsonPhys20-Chap08 7/24/08 2:22 PM Page 393 8-1 QuickLab 8-1 QuickLab Fundamental Properties of Wave Motion Problem To determine properties of waves in a ripple tank. Materials ripple tank and apparatus for its operation dowel ( 1.5 cm in diameter) 2 stopwatches ruler, two paper clips light and stand to project waves onto screen screen (a large sheet of newsprint works well) Procedure 1 Set up the ripple tank as shown in Figure 8.2. The water should be about 1 cm deep. Make sure that energy-absorbing buffers are placed around the edge of the tank to prevent unwanted reflections. Check your assembly with your instructor. light source Figure 8.2 paper screen 2 (a) Place a tiny spot of paper in the middle of the ripple tank. (b) Dip the end of your finger once into the water about the middle of the ripple tank to create a single, circular, wave front. Observe the speck of paper as the wave front passes it. 3 (a) On the screen, place the two paper clips at a measured distance apart, 30–40 cm. (b) Position your finger so that its shadow is over one of the paper clips and generate another single wave front. (c) Using a stopwatch, measure the time for the wave to travel from one paper clip to the other. Record the distance and time. Calculate the speed of the wave. Do a few trials for accuracy. 4 (a) Place the dowel horizontally in the water near one edge of the tank. Tap the dowel gently and observe the wave front. Sketch and describe the motion. (b) Position the paper clips in the wave’s path and measure the speed of the straight wave front. CAUTION: Use care with ripple tanks. It is easy to break the glass bottom or to spill water. This is a serious hazard in an area where electricity is being used. Vibrating the tank will generate unwanted waves. Questions 1. When a wave front passes the speck of paper, what motion does the paper make? Does it move in the same or the opposite direction to the motion of the wave front? What does that tell you about the motion of the water as the wave moves through it? 2. On your sketches, draw several vector arrows along the fronts to indicate the direction in which they are moving. What is the angle between the line of the wave front and its motion? In Procedure 4(a), what is the angle between the edge of the dowel and the direction of the motion of the wave front? Sketch what you observe. Describe the motion. 3. Which wave front moves faster, the circular wave front or the straight wave front? Think About It 1. What differences and similarities are there between the ways energy is transmitted by waves and by matter? e SIM Find out more about waves in ripple tanks. Go to www.pearsoned.ca/school/ 2. What assumptions must be made to use water waves as a model for physicssource. sound waves? Discuss your answers in a small group and record them. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. Chapter 8 Mechanical waves transmit energy in a variety of ways. 393 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 394 8.1 The Properties of Waves info BIT In December of 2004, an earthquake near the island of Sumatra set off a tsunami that is estimated to have had more than 2 petajoules (1015 J) of energy. This tsunami, the most powerful in recorded history, took over 225 000 lives and did untold billions of dollars in damage to the economies and the environments of the countries that border on the Indian Ocean. info BIT On the day of the tsunami of 2004 that devastated Phuket, Thailand, people travelling in a ferry in deep water offshore from Phuket felt only a greater than usual swell as the wave passed them by. Figure 8.3 Surfers use a wave’s energy to speed their boards across the water. medium: material, for example, air or water through which waves travel; the medium does not travel with the wave wave: disturbance that moves outward from its point of origin, transferring energy through a medium by means of vibrations equilibrium position: rest position or position of a medium from which the amplitude of a wave can be measured crest: region where the medium rises above the equilibrium position trough: region where the medium is lower than the equilibrium position When a surfer catches a wave, many people assume that the forward motion of the surfer is the result of the forward motion of the water in the wave. However, experimental evidence indicates that in a deep-water wave the water does not, in general, move in the direction of the wave motion. In fact, the surfer glides down the surface of the wave just as a skier glides down the surface of a ski hill. Like the skier, the surfer can traverse across the face of the hill as well as slide down the hill. But, unlike the ski hill, the water in the wave front is constantly rising. So, even though the surfer is sliding down the front of the wave he never seems to get much closer to the bottom of the wave. It is a common misconception that the water in a wave moves in the direction in which the waves are travelling. This may be because waves arriving at the shoreline move water to and fro across the sand. As you will see, this movement is a feature of the interaction of the wave with the sloping shoreline rather than the actual motion of the wave itself. In deep water, there is only very limited lateral motion of water when a wave moves past a particular point. 394 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 395 Waves and Wave Trains When a stone is thrown into a still pond or lake, a ripple moves outward in ever-enlarging concentric circles (Figure 8.4). The water is the transporting medium of the wave and the undisturbed surface of the water is known as the wave’s equilibrium position. Regions where the water rises above the equilibrium position are called crests and regions where the water is lower than its equilibrium position are called troughs. In the crest or trough, the magnitude of greatest displacement from the equilibrium is defined as the waves’ amplitude (A). A complete cycle of a crest followed by a trough is called a wavelength; |
its symbol is the Greek letter lambda, (Figure 8.5). λ A crest A equilibrium position λ trough Figure 8.5 Properties of a wave wavelength wavelength A wave front moving out from the point of origin toward a barrier is called an incident wave. A wave front moving away from the barrier is called a reflected wave, while a series of waves linked together is a wave train. The concept of a wave train implies a regular repetition of the motion of the medium through which the wave travels. As a result, many parts of the medium are moving in a motion that is identical to the motion of other points on the wave train. At these points, the waves are said to be in phase (Figure 8.6). λ B λ D E A C λ A and B are in phase C, D, and E are in phase Figure 8.6 In-phase points along a wave train have identical status relative to the medium and are separated by one wavelength, λ. Instead of creating individual pulses by hand in a ripple tank, you may use a wave generator to create a continuous series of crests and troughs forming a wave train. Wave generators can act as a point source similar to the tip of a finger, or as a straight line source, similar to a dowel. In 8-1 QuickLab you measured the speed of a single pulse by observing its motion. However, because it is impossible to keep track of a single wave in a wave train, to measure the speed of a wave train requires a greater understanding of the properties of waves. crests crests troughs troughs Figure 8.4 Many of the terms used to describe wave motions come from the observation of waves on the surface of water. e WEB To learn more about experiments using ripple tanks, follow the links at www.pearsoned.ca/school/ physicssource. amplitude: the distance from the equilibrium position to the top of a crest or the bottom of a trough wavelength: the distance between two points on a wave that have identical status. It is usually measured from crest to crest or from trough to trough. wave front: an imaginary line that joins all points reached by the wave at the same instant incident wave: a wave front moving from the point of origin toward a barrier reflected wave: a wave front moving away from a barrier wave train: a series of waves forming a continuous series of crests and troughs point source: a single point of disturbance that generates a circular wave Chapter 8 Mechanical waves transmit energy in a variety of ways. 395 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 396 8-2 Inquiry Lab 8-2 Inquiry Lab Wave Trains in a Ripple Tank, Part 1: Reflecting Waves In this ripple tank experiment, the properties of a twodimensional wave train are analyzed. Question How do the incident and reflected wave trains interact when wave trains reflect from a straight barrier? Materials and Equipment ripple tank, including the apparatus for its operation straight barrier wave generators (point-source and straight-line) light and stand to project waves onto screen screen (a large sheet of newsprint works well) Variables In this experiment you are to observe the directions of motion of the incident waves and reflected waves and how these directions are related to each other. Other variables to be observed are the interactions that occur when the incident and reflected wave trains move in different directions through the same point in the ripple tank. As you observe the wave motions you should identify which are the controlled variables, manipulated variables, and responding variables. General Procedure 1 (a) Set up the ripple tank as shown in Figure 8.7. (b) When using motorized wave generators, it is important that the generator just barely contacts the surface of the water. It should never touch the tank during operation. Check with your instructor to make sure that your apparatus is properly assembled. CAUTION: Use care with ripple tanks. It is easy to break the glass bottom or to spill water. This is a serious hazard in an area where electricity is being used. Vibrating the tank will generate unwanted waves that interfere with the desired observations. The wave generator should never touch the tank during operation. Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork light source generator motor generator support point sources To power supply paper screen Figure 8.7 Procedure 1 (a) Place the point-source wave generator at one edge of the ripple tank and the straight barrier at the other edge. The shadow of both the barrier and the source should be visible on the screen. (b) Use the point-source wave generator to create a continuous wave train in the ripple tank. Observe what happens to the incident wave train when it meets the reflected wave train. (c) Make a sketch of your observations. Wave trains are a bit tricky to observe at first. Discuss what you see with your team members. When you have reached a consensus, write a brief description of your observations. On your sketch, place vector arrows along an incident and a reflected wave front to indicate the direction and speed of their motions. 2 (a) Set up the straight-line wave generator at one edge of the ripple tank. Place the barrier at the other edge parallel to the generator. The shadows of both the generator and the barrier should be visible on the screen. (b) Start the generator to create a continuous wave train. Observe what happens when the reflected wave train moves back through the incident wave train. Draw diagrams and write a description of the observations. Again, draw vector arrows along incident and reflected wave fronts to indicate their relative velocities. 396 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 397 3 Move the barrier so that it is at an angle of about 30° 2. (a) When the barrier is parallel to the straight wave to the generator and repeat step 2 (b). 4 Set the barrier so that the angle between it and the generator is about 60º and repeat step 2 (b). Analysis 1. (a) When the incident wave train created by the point-source generator is passing through the reflected wave train, what happens to the waves in the region where they overlap? (b) Can you see the direction of the motion for both the incident and reflected wave trains? generator, what pattern do you observe when the reflected waves are moving back through the incident waves? (b) In which direction does the pattern seem to be moving? Can you see the direction of the motion for both the incident and reflected wave trains? 3. Answer question 2 for the set-up when the barrier is at an angle to the straight wave generator. 4. In all cases above, how does the spacing of the waves in the reflected wave train compare to the spacing of the waves in the incident wave train? Waves and Rays When waves in a ripple tank are viewed from above (Figure 8.8) the wave fronts appear as a set of bright and dark bands (crests and troughs). When we draw wave trains as seen from above, we use a line to represent a wave front along the top of a crest. The point halfway between two lines is the bottom of a trough. A series of concentric circles represents the wave train generated by a point source. ray: a line that indicates only the direction of motion of the wave front at any point where the ray and the wave intersect rays wavelength wave source crests troughs velocity vectors Figure 8.8 View of a ripple tank from above Figure 8.9 A point source generates waves that move outward as concentric circles with the source at their centre Waves are in constant motion. At all points on a wave front, the wave is moving at right angles to the line of the crest. There are two ways to indicate this (Figure 8.9). You could draw a series of vector arrows at right angles to the wave front with their length indicating the speed of the wave. Or, you could draw rays, lines indicating only the direction of motion of the wave front at any point where the ray and the wave front intersect. The rays in Figure 8.9 are called diverging rays since they spread out as they move away from the origin. When rays diverge, it indicates that the energy given to the wave at its source is being spread over a larger and larger area. This is why, as sound moves away from a point source such as a bell, the volume decreases with the square of the distance. e WEB To learn more about the mathematical relationship between the volume of sound and the distance from the source, follow the links at www.pearsoned.ca/ school/physicssource. Chapter 8 Mechanical waves transmit energy in a variety of ways. 397 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 398 light rays trough crest water ripple tank bottom paper screen bright areas under crests dark areas under troughs Figure 8.10 Sketch of a wave showing light refracting through a crest and a trough When waves in a ripple tank are projected onto a screen below, the wave fronts appear as a set of bright and dark bands. It may seem logical that the light and dark bands seen on the screen below the ripple tank result from the differences in water depth between the crests and the troughs. But that difference is only about a millimetre and cannot account for the high contrast in light seen on the screen. In fact, a crest acts like a converging lens to concentrate the light, creating a bright bar. A trough acts like a diverging lens to spread the light out, making the area under the trough darker (Figure 8.10). You will learn more about light refraction in Unit VII of this course. Reflection of a Wave Front When a wave front is incident on a straight barrier, it reflects. The direction the wave travels after reflection depends on the angle between the incident wave front and the barrier. A circular wave front, as generated by a point source, S, produces the simplest reflection pattern to explain. In this case, the reflected wave follows a path as if it had been generated by an imaginary point source S, at a position behind the barrier identical to that of the actual point sour |
ce in front of the barrier (Figure 8.11). Now consider an incident wave front created by a straight wave generator (Figure 8.12). The straight wave front also reflects as if the reflected wave had been generated by an imaginary generator located behind the barrier. The position of the imaginary generator behind the barrier is equivalent to the position of the real generator in front of the barrier. The incident wave front and the reflected wave front are travelling in different directions, but the angle between the incident wave front and the barrier must be identical to the angle between the reflected wave front and the barrier. e SIM Find out more about the ways waves reflect. Go to www.pearsoned.ca/school/ physicssource. S imaginary source rays from the imaginary source reflecting surface of barrier S real source imaginary incident wave generated by S reflected portion of incident wave incident wave generated by S imaginary straight wave generator reflected wave front real straight wave generator incident wave from imaginary generator reflecting surface of barrier incident wave front Figure 8.11 When circular waves reflect from a straight barrier, the reflected waves seem to be moving away from an imaginary source. Figure 8.12 When straight waves reflect from a straight barrier, the angle between the reflected wave front and the barrier must be equal to the angle between the incident wave front and the barrier. 398 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 399 M I N D S O N Waves Can Have Curls Too When waves travel in deep water, their shape is similar to the waves in a ripple tank. But as waves near the shoreline they change shape and develop what is known as a curl in which the top of the wave falls in front of the wave. Recall Figure 8.3 on page 394. Explain the causes of a wave’s curl in terms of its motion. 8-3 Inquiry Lab 8-3 Inquiry Lab Wave Trains in a Ripple Tank, Part 2: Wave Speed and Wavelength Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork In this ripple tank experiment, the properties of a twodimensional wave train are further analyzed. Question What effect does a change in speed have on wave trains? Materials and Equipment ripple tank, including the apparatus for its operation wave generators (point-source and straight-line) light and stand to project waves onto screen screen (a large sheet of newsprint works well) two small blocks of wood about 8 mm thick Variables In this lab you will be observing how water depth affects the properties of waves. The variables that might be affected by changes in the depth are speed, frequency, wavelength, and direction. As you make your observations, consider which of the variables are controlled variables, manipulated variables, and responding variables. General Procedure 1 (a) Set up the ripple tank as shown in Figure 8.7 (page 396). (b) When using motorized wave generators, it is important that the generator just barely contacts the surface of the water. It should never touch the tank during operation. Check with your instructor to make sure that your apparatus is properly assembled. CAUTION: Use care with ripple tanks. It is easy to break the glass bottom or to spill water. This is a serious hazard in an area where electricity is being used. Vibrating the tank will generate unwanted waves that interfere with the desired observations. The wave generator should never touch the tank during operation. Procedure 1 Place small pads (about 8 mm thick) under the legs along one edge of the ripple tank so that the water gets shallower toward that edge. The water should be less than 1 mm deep at the shallow edge. 2 (a) Near the deep edge, create a wave train using the point-source generator. (b) Observe what happens to the wave fronts as the wave train moves toward the shallow edge. Discuss your observations with your team members. Sketch and briefly describe your observations. (c) Place vector arrows along several of the wave fronts to indicate the direction and speed of the wave fronts as they move into shallow water. 3 (a) Set up the straight-line wave generator at the deep edge of the tank. (b) Turn on the generator and observe the wave train as it moves into the shallow water. (c) Sketch your observations and describe the motion of the wave train as it moves into shallow water. Use vector arrows along the wave fronts to assist you in your descriptions. 4 (a) Now place the pads under the legs of the ripple tank on an edge that is at a right angle to the position of the straight-line wave generator. (b) Use the straight-line wave generator to create a wave train. (c) Observe the wave train as it travels across the tank. Discuss your observations with your team members. Sketch and write a brief description of what you saw to accompany your sketch. Use vector arrows drawn along several of the wave fronts to indicate their relative velocity as they move into the shallow water. Chapter 8 Mechanical waves transmit energy in a variety of ways. 399 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 400 Analysis 2. What do you think causes the observed changes? 1. For each of the trials, when the waves moved from deep to shallow edge (or vice versa), comment on the kinds of changes you observed. 3. When the straight wave fronts moved across the tank at right angles to the change in the depth of the water, was the shape of the wave front affected? (a) Were the wavelengths of the incident waves affected as they moved into shallow water? 4. What properties of waves are affected as the waves move from deep to shallow water? (b) Was the shape of the wave fronts affected as they entered shallow water? (c) If so, how did the shape of the wave fronts change as they changed speed? 5. When a water wave moves toward a beach, how would the change in the depth of the water affect the motion of the wave? 8.1 Check and Reflect 8.1 Check and Reflect Knowledge 1. If a wave pattern is created by a point source, what is the nature of the ray diagram that would represent the wave fronts? 2. When a wave front reflects from a barrier, what is the relationship between the direction of the motions of the incident and reflected wave fronts? Applications 3. The sketch shows a ray diagram that represents the motion of a set of wave fronts. If you were observing these wave fronts in a ripple tank, describe what you would see. rays 4. Draw a diagram of a set of straight wave fronts that are incident on a straight barrier such that the angle between the wave fronts and the barrier is 40˚. Draw the reflected wave fronts resulting from this interaction. How do the properties (speed, wavelength, and amplitude) of the reflected wave compare with the properties of the incident wave? Use a wavelength of about 1 cm in your diagram. Extensions 5. Reflection of light is the essence of how we use mirrors to see images. What does the reflection of waves in a ripple tank tell you about the formation of images? Hint: Think of where the reflected waves in the ripple tank seem to originate. 6. When a sound travels in water, the speed of the sound depends on the temperature of the water. If the sonar ping emitted by a submarine has a wavelength of 2.50 m, what happens to that wavelength when it enters a region where sound travels faster? e TEST To check your understanding of the properties of waves, follow the eTest links at www.pearsoned.ca/ school/physicssource. 400 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 401 8.2 Transverse and Longitudinal Waves Did you ever have a Slinky™ toy when you were a child? When a Slinky™ is stretched out along the floor and oscillated from side to side across its axis (centre line), or forward and back along its axis, mechanical waves are transmitted along its length. The sideways oscillations set up a transverse wave while those along the axis set up a longitudinal wave as shown in Figure 8.13. In this section we will consider the characteristics of such waves. Transverse Pulses info BIT The ever-popular Slinky™ was invented in 1945 by Richard James, a naval engineer working on tension springs. The name comes from the Swedish for “sleek” or “sinuous.” Each Slinky™ is made from 80 feet (24.384 m) of wire. A pulse moving through a spring is a good introduction to the way a wave moves through a medium. An ideal spring is one that allows a pulse to travel through it without loss of energy. By definition, a pulse is just the crest or the trough of a wave; its length is one-half a wavelength. The spring provides a medium in which the motion of a pulse can be observed from the side. Initially, the spring is in its equilibrium position. When you flip the spring sharply to the side and back, the motion of your hand sets up a series of sequential motions in the coils of the spring. Each coil imitates, in turn, the motion of the hand. This results in a transverse pulse (Figure 8.14) that moves along the spring. As a pulse moves along a spring, the coils of the spring move at right angles to the direction of the pulse’s motion. Compare v hand and v pulse in Figure 8.14. At the front of the pulse, the coils are moving away from the spring’s equilibrium position toward the point of maximum displacement from the equilibrium. In the trailing edge of the pulse, the coils are moving back toward the equilibrium position. Hand starts at equilibrium position of spring. Front of pulse starts to move along spring. Hand is at maximum amplitude. Hand continues to move up. vhand 0 vpulse vhand 0 v A Hand continues to move down. Figure 8.13 (a) A transverse pulse (b) A longitudinal pulse Arrows indicate the direction of the medium. The pulses are moving through the springs toward the bottom of the page. pulse: a disturbance of short duration in a medium; usually seen as the crest or trough of a wave e WEB To learn more about the forces opera |
ting in an oscillating spring, follow the links at www.pearsoned.ca/school/ physicssource. amplitude A As the hand moves toward the equilibrium position, the amplitude of the pulse moves along the spring. Pulse is complete when the hand is at equilibrium position. Figure 8.14 When you move your hand you set up a sequence in which the coils of the spring imitate the motion of your hand. This creates a moving pulse. l vhand 0 Chapter 8 Mechanical waves transmit energy in a variety of ways. 401 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 402 Energy Changes During the Movement of a Pulse Along the pulse, energy is stored in the form of both elastic potential energy and kinetic energy. As a section of the spring moves from the equilibrium position to the top of the pulse, that section has both kinetic energy (it is moving sideways relative to the direction of the pulse) and elastic potential energy (it is stretched sideways). At the point on the pulse where the displacement is greatest the coils are, for an instant, motionless. Then, the tension in the spring returns the coils to their equilibrium position. No blurring. This indicates: Ek Ep 0 maximum 0 0 Ek Ep Figure 8.15 A transverse pulse is generated when a spring is given a sharp flip to the side. Arrows indicate the direction of motion of the coils. Can you determine which way the pulse is moving? In Figure 8.15 the blurring on the front and back segments of the pulse indicates the transverse motion and the presence of kinetic energy as well as elastic potential energy. At the top, there is no blurring as the coils are temporarily motionless. At that instant that segment of the spring has only elastic potential energy. As it returns to its equilibrium position, the segment has, again, both kinetic and potential energies. The energy in a pulse moves along the spring by the sequential transverse motions of the coils. Recall from section 6.3 that a pendulum, along the arc of its path, has both kinetic and potential energy, but at the point where the pendulum’s displacement is greatest, all the energy is in the form of potential energy. Thus, the energy of an oscillating pendulum is equivalent to its potential energy at the point where its displacement is greatest. Similarly, the amplitude of the wave in an experimental spring can be used to determine the quantity of energy that is stored in the pulse. Concept Check You generate a pulse in a Slinky™ stretched out on the floor. If you wish to, you could give the next pulse more energy. How would you do that? info BIT When considering sound, amplitude determines loudness. 402 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 403 8-4 Inquiry Lab 8-4 Inquiry Lab Pulses in a Spring, Part 1: Pulses in an Elastic Medium Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork In this experiment, you will study how a pulse moves through a medium. Question What are the mechanics by which pulses move through a medium? Variables The measured properties of a pulse include its amplitude (A), pulse length (l ), period (t), and speed (v). 5 On your sketch of the pulse, label the following parts: — The amplitude (A) is the perpendicular distance from the equilibrium position of the spring to the top of the pulse. — The pulse length (l) is the distance over which the spring is distorted from its equilibrium position. NOTE: When you stand at the side of the spring, the pulse seems to move past you very quickly, almost as a blur. Watching from the end of the spring may make it easier to observe the details of the motion. Materials and Equipment 6 Make a longitudinal pulse by moving your hand light spring metre-stick or measuring tape stopwatch masking tape CAUTION: A stretched spring stores considerable amounts of elastic potential energy. Be careful not to release the end of a spring while it is stretched. When collapsing a spring, have the person holding one end walk the spring slowly toward the other end. If you allow the spring to gently unwind as you are walking you will prevent the spring from tying itself into a knot. Procedure 1 Have one team member hold the end of the spring while another stretches it until it is moderately stretched (about 5–6 m). 2 Place strips of masking tape on the floor at either end of the spring to mark this length. Near the middle of the spring, attach a strip of tape about 5 cm long to one of the coils. 3 Have one of the people holding the spring generate a transverse pulse. Generate the pulse by moving your hand sharply to one side (about 60–75 cm) and back to its original position. This is a transverse pulse since its amplitude is perpendicular to the direction of its motion. 4 Sketch the pulse. Indicate the motions of the pulse and the coils using vector arrows. Observe the motion of the tape at the middle of the spring to assist in these observations. Generate more pulses until you understand the nature of the motion of the pulse in the spring. sharply toward the person holding the spring at the other end, and then back to its original position. This pulse is called a longitudinal pulse, because its amplitude is along the direction of its motion. Repeat the pulse a few times to determine the nature of the motion of the spring as the pulse moves through it. Sketch and describe the motion of the coils as the pulse moves along the spring. Analysis 1. What determines the amplitude of the transverse pulse? The longitudinal pulse? 2. Does the pulse change shape as it moves along the spring? If so, what causes the change in shape of the pulse? Would you expect the pulse shape to change if this were an isolated system? 3. How is the reflected pulse different from the incident pulse? If this were an isolated system, how would the reflected pulse differ from the incident pulse? 4. Describe the motion of the strip of tape at the middle of the spring as the pulse passes it. Does the tape move in the direction of the pulse? 5. How does the motion of the medium relate to the motion of the pulse? 6. How does the pulse transfer energy from one end of the spring to the other? 7. The motion of a pulse in the spring requires you to make assumptions about the motion of an ideal pulse. What assumptions must you make to create a model of how a transverse pulse moves through an elastic medium? Chapter 8 Mechanical waves transmit energy in a variety of ways. 403 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 404 v Reflection of Pulses from a Fixed Point incident pulse reflected pulse Figure 8.16 Reflection from the fixed end of a spring causes the pulse to be inverted. e SIM Find out about similarities and differences between transverse and longitudinal waves. Go to www.pearsoned.ca/ school/physicssource. When a pulse (or wave) is generated in a spring it soon arrives at the other end of the spring. If that end is held in place, the total pulse reflects from the end and travels back toward the source. The reflected pulse is always inverted relative to the incident pulse (Figure 8.16). In an ideal medium, the other properties of the pulse (amplitude, length, and speed) are unaffected by reflection. These properties of the reflected pulse are identical to those of the incident pulse. When a wave train is generated in the spring, the crests of the incident wave are reflected as troughs while the troughs of the incident wave are reflected as crests. v Longitudinal Waves If, instead of moving your hand across the line of the spring, you give the spring a sharp push along its length, you will observe that a pulse moves along the spring. This pulse is evidence of a longitudinal wave. The pulse is seen as a region where the coils are more tightly compressed followed by a region where the coils are more widely spaced. These two regions are called, respectively, a compression and a rarefaction and correspond to the crest and trough in a transverse wave. In the case of a longitudinal wave, the coils of the spring oscillate back and forth parallel to the direction of the motion of the wave through the medium (Figure 8.17). But, as with transverse waves, once the wave has passed through, the medium returns to its original position. Once again, energy is transmitted through the medium without the transmission of matter. Figure 8.17 Longitudinal waves are formed when the source oscillates parallel to the direction of the wave motion. spring in equilibrium position Project LINK What aspect of your seismograph will relate to the ideas of compression and rarefaction in a longitudinal wave? longitudinal wave in the spring vwave vcoils vcoils vcoils vcoils rarefaction rarefaction compression compression wavelength 404 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 405 8-5 Inquiry Lab 8-5 Inquiry Lab Pulses in a Spring, Part 2: Speed, Amplitude, and Length Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork In this experiment, you will study the speed, amplitude, and length of pulses. You will set up the experiment similarly to 8-4 Inquiry Lab on page 403. Question What is the relationship between the amplitude, length, and speed of a pulse? Variables The measured properties of a pulse include its amplitude (A), pulse length (l ), period (t), and speed (v). Materials and Equipment light spring stopwatch metre-stick or measuring tape masking tape Procedure 1 (a) Measure the speed of a transverse pulse as it moves along the spring. Have the person creating the pulse “count down” so that team members with stopwatches can time the pulse as it moves toward the other end. Measure the time from the instant the front edge of the pulse leaves the hand of the person generating it until the front edge arrives at the hand at the other end. Do this a few times to establish a consistent value. Record your results. Use the time and the di |
stance between the hands to calculate the speed of the pulse. (b) Generate pulses by moving your hand to the side and back at different speeds (more quickly or more slowly). Measure the speed of each of these pulses. 2 Have the person holding one end of the spring move so that the spring is stretched about 1 m farther. (Do not overstretch the spring.) Generate a pulse and measure the speed of the pulse in the spring at this higher tension. Carefully walk the spring back to the length used initially. 3 Make a transverse pulse by moving your hand a different distance sideways. Try to keep the time used to make the pulse the same as before. Repeat this a few times to observe changes in the pulse. Record your observations. 4 Now make several transverse pulses by moving your hand to a given amplitude but change the speed at which you move. Repeat a few times and record your observations. Analysis 1. Does the speed at which you moved your hand to generate a pulse affect the speed of the pulse? 2. When the spring was stretched to a greater length, what happened to the speed of the pulse? 3. What controls the amplitude (A) of the pulse? Can you create pulses with equal lengths but different amplitudes? 4. What controls the length (l ) of the pulse? Can pulses of equal amplitudes have different lengths? 5. What is the relationship between the length of the pulse and the speed (v) of the pulse in the medium? 6. Does the length of a pulse affect its larger amplitude or vice versa? Explain why or why not. 7. Does the energy in a pulse seem to depend on its amplitude or its length? Give reasons for your decision. Consider what changes occur as the pulse moves through the spring. 8. What determines the speed, the length, and the amplitude of the pulse? 9. What aspect of wave motion in water can you simulate by changing the tension in a spring? 10. What do your findings for the relationship of the amplitudes and lengths of pulses in springs tell you about the relationship between the amplitudes and wavelengths of waves in water? 11. Sound is often referred to as a wave. What aspect of a sound would relate to (a) the amplitude and (b) the wavelength of its waves? Chapter 8 Mechanical waves transmit energy in a variety of ways. 405 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 406 Pulse Length and Speed Figure 8.18 The length (l) of the pulse depends on the speed (v) of the pulse and the time (t) taken to complete the pulse. e WEB To learn more about the way the structures of the human ear transfer sound waves, follow the links at www.pearsoned.ca/school/ physicssource. info BIT When a wave moves across the surface of water, the water moves between crests and troughs by localized circular motions. This local circular motion moves water back and forth between a trough and the adjacent crest. direction of motion of surface vwave vmedium The speed of the pulse depends on the medium. If you stretch the spring so that the tension increases, then the speed of the pulse increases. Relaxing the tension causes the speed to decrease. The speed of the pulse in the spring also determines the length of the pulse. vhand 0 Hand starts at equilibrium position of spring. vhand 0 v A Front of pulse starts to move along spring. vpulse Hand is at maximum amplitude. amplitude A Pulse is complete when the hand is at equilibrium position. l vhand 0 The instant you start to move your hand to generate a pulse, the disturbance begins to move along the spring at a constant speed, v. Assume that the time it takes to move your hand to create the complete pulse is t. By the time your hand returns the spring to its equilibrium position, the front of the pulse will have travelled a distance d, which can be defined as the length l of the pulse (Figure 8.18). Remember d v and, therefore t d vt l vt Waves and the Medium A solid such as a spring is an elastic medium and can store elastic potential energy by stretching longitudinally or transversely. Typically, the way that fluids (liquids and gases) store elastic potential energy is by being compressed. Therefore, waves within fluids are typically longitudinal waves, known as pressure waves. This is the principle used in engines and aerosol sprays. As compressions and rarefactions move through a fluid, the motion of the molecules in the fluid is very similar to the motion of the coils when a longitudinal wave moves through a spring. For water to transmit energy as a transverse wave, the waves must be displaced vertically, but in liquids the vertical displacement cannot be a form of elastic potential energy. Thus, transverse waves can be transmitted only at the surface of water, or other liquids, where the waves are the result of gravitational potential energy rather than elastic potential energy. motion of water within wave Figure 8.19 406 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 407 M I N D S O N Wave Motion in Fluids Water can transmit both transverse (surface) waves and longitudinal (internal) waves such as sound. We know that sound waves in gases are longitudinal waves. Is it possible to create a transverse wave in a gas? Why or why not? Consider how transverse waves are created in liquids. Example 8.1 To create a pulse in a fixed ideal spring, you move your hand sideways a distance of 45 cm from the equilibrium position. It takes 0.80 s from the time you begin to move your hand until it returns the spring to its equilibrium position. If the pulse moves at a speed of 2.5 m/s, calculate the length of the pulse and describe the incident pulse and reflected pulse that pass through the midpoint of the spring. Practice Problems 1. A pulse is generated in a spring where it travels at 5.30 m/s. (a) If the time to generate the pulse is 0.640 s, what will be its length? (b) How does the speed of the pulse affect its amplitude? 2. A pulse moves along a spring at a speed of 3.60 m/s. If the length of the pulse is 2.50 m, how long did it take to generate the pulse? 3. A pulse that is 1.80 m long with an amplitude of 0.50 m is generated in 0.50 s. If the spring, in which this pulse is travelling, is 5.0 m long, how long does it take the pulse to return to its point of origin? 4. A spring is stretched to a length of 6.0 m. A pulse 1.50 m long travels down the spring and back to its point of origin in 3.6 s. How long did it take to generate the pulse? Answers 1. (a) 3.39 m (b) It does not; they are independent. 2. 0.694 s 3. 2.8 s 4. 0.45 s Given A 45 cm 0.45 m t 0.80 s v 2.5 m/s Required (a) length of the pulse (b) description of incident pulse passing the midpoint of the spring (c) description of reflected pulse passing the midpoint of the spring Analysis and Solution (a) The length of the pulse can be found using l vt. l vt (0.80 s) 2.5 m s 2.0 m (b) The spring is defined as an ideal spring, so the amplitude of the pulse is constant. The amplitude at all points on the spring will be the same as at the source. Therefore, A 0.45 m At the midpoint of the spring, the amplitude of the incident pulse is 0.45 m, its length is 2.0 m, and its speed is 2.5 m/s. (c) Reflection inverts the pulse but does not change any of its properties. The reflected pulse is identical to the incident pulse except that it is inverted relative to the incident pulse. Paraphrase and Verify (a) The length of the pulse is equal to 2.0 m. (b) In an ideal spring the amplitude of the pulse is constant. (c) When pulses are reflected from a fixed end of a spring they are inverted. Chapter 8 Mechanical waves transmit energy in a variety of ways. 407 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 408 direction of pulse motion Waves Are a Form of Simple Harmonic Motion path of hand motion equilibrium position of spring Figure 8.20 Simple harmonic motion generates a wave train in the form of a sine curve. spring If you move your hand from side to side in simple harmonic motion, as indicated in Figure 8.20, transverse waves are generated in the spring. When a transverse wave moves through a medium, the motion of the medium may seem, at first, quite complex. In a transverse wave, each segment of the medium simply oscillates in simple harmonic motion about its equilibrium position in the direction perpendicular to the direction of the wave motion. This simple harmonic motion is transferred sequentially from one segment of the medium to the next to produce the motion of a continuous wave. Universal Wave Equation Pulses provide a useful tool to introduce the nature of waves. However, in nature, sound and light are wave phenomena rather than pulses. In this section, we will begin to shift the emphasis to the properties of waves. Whereas the letter l is used to indicate the length of a pulse, the Greek letter lambda, λ, is used to indicate wavelength. The terms crest and trough come from the description of water waves but are used throughout wave studies. For a water wave, the crest occurs where the medium is displaced above the equilibrium position, while a trough is the region displaced below the equilibrium position. However, for media such as springs, the terms crest and trough merely refer to two regions in the medium that are displaced to opposite sides of the equilibrium position (Figure 8.20). Other variables used in wave studies (frequency–f, period–T, amplitude–A) come from and have the same meanings as in your study of simple harmonic motion in section 7.2. The period (T) is the time taken to generate one complete wavelength. Since two pulses join to create one wave, the period for a wave is twice the time required to generate a pulse. Therefore, the wavelength of a wave is twice the length of a pulse. With this in mind, the relationship between wavelength, speed, and period is the same for waves as it is for pulses. That is, λ vT rather than l vt. For periodic motion, T 1 . f The equation for wavelength now can be written as λ v f or v f λ . The latter form is known as the universal wave equation. 408 Unit |
IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 409 Constant Frequency, Speed, and Wavelength In 8-3 Inquiry Lab, you investigated what happened to a wave train as it moved from deep to shallow water. Changes occurred because the speed in shallow water was slower than it was in deep water. Since the frequency of the waves as they moved from deep to shallow water was unchanged, the reduction in speed was, as predicted by the universal wave equation, accompanied by a reduction in wavelength (Figure 8.21). For a constant frequency, the ratio of the velocities is the same as the ratio of the wavelengths. λ 2 v2 boundary between deep and shallow water straight wave generator λ 1 v1 shallow water (slow speed) wave fronts deep water (faster speed) ripple tank Figure 8.21 When the frequency is constant, a change in speed results in a change in wavelength. When waves change speed, they often change direction as well. You will study this phenomenon further in Unit VII. Example 8.2 To generate waves in a stretched spring, you oscillate your hand back and forth at a frequency of 2.00 Hz. If the speed of the waves in the spring is 5.40 m/s, what is the wavelength? Given v 5.40 m/s f 2.00 Hz Required wavelength Analysis and Solution The variables (v, f, λ) are related by the universal wave equation. v f λ v λ f m 5.40 s 2.00 Hz 5.40 m s 2.00 1 s 2.70 m Paraphrase and Verify The wavelength is 2.70 m. Practice Problems 1. Orchestras use the note with a frequency of 440 Hz (“A” above middle “C”) for tuning their instruments. If the speed of sound in an auditorium is 350 m/s, what is the length of the sound wave generated by this frequency? 2. A submarine sonar system sends a burst of sound with a frequency of 325 Hz. The sound wave bounces off an underwater rock face and returns to the submarine in 8.50 s. If the wavelength of the sound is 4.71 m, how far away is the rock face? 3. A fisherman anchors his dinghy in a lake 250 m from shore. The dinghy rises and falls 8.0 times per minute. He finds that it takes a wave 3.00 min to reach the shore. How far apart are the wave crests? Answers 1. 0.795 m 2. 6.51 km 3. 10 m Chapter 8 Mechanical waves transmit energy in a variety of ways. 409 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 410 M I N D S O N Wavelength, Frequency, and Speed • Walk side by side with a partner at • With both students keeping their the same speed. One student should take long steps while the other takes very short steps. 1. If the two students maintain their pace, what is the relationship between the frequency and the length of their steps? steps the same length as in the first trial, walk so that your steps are in phase (take steps at the same time). 2. When the two students walk in phase, what is the effect of taking shorter steps? What is the relationship between speed and step length? 8.2 Check and Reflect 8.2 Check and Reflect Knowledge 1. Explain the relationship between the motion of a transverse wave and the motion of the medium through which it moves. 2. Explain how the medium moves when a longitudinal wave passes through it. 3. What is the difference between a transverse and a longitudinal wave? 4. What determines the amount of energy stored in a wave? Applications 5. Sound waves travel through seawater at about 1500 m/s. What frequency would generate a wavelength of 1.25 m in seawater? 6. Temperature changes in seawater affect the speed at which sound moves through it. A wave with a length of 2.00 m, travelling at a speed of 1500 m/s, reaches a section of warm water where the speed is 1550 m/s. What would you expect the wavelength in the warmer water to be? 7. A speaker system generates sound waves at a frequency of 2400 Hz. If the wave speed in air is 325 m/s, what is the wavelength? 8. When you generate a wave in a spring, what is the relationship between the frequency, wavelength, and amplitude? 410 Unit IV Oscillatory Motion and Mechanical Waves 9. Two tuning forks are generating sound waves with a frequency of 384 Hz. The waves from one tuning fork are generated in air where the speed of sound is 350 m/s. The other tuning fork is generating sound under water where the speed of sound is 1500 m/s. Calculate the wavelength for the sound (a) in air, and (b) in water. (c) Would you hear the same musical note under water as you did in air? Extensions 10. A radio speaker generates sounds that your eardrum can detect. What does the operation of the speaker and your eardrum suggest about the nature of sound waves? How does the nature of the medium (air) through which sound travels support your assumptions? radio speaker direction of oscillation e TEST eardrum direction of oscillation To check your understanding of transverse and longitudinal waves, follow the eTest links at www.pearsoned.ca/school/physicssource. 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 411 8.3 Superposition and Interference Superposition of Pulses and Interference When waves travel through space it is inevitable that they will cross paths with other waves. In nature, this occurs all the time. Imagine two people who sit facing each other and are speaking at the same time. As each person’s sound waves travel toward the other person they must meet and pass simultaneously through the same point in space (Figure 8.22). Still, both people are able to hear quite plainly what the other person is saying. The waves obviously were able to pass through each other so that they reached the other person’s ears unchanged. How waves interact when they cross paths is well understood. When you observe two waves crossing in the ripple tank, things happen so quickly that it is difficult to see what is happening. Still, it is plain that the waves do pass through each other. By sending two pulses toward each other in a spring, it is easier to analyze the events. It is helpful to imagine that the spring in which the pulses are travelling is an ideal, isolated system. The pulses then travel without loss of energy. First, consider two upright pulses moving through each other. When two pulses pass through the same place in the spring at the same time, they are said to interfere with each other. In the section of the spring where interference occurs, the spring takes on a shape that is different from the shape of either of the pulses individually (Figure 8.23). vA pulse A vB pulse B actual position of spring region of pulse overlap pulse A pulse B pulse A x x y y pulse B z z pulse A pulse B Region of overlap Original position of pulse A Original position of pulse B Resultant sound waves from girl sound waves from boy Figure 8.22 When two people talk simultaneously, each person’s sound waves reach the other person’s ears in their original form. interference: the effect of two pulses (or two waves) crossing within a medium; the medium takes on a shape that is different from the shape of either pulse alone Figure 8.23 When two upright pulses move through each other, the displacement of the resultant pulse is the sum of the displacements of pulse A and pulse B. If at any point in the region of overlap, the displacement of one pulse, shown here as x, y, and z, is added to the displacement of the other, the displacement of the resultant pulse is increased. This is called constructive interference. Chapter 8 Mechanical waves transmit energy in a variety of ways. 411 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 412 principle of superposition: the displacement of the combined pulse at each point of interference is the sum of the displacements of the individual pulses constructive interference: the overlap of pulses to create a pulse of greater amplitude destructive interference: the overlap of pulses to create a pulse of lesser amplitude e WEB Find out more about superposition of pulses. Follow the links at www.pearsoned.ca/ school/physicssource. The new shape that the spring takes on is predicted by the principle of superposition. This principle, based on the conservation of energy, makes it quite easy to predict the shape of the spring at any instant during which the pulses overlap. The displacement of the combined pulse at each point of interference is the algebraic sum of the displacements of the individual pulses. In Figure 8.23 the two pulses have different sizes and shapes and are moving in opposite directions. The displacement of a pulse is positive for crests and negative for troughs. Since in Figure 8.23 both displacements are positive, at any point where the two pulses overlap, the displacement of the resultant pulse is greater than the displacements of the individual pulses. When pulses overlap to create a pulse of greater amplitude, the result is constructive interference (Figure 8.24). Now consider the case when an inverted pulse meets an upright pulse. The displacement of the inverted pulse is a negative value. When the displacements of these pulses are added together, the displacement of the resultant pulse is smaller than the displacement of either pulse. When pulses that are inverted with respect to each other overlap to create a pulse of lesser amplitude, the result is destructive interference (Figure 8.25). resultant pulse pulse A pulse B Figure 8.24 Constructive interference pulse A Figure 8.25 Destructive interference resultant pulse pulse B 412 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 413 Figure 8.26 shows a special case of destructive interference. Two pulses that have the same shape and size are shown passing through each other. Because the pulses are identical in shape and size, their displacements at any position equidistant from the front of each pulse are equal in magnitude but opposite in sign. At the point where the two pulses meet, the sum of their displacements will always be zero. At the instant when these two pulses exactly overlap, the displacement at all points is zero and the pulses disappear. The resultant is a flat line. Immediately foll |
owing this instant, the pulses reappear as they move on their way. The Inversion of Reflected Pulses in a Fixed Spring The principle of superposition explains why pulses are inverted when they reflect from the fixed end of a spring (Figure 8.27). Because the end of the spring is fixed in place, at that point the sum of the displacements of the incident pulse and the reflected pulse must always be zero. Thus, at the point of reflection, the displacement of the reflected pulse must be the negative of the incident pulse. Hence, the reflected pulse must be inverted relative to the incident pulse. info BIT Since, at the point of reflection in the spring the system is basically an isolated system, all the energy in the incident pulse must be carried away by the reflected pulse. actual position of spring original incident pulse reflected segment of incident pulse Pulse arrives at fixed end of the spring. vP vP actual position of spring vA pulse A point where pulses meet pulse B vB region of overlap pulse B pulse B vB pulse A x x y y z z pulse A vA pulse A pulse B pulse A vA pulse B vB Region of overlap Original position of pulse A Original position of pulse B Resultant Figure 8.26 When pulses that are inverted with respect to each other overlap, the displacement of one pulse is reduced by the displacement of the other pulse. At any point in the region of overlap, the displacement of Pulse B, shown here as x, y, and z, reduces the displacement of Pulse A to produce the resultant. This is called destructive interference. vP vP Reflected pulse leaves fixed end of the spring. Figure 8.27 If the end of the spring is fixed, the reflected pulse must be inverted relative to the incident pulse. Chapter 8 Mechanical waves transmit energy in a variety of ways. 413 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 414 8-6 Inquiry Lab 8-6 Inquiry Lab Interference of Waves Questions 1 What happens when two pulses pass through the same point in a medium? 2 How can two waves, moving in opposite directions, exist simultaneously in the same space? 3 What causes a standing wave? Materials light spring heavy spring masking tape stopwatch tape measure or metre-stick CAUTION: A stretched spring stores considerable amounts of elastic energy. Be careful not to release the end of a spring while it is stretched. When collapsing a spring, have the person holding one end walk the spring slowly toward the other end. Allow the spring to gently unwind as you are walking to prevent the spring from tying itself into a knot. Variables Part 1: In this part you are concerned with the amplitudes and lengths of the pulses. Observe these variables before, during, and after the period in which they interfere with each other. Part 2: In this part you will explore the relationship between the frequency and the standing wave pattern generated in a spring. From the structure of the standing wave pattern and the length of the spring, the wavelength and the speed of the standing wave are easily calculated. For both parts, identify which are the controlled variables, manipulated variables, and responding variables. Part 1: Superposition and Interference of Pulses 1 (a) Place two parallel strips of tape on the floor about 5 m apart. Measure and record the distance between them. Use these tapes to maintain a constant length for the spring while it is stretched. Attach a third strip of tape about 5 cm long to one of the coils near the middle of the spring as a marker. Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork (b) Have the team member holding one end of the spring generate a transverse pulse. When this pulse reaches the fixed end of the spring, have the same team member generate a second similar pulse. Try to generate the second incident pulse so that it meets the reflection of the first pulse at the strip of tape near the middle of the spring. Focus on the nature of the spring’s motion while the pulses interact. This complex interaction occurs quite quickly and may need to be repeated a few times until you are confident that you can see what is happening. Discuss the observations with your team members. (c) Record your observations in sketches and writing. 2 (a) Again, have one team member generate two pulses. This time, however, generate the second pulse so that it is on the opposite side of the spring (i.e., inverted) to the first pulse. The second pulse will now be on the same side of the spring as the reflected pulse. Again, time the pulses so that they meet near the centre of the spring. (b) Observe how these pulses interact when they meet at the centre of the spring. Discuss what you think is happening with the other members of your team. Analysis 1. When pulses on opposite sides of the spring meet, does the amplitude increase or decrease in the region of overlap? 2. When pulses on the same side of the spring meet, does the amplitude increase or decrease in the overlap region? Part 2: Standing Waves Procedure 1 (a) Have a team member at one end of the spring create a double wave (a series of four pulses) by oscillating his or her hand back and forth twice across the spring’s equilibrium position. (b) Observe what happens as this wave travels back and forth along the spring. Pay particular attention to what happens when the reflected portion of the wave is passing through the incident wave. Discuss your observations with your lab team to come to a consensus on what is occurring. 414 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 415 (c) Record your observations. Keep in mind what Analysis you observed when the pulses crossed in Part 1 of the lab. 2 (a) Now create a steady wave train by moving your hand back and forth. Try to find the frequency such that the spring oscillates in two segments about its midpoint. If, at first, there are more than two segments, then reduce the frequency slightly. If, at first, the spring is oscillating as only one segment, then increase the frequency until the second segment appears. Once the spring begins to oscillate as two segments, maintain that frequency. (b) Measure and record the frequency of oscillation by timing ten oscillations. Since you know the length of the spring (the distance between the tapes you placed on the floor), record the length of a wave for this frequency. Each half of the spring is a pulse so that, in this mode, the wavelength is equal to the length of the spring. (c) Record the data obtained in step 2(b) in a table. Use column headings: trial number, number of segments, frequency, wavelength, and speed. 3 (a) Begin with the frequency at which the spring oscillates in two halves and gradually increase the frequency. (b) Describe what happens when you try to maintain a slight increase in the frequency. Keep increasing the frequency until a new oscillation pattern is established. Measure the frequency for this pattern. Record your results in your table of data. (c) Starting from this frequency, gradually increase the frequency until a new pattern of oscillation is found. Once the new wave pattern is established, measure its frequency and record your measurements in your data table. 1. When you created a sustained wave so that the spring oscillated as a stable pattern, in which direction did the waves move? Why do you think that is the case? Does this tell you why this pattern is known as a standing wave? 2. (a) Two segments of a standing wave are equal to one wavelength. For each trial recorded in the table of data, calculate the wavelength of the standing wave. (b) For each trial, use the universal wave equation to calculate the speed of the waves in the spring. 3. To what does the speed of a standing wave refer? 4. Express the frequencies, for the different trials recorded in your data table, as ratios using simple whole numbers. Compare these ratios to the number of segments in which the spring oscillates for each trial. NOTE: The parts of a standing wave that remain motionless are called nodes or nodal points. The midpoints of the parts that oscillate back and forth are called antinodes. Each segment that contains an antinode is simply a pulse or one-half a wavelength. In a standing wave two adjacent segments are required to complete one wavelength. 5. Once a standing wave is established in the spring, what do you notice about the amplitude of the oscillations you use to sustain the wave compared with the amplitude of the antinodes? What explanation might exist for the difference in these two amplitudes? 6. Beginning at the fixed end of the spring, describe the locations of the nodes (points that remain motionless) and antinodes (midpoints of the parts that oscillate back and forth) along the spring in terms of wavelength. 7. How does the principle of superposition explain what must be happening at the antinodes of a standing wave? 4 If time permits, change to the heavier spring and repeat steps 2 and 3. 8. What relationship exists between the wavelength of a standing wave and the frequency creating the wavelength? CAUTION: Be very careful not to accidentally release the heavy spring while it is stretched. It will contain a large quantity of elastic potential energy and may seriously injure someone. To relax the tension in the spring, walk one end of the spring slowly toward the other end. 9. Go back to your observations in Part 1 of 8-2 Inquiry Lab. When a train of straight waves parallel to the barrier was reflected back through the incident wave train, did you observe a standing wave? 10. If you could generate a standing wave for sound, what do you think would be the nature of the sound at the location of an antinode? a node? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Chapter 8 Mechanical waves transmit energy in a variety of ways. 415 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 416 M I N D S O N Total Destruction? At the instant when two pulses “co |
mpletely destroy” each other, the spring is in its equilibrium position. How is it possible for the two pulses to reappear as if from nothing? Where does the energy in the pulses go when the sum of the amplitudes is zero? Hint: It might help to think of the spring in terms of a system. Standing Waves and Resonance When two wave trains with identical wavelengths and amplitudes move through each other (Figure 8.28), the resulting interference pattern can be explained by using the principle of superposition. When crests from the two waves or troughs from the two waves occupy the same point in the medium, the waves are in phase. Waves that are in phase produce constructive interference. When a crest from one wave occupies the same point in the medium as a trough from a second wave, we say that these waves are out of phase. Out-of-phase waves produce destructive interference. As the two wave trains pass through each other in opposite directions, they continually shift in and out of phase to produce a wave that seems to oscillate between fixed nodes, rather than move through the medium. wave A wave B wave A B vA vB A λ1 a) (b) (c) (d) (e) (f) actual position of medium at area where waves overlap constructive interference points at which only destructive interference occurs e MATH To graphically analyze the superposition of waves that are in or out of phase, visit www.pearsoned.ca/school/ physicssource. e WEB Find out more about the superposition of waves. Follow the links at www.pearsoned.ca/school/ physicssource. Figure 8.28 The diagrams show how waves travelling in opposite directions interfere as they move through each other. 416 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 417 • (a) Point A is the initial point of contact between the two wave trains shown in blue and purple. The crest from the purple wave train and the trough from the blue wave train arrive at point A at the same instant. • (b) The two identical waves have moved a distance of 1 λ in opposite 4 directions. This overlap results in destructive interference and the spring is flat in the region of overlap. The position of the spring where the two waves overlap, the resultant, is shown in red. • (c) Each wave has moved a further 1 λ along the spring. Now the 4 waves are exactly in phase and constructive interference occurs. The regions to the left and right of point A show a crest and a trough, respectively, with displacement of the resultant being twice that of the blue or purple waves. • Every time the wave trains move a further 1 λ along the spring, the 4 interference changes from constructive to destructive and vice versa. At point A, only destructive interference occurs. The magnitudes of the displacements of the waves arriving at point A are always equal but opposite in sign. As the waves continue to move in opposite directions, the nature of the interference continually changes. However, at point A and every 1 λ from point A, there are points at which only destructive 2 interference occurs. These are called nodal points or nodes. Between the nodes, the spring goes into a flip-flop motion as the interference in these areas switches from constructive (crest crossing crest) to destructive (crest crossing trough) and back to constructive interference (trough crossing trough). The midpoints of these regions on the spring are called antinodes. The first antinode occurs at a distance λ on either side of A, and then at every 1 of 1 λ after that point. Because 2 4 the wave seems to oscillate around stationary nodes along the spring, it is known as a standing wave. Standing waves are also seen in nature; an example is shown in Figure 8.29. Standing Waves in a Fixed Spring When you generate a wave train in a spring that is fixed at one end, the reflected wave train must pass back through the incident wave train. These two wave trains have identical wavelengths and nearly identical amplitudes. For incident and reflected wave trains, the initial point of contact is by definition the fixed point at which reflection occurs. This means that the endpoint of the spring is always a nodal point and, as shown in Figure 8.30, nodes occur every 1 λ from that point with antinodes 2 between them. λ1 2 λ1 2 λ1 2 λ1 2 λ1 2 λ1 2 node: a point on a spring or other medium at which only destructive interference occurs; a point that never vibrates between maximum positive amplitude and maximum negative amplitude; in a standing wave nodes occur at intervals of 1 λ 2 antinode: a point in an interference pattern that oscillates with maximum amplitude; in a standing wave antinodes occur at intervals of 1 λ 2 standing wave: a condition in a spring or other medium in which a wave seems to oscillate around stationary points called nodes. The wavelength of a standing wave is the distance between alternate nodes or alternate antinodes. Figure 8.29 Standing waves occur in nature. This photograph shows a standing wave in a stream crossing a sandy beach in Scotland. e WEB Find out about the details of a standing wave in a spring. Follow the links at www.pearsoned.ca/school/ physicssource. Figure 8.30 In a spring with a fixed end, a standing wave must contain a whole number of antinodes. Nodes occur every half-wavelength from the ends. Antinodes oscillate between shown positions. Nodes remain motionless. antinode Chapter 8 Mechanical waves transmit energy in a variety of ways. 417 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 418 e WEB To learn more about the Tacoma Narrows Bridge collapse, follow the links at www.pearsoned.ca/ school/physicssource. Figure 8.31 Resonance, caused by wind, set up a standing wave that destroyed the Tacoma Narrows Bridge. resonance: an increase in the amplitude of a wave due to a transfer of energy in phase with the natural frequency of the wave e WEB To learn more about the giant shock absorbers added to the Millennium Bridge, follow the links at www.pearsoned.ca/ school/physicssource. Figure 8.32 The tone produced when you blow across the top of an open bottle depends on the length of the air column. Resonant Frequencies When a standing wave is present in a spring, the wave reflects from both ends of the spring. There must be a nodal point at both ends with an integral number of antinodes in between. The spring “prefers” to oscillate at those frequencies that will produce a standing wave pattern, called the resonant frequencies for the spring. When the generator is oscillating at a resonant frequency, the energy is added to the spring in phase with existing oscillations. This reinforces and enhances the standing wave pattern. The added energy works to construct waves with ever-larger amplitudes. If the generator is not oscillating at a resonant frequency of the medium, the oscillations tend to destroy the standing wave motion. Amplitude and Resonance Perhaps the most impressive display of a standing wave occurred when resonance set up a standing wave in the bridge across the Tacoma Narrows in the state of Washington (Figure 8.31). Opened in November 1940, the bridge was in operation only a few months before resonance ripped it apart. More recently, in June 2000, the newly opened Millennium Bridge in London had to be closed for modifications when the footsteps of pedestrians set up resonance patterns. Anyone who has ever “pumped up” a swing has used the principle of resonance. To increase the amplitude of its motion, the swing must be given a series of nudges in phase with its natural frequency of motion. Each time the swing begins to move forward, you give it a little push. Since these little pushes are produced in resonance with the swing’s natural motion, they are added to its energy and the amplitude increases. If you pushed out of phase with its natural motion, the swing would soon come to rest. Concept Check Why does it take so little energy to sustain a standing wave in a spring? Concept Check Resonating Air Columns All wind instruments use the principle of resonance to produce music. The simplest example of resonance in music is the note produced when you blow over the top of a bottle (Figure 8.32). Blowing across the top of the bottle oscillates the air in the bottle and generates a standing wave. This standing wave is like the waves travelling in a spring, but unlike a spring that is fixed at both ends, the air column is fixed only at the end where reflection occurs and is free to oscillate at the open end. The resonant frequency of the note produced depends on the length of the air column because, to resonate, the standing wave must have a node at the closed end of the bottle and an antinode at the open end (Figure 8.33). 418 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 419 Closed-Pipe or Closed-Tube Resonance NR R resonance heard NR no resonance heard When a wave source is held at the open end of a pipe, it sends down a wave that reflects from the closed end of the pipe and establishes a standing wave pattern. The sound one hears depends on the length of the air column in the pipe relative to the length of the standing wave. If an antinode occurs at the open end of the pipe (Figure 8.33 (a) and (c)), a point of resonance (resulting from constructive interference) occurs at the open end of the pipe and the sound appears to be amplified. This phenomenon is known as closed-pipe or closed-tube resonance. However, if the open end of the pipe coincides with the position of a node (destructive interference), then almost no sound can be heard because the source (tuning fork) and the standing wave are out of phase (Figure 8.33 (b) and (d)). (b) (a) R NR R (c) (d) Figure 8.33 Resonance series. A tuning fork sets up a standing wave in the air column. The volume of the sound one hears will vary depending on whether there is an antinode, (a) and (c), or a node, (b) and (d), at the end of the pipe. Nodes and Antinodes in Closed-Pipe Resonance In the air column, node |
s are located every half-wavelength from the end at which the wave is reflected, just as they are in a standing wave in a spring. If the pipe length is equal to any multiple of 1 λ, there will 2 be a node at the upper end of the pipe, and destructive interference λ, 4 λ, 3 λ, 2 will occur. Thus, when the air column is 1 λ, … in length, 2 2 2 2 little or no sound will be heard. Antinodes in the air column are located one quarter-wavelength from the end of the pipe where reflection occurs, and then every halfwavelength from that point. Thus, resonance is heard when the pipe is λ, 5 λ, 3 1 λ, … long. When resonance is heard for an air column closed 4 4 4 at one end, we know that the open end of the column coincides with the location of one of the antinodes. This information can be used to measure the wavelength of sound in gases. If the frequency of the sound is known, then the wavelength can be used to calculate the speed of sound in the gas. Concept Check Is the volume of a sound related to speed, wavelength, amplitude, or frequency of the wave? What evidence is there to support your answer? Example 8.3 A tuning fork with a frequency of 384 Hz is held above an air column. As the column is lengthened, a closed-pipe resonant point is found when the length of the air column is 67.5 cm. What are possible wavelengths for this data? If the speed of sound is known to be slightly greater than 300 m/s, what is (a) the actual wavelength, and (b) the actual speed of sound? Chapter 8 Mechanical waves transmit energy in a variety of ways. 419 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 420 Practice Problems 1. A tuning fork of frequency 512 Hz is used to generate a standing wave pattern in a closed pipe, 0.850 m long. A strong resonant note is heard indicating that an antinode is located at the open end of the pipe. (a) What are the possible wavelengths for this note? (b) Which wavelength will give the most reasonable value for the calculation of the speed of sound in air? 2. A tuning fork with a frequency of 256 Hz is held above a closed air column while the column is gradually increased in length. At what lengths for this air column would the first 4 resonant points be found, if the speed of sound is 330 m/s? 3. A standing wave is generated in a spring that is stretched to a length of 6.00 m. The standing wave pattern consists of three antinodes. If the frequency used to generate this wave is 2.50 Hz, what is the speed of the wave in the spring? 4. When a spring is stretched to a length of 8.00 m, the speed of waves in the spring is 5.00 m/s. The simplest standing wave pattern for this spring is that of a single antinode between two nodes at opposite ends of the spring. (a) What is the frequency that produces this standing wave? (b) What is the next higher frequency for which a standing wave exists in this spring? Answers 1. (a) 3.40 m @ 1.74 103 m/s; 1.13 m @ 580 m/s; 0.680 m @ 348 m/s; 0.486 m @ 249 m/s (b) 0.680 m 2. 0.322 m, 0.967 m, 1.61 m, 2.26 m 3. 10.0 m/s 4. 0.313 Hz, 0.625 Hz Given f 384 Hz l 67.5 cm 0.675 m Required wavelength and speed of sound Analysis and Solution λ, 5 λ, 3 The resonant point might represent 1 λ,…, etc., for 4 4 4 this tuning fork. Assume that 67.5 cm is the first resonant point; that means 67.5 cm is 1 λ. Calculate the wavelength 4 and the speed of sound from that data. Assume that l 1 λ. Therefore, 4 λ 4l v fλ 4(0.675 m) 2.70 m (384 Hz)(2.70 m) 1037 m/s 1.04 103 m/s This value is larger than the speed of sound in air. If the speed of sound is not of the proper order of magnitude, then assume that the resonant point is the second point of resonance and that 67.5 cm is 3 λ. Calculate the wavelength 4 and the speed of sound from that data. Assume that l 3 λ. Therefore, 4 l λ 4 3 75 m) (384 Hz)(0.900 m) 4(0.6 3 v fλ 0.900 m 345.6 m/s 346 m/s This is a reasonable speed for sound in air. Complete the analysis by assuming that l 5 λ. Therefore, 4 l λ 4 5 75 m) (384 Hz)(0.540 m) 4(0.6 5 v fλ 0.540 m 207.4 m/s 207 m/s This value is less than the speed of sound in air. Paraphrase and Verify The calculations for the speed of sound indicate that the data must have been for the second point of resonance. This assumption gives the speed for sound of 346 m/s. The assumption that the pipe length is for the first resonant point results in a speed about three times that of sound. The assumption that the pipe length is for the third resonant point produces a speed less than 300 m/s. 420 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 421 8-7 Inquiry Lab 8-7 Inquiry Lab Measuring the Speed of Sound Using Closed-pipe Resonance Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork When a sound wave travels down a closed pipe, the incident wave reflects off the end of the pipe and back toward the source. The interaction of the incident and reflected waves sets up an interference pattern inside the pipe, known as a standing wave. This standing wave can be used to determine the wavelength of the sound. Problem What is the speed of sound in air? Variables The universal wave equation relates the speed (v) of a wave to its frequency (f ) and wavelength (). The wavelength is determined from the length of the pipe (l ) and the number of the resonant point as counted from the reflecting surface. Materials and Equipment tuning forks and tuning fork hammer or an audio frequency generator glass or plastic pipe tall cylinder Procedure 1 Assemble the apparatus as shown in Figure 8.34. metrestick tuning fork open-ended pipe tall cylinder water Figure 8.34 2 Place the pipe in the water-filled cylinder so that the column of air in the pipe is quite short. 3 Strike the tuning fork with the hammer. 4 Hold the tuning fork as shown over the end of the pipe and lift the pipe slowly so that the length of the column of air in the pipe increases. 5 As you approach a point where the volume of the sound increases, move the pipe slowly up and down to find the point where the resonance is greatest. Strike the tuning fork as often as necessary to maintain the sound source. 6 Determine the length of the column of air that gives the greatest resonance and record it in a table like Table 8.1. Measure the length from the surface of the water to the location of the tuning fork. 7 Beginning with the column of air at the previously recorded length, gradually increase the length until you have determined the length of the air column that gives the next point of resonance. Record this length. 8 Repeat step 7 to find the length of the column for the third resonant point. Table 8.1 Column Length and Resonance Frequency f (Hz) Length of column at first resonant point l1 (m) Length of column at second resonant point l2 (m) Length of column at third resonant point l3 (m) Analysis 1. In terms of wavelength, how far is each of the first three resonant points from the reflecting surface of the water at the bottom of the air column? 2. Calculate the wavelength of the sound from the tuning fork for each resonant point. Record your answers in a table similar to Table 8.2. Calculate the speed of sound for each of the wavelengths. 3. When you calculate the wavelength for different resonant points, do the answers agree? If not, what might cause the differences? 4. Why should you start with a short column of air and increase its length if you are to be sure that you have correctly determined the wavelength? Chapter 8 Mechanical waves transmit energy in a variety of ways. 421 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 422 ▼ Table 8.2 Resonant Points, Wavelength, and Speed of Sound First Resonant Point Second Resonant Point Third Resonant Point Frequency f (Hz) Wavelength 4l (m) Speed v (m/s) Wavelength 4l/3 (m) Speed v (m/s) Wavelength 4l/5 (m) Speed v (m/s) 5. Why should you measure the length of the column from the reflecting surface to the tuning fork rather than to the top end of the pipe? 6. What is the speed of sound at room temperature? 7. Investigate the effect that air temperature has on the speed of sound. Use hot water or ice water to modify the temperature of the air in the column. Compare the measured speed of sound for at least three temperatures. Suspend a thermometer down the pipe to determine the temperature of the air in the pipe. Plot a graph of the measured speed of sound versus the temperature. Does the graph suggest a linear relationship? Can you use the graph to predict the speed of sound at other temperatures? 8. An alternative technique to determine the speed of sound is to measure the time for an echo to return to you. Stand a measured distance from a wall or other surface that reflects sound. Create an echo by striking together two hard objects, such as metal bars, or, perhaps beating a drum. Listen for the echo. Once you have established an approximate time for the echo to return, strike the bars in a rhythm so that they are in phase with the echo. Have a team member count the number of beats in 1 min. The period of this frequency is the time required for the sound to travel to the wall and back. Use that data to calculate the speed of the sound. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. fundamental frequency: the lowest frequency produced by a particular instrument; corresponds to the standing wave having a single antinode, with a node at each end of the string Music and Resonance Complex modes of vibration give instruments their distinctive sounds and add depth to the musical tones they create. A string of a musical instrument is simply a tightly stretched spring for which the simplest standing wave possible is a single antinode with a node at either end. For this pattern, the length of the string equals one-half a wavelength and the frequency produced is called the fundamental frequency (Figure 8.35(a)). fundamental frequency equilibrium position info BIT Assume that the fundamental frequ |
ency is f . In physics and in music, the frequency 2f is called the first overtone; 3f is the second overtone, and so on. These frequencies are said to form a harmonic series. Thus, physicists may also refer to the fundamental frequency (f ) as the first harmonic, the frequency 2f as the second harmonic, the frequency 3f as the third harmonic, and so on. Figure 8.35 (a) standing wave with an antinode at the centre of the string. The fundamental frequency of a vibrating string oscillates as a This is the lowest frequency produced by a particular instrument. But other standing wave patterns can exist in the string at the same time as it oscillates at its fundamental frequency. By plucking or bowing a string nearer its end than its middle, the string is encouraged to vibrate with multiple frequencies. The frequencies above the fundamental 422 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 423 frequency that may exist simultaneously with the fundamental frequency are called overtones. Figures 8.35 (b) and (c) show the shape of a string vibrating in its first and second overtones, respectively. Figure 8.36 shows a violinist bowing and fingering the strings of her violin to produce notes. 1st overtone without the fundamental frequency equilibrium position fundamental frequency 1st overtone with the fundamental frequency equilibrium position Figure 8.36 The violinist’s fingering technique changes the length of the string and thus changes the fundamental frequency of vibration. overtone: any frequency of vibration of a string that may exist simultaneously with the fundamental frequency The first overtone has the form of a standing wave with two Figure 8.35 (b) antinodes. A node exists at the midpoint of the string. The lower portion of the diagram shows a string vibrating with both the fundamental frequency and the first overtone simultaneously. 2nd overtone without the fundamental frequency equilibrium position fundamental frequency 2nd overtone with the fundamental frequency equilibrium position Figure 8.35(c) The lower portion of the diagram shows a string vibrating with both the fundamental frequency and second overtone. The vibration that produces the second overtone has three antinodes. The actual form of a vibrating string can be very complex as many overtones can exist simultaneously with the fundamental frequency. The actual wave form for a vibrating string is the result of the constructive and destructive interference of the fundamental wave with all the existing overtones that occur in the string. For example, Figure 8.37 shows the wave trace on an oscilloscope for the sound of a violin. Figure 8.37 The interference of the fundamental frequency with the overtones produced by a bowed string creates the wave form that gives the violin its unique sound. The wavelength of the fundamental frequency is the distance between the tall sharp crests. Chapter 8 Mechanical waves transmit energy in a variety of ways. 423 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 424 Tuning a Stringed Instrument Tuning a stringed instrument involves several principles of physics. The universal wave equation, v fλ, indicates that the frequency of a sound wave is directly proportional to the speed of the sound and inversely proportional to its wavelength. The wavelength for the fundamental frequency of the standing wave in a string is fixed at twice the length of the string, but the speed of a wave in a string increases with tension. Thus, if the wavelength does not change, the frequency at which a string vibrates must increase with tension. Changing the tension in the string is known as tuning (Figure 8.38). Wind Instruments Wind instruments produce different musical notes by changing the length of the air columns (Figure 8.39). In 8-7 Inquiry Lab you used a closed pipe and saw that for resonance to occur, a node must be present at the closed end while an antinode is created at the open end. For a closed pipe, the longest wavelength that can resonate is four times as long as the pipe (Figure 8.40). If the pipe is open at both ends, then the wavelengths for which resonance occurs must have antinodes at both ends of the open pipe or open tube (Figure 8.41). The distance from one antinode to the next is one-half a wavelength; thus, the longest wavelength that can resonate in an open pipe is twice as long as the pipe. antinode antinode l λ1 4 l λ1 2 node antinode Figure 8.40 In a closed pipe, the longest possible resonant wavelength is four times the length of the pipe. Figure 8.41 In an open pipe, the longest possible resonant wavelength is twice the length of the pipe. Wind instruments are generally open pipes. The wavelength of the resonant frequency will be decided by the length of the pipe (Figure 8.42). In a clarinet or oboe, for example, the effective length of the pipe is changed by covering or uncovering holes at various lengths down the side of the pipe. The strongest or most resonant frequency will be the wave whose length is twice the distance from the mouthpiece to the first open hole. Overtones are also generated but the note you hear is that with the longest wavelength. As with stringed instruments, the overtones contribute to the wind instrument’s characteristic sound. If the speed of sound in air never varied, then a given wavelength would always be associated with the same frequency. But the speed of sound changes slightly with air temperature and pressure. Thus, in the case of resonance in a pipe, the length of the pipe must be increased or decreased as the speed of sound increases or decreases to ensure that the frequency is that of the desired note. Figure 8.38 Tuning a guitar e LAB For a probeware activity, go to www.pearsoned.ca/ school/physicssource. Figure 8.39 The trumpeter produces different notes by opening valves to change the instrument’s overall pipe length. Figure 8.42 A variety of wind instruments e WEB To learn how and why wind instruments are affected by temperature, follow the links at www.pearsoned.ca/school/ physicssource. 424 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 425 An Interference Pattern from Two In-phase Point Sources Interference patterns carry information about the waves that create them. For this reason, the patterns are often used to determine the properties of the waves. One of the most interesting interference patterns results from waves generated by two point sources that are in phase. Remember that wave sources are in phase if they generate crests at the same time. The ripple tank photograph (Figure 8.43) shows the interference pattern generated by two in-phase point sources that are separated in space. This pattern is the result of constructive and destructive interference as the waves cross. Generally crests appear bright and troughs appear dark. However, in areas where destructive interference occurs, there appear to be fuzzy lines (such as the line indicated by Q1) that seem to radiate approximately from the midpoint between the sources. While the pattern may appear to be complex, its explanation is fairly simple. P1 Q1 R1 S1 P2 S2 Figure 8.43 The interference pattern generated by two in-phase point sources in a ripple tank. The distance between the sources is 3. Individually, point sources generate waves that are sets of expanding concentric circles. As the crests and troughs from each source move outward, they cross through each other. As with all waves, when the crests from one source overlap crests from the other source (or troughs overlap troughs), constructive interference occurs. In these regions there is increased contrast (as indicated by P1 and R1). At locations where the crests from one source overlap troughs from the other source, destructive interference occurs. In these regions, contrast is reduced. Because the sources oscillate in phase, the locations where constructive and destructive interference occur are at predictable, fixed points. Like standing waves in a spring, the positions of the nodes and antinodes depend on the wavelength and the distance between the sources. Can you identify the regions of constructive and destructive interference in Figure 8.43 above? info BIT Common effects of interference patterns result in the “hot” and “cold” spots for sound in an auditorium. In 2005 renovations were completed for the Jubilee Auditoriums in Edmonton and Calgary. During renovations, the auditoriums were retuned to improve their acoustic properties. PHYSICS INSIGHT An interference pattern for sound can result if two loudspeakers, at an appropriate distance apart, are connected to the same audio frequency generator. When the sound waves diverging from the speakers overlap and interfere, regions of loud sound (maxima) and regions of relative quiet (minima) will be created. interference pattern: a pattern of maxima and minima resulting from the interaction of waves, as crests and troughs overlap while the waves move through each other Chapter 8 Mechanical waves transmit energy in a variety of ways. 425 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 426 The pattern in Figure 8.43 can be reproduced by drawing sets of concentric circles about two point sources where each circle represents the crest of a wave front (Figure 8.44). In Figure 8.43 the distance (d) between the sources is equal to three wavelengths (3λ). This can be shown by counting the wavelengths between S1 and S2 in Figure 8.44. P1 Q1 R1 central maximum first order minimum first order maximum minima maxima 5λ 5λ 4 λ1 2 4λ 3λ S1 P2 Q2 R2 S2 Figure 8.44 The interference pattern for two in-phase point sources results from the overlap of two sets of concentric circles. In this diagram, the centres of the circles are three wavelengths apart. Maxima, Minima, and Phase Shifts The central maximum is a line of antinodes. In Figure 8.44, the line P1P2 is the perpendicular bisector of the line S1S2. By definition, every point |
on P1P2 is equidistant from the points S1 and S2. Thus, crests (or troughs) generated simultaneously at S1 and S2 must arrive at P1P2 at the same time, resulting in constructive interference. Along the line P1P2 only antinodes are created. The line of antinodes along P1P2 is called the central maximum. A nodal line, or minimum, marks locations where waves are exactly out of phase. A little to the right of the central maximum is the line Q1Q2. If you follow this line from one end to the other you will notice that it marks the locations where the crests (lines) from S1 overlap the troughs (spaces) from S2 and vice versa. Waves leave the sources in phase, but all points on Q1Q2 are a one-half wavelength farther from S1 than they are from S2. Thus, at any point on Q1Q2, the crests from S1 arrive one-half a wavelength later than the crests from S2. This means they arrive at the same time as troughs from S2. The greater distance travelled by waves from S1 produces what is called a one-half wavelength phase shift. Waves that began in phase arrive at points on Q1Q2 exactly out of phase. Thus, at every point on Q1Q2 destructive interference occurs. The line, Q1Q2, is known as a nodal line or a minimum. maximum: a line of points linking antinodes that occur as the result of constructive interference between waves minimum: a line of points linking nodes that occur as the result of destructive interference between waves phase shift: the result of waves from one source having to travel farther to reach a particular point in the interference pattern than waves from the other source 426 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 427 e WEB To learn more about two-point interference systems, follow the links at www.pearsoned.ca/school/ physicssource. A first order maximum is the result of a one wavelength phase shift. Moving farther right, another region of constructive interference occurs. To arrive at any point on R1R2, crests from S1 travel exactly one wavelength farther than crests from S2. Crests from S1 arrive at points on R1R2 at the same time as crests from S2 that were generated one cycle later. This one-wavelength phase shift means that all waves arriving at any point on R1R2 are still in phase. The line of antinodes resulting from a one-wavelength shift is known as a first order maximum. An identical first order maximum exists on the left side. The interference pattern is symmetrical about the central maximum. Phase shifts equal to whole wavelengths produce maxima. Moving farther outward from the central maximum, you pass through lines of destructive and constructive interference (minima and maxima). Each region is the result of a phase shift produced when waves travel farther from one source than the other. When the phase shift equals a whole number of wavelengths (0λ, 1λ, 2λ, . . .), the waves arrive in phase, producing antinodes and resulting in the central, first, second, and third order maxima, etc. In Figure 8.44, since the sources are 3λ apart, the greatest phase shift possible is three wavelengths. This produces the third order maximum directly along the line of S1S2. Phase shifts equal to an odd number of half-wavelengths produce minima. When the phase shift equals an odd number of half-wavelengths 1 λ, 2 λ, … the waves arrive out of phase, producing a nodal line or λ, 5 3 2 2 minimum. In Figure 8.44, the greatest phase shift to produce destructive interference is one-half wavelength less than the three-wavelength λ 5 separation of the sources, or 3λ 1 λ. Because the sources are three 2 2 wavelengths apart, there are exactly three maxima and three minima to the right and to the left of the central maximum. 8-8 Design a Lab 8-8 Design a Lab Interference Patterns and In-phase Sound Sources The Question Do interference patterns exist for two in-phase sound sources? Design and Conduct Your Investigation An audio frequency generator and two speakers can be used to create an interference pattern for sound. Design a set-up that will enable you to measure the wavelength of sound of known frequencies. If electronic equipment (probeware or waveport) is available, design lab 8-8 to incorporate this equipment. Measure the wavelengths using several maxima and minima to compare measurements. Which type of line gives the best results? How well do the results from this experiment compare with the results from measuring wavelengths using closed-pipe resonance? Chapter 8 Mechanical waves transmit energy in a variety of ways. 427 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 428 8.3 Check and Reflect 8.3 Check and Reflect Knowledge 1. What is meant by the term interference? 2. For a standing wave, what is the relationship between the amplitude of an antinode and the amplitude of the waves that combine to create the standing wave? 3. In terms of the length of an air column, what is the longest standing wavelength that can exist in an air column that is (a) closed at one end and (b) open at both ends? 4. An air column is said to be closed if it is closed at one end. Consider a pipe of length (l). For a standing wave in this pipe, what are the lengths of the three longest wavelengths for which an antinode exists at the open end of the pipe? 5. What does it mean to say that two wave generators are in phase? What does it mean to say that two waves are in phase? Applications 6. Two pulses of the same length (l) travel along a spring in opposite directions. The amplitude of the pulse from the right is three units while the amplitude of the pulse from the left is four units. Describe the pulse that would appear at the moment when they exactly overlap if (a) the pulses are on the same side of the spring and (b) the pulses are on opposite sides of the spring. 7. A standing wave is generated in a closed air column by a source that has a frequency of 768 Hz. The speed of sound in air is 325 m/s. What is the shortest column for which resonance will occur at the open end? 8. Draw the interference pattern for two inphase point sources that are 5 apart, as follows. Place two points, S1 and S2, 5 cm apart near the centre of a sheet of paper. Using each of these points as a centre, draw two sets of concentric circles with increasing radii of 1 cm, 2 cm, 3 cm, . . . , until you reach the edge of the paper. On the diagram, draw solid lines along maxima and dotted lines along minima. Label the maxima according to their order. Explain why there are five minima on either side of the central maximum. 9. An interference pattern from two in-phase point sources is generated in a ripple tank. On the screen, a point on the second order maximum is measured to be 8.0 cm from one point source and 6.8 cm from the other source. What is the wavelength of this pattern? Extension 10. Do pipe organs, such as those found in churches and concert halls, use closed or open pipes to produce music? What is the advantage of using a real pipe organ as opposed to an electronic organ that synthesizes the sound? e TEST To check your understanding of superposition and interference of pulses and waves, follow the eTest links at www.pearsoned.ca/school/physicssource. 428 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 429 info BIT It is common to see police officers using radar to measure the speed of cars on the highway. The radar gun emits waves that are reflected back to it from an oncoming car. A computer in the gun measures the change in frequency and uses that change to calculate the speed of the car. reflected wave λ transmitted RADAR wave moving car λ v radio transmitter Figure 8.45 8.4 The Doppler Effect Have you ever stood at the side of a road and listened to the cars pass? If you listen carefully, you will detect a very interesting phenomenon. At the instant a car passes you, the sound it makes suddenly becomes lower in pitch. This phenomenon was explained by an Austrian physicist named Christian Doppler (1803–1853). Doppler realized that the motion of the source affected the wavelength of the sound. Those waves that moved in the same direction as the source was moving were shortened, making the pitch of the sound higher. Moving in the direction opposite to the motion of the source, the sound waves from the source were lengthened, making the pitch lower. Wavelength and Frequency of a Source at Rest Assume that the frequency of a source is 100 Hz and the speed of sound is 350 m/s (Figure 8.46). According to the universal wave equation, if this source is at rest, the wavelength of the sound is 3.50 m. v fλ λ v f 350 m s 0 10 s 3.50 m vw 350 m/s λ 3.50 m Figure 8.46 When a wavelength of 3.50 m travels toward you at a speed of 350 m/s, you hear sound that has a frequency of 100 Hz (diagram not to scale). You hear the sound at a frequency of 100 Hz because at a speed of 350 m/s, the time lapse between crests that are 3.50 m apart is 1/100 s. If, however, the wavelengths that travel toward you were 7.0 m long, the time lapse between successive crests would be 1/50 s, a frequency equal to 50 Hz. v f λ 350 m s 7.0 m 50 1 s 50 Hz Chapter 8 Mechanical waves transmit energy in a variety of ways. 429 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 430 Wavelength and Frequency of a Moving Source 0.7 m A B C D 4.2 m Figure 8.47 When a sound source moves toward you, the wavelengths in the direction of the motion are decreased. e SIM Research examples of shock waves and the variation of wavelength with a moving source. Go to www.pearsoned.ca/ school/physicssource. direction of motion wave front generated by source at position A by source at position B by source at position C by source at position D Imagine that the source generating the 100-Hz sound is moving toward you at a speed of 70 m/s. Assume that the source is at point A (Figure 8.47) when it generates a crest. While the first crest moves a distance of 3.5 m toward you, the source also moves toward you. The distance the source moves |
while it generates one wavelength is the distance the source travels in 1/100 s at 70 m/s, or 0.7 m. Because of the motion of the source, the next crest is generated (at point B) only 2.8 m behind the first crest. As long as the source continues at the speed of 70 m/s toward you, the crests travelling in your direction will be only 2.8 m apart. Hence, for a car moving toward you, the sound waves emitted by the car will be “squashed together” and thus reach you more frequently than if the car were stationary. 2.8 m If waves that are 2.8 m long travel toward you at a speed of 350 m/s, then the frequency of the sound arriving at your ear will be 125 Hz. The pitch of the sound that you hear will have been increased because the source is moving toward you. v fλ v f λ 350 m s 2.80 m 125 1 s 125 Hz At the same time, along a line in the direction opposite to the motion of the source, the wavelengths are increased by the same amount that the waves in front of the source are shortened. For the 100-Hz sound source moving at 70 m/s, the waves behind the source are increased by 0.7 m, to a length of 4.2 m. The time lapse between these crests, which are 4.2 m apart and travelling at 350 m/s, is 0.012 s. Therefore, the perceived frequency in the direction opposite to the motion of the source is about 83 Hz. The pitch of the sound has been lowered. Analysis of the Doppler Effect If the velocity of the sound waves in air is vw, then the wavelength (λ s) that a stationary source(s) with a frequency of fs generates is given by λ s vw fs. The key to this Doppler’s analysis is to calculate the distance the source moves in the time required to generate one wavelength (the period (Ts) of the source). If the source is moving at speed vs, then in the period (Ts) the source moves a distance (ds) that is given by ds vsTs . Since, by definition 430 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/28/08 9:13 AM Page 431 Ts 1fs, then ds vs fs. Sources Moving Toward You For sources that are moving toward you, ds is the distance by which the wavelengths are shortened. Subtracting ds from λ s gives the lengths of the waves (λ d) that reach the listener. Therefore, ds. s and ds by their equivalent forms gives λ λ d s Replacing v v s w fs f s λ d λ d vs ) (vw fs This is the apparent wavelength (Doppler wavelength) of the sound generated by a source that is moving toward you at a speed vs. Dividing the speed of the waves (vw) by the Doppler wavelength (λ d) produces the Doppler frequency (fd) of the sound that you hear as the source approaches you. Therefore, v w λ d fd vw vs vw f s f s vw fs vw vs v w vw vs is the Doppler frequency when the source is approaching the listener. Sources Moving Away from You If the source is moving away from the listener, the value of ds is added to the value of λ s, giving λ λ ds. s d s and ds by their equivalent forms and complete the If you replace λ development to find fd, it is easy to see that the Doppler frequency for a sound where the source moves away from the listener is given by fs v w vw vs fd General Form of the Doppler Equation The equations for the Doppler effect are usually written as a single equation of the form fs v w vw vs fd PHYSICS INSIGHT When the distance between you and the source is decreasing, you must subtract to calculate the Doppler effect on frequency and wavelength. Chapter 8 Mechanical waves transmit energy in a variety of ways. 431 08-PearsonPhys20-Chap08 7/28/08 9:17 AM Page 432 Concept Check If you are travelling in your car beside a train that is blowing its whistle, is the pitch that you hear for the whistle higher or lower than the true pitch of the whistle? Explain. Example 8.4 A train is travelling at a speed of 30.0 m/s. Its whistle generates a sound wave with a frequency of 224 Hz. You are standing beside the tracks as the train passes you with its whistle blowing. What change in frequency do you detect for the pitch of the whistle as the train passes, if the speed of sound in air is 330 m/s? Practice Problems 1. You are crossing in a crosswalk when an approaching driver blows his horn. If the true frequency of the horn is 264 Hz and the car is approaching you at a speed of 60.0 km/h, what is the apparent (or Doppler) frequency of the horn? Assume that the speed of sound in air is 340 m/s. 2. An airplane is approaching at a speed of 360 km/h. If you measure the pitch of its approaching engines to be 512 Hz, what must be the actual frequency of the sound of the engines? The speed of sound in air is 345 m/s. 3. An automobile is travelling toward you at a speed of 25.0 m/s. When you measure the frequency of its horn, you obtain a value of 260 Hz. If the actual frequency of the horn is known to be 240 Hz, calculate vw, the speed of sound in air. 4. As a train moves away from you, the frequency of its whistle is determined to be 475 Hz. If the actual frequency of the whistle is 500 Hz and the speed of sound in air is 350 m/s, what is the train’s speed? Answers 1. 278 Hz 2. 364 Hz 3. 325 m/s 4. 18.4 m/s Given fs vw vs 224 Hz 330 m/s 30.0 m/s Required (a) Doppler frequency for the whistle as the train approaches (b) Doppler frequency for the whistle as the train moves away (c) change in frequency Analysis and Solution Use the equations for Doppler shifts to find the Doppler frequencies of the whistle. (a) For the approaching whistle, (b) For the receding whistle, fs v w vs vw fd fd m 330 s m m 30.0 330 s s fs v w vs fd fd 224 Hz vw m 330 s m m 30.0 330 s s 224 Hz 224 Hz 330 m s 300 m s 246.4 Hz 246 Hz 224 Hz 330 m s 360 m s 205.3 Hz 205 Hz (c) The change in pitch is the difference in the two frequencies. Therefore, the pitch change is f 246.4 Hz 205.3 Hz 41.1 Hz Paraphrase and Verify As the train passes, the pitch of its whistle is lowered by a frequency of about 41.1 Hz. 432 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 433 e WEB To learn more about possible health effects of sonic booms at close range, and the recent concerns of the Innu Nation, follow the links at www.pearsoned.ca/ school/physicssource. The Sound Barrier Jet planes are not allowed to break or exceed the sound barrier in the airspace over most cities. When an object travels at speeds at, or greater than, the speed of sound, it creates a sonic boom. The boom is the result of the shock wave created by the motion of the object. Bow Waves A boat moving through water produces a bow wave. The crest of the wave moves sideways away from the object, producing the wave’s characteristic V-shape. For an airplane moving through the fluid medium of the atmosphere, a V-shaped bow wave, or pressure wave, travels outward at the speed of sound (Figure 8.48). If the speed of the airplane is less than the speed of sound, the bow wave produced at any instant lags behind the bow wave produced just an instant earlier. The bow wave carries energy away from the plane in a continuous stream (Figure 8.49(a)). Sonic Boom However, for an airplane travelling at the speed of sound, the bow wave and the airplane travel at the same speed. Instant by instant, crests of the bow wave are produced at the same location as the crest of the bow wave produced by the plane an instant earlier (Figure 8.49). The energy stored in the bow wave becomes very intense. To the ear of an observer, crests of successively produced bow waves arrive simultaneously in what is known as a sonic boom. In early attempts to surpass the speed of sound, many airplanes were damaged. At the speed of sound, there is a marked increase in drag and turbulence. This effect damaged planes not designed to withstand it. A reporter assumed the increased drag acted like a barrier to travelling faster than sound and coined the term sound barrier. Mathematically, from the arguments presented above, the Doppler wavelength is given by λ d vs) (vw . f s Figure 8.48 When conditions are right, the change in pressure produced by the airplane’s wings can cause sufficient cooling of the atmosphere so that a cloud forms. The extreme conditions present when a jet is travelling near the speed of sound often result in the type of cloud seen in this photo. (a) Slower than speed of sound: Pressure waves move out around plane. (b) At speed of sound: Pressure waves at nose form a shock wave. (c) At supersonic speed: Shock waves form a cone, resulting in a sonic boom. Figure 8.49 As an airplane accelerates from subsonic to supersonic speeds, the changing relationship between the plane and the bow waves or pressure waves results in a sonic boom. Chapter 8 Mechanical waves transmit energy in a variety of ways. 433 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 434 vw, which If a plane is travelling at the speed of sound, then vs means that for any sound produced by the jet, the Doppler wavelength in the direction of the jet’s motion is zero. Even if the plane’s speed is greater than that of sound, the bow waves still combine to form a shock wave. In this way a sonic boom can be heard for any object, such as a rifle bullet, that has a supersonic speed. THEN, NOW, AND FUTURE Ultrasound While impressive given the technology available at the time, the results of early attempts at using ultrasound in medicine were of poor quality. Initially, ultrasound images from within a body were very blurry and two-dimensional. By today’s standards, the technology was extremely crude and there was virtually no scientific understanding of how the sound would behave when it encountered different types of tissue. Today, computers have made it possible to form three-dimensional images that can be rotated so that you can see all sides. Doppler ultrasound is used to detect blood flow through an organ. Today, 4-D ultrasound (time is the 4th dimension) is a real-time 3-D image that moves. 1. What are the advantages and disadvantages of ultrasound imaging compared with other imaging techniques such as CT scans and MRI? Figure 8.50 A 3-D ultrasound picture of a d |
eveloping fetus 8.4 Check and Reflect 8.4 Check and Reflect Knowledge 1. What causes the Doppler effect? 2. Two sound sources have the same frequency when at rest. If they are both moving away from you, how could you tell if one was travelling faster than the other? 3. Explain the cause of a sonic boom. Applications 4. The siren of a police car has a frequency of 660 Hz. If the car is travelling toward you at 40.0 m/s, what do you perceive to be the frequency of the siren? The speed of sound in air is 340 m/s. 5. A police car siren has a frequency of 850 Hz. If you hear this siren to have a frequency that is 40.0 Hz greater than its true frequency, what was the speed of the car? The speed of sound is 350 m/s. 6. A jet, travelling at the speed of sound (Mach 1), emits a sound wave with a frequency of 1000 Hz. Use the Doppler effect equations to calculate the frequency of this sound as the jet first approaches you, then moves away from you. Explain what these answers mean in terms of what you would hear as the jet moved toward, then past, you. Extensions 7. Astronomers have shown that the colour of light from distant stars is shifted from the blue end toward the red end of the spectrum. This is known as red shift. Astronomers realized that since light energy is transmitted as a wave, the red shift was the result of the Doppler effect applied to light. What does the red shift indicate about the motions of the star, which emits light that we see as a red shift? Hint: Investigate the relationship between the frequency and colour for light. e TEST To check your understanding of the Doppler effect, follow the eTest at www.pearsoned.ca/school/ physicssource. 434 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 435 CHAPTER 8 SUMMARY Key Terms and Concepts medium wave equilibrium position crest trough amplitude wavelength wave front incident wave reflected wave wave train point source ray pulse interference principle of superposition constructive interference destructive interference node antinode standing wave resonance closed pipe fundamental frequency overtone open pipe interference pattern maximum minimum phase shift Key Equations λ vT v f λ fd v fs w vw vs Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to produce a full summary of the chapter. wave trains travel as Waves which have which can be transmit from moving sources produce the or stored in the longitudinal amplitude wavelength The source of waves can move either or away from the observer which causes the apparent frequency wavelength frequency to decrease increase of the or troughs strings open pipes which can overlap to which is explained using the This results in interference patterns known as or principle of superposition to describe resulting in caused by and destructive interference with producing in antinodes and Chapter 8 Mechanical waves transmit energy in a variety of ways. 435 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 436 CHAPTER 8 REVIEW Knowledge 1. (8.1) (a) When a wave moves on water, what is the nature of the motion of the water within the wave? (b) What is the relationship between the direction of an incident wave and a reflected wave? (c) If you were able see the sound waves emitted when a tuning fork was struck, what would you record as your observation? 2. (8.2) (a) What affects the speed of a water wave? (b) What is the nature of the motion of the medium when a longitudinal wave moves through it? (c) Describe how the speed of a wave affects its wavelength and its amplitude. (d) Explain why waves are considered a form of Simple Harmonic Motion. (e) If speed is constant, how does wavelength vary with frequency? 3. (8.3) (a) Describe the conditions required to produce constructive and destructive interference in waves. (b) Describe how the principle of superposition applies to what happens when two pulses of identical length and amplitude interfere to produce no apparent pulse. (c) Define node, antinode, and standing wave. (d) In terms of the wavelength of the waves that have combined to form a standing wave, describe the position of the nodes and antinodes as you move away from the fixed end of a spring. (e) Why can a standing wave be generated only by what is defined as resonant frequency? 4. (8.4) (a) Does the Doppler effect apply only to sound or can it apply to any form of wave motion? Explain. (b) How are the waves in the direction of a source’s motion affected as the speed of the source increases? Applications 5. The speed of a wave in a spring is 15.0 m/s. If the length of a pulse moving in the spring is 2.00 m, how long did it take to generate the pulse? Why don’t we talk about the frequency for a pulse? 436 Unit IV Oscillatory Motion and Mechanical Waves 6. Waves are generated by a straight wave generator. The waves move toward and reflect from a straight barrier. The angle between the wave front and the barrier is 30°. Draw a diagram that shows what you would observe if this occurred in a ripple tank. Use a line drawn across the middle of a blank sheet of paper to represent the barrier. Draw a series of wave fronts about 1 cm apart intersecting the barrier at 30°. Use a protractor to make sure the angle is correct. Now draw the reflected waves. Draw a ray to indicate the motion of the incident wave front and continue this ray to indicate the motion of the reflected wave front. Hint: Draw the reflected ray as if it had a new source. 7. A ripple tank is set up so that the water in it is 0.7 cm deep. In half of the tank, a glass plate is placed on the bottom to make the water in that half shallower. The glass plate is 0.5 cm thick. Thus, the tank has a deep section (0.7 cm) and a shallow section (0.2 cm). In the deep section the wave velocity is 15.0 cm/s while in the shallow section the velocity is 10.0 cm/s. Straight waves, parallel to the edge of the glass, move toward the line between the deep and shallow sections. If the waves have a frequency of 12.0 Hz, what changes in wavelength would you observe as they enter the shallow section? What would happen to the direction of the motion? 8. A ripple tank is set up as described in question 7. For this ripple tank you measure the speed of the waves to be 12.0 cm/s and 9.0 cm/s in the deep and shallow sections, respectively. If waves in the deep section that are 11.5 cm long cross over to the shallow section, what would be the wavelength in the shallow section? 9. The term ultrasound means the frequency is higher than those that our ears can detect (about 20 kHz). Animals can often hear sounds that, to our ears, are ultrasound. For example, a dog whistle has a frequency of 22 kHz. If the speed of sound in air is 350 m/s, what is the wavelength of the sound generated by this whistle? 10. A spring is stretched to a length of 7.0 m. A frequency of 2.0 Hz generates a standing wave in the spring that has six nodes. (a) Sketch the standing wave pattern for the spring. (b) Calculate the velocity of the wave. 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 437 11. The figure shows two waves that occupy the same point in space. Copy the sketch onto a sheet of paper using the dimensions indicated. Draw the wave that results from the interference of these two waves. 20. When a police car is at rest, the wavelength of the sound from its siren is 0.550 m. If the car is moving toward you at a speed of 120 km/h, what is the frequency at which you hear the siren? Assume that the speed of sound is 345 m/s. 3.5 cm 5.0 cm A 1.5 cm A A 10 cm A A 12. If a frequency of 1.5 Hz generates a standing wave in a spring that has three antinodes, (a) what frequency generates a standing wave with five antinodes in the same spring, and (b) what is the fundamental frequency for this spring? 13. A violin string is 33.0 cm long. The thinnest string on the violin is tuned to vibrate at a frequency of 659 Hz. (a) What is the wave velocity in the string? (b) If you place your finger on the string so that its length is shortened to 28.0 cm, what is the frequency of the note that the string produces? 14. (a) What is the shortest closed pipe for which resonance is heard when a tuning fork with a frequency of 426 Hz is held at the open end of the pipe? The speed of sound in air is 335 m/s. (b) What is the length of the next longest pipe that produces resonance? 21. If the speed of sound in air is 350 m/s, how fast must a sound source move toward you if the frequency that you hear is twice the true frequency of the sound? What frequency would you hear if this sound source had been moving away from you? Extensions 22. Describe an arrangement that you might use if you wanted to create an interference pattern similar to the one in Figure 8.44 on page 426 by using sound waves that have a frequency of 512 Hz. Except for standing waves in strings or pipes, why do you think that we do not often find interference patterns in nature? 23. Explain why the number of maxima and minima in the interference pattern generated by two inphase point sources depends on the ratio of the distance between the sources to the wavelength. e TEST To check your understanding of waves and wave motion, follow the eTest links at www.pearsoned.ca/ school/physicssource. Consolidate Your Understanding 15. Draw the interference pattern generated by two inphase point sources that are four wavelengths apart. Answer each of the following questions in your own words. Provide examples to illustrate your explanation. 16. In the interference pattern for two in-phase point sources, a point on a second order maximum is 2.8 cm farther from one source than the other. What is the wavelength generated by these sources? 17. The horn on a car has a frequency of 290 Hz. If the speed of sound in air is 340 m/s and the car is moving toward you at a speed of 72.0 km/h, what is the apparent frequency of the sound? 18. How fast is a sound source moving towa |
rd you if you hear the frequency to be 580 Hz when the true frequency is 540 Hz? The speed of sound in air is 350 m/s. Express your answer in km/h. 19. If the speed of sound in air is 350 m/s, how fast would a sound source need to travel away from you if the frequency that you hear is to be onehalf the true frequency? What would you hear if this sound source had been moving toward you? 1. What are the advantages and disadvantages of using a spring as a model for wave motion? 2. What are the conditions for which a standing wave pattern is generated? Why are standing waves not often seen in nature? 3. Explain how the energy in a wave is transmitted from one place to another. 4. Describe what is meant by the principle of superposition. How does this principle explain standing waves? 5. What is meant by resonance? Think About It Review your answers to the Think About It questions on page 393. How would you answer each question now? Chapter 8 Mechanical waves transmit energy in a variety of ways. 437 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 438 UNIT IV PROJECT Earthquakes Scenario The tsunami that swept coastal regions of the Indian Ocean on December 26, 2004, was set off by an earthquake centred off the coast of the island of Sumatra in Indonesia. Seismographs around the world identified the location and strength of the earthquake. It was determined that the earthquake rated about 9.0–9.3 on the Richter scale. You have been asked by your government to make a presentation on the seismology of earthquakes. Your challenge is threefold. • First: you are to explain the nature of earthquake shock waves, their movement through Earth, and how the location of the earthquake epicentre is identified by seismographs around the world. • Second: you are to explain how the intensity of earthquakes is measured. This means that you must explain what the Richter scale is, and how it is used to rate earthquake intensity. • Third: you are to demonstrate the operation of a seismograph. Planning Your team should consist of three to five members. Choose a team manager and a record keeper. Assign other tasks as they arise. The first task is to decide the structure of your presentation and the research questions you will need to investigate. Questions you will need to consider are: How will you present the information to your audience? What are the resources at your disposal? Do you have access to computers and presentation programs such as PowerPoint®? Which team members will design, build, and demonstrate the model seismograph? Brainstorm strategies for research and create a schedule for meeting the deadlines for all phases of the project. Where is your team going to look for the information necessary to complete the project? What types of graphics will be most effective to assist your presentation? How will you best demonstrate the function of your seismograph? Your final report should include written, graphic, and photographic analyses of your presentation. Assessing Results Assess the success of your project based on a rubric* designed in class that considers: research strategies thoroughness of the experimental design effectiveness of the experimental technique effectiveness of the team’s public presentation Materials • materials, as needed, for the construction of your model seismograph Procedure 1 Research the nature of the shock waves set off by an earthquake. Be alert to Internet sites that may contain unreliable or inaccurate information. Make sure that you evaluate the reliability of the sources of information that you use for your research. If you gather information from the Internet, make sure you identify who sponsors the site and decide whether or not it is a reputable source of information. Maintain a list of your references and include it as an appendix to your report. Use graphics to explain how the shock waves move through Earth, and how seismologists locate the epicentre of an earthquake. 2 Research the history of the Richter scale and its use in identifying the intensity of an earthquake. 3 Design and build your model of a seismograph. Decide how you will demonstrate its use in your presentation. 4 Prepare an audio-visual presentation that would inform your audience on the nature of earthquakes and how they are detected. Thinking Further Write a short appendix (three or four paragraphs) to your report to suggest steps that governments might take to make buildings safer in earthquake zones. Answer questions such as: What types of structures are least susceptible to damage by earthquakes? What types are most susceptible? *Note: Your instructor will assess the project using a similar assessment rubric. 438 Unit IV Oscillatory Motion and Mechanical Waves 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 439 UNIT IV SUMMARY Unit Concepts and Skills: Quick Reference Concepts Chapter 7 Period Frequency Spring constant Summary Resources and Skill Building Oscillatory motion requires a set of conditions. 7.1 Period and Frequency Period is the time for one complete cycle, measured in seconds (s). If the period of each cycle remains constant, the object is moving with oscillatory motion. Frequency is the number of cycles per second, measured in Hertz (Hz). 7.2 Simple Harmonic Motion The spring constant is the amount of force needed to stretch or compress the spring 1 m and is measured in N/m. It can also be thought of as the stiffness of a spring. QuickLab 7-1; Inquiry Lab 7-2; Minds On; Figures 7.4, 7.5 QuickLab 7-1; Inquiry Lab 7-2; Example 7.1; Figure 7.5 QuickLab 7-3; Examples 7.2–7.4 Hooke’s law Hooke’s law states that the deformation of an object is proportional to the force causing it. Figures 7.9–7.16 Simple harmonic motion SHM refers to anything that moves with uniform oscillatory motion and conforms to Hooke’s law. Figures 7.19–7.23; eSIM Pendulum motion The pendulum is a simple harmonic oscillator for angles less than 15°. Figures 7.25–7.27; Example 7.5; Inquiry Lab 7-4; Table 7.5 Figures 7.44, 7.45; Then, Now, and Future QuickLab 8-1; Inquiry Lab 8-2; Inquiry Lab 8-3 Acceleration of a mass-spring system Relationship between acceleration and velocity of a mass-spring system Period of a mass-spring system Period of a pendulum Resonance Forced frequency Resonance effects on buildings and bridges 7.3 Position, Velocity, Acceleration, and Time Relationships The acceleration of a mass-spring system depends on displacement, mass, and the spring constant, and it varies throughout the motion of the mass-spring system. Figure 7.28 The acceleration and velocity of a mass-spring system are continually changing. The velocity of a mass-spring system is determined by its displacement, spring constant, and mass. Figures 7.29–7.33; Example 7.6 The period of a mass-spring oscillator is determined by its mass and spring constant, but not its amplitude. Figures 7.35–7.37; Example 7.7 A pendulum’s period is determined by its length and the gravitational field strength, but not the mass of the bob. Figures 7.39, 7.40; eSIM; Example 7.8 7.4 Applications of Simple Harmonic Motion Resonance is the natural frequency of vibration of an object. Figure 7.41; QuickLab 7-5 Forced frequency is the frequency at which an external force is applied to an oscillating object. Figure 7.41; QuickLab 7-5 Bridges and buildings can resonate due to the force of the wind. Chapter 8 Mechanical waves transmit energy in a variety of ways. Wave properties may be qualitative or quantitative. 8.1 The Properties of Waves Waves have many properties that can be used to analyze the nature of the wave and the way it behaves as it moves through a medium. Some of these properties are qualitative (crest, trough, wave front, medium, incident wave, reflected wave, wave train) while others are quantitative (amplitude, wavelength, frequency, wave velocity). Universal wave equation 8.2 Transverse and Longitudinal Waves Waves can move through a medium either as transverse or longitudinal waves. The relationship among the frequency, wavelength, and wave velocity is given by the universal wave equation. Inquiry Lab 8-4; Inquiry Lab 8-5; Example 8.1; Example 8.2 Interference patterns may result when more than one wave moves through a medium. 8.3 Superposition and Interference When two or more waves travel in different directions through the same point in space, their amplitudes combine according to the principle of superposition. Depending on the properties of the waves, they may form an interference pattern. Interference patterns can often be used to determine the properties of the waves from which they are formed. Inquiry Lab 8-6; Example 8.3; Inquiry Lab 8-7; Design a Lab 8-8 Doppler effect Sonic boom 8.4 The Doppler Effect When a sound source moves either toward or away from a sensor (ear or microphone), the frequency of the sound that is detected will be different from the frequency emitted by the source. When an object is travelling at the speed of sound, it creates a shock wave known as a sonic boom. Example 8.4 Figure 8.49 Unit IV Oscillatory Motion and Mechanical Waves 439 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 440 UNIT IV REVIEW Vocabulary 1. Use your own words to define the following terms, concepts, principles, or laws. Give examples where appropriate. amplitude antinodes closed-pipe air column constructive interference crest destructive interference diverging Doppler effect equilibrium forced frequency frequency fundamental frequency Hooke’s law incident wave in phase interference longitudinal wave maximum mechanical resonance medium minimum nodes or nodal points open-pipe air column oscillation oscillatory motion overtone period phase shift principle of superposition pulse ray reflected wave resonance resonant frequency restoring force shock wave simple harmonic motion simple harmonic oscillator sonic boom sound barrier spring constant standing waves transverse wave trough two-point-source interference pattern 440 Unit IV Oscillatory Motion and Mechanical Waves w |
ave wave front wave train wave velocity wavelength Knowledge CHAPTER 7 2. How are the units of frequency and period similar? How are they different? 3. The SI unit for frequency is Hz. What are two other accepted units? 4. For any simple harmonic oscillator, in what position is (a) the velocity zero? (b) the restoring force the greatest? 5. Why doesn’t a pendulum act like a simple harmonic oscillator for large amplitudes? 6. The equation for Hooke’s law uses a negative sign (F kx). Why is this sign necessary? 7. Aboriginal bows used for hunting were made from wood. Assuming the wood deforms according to Hooke’s law, explain how you would go about measuring the spring constant of the wood. 8. Suppose the same pendulum was tested in both Calgary and Jasper. In which location would you expect the pendulum to oscillate more slowly? Explain. 9. Explain why the sound from one tuning fork can make a second tuning fork hum. What conditions must be necessary for this to happen? 10. A pendulum in a clock oscillates with a resonant frequency that depends on several factors. From the list below, indicate what effect (if any) the following variables have on the pendulum’s resonant frequency. (a) length of pendulum arm (b) latitude of clock’s position (c) longitude of clock’s position (d) elevation (e) restoring force 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 441 CHAPTER 8 11. The diagram below shows waves in two springs. For each of the springs, how many wavelengths are shown? 12. Sound waves, travelling through air, are reflected from the wall of a building. Describe how the reflection affects the speed, (a) the wavelength, (b) (c) the amplitude, and (d) the direction of a wave train. 13. Points of zero displacement on a transverse wave have the greatest kinetic energy. Which points on a longitudinal wave have the greatest kinetic energy? 14. How is the shape of a circular wave front changed when it reflects from a straight barrier? 15. What aspect of a pulse determines the amount of energy it transfers? 16. When water waves enter a region where they travel slower, what happens to the (a) frequency, (b) wavelength, and (c) direction of the waves? 17. In the interference pattern from two in-phase point sources, what name is given to a line along which destructive interference occurs? 18. What determines the speed at which a wave travels through a spring? 19. What causes a standing wave in a spring? 20. Draw a transverse wave train that consists of two wavelengths. On your diagram, label the equilibrium position for the medium, a crest, a trough, the amplitude, a wavelength, and the direction of the wave velocity. Along the wavelength that you identified above, draw several vector arrows to indicate the direction of the motion of the medium. 21. Why does moving your finger along the string of a violin alter the note that it produces? 22. What property of the sound produced by a tuning fork is affected by striking the tuning fork with different forces? What does that tell you about the relationship between the properties of the sound and the sound wave created by striking the tuning fork? 23. When two in-phase point sources generate an interference pattern, what conditions are required to create (a) the central maximum and (b) a second order maximum? 24. In terms of the length of an open pipe, what is the longest wavelength for which resonance can occur? 25. You are walking north along a street when a police car with its siren on comes down a side street (travelling east) and turns northward on the street in front of you. Describe what you would hear, in terms of frequency of the sound of its siren, before and after the police car turns. 26. What is the relationship between frequency, wavelength, and wave velocity? 27. Why does the frequency of a sound source that is moving toward you seem to be higher than it would be if the source were at rest? Applications 28. Determine the force necessary to stretch a spring (k 2.55 N/m) to a distance of 1.20 m. 29. A musician plucks a guitar string. The string has a frequency of 400.0 Hz and a spring constant of 5.0 104 N/m. What is the mass of the string? 30. When a pendulum is displaced 90.0° from the vertical, what proportion of the force of gravity is the restoring force? 31. While performing a demonstration to determine the spring constant of an elastic band, a student pulls an elastic band to different displacements and measures the applied force. The observations were recorded in the table below. Plot the graph of this data. Can the spring constant be determined? Why or why not? Displacement (m) 0.1 0.2 0.3 0.4 0.5 0.6 Force (N) 0.38 1.52 3.42 6.08 9.5 13.68 32. A force of 40.0 N is required to move a 10.0-kg horizontal mass-spring system through a displacement of 80.0 cm. Determine the acceleration of the mass when its displacement is 25.0 cm. Unit IV Oscillatory Motion and Mechanical Waves 441 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 442 33. Use the following table to determine the spring constant of a spring. Displacement (cm) Force (mN) 2.5 5.0 7.5 10.0 12.5 10.0 21.0 31.0 39.0 49.0 34. A 50.0-g mass oscillates on the end of a vertical mass-spring system (k 25.0 N/m) with a maximum acceleration of 50.0 m/s2. (a) What is its amplitude of vibration? (b) What is the maximum velocity of the mass? 35. A bee’s wing has a mass of 1.0 105 kg and makes one complete oscillation in 4.5 103 s. What is the maximum wing speed if the amplitude of its motion is 1.10 cm? 36. A skyscraper begins resonating in a strong wind. A tuned mass damper (m 10.0 t) at the top of the building moves through a maximum displacement of 1.50 m in the opposite direction to dampen the oscillations. If the mass damper is attached to a horizontal spring and has a maximum speed of 1.40 m/s, what is the period of its oscillations? 37. A branch at the top of a tree sways with simple harmonic motion. The amplitude of motion is 0.80 m and its speed is 1.5 m/s in the equilibrium position. What is the speed of the branch at the displacement of 0.60 m? 38. A tuned mass damper at the top of a skyscraper is a mass suspended from a thick cable. If the building sways with a frequency of 0.125 Hz, what length must the cable supporting the weight be to create a resonance in the damper? 39. When a wave slows down, what property of the wave is not affected? What effect does this have on the other properties of the wave? Explain. 40. Explain how a wave can transmit energy through a medium without actually transmitting any matter. 41. A light wave is transmitted through space at 3.00 108 m/s. If visible light has wavelengths ranging from about 4.30 107 m to 7.50 107 m long, what range of frequencies are we able to see? 42. Radio waves travel at the speed of light waves (3.00 108 m/s). If your radio is tuned to a station broadcasting at 1250 kHz, what is the length of the waves arriving at the radio antenna? 442 Unit IV Oscillatory Motion and Mechanical Waves 43. A pendulum oscillates with a period of 0.350 s. Attached to the pendulum is a pen that marks a strip of paper on the table below the pendulum as it oscillates. When the strip of paper is pulled sideways at a steady speed, the pen draws a sine curve on the paper. What will be the wavelength of the sine curve if the speed of the paper is 0.840 m/s? 44. A submarine sends out a sonar wave that has a frequency of 545 Hz. If the wavelength of the sound is 2.60 m, how long does it take for the echo to return when the sound is reflected from a submarine that is 5.50 km away? 45. A wire is stretched between two points that are 3.00 m apart. A generator oscillating at 480 Hz sets up a standing wave in the wire that consists of 24 antinodes. What is the velocity at which waves move in this wire? 46. A spring is stretched to a length of 5.40 m. At that length the speed of waves in the spring is 3.00 m/s. If a standing wave with a frequency 2.50 Hz (a) were generated in this spring, how many nodes and antinodes would there be along the spring? (b) What is the next lower frequency for which a standing wave pattern could exist in this spring? 47. The second string on a violin is tuned to the note D with a frequency of 293 Hz. This is the fundamental frequency for the open string, which is 33.0 cm long. (a) What is the speed of the waves in the string? If you press on the string with your finger so (b) that the oscillating portion of the string is 2/3 the length of the open string, what is the frequency of the note that is created? 48. An audio frequency generator set at 154 Hz is used to generate a standing wave in a closed-pipe resonator, where the speed of sound is 340 m/s. (a) What is the shortest air column for which resonance is heard? (b) What is the next longer column length for which resonance is heard? 49. A submarine’s sonar emits a sound with a frequency of 875 Hz. The speed of sound in seawater is about 1500 m/s. If you measure the frequency of the sound to be 870 Hz, what is the velocity of the submarine? 50. A police car is travelling at a speed of 144 km/h. It has a siren with a frequency of 1120 Hz. Assume that the speed of sound in air is 320 m/s. (a) If the car is moving toward you, what frequency will you hear for the siren? If the car had been moving away from you at the same speed, what frequency would you have heard? (b) 08-PearsonPhys20-Chap08 7/24/08 2:23 PM Page 443 Extensions 51. What generalization can be made about the frequency of vibration with regard to the mass for a mass-spring system? (Assume all other qualities remain constant.) 52. An alien crash-lands its spaceship on a planet in our solar system. Unfortunately, it is unable to tell which planet it is. From the wreckage of the spaceship the alien constructs a 1.0-m-long pendulum from a piece of wire with four metal nuts on the end. If this pendulum swings with a period of 3.27 s, on which planet did the alien land: Mercury, Venus, or Earth? 53. Use a compass to draw a simulation of the wave pat |
tern generated by two in-phase point source generators that are 3.5 wavelengths apart. Near the middle of the page, place two points (S1 and S2) 3.5 cm apart to represent the positions of the sources. Draw wavelengths 1.0 cm long by drawing concentric circles that increase in radii by 1.0-cm increments. Locate on the diagram all the maxima and minima that are generated by this set-up and draw lines to indicate their positions. How does this pattern differ from the one in Figure 8.44 on page 426? Explain why these differences occur. 59. Outline a procedure that you could use to determine the mass of a horizontal mass-spring system without measuring the mass on a scale. 60. A student wants to determine the mass of the bob on a pendulum but only has access to a stopwatch and a ruler. She decides to pull the pendulum bob back through a displacement of 10° and time 20 complete oscillations. Will it be possible to determine the mass from the data gathered? Explain. 61. Construct a concept map for the simple harmonic motion of a pendulum. Include the following terms: period, displacement, restoring force, velocity, length, and gravitational field strength. 62. In a paragraph, explain why Huygens’s pendulum clock was a revolution in clock making and what the limitations were in its design. Be sure to use these terms: pendulum length, resonant frequency, forced frequency, and gravitational field strength. 63. Research the term “red shift” as used in astronomy. Prepare a report on the importance of red shift to our understanding of the nature of the universe. 54. In a stereo system, there are two speakers set at some distance apart. Why does a stereo system not result in an interference pattern? 64. Describe how to use springs to explore what happens to pulses transmitted from one medium to another in which the wave speed is different. 55. If a sound source is at rest, the frequency you hear and the actual frequency are equal. Their 1). If the sound source ratio equals one (fd/fs moves toward you at an ever-increasing speed, this frequency ratio also increases. Plot a graph for the ratio of the frequencies vs. the speed of the sound source as the speed of the source increases from zero to Mach 1. What is the value of the ratio when the speed of the source is Mach 1? Skills Practice 56. Use a graphing calculator or another suitable means to plot a graph of period against frequency. What type of relationship is this? 57. Outline an experimental procedure that you could perform to determine the spring constant of a vertical mass-spring system. 58. Sketch a diagram of a horizontal mass-spring system in three positions: at both extremes of its motion, and in its equilibrium position. In each diagram, draw vector arrows representing the restoring force, velocity, and acceleration. State whether these are at a maximum or a minimum. 65. Explain to someone who has not studied physics the differences in the ways objects and waves transport energy between points on Earth. Self-assessment 66. Identify a concept or issue that you studied in this unit and would like to learn more about. 67. Learning often requires that we change the way we think about things. Which concept in this unit required the greatest change in your thinking about it? Explain how your thinking changed. 68. Which of the concepts in this unit was most helpful in explaining to you how objects interact? e TEST To check your understanding of oscillatory motion and waves, follow the eTest links at www.pearsoned.ca/school/physicssource. Unit IV Oscillatory Motion and Mechanical Waves 443 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 444 U N I T V Momentum Momentum and Impulse and Impulse Many situations and activities in the real world, such as snowboarding, involve an object gaining speed and momentum as it moves. Sometimes two or more objects collide, such as a hockey stick hitting a puck across the ice. What physics principles apply to the motion of colliding objects? How does the combination of the net force during impact and the interaction time affect an object during a collision? 444 Unit V 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 445 Unit at a Glance C H A P T E R 9 The momentum of an isolated system of interacting objects is conserved. 9.1 Momentum Is Mass Times Velocity 9.2 Impulse Is Equivalent to a Change in Momentum 9.3 Collisions in One Dimension 9.4 Collisions in Two Dimensions Unit Themes and Emphases • Change and Systems • Science and Technology Focussing Questions In this study of momentum and impulse, you will investigate the motion of objects that interact, how the velocity of a system of objects is related before and after collision, and how safety devices incorporate the concepts of momentum and impulse. As you study this unit, consider these questions: • What characteristics of an object affect its momentum? • How are momentum and impulse related? Unit Project An Impulsive Water Balloon • By the time you complete this unit, you will have the skills to design a model of an amusement ride that is suitable for a diverse group of people. You will first need to consider acceptable accelerations that most people can tolerate. To test your model, you will drop a water balloon from a height of 2.4 m to see if it will remain intact. e WEB Research the physics concepts that apply to collisions in sports. How do athletes apply these concepts when trying to score goals for their team? How do they apply these concepts to minimize injury? Write a summary of your findings. Begin your search at www.pearsoned.ca/school/physicssource. Unit V Momentum and Impulse 445 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 446 C H A P T E R 9 Key Concepts In this chapter, you will learn about: impulse momentum Newton’s laws of motion elastic and inelastic collisions Learning Outcomes When you have completed this chapter, you will be able to: Knowledge define momentum as a vector quantity explain impulse and momentum using Newton’s laws of motion explain that momentum is conserved in an isolated system explain that momentum is conserved in one- and twodimensional interactions compare and contrast elastic and inelastic collisions Science, Technology, and Society explain that technological problems lend themselves to multiple solutions 446 Unit V The momentum of an isolated system of interacting objects is conserved. Most sports involve objects colliding during the play. Hockey checks, curling takeouts, football tackles, skeet shooting, lacrosse catches, and interceptions of the ball in soccer are examples of collisions in sports action. Players, such as Randy Ferbey, who are able to accurately predict the resulting motion of colliding objects have a better chance of helping their team win (Figure 9.1). When objects interact during a short period of time, they may experience very large forces. Evidence of these forces is the distortion in shape of an object at the moment of impact. In hockey, the boards become distorted for an instant when a player collides with them. Another evidence of these forces is a change in the motion of an object. If a goalie gloves a shot aimed at the net, you can see how the impact of the puck affects the motion of the goalie’s hand. In this chapter, you will examine how the net force on an object and the time interval during which the force acts affect the motion of the object. Designers of safety equipment for sports and vehicles use this type of analysis when developing new safety devices. In a system of objects, you will also investigate how their respective velocities change when the objects interact with each other. Figure 9.1 Sports such as curling involve applying physics principles to change the score. Randy Ferbey, originally from Edmonton, won the Brier (Canadian) Curling Championship six times, and the World Curling Championship four times. 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 447 9-1 QuickLab 9-1 QuickLab Predicting Angles After Collisions Problem How do the masses of two objects affect the angle between their paths after they collide off centre? Materials pennies and nickels with smooth, circular outer edges marking devices for the paths (paper, tape, pencil, ruler) stack of books protractor (optional) Procedure 1 Set up the books and paper as shown in Figure 9.2. Open the cover of the book at the top of the stack for backing. Tape the paper securely to the lab bench. 2 Position one penny at the bottom of the ramp. Mark its initial position by drawing an outline on the paper. 3 Place the incoming penny at the top of the ramp as shown in Figure 9.2. Mark its initial position. 4 Predict the path each coin will take after they collide off centre. Lightly mark the predicted paths. 5 Send the coin down the ramp and mark the position of each coin after collision. Observe the relative velocities of the coins to each other both before and after collision. initial position of penny tape stack of books for ramp paper taped in position identical penny at bottom of ramp tape Figure 9.2 Think About It before after direction of motion angle between two coins after collision Figure 9.3 6 Determine if the angle between the paths after collision is less than 90, 90, or greater than 90 (Figure 9.3). 7 Repeat steps 5 and 6, but have the incoming coin collide at a different contact point with the coin at the bottom of the ramp. 8 Repeat steps 2 to 7 using a penny as the incoming coin and a nickel at the bottom of the ramp. 9 Repeat steps 2 to 7 using a nickel as the incoming coin and a penny at the bottom of the ramp. Questions 1. What was the approximate angle formed by the paths of the two coins after collision when the coins were (a) the same mass? (b) of different mass? 2. Describe how the speeds of the two coins changed before and after collision. 3. How can you predict which coin will move faster after collision? 1. Under what circumstances could an object initially at rest be struck and move at a greater speed after collision than the incoming object |
? 2. Under what circumstances could a coin in 9-1 QuickLab rebound toward the ramp after collision? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 447 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 448 info BIT A tragic avalanche occurred during the New Year’s Eve party in the Inuit community of Kangiqsualujjuaq, formerly in Quebec and now part of Nunavut. At 1:30 a.m. on January 1, 1999, snow from the nearby 365-m mountain slope came cascading down, knocking out a wall and swamping those inside the gymnasium at the party. The snow on the mountain was initially about 1 m thick. After the avalanche was over, the school was covered with up to 3 m of snow. 9.1 Momentum Is Mass Times Velocity Snow avalanches sliding down mountains involve large masses in motion. They can be both spectacular and catastrophic (Figure 9.4). Unbalanced forces affect the motion of all objects. A mass of snow on the side of a mountain experiences many forces, such as wind, friction between the snow and the mountain, a normal force exerted by the mountain on the snow, and gravity acting vertically downward. Skiers and animals moving along the mountain slope also apply forces on the mass of snow. When a large mass of snow becomes dislodged and slides down a mountain slope due to gravity, it not only gains speed but also more mass as additional snow becomes dislodged along the downward path. info BIT Most avalanches occur on slopes that form an angle of 30 to 45 with the horizontal, although they can occur on any slope if the right conditions exist. In North America, a large avalanche may release about 230 000 m3 of snow. Figure 9.4 When the risk of an avalanche seems imminent, ski patrols reduce the mass of snow along a mountain slope by forcing an avalanche to take place. They do this by targeting large masses of snow with guns or explosives to dislodge the snow. 448 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 449 momentum: product of the mass of an object and its velocity Momentum Is a Vector Quantity All objects have mass. The momentum, p, of an object is defined as the product of the mass of the object and its velocity. Since momentum is the product of a scalar (mass) and a vector (velocity), momentum is a vector quantity that has the same direction as the velocity. p mv Momentum has units of kilogram-metres per second (kgm/s). When you compare the momenta of two objects, you need to consider both the mass and the velocity of each object (Figure 9.5). Although two identical bowling balls, A and B, have the same mass, they do not necessarily have the same momentum. If ball A is moving very slowly, it has a very small momentum. If ball B is moving much faster than ball A, ball B’s momentum will have a greater magnitude than ball A’s because of its greater speed. Figure 9.5 The bowling ball in both photos is the same. However, the bowling ball on the left has less momentum than the ball on the right. What evidence suggests this? In real life, almost no object in motion has constant momentum because its velocity is usually not constant. Friction opposes the motion of all objects and eventually slows them down. In most instances, it is more accurate to state the instantaneous momentum of an object if you can measure its instantaneous velocity and mass. Concept Check How would the momentum of an object change if (a) the mass is doubled but the velocity remains the same? (b) the velocity is reduced to 1 of its original magnitude? 3 (c) the direction of the velocity changes from [E] to [W]? e WEB Switzerland has a long history of studying avalanches. Find out what causes an avalanche. What physical variables do avalanche experts monitor? What models are scientists working on to better predict the likelihood and severity of avalanches? Begin your search at www.pearsoned.ca/school/ physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 449 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 450 Relating Momentum to Newton’s Second Law e WEB Research how momentum applies to cycling and other sports. Write a brief report of your findings. Begin your search at www.pearsoned.ca/school/ physicssource. The concept of momentum can be used to restate Newton’s second law. From Unit II, Newton’s second law states that an external non-zero net force acting on an object is equal to the product of the mass of the object ma. Acceleration is defined as the rate of and its acceleration, F net v change of velocity. For constant acceleration, a i. If you or t t i for a in Newton’s second law, you get substitute t v v v v f f F net f ma v v i m t mv mv i t f The quantity mv is momentum. So the equation can be written as F net p p i t f p where F net is constant t Written this way, Newton’s second law relates the net force acting on an object to its rate of change of momentum. It is interesting to note that Newton stated his second law of motion in terms of the rate of change of momentum. It may be worded as: An external non-zero net force acting on an object is equal to the rate of change of momentum of the object. F net p where F net is constant t e SIM For a given net force, learn how the mass of an object affects its momentum and its final velocity. Follow the eSim links at www.pearsoned.ca/school/ physicssource. This form of Newton’s law has some major advantages over the way it was ma only applies to situations where written in Unit II. The equation F net the mass is constant. However, by using the concept of momentum, it is possible to derive another form for Newton’s second law that applies to situations where the mass, the velocity, or both the mass and velocity are changing, such as an accelerating rocket where the mass is decreasing as fuel is being burned, while the velocity is increasing. In situations where the net force changes over a time interval, the average net force is equal to the rate of change of momentum of the object. F netave p t 450 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 451 In Example 9.1, a person in a bumper car is moving at constant velocity. Since both the person and the car move together as a unit, both objects form a system. The momentum of the system is equal to the total mass of the system times the velocity of the system. Example 9.1 W E v Figure 9.6 A 180-kg bumper car carrying a 70-kg driver has a constant velocity of 3.0 m/s [E]. Calculate the momentum of the cardriver system. Draw both the velocity vector and the momentum vector. Given mc 180 kg md 70 kg v 3.0 m/s [E] Required momentum of system ( p) velocity and momentum vector diagrams Analysis and Solution The driver and bumper car are a system because they move together as a unit. Find the total mass of the system. mT mc md 180 kg 70 kg 250 kg The momentum of the system is in the direction of the velocity of the system. So use the scalar form of p mv to find the magnitude of the momentum. p mTv (250 kg)(3.0 m/s) 7.5 102 kgm/s Draw the velocity vector to scale (Figure 9.7). W E 1.0 m/s v 3.0 m/s Figure 9.7 Practice Problems 1. A 65-kg girl is driving a 535-kg snowmobile at a constant velocity of 11.5 m/s [60.0 N of E]. (a) Calculate the momentum of the girl-snowmobile system. (b) Draw the momentum vector for this situation. 2. The combined mass of a bobsled and two riders is 390 kg. The sled-rider system has a constant momentum of 4.68 103 kgm/s [W]. Calculate the velocity of the sled. Answers 1. (a) 6.90 103 kgm/s [60.0 N of E] (b) N 3 kgm/s p 6.90 10 60.0° 2. 12.0 m/s [W] 2000 kgm/s E Chapter 9 The momentum of an isolated system of interacting objects is conserved. 451 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 452 Draw the momentum vector to scale (Figure 9.8). W E 100 kgm/s p 7.5 102 kgm/s Figure 9.8 Paraphrase The momentum of the car-driver system is 7.5 102 kgm/s [E]. Using Proportionalities to Solve Momentum Problems Example 9.2 demonstrates how to solve momentum problems using proportionalities. In this example, both the mass and velocity of an object change. Example 9.2 An object has a constant momentum of 2.45 102 kgm/s [N]. Determine the momentum of the object if its mass decreases to 1 of its original value 3 and an applied force causes the speed to increase by exactly four times. The direction of the velocity remains the same. Explain your reasoning. Practice Problems 1. Many modern rifles use bullets that have less mass and reach higher speeds than bullets for older rifles, resulting in increased accuracy over longer distances. The momentum of a bullet is initially 8.25 kgm/s [W]. What is the momentum if the speed of the bullet increases by a factor of 3 and 2 its mass decreases by a factor of 3 ? 4 2. During one part of the liftoff of a model rocket, its momentum increases by a factor of four while its mass is halved. The velocity of the rocket is initially 8.5 m/s [up]. What is the final velocity during that time interval? Answers 1. 9.28 kgm/s [W] 2. 68 m/s [up] 452 Unit V Momentum and Impulse Analysis and Solution From the equation p mv, p m and p v. Figure 9.9 represents the situation of the problem. before after N S v 4 v Figure 9.9 p 2.45 102 kgm/s [N] p ? p 1 m 3 and p 4v Calculate the factor change of p. 4 4 1 3 3 Calculate p. p 4 4 (2.45 102 kgm/s) 3 3 3.27 102 kgm/s The new momentum will be 3.27 102 kgm/s [N]. 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 453 9.1 Check and Reflect 9.1 Check and Reflect Knowledge 1. (a) Explain, in your own words, the concept of momentum. (b) State the SI units of momentum. 2. Explain why momentum is a vector quantity. 3. How is momentum related to Newton’s second law? 4. Explain why stating Newton’s second law in terms of momentum is more useful than stating it in terms of acceleration. 5. Explain, in your own |
words, the difference between momentum and inertia. 6. Provide three examples of situations in which (a) velocity is the dominant factor affecting the momentum of an object (b) mass is the dominant factor affecting the momentum of an object. 12. Draw a momentum vector diagram to represent a 425-g soccer ball flying at 18.6 m/s [214]. 13. At what velocity does a 0.046-kg golf ball leave the tee if the club has given the ball a momentum of 3.45 kgm/s [S]? 14. (a) A jet flies west at 190 m/s. What is the momentum of the jet if its total mass is 2250 kg? (b) What would be the momentum of the jet if the mass was 3 of its original 4 value and the speed increased to 6 of 5 its original value? 15. The blue whale is the largest mammal ever to inhabit Earth. Calculate the mass of a female blue whale if, when alarmed, it swims at a velocity of 57.0 km/h [E] and has a momentum of 2.15 106 kgm/s [E]. Applications Extensions 7. A Mexican jumping bean moves because an insect larva inside the shell wiggles. Would it increase the motion to have the mass of the insect greater or to have the mass of the shell greater? Explain. 8. What is the momentum of a 6.0-kg bowling ball with a velocity of 2.2 m/s [S]? 9. The momentum of a 75-g bullet is 9.00 kgm/s [N]. What is the velocity of the bullet? 10. (a) Draw a momentum vector diagram for a 4.6-kg Canada goose flying with a velocity of 8.5 m/s [210]. (b) A 10.0-kg bicycle and a 54.0-kg rider both have a velocity of 4.2 m/s [40.0 N of E]. Draw momentum vectors for each mass and for the bicycle-rider system. 11. A hockey puck has a momentum of 3.8 kgm/s [E]. If its speed is 24 m/s, what is the mass of the puck? 16. A loaded transport truck with a mass of 38 000 kg is travelling at 1.20 m/s [W]. What will be the velocity of a 1400-kg car if it has the same momentum? 17. Summarize the concepts and ideas associated with momentum using a graphic organizer of your choice. See Student References 4: Using Graphic Organizers on pp. 869–871 for examples of different graphic organizers. Make sure that the concepts and ideas are clearly presented and appropriately linked. e TEST To check your understanding of momentum, follow the eTest links at www.pearsoned.ca/school/ physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 453 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 454 info BIT Legendary stunt person Dar Robinson broke nine world records and made 21 “world firsts” during his career. One world record was a cable jump from the CN Tower in 1980 for the film The World’s Most Spectacular Stuntman. While tied to a 3-mm steel cable, Robinson jumped more than 366 m and stopped only a short distance above the ground. 9.2 Impulse Is Equivalent to a Change in Momentum Stunt people take the saying, “It isn’t the fall that hurts, it’s the sudden stop at the end,” very seriously. During the filming of a movie, when a stunt person jumps out of a building, the fall can be very dangerous. To minimize injury, stunt people avoid a sudden stop when landing by using different techniques to slow down more gradually out of sight of the cameras. These techniques involve reducing the peak force required to change their momentum. Sometimes stunt people jump and land on a net. Other times, they may roll when they land. For more extreme jumps, such as from the roof of a tall building, a huge oversized, but slightly under-inflated, air mattress may be used (Figure 9.10). A hidden parachute may even be used to slow the jumper to a safer speed before impact with the surface below. Despite all these precautions, injuries occur as stunt people push the limits of what is possible in their profession. Designers of safety equipment know that a cushioned surface can reduce the severity of an impact. Find out how the properties of a landing surface affect the shape of a putty ball that is dropped from a height of 1 m by doing 9-2 QuickLab. Figure 9.10 The thick mattress on the ground provides a protective cushion for the stunt person when he lands. Why do you think the hardness of a surface affects the extent of injury upon impact? 454 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 455 9-2 QuickLab 9-2 QuickLab Softening the Hit Problem How is the change in the shape of a putty ball upon impact related to the structure of the landing surface? Materials putty-type material closed cell foam or felt pad urethane foam pad or pillow waxed paper or plastic wrap metre-stick Procedure 1 Choose three surfaces of varying softness onto which to drop a putty ball. One of the surfaces should be either a lab bench or the floor. Cover each surface with some waxed paper or plastic wrap to protect it. 2 Knead or work the putty until you can form three pliable balls of equal size. 3 Measure a height of 1 m above the top of each surface. Then drop the balls, one for each surface (Figure 9.11). 4 Draw a side-view sketch of each ball after impact. 1 m 1 m Figure 9.11 Questions 1. Describe any differences in the shape of the putty balls after impact. 2. How does the amount of cushioning affect the deformation of the putty? 3. Discuss how the softness of the landing surface might be related to the time required for the putty ball to come to a stop. Justify your answer with an analysis involving the kinematics equations. Force and Time Affect Momentum In 9-2 QuickLab, you found that the softer the landing surface, the less the shape of the putty ball changed upon impact. The more cushioned the surface, the more the surface became indented when the putty ball collided with it. In other words, the softer and more cushioned landing surface provided a greater stopping distance for the putty ball. Suppose you label the speed of the putty ball at the instant it touches the landing surface vi, and the speed of the putty ball after the impact vf. 0. So the greater the For all the landing surfaces, vi was the same and vf stopping distance, the longer the time required for the putty ball to stop (Figure 9.12). In other words, the deformation of an object is less when the stopping time is increased. vf 0 for surfaces A and B viA viB tB tA viA dA viB dB harder landing surface (A) more cushioned landing surface (B) Figure 9.12 The stopping distance of the putty ball was greater for the more cushioned landing surface (B). So the time interval of interaction was greater on surface B. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 455 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 456 PHYSICS INSIGHT To visualize the effect of how Fnet and t can vary but p remains the same, consider the effect of changing two numbers being multiplied together to give the same product. 3 12 36 6 6 36 18 2 36 As the first number increases, the second number decreases in order to get the same product. Project LINK How will the net force and time interval affect the water balloon when it is brought to a stop? What types of protective material will you use to surround the water balloon and why? Apart from the different stopping times, what other differences were there between the drops that would have affected the shape of the putty ball upon impact? The answer to this question requires looking at Newton’s second law written in terms of momentum. From the previous section, the general form of Newton’s second law states that the rate of change of momentum is equal to the net force acting on an object. p t F net If you multiply both sides of the equation by t, you get F net t p For all the landing surfaces, since m, vi (at the first instant of impact), and vf (after the impact is over) of the putty ball were the same, pi was the t p, so the 0. So p was the same for all drops. But F same and pf net product of net force and stopping time was the same for all drops. From Figure 9.12 on page 455, the stopping time varied depending on the amount of cushioning provided by the landing surface. If the stopping time was short, as on a hard landing surface, the magnitude of the net force acting on the putty ball was large. Similarly, if the stopping time was long, as on a very cushioned landing surface, the magnitude of the net force acting on the putty ball was small. This analysis can be used to explain why the putty ball became more deformed when it landed on a hard surface. To minimize changes to the shape of an object being dropped, it is important to minimize Fnet required to stop the object, and this happens when you maximize t (Figure 9.13). It is also important to note where net acts on a large area, the result of the impact will have a F net acts. If F net acts on only one different effect on the shape of the object than if F small part on the surface of the object. (a) direction of motion (b) direction of motion concrete p Fnet t p Fnet t bed of straw Figure 9.13 Identical eggs are dropped from a height of 2 m onto a concrete floor or a pile of straw. Although p is the same in both situations, the magnitude of the net force acting on the egg determines whether or not the egg will break. Concept Check In 9-2 QuickLab, was the momentum of the putty ball at the first instant of impact zero? Explain your reasoning. 456 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 457 Impulse Is the Product of Net Force and Interaction Time impulse: product of the net force on an object and the time interval during an interaction. Impulse causes a change in the momentum of the object. e SIM Learn how the mass and acceleration of an object affect its change in momentum. Follow the eSim links at www.pearsoned.ca/school/ physicssource. t p, the product of net force and interaction time In the equation F net is called impulse. Impulse is equivalent to the change in momentum that an object experiences during an interaction. Every time a net force acts on an object, the object is provided with an impulse because the force is applied for a specific length of time. F net If you substitute the definitio |
n of momentum, p mv, the equation t p becomes t (mv) F net If m is constant, then the only quantity changing on the right-hand side of the equation is v. So the equation becomes F net t mv So impulse can be calculated using either equation: F net t p or F net t mv The unit of impulse is the newton-second (Ns). From Unit II, a newton is defined as 1 N 1 kgm/s2. If you substitute the definition of a newton in the unit newton-seconds, you get (s) 1 Ns 1 m kg 2 s m kg 1 s which are the units for momentum. So the units on both sides of the impulse equation are equivalent. Since force is a vector quantity, impulse is also a vector quantity, and the impulse is in the same direction as the net force. To better understand how net force and interaction time affect the change in momentum of an object, do 9-3 Design a Lab. 9-3 Design a Lab 9-3 Design a Lab Providing Impulse The Question What is the effect of varying either the net force or the interaction time on the momentum of an object? Design and Conduct Your Investigation State a hypothesis to answer the question using an “if/then” statement. Then design an experiment to measure the change in momentum of an object. First vary Fnet, then repeat the experiment and vary t instead. List the materials you will use, as well as a detailed procedure. Check the procedure with your teacher and then do the investigation. To find the net force, you may need to find the force of friction and add it, using vectors, to the applied force. The force of kinetic friction is the minimum force needed to keep an object moving at constant velocity once the object is in motion. Analyze your data and form conclusions. How well did your results agree with your hypothesis? Compare your results with those of other groups in your class. Account for any discrepancies. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 457 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 458 Example 9.3 demonstrates how, for the same impulse, varying the interaction time affects the average net force on a car during a front-end collision (the net force on the car is not constant). Example 9.3 W E Practice Problems 1. Two people push a car for 3.64 s with a combined net force of 200 N [S]. (a) Calculate the impulse provided to the car. (b) If the car has a mass of 1100 kg, what will be its change in velocity? 2. A dog team pulls a 400-kg sled that has begun to slide backward. In 4.20 s, the velocity of the sled changes from 0.200 m/s [backward] to 1.80 m/s [forward]. Calculate the average net force the dog team exerts on the sled. Answers 1. (a) 728 Ns [S], (b) 0.662 m/s [S] 2. 190 N [forward] v Figure 9.14 To improve the safety of motorists, modern cars are built so the front end crumples upon impact. A 1200-kg car is travelling at a constant velocity of 8.0 m/s [E] (Figure 9.14). It hits an immovable wall and comes to a complete stop in 0.25 s. (a) Calculate the impulse provided to the car. (b) What is the average net force exerted on the car? (c) For the same impulse, what would be the average net force exerted on the car if it had a rigid bumper and frame that stopped the car in 0.040 s? Given m 1200 kg (a) and (b) t 0.25 s (c) t 0.040 s v i 8.0 m/s [E] info BIT Some early cars were built with spring bumpers that tended to bounce off whatever they hit. These bumpers were used at a time when people generally travelled at much slower speeds. For safety reasons, cars today are built to crumple upon impact, not bounce. This results in a smaller change in momentum and a reduced average net force on motorists. The crushing also increases the time interval during the impulse, further decreasing the net force on motorists. Required (a) impulse provided to car (b) and (c) average net force on car ( F netave) Analysis and Solution When the car hits the wall, the final velocity of the car is zero. v f 0 m/s During each collision with the wall, the net force on the car is not constant, but the mass of the car remains constant. (a) Use the equation of impulse to calculate the impulse provided to the car. f F netave v i) t mv m(v (1200 kg)[0 (8.0 m/s)] (1200 kg)(8.0 m/s) 9.6 103 kgm/s impulse 9.6 103 Ns [W] 458 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 459 netave. For (b) and (c), substitute the impulse from part (a) and solve for F F netave t 9.6 103 Ns 03 Ns 9.6 1 t F netave (b) F netave 9.6 0 .25 3 Ns s 10 (c) F netave m kg 3.8 104 2 s 3.8 104 N 9.6 04 0. 3 Ns 1 0 0 s m kg 2.4 105 2 s 2.4 105 N PHYSICS INSIGHT netave is in the opposite F direction to the initial momentum of the car, because from Newton’s third law, the wall is exerting a force directed west on the car. F netave 3.8 104 N [W] F netave 2.4 105 N [W] Paraphrase and Verify (a) The impulse provided to the car is 9.6 103 Ns [W]. The average net force exerted by the wall on the car is (b) 3.8 104 N [W] when it crumples, and (c) 2.4 105 N [W] when it is rigid. The change in momentum is the same in parts (b) and (c), but the time intervals are different. So the average net force is different in both netave on the car with the rigid frame is situations. The magnitude of F more than 6 times greater than when the car crumples. Impulse Can Be Calculated Using a Net Force-Time Graph One way to calculate the impulse provided to an object is to graph the net force acting on the object as a function of the interaction time. Suppose a net force of magnitude 30 N acts on a model rocket for 0.60 s during liftoff (Figure 9.15). From the net force-time graph in Figure 9.16, the product t is equal to the magnitude of the impulse. But this product is also Fnet the area under the graph. Magnitude of Net Force vs. Interaction Time for a Model Rocket Fnet ) 50 40 30 20 10 0 0 0.10 0.20 0.30 Time t (s) 0.40 0.50 0.60 Figure 9.15 What forces act on the rocket during liftoff? Figure 9.16 Magnitude of net force as a function of interaction time for a model rocket. The area under the graph is equal to the magnitude of the impulse provided to the rocket. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 459 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 460 The magnitude of the impulse provided to the rocket is magnitude of impulse Fnet t (30 N)(0.60 s) 18 Ns In other words, the area under a net force-time graph gives the magnitude of the impulse. Note that a net force acting over a period of time causes a change in momentum. When Fnet is not constant, you can still calculate the impulse by finding the area under a net force-time graph. Figure 9.17 shows the magnitude of the net force exerted by a bow on an arrow during the first part of its release. The magnitude of the net force is greatest at the beginning and decreases linearly with time Magnitude of Net Force vs. Interaction Time for an Arrow Shot with a Bow 200 150 100 50 0 0 10 20 30 Time t (ms) 40 50 Figure 9.17 Magnitude of net force as a function of interaction time for an arrow shot with a bow. In this case, the area under the graph could be divided into a rectangle and a triangle or left as a trapezoid (Figure 9.18). So the magnitude of the impulse provided to the arrow is 1 (a b)(h) magnitude of impulse 2 1 (100 N 200 N)(0.050 s) 2 7.5 Ns Sometimes two net force-time graphs may appear different but the magnitude of the impulse is the same in both cases. Figure 9.19 (a) shows a graph where Fnet is much smaller than in Figure 9.19 (b). The value of t is different in each case, but the area under both graphs is the same. So the magnitude of the impulse is the same in both situations. (a 30 25 20 15 10 5 0 0 Magnitude of Net Force vs. Interaction Time (b) 2.00 4.00 6.00 Time t (s 30 25 20 15 10 5 0 0 Magnitude of Net Force vs. Interaction Time 2.00 4.00 6.00 Time t (s) Figure 9.19 What other graph could you draw that has the same magnitude of impulse? info BIT The area of a trapezoid is equal to 1 (a b)(h). 2 a h b Figure 9.18 460 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 461 Effect of a Non-linear Net Force on Momentum In real life, many interactions occur during very short time intervals (Figure 9.20). If you tried to accurately measure the net force, you would find that it is difficult, if not impossible, to do. In addition, the relationship between Fnet and t is usually non-linear, because Fnet increases from zero to a very large value in a short period of time (Figure 9.21). Magnitude of Net Force vs. Interaction Time Fnetave ) Time t (s) Figure 9.21 The average net force gives some idea of the maximum instantaneous net force that an object actually experienced during impact. Figure 9.20 When a baseball bat hits a ball, what evidence demonstrates that the force during the interaction is very large? What evidence demonstrates that the force on the ball changes at the instant the ball and bat separate? From a practical point of view, it is much easier to measure the interaction time and the overall change in momentum of an object during an interaction, rather than Fnet. In this case, the equation of Newton’s second law expressed in terms of momentum is F netave p t and the equation of impulse is F netave t p or F netave t mv In all the above equations, F netave on the object during the interaction. represents the average net force that acted In Example 9.4, a golf club strikes a golf ball and an approximation of the net force-time graph is used to simplify the calculations for impulse. In reality, the net force-time graph for such a situation would be similar to that shown in Figure 9.21. info BIT The fastest recorded speed for a golf ball hit by a golf club is 273 km/h. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 461 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 462 Example 9.4 A golfer hits a long drive sending a 45.9-g golf ball due east. Figure 9.22 shows an approximation of the net force as a function of t |
ime for the collision between the golf club and the ball. (a) What is the impulse provided to the ball? (b) What is the velocity of the ball Magnitude of Net Force vs. Interaction Time for a Golf Ball Being Hit by a Golf Club 6000 5000 4000 3000 2000 1000 ) .2 0.6 0.4 Time t (ms) 0.8 1.0 1.2 Figure 9.22 W E at the moment the golf club and ball separate? Given m 45.9 g ti 1.1 ms tf Fneti 5000 N Fnetf Fnetmax 0.1 ms 0 N 0 N Required (a) impulse provided to ball (b) velocity of ball after impact (v f ) Analysis and Solution The impulse and velocity after impact are in the east direction since the golfer hits the ball due east. (a) t tf ti 1.1 ms 0.1 ms 1.0 ms or 1.0 103 s Fnet Figure 9.23 magnitude of impulse area under net force-time graph 1 (t)(Fnetmax) 2 1 (1.0 103 s)(5000 N) 2 2.5 Ns impulse 2.5 Ns [E] (b) Impulse is numerically equal to mv or m(v v i). f i But v 0 m/s So, impulse m (v 2.5 Ns mv f 0) v f f Ns 2.5 m m kg s 2.50 2 s g (45.9 g) k 1 g 0 0 10 54 m/s Paraphrase (a) The impulse provided to the ball is 2.5 Ns [E]. (b) The velocity of the ball after impact is 54 m/s [E]. Practice Problems 1. (a) Draw a graph of net force as a function of time for a 0.650-kg basketball being shot. During the first 0.15 s, Fnet increases linearly from 0 N to 22 N. During the next 0.25 s, Fnet decreases linearly to 0 N. (b) Using the graph, calculate the magnitude of the impulse provided to the basketball. (c) What is the speed of the ball when it leaves the shooter’s hands? 2. (a) A soccer player heads the ball with an average net force of 21 N [W] for 0.12 s. Draw a graph of the average net force on the ball as a function of time. Assume that Fnetave is constant during the interaction. (b) Calculate the impulse provided to the soccer ball. (c) The impulse changes the velocity of the ball from 4.0 m/s [E] to 2.0 m/s [W]. What is the mass of the ball? Answers 1. (a 25 20 15 10 5 0 Magnitude of Net Force vs. Interaction Time for a Basketball Being Shot 0 0.10 0.20 0.30 Time t (s) 0.40 0.50 (b) 4.4 Ns, (c) 6.8 m/s 2. (a 30 25 20 15 10 5 0 Magnitude of Average Net Force vs. Interaction Time for a Soccer Ball Being Hit 0 0.03 0.06 Time t (s) 0.09 0.12 (b) 2.5 Ns [W], (c) 0.42 kg 462 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 463 The Design of Safety Devices Involves Varying Fnetave and t Many safety devices are based on varying both the average net force acting on an object and the interaction time for a given impulse. Suppose you attached a sled to a snowmobile with a rope hitch. As long as the sled is accelerating along a horizontal surface or is being pulled uphill, there is tension in the rope because the snowmobile applies a force on the sled (Figure 9.24). If the driver in Figure 9.24 (a) brakes suddenly to slow down, the momentum of the snowmobile changes suddenly. However, the sled continues to move in a straight line until friction eventually slows it down to a stop. In other words, the only way that the momentum of the f acts for a long enough period of time. sled changes noticeably is if F (a) (b FT 0 a FT 0 θ Figure 9.24 (a) A snowmobile accelerating along a horizontal surface, and (b) the same snowmobile either moving at constant speed or accelerating uphill. In both (a) and (b), the tension in the rope is not zero. Suppose the snowmobile driver is heading downhill and applies the brakes suddenly as in Figure 9.25 (a). F g will cause the sled to accelerate downhill as shown in Figure 9.25 (b). The speed of the sled could become large enough to overtake the snowmobile, bump into it, or tangle the rope. (a) n o f c ti o n o ti o m d ir e FT 0 θ (b) a Ff on m1 m1 Fg θ a m2 Ff on m2 Fg θ Figure 9.25 (a) The snowmobile is braking rapidly, and the tension in the rope is zero. (b) The free-body diagrams for the snowmobile and sled only show forces parallel to the incline. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 463 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 464 The driver can change the momentum of the snowmobile suddenly by using the brakes. But, as before, the only way that the momentum of the f acts for a long enough time interval. sled can eventually become zero is if F With experience, a driver learns to slow down gradually so that a towed sled remains in its proper position. Some sleds are attached to snowmobiles using a metal tow bar, which alleviates this problem (Figure 9.26). Since the tow bar can never become slack like a rope, the sled always remains a fixed distance from the snowmobile. Tow bars usually have a spring mechanism that increases the time during which a force can be exerted. So if the driver brakes or changes direction suddenly, the force exerted by the snowmobile on the sled acts for a longer period of time. Compared to a towrope, the spring mechanism in the tow bar can safely cause the momentum of the sled to decrease in a shorter period of time. Figure 9.26 A rigid tow bar with a spring mechanism provides the impulse necessary to increase or decrease the momentum of a towed sled. e WEB During takeoff, the magnitude of Earth’s gravitational field changes as a rocket moves farther away from Earth’s surface. The mass of a rocket also changes because it is burning fuel to move upward. Research how impulse and momentum apply to the design and function of rockets and thrust systems. Write a brief report of your findings, including diagrams where appropriate. Begin your search at www.pearsoned.ca/ school/physicssource. Safety devices in vehicles are designed so that, for a given impulse such as in a collision, the interaction time is increased, thereby reducing the average net force. This is achieved by providing motorists with a greater distance to travel, which increases the time interval required to stop the motion of the motorist. Three methods are used to provide this extra distance and time: • The dashboard is padded and the front end of the vehicle is designed to crumple. • The steering column telescopes to collapse, providing an additional 15–20 cm of distance for the driver to travel forward. • The airbag is designed to leak after inflation so that the fully inflated bag can decrease in thickness over time from about 30 cm to about 10 cm. In fact, an inflated airbag distributes the net force experienced during a collision over the motorist’s chest and head. By spreading the force over a greater area, the magnitude of the average net force at any one point on the motorist’s body is reduced, lowering the risk of a major injury. 464 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 465 A similar reasoning applies to the cushioning in running shoes and the padding in helmets and body pads used in sports (Figure 9.27). For a given impulse, all these pieces of equipment increase the interaction time and decrease the average net force. (a) (b) Figure 9.27 Padding in sports equipment helps reduce the risk of major injuries, because for a given impulse, the interaction time is increased and the average net force on the body part is reduced. (a) Team Canada in the World Women’s hockey tournament in Sweden, 2005. (b) Calgary Stampeders (in red) playing against the B.C. Lions in 2005. The effect of varying the average net force and the interaction time can be seen with projectiles. A bullet fired from a pistol with a short barrel does not gain the same momentum as another identical bullet fired from a rifle with a long barrel, assuming that each bullet experiences the same average net force (Figure 9.28). In the gun with the shorter barrel, the force from the expanding gases acts for a shorter period of time. So the change in momentum of the bullet is less. t p t p Figure 9.28 For the same average net force on a bullet, a gun with a longer barrel increases the time during which this force acts. So the change in momentum is greater for a bullet fired from a long-barrelled gun. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 465 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 466 Improved Sports Performance Involves Varying Fnetave and t In baseball, a skilled pitcher knows how to vary both the net force acting on the ball and the interaction time, so that the ball acquires maximum velocity before it leaves the pitcher’s hand (Figure 9.29). To exert the maximum possible force on the ball, a pitcher uses his arms, torso, and legs to propel the ball forward. To maximize the time he can exert that force, the pitcher leans back using a windup and then takes a long step forward. This way, his hand can be in contact with the ball for a longer period of time. The combination of the greater net force and the longer interaction time increases the change in momentum of the ball. Figure 9.29 When a pitcher exerts a force on the ball during a longer time interval, the momentum of a fastball increases even more. In sports such as hockey, golf, and tennis, coaches emphasize proper “follow through.” The reason is that it increases the time during which the puck or ball is in contact with the player’s stick, club, or racquet. So the change in momentum of the object being propelled increases. A similar reasoning applies when a person catches a ball. In this case, a baseball catcher should decrease the net force on the ball so that the ball doesn’t cause injury and is easier to hold onto. Players soon learn to do this by letting their hands move with the ball. For the same impulse, the extra movement with the hands results in an increased interaction time, which reduces the net force. This intentional flexibility when catching is sometimes referred to as having “soft hands,” and it is a great compliment to a football receiver. Hockey goalies allow their glove hand to fly back when snagging a puck to reduce the impact and allow them a better chance of keeping the puck in their glove. Boxers are also taught to “roll with the punch,” because if they move backward when hit, it increases the interaction time an |
d decreases the average net force of an opponent’s blow. 466 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 467 9.2 Check and Reflect 9.2 Check and Reflect Knowledge 1. (a) What quantities are used to calculate impulse? (b) State the units of impulse. 2. How are impulse and momentum related? 3. What graph could you use to determine the impulse provided to an object? Explain how to calculate the impulse using the graph. 4. What is the effect on impulse if (a) the time interval is doubled? (b) the net force is reduced to 1 of its 3 original magnitude? 5. Even though your mass is much greater than that of a curling stone, it is dangerous for a moving stone to hit your feet. Explain why. Applications 6. Using the concept of impulse, explain how a karate expert can break a board. 7. (a) From the graph below, what is the magnitude of the impulse provided to a 48-g tennis ball that is served due south? (b) What is the velocity of the ball when the racquet and ball separate? Magnitude of Net Force vs. Interaction Time for a Tennis Ball Being Hit by a Racquet ) 1000 900 800 700 600 500 400 300 200 100 0 0.0 1.0 2.0 3.0 Time t (ms) 4.0 5.0 6.0 9. During competitive world-class events, a four-person bobsled experiences an average net force of magnitude 1390 N during the first 5.0 s of a run. (a) What will be the magnitude of the impulse provided to the bobsled? (b) If the sled has the maximum mass of 630 kg, what will be the speed of the sled? 10. An advertisement for a battery-powered 25-kg skateboard says that it can carry an 80-kg person at a speed of 8.5 m/s. If the skateboard motor can exert a net force of magnitude 75 N, how long will it take to attain that speed? 11. Whiplash occurs when a car is rear-ended and either there is no headrest or the headrest is not properly adjusted. The torso of the motorist is accelerated by the seat, but the head is jerked forward only by the neck, causing injury to the joints and soft tissue. What is the average net force on a motorist’s neck if the torso is accelerated from 0 to 14.0 m/s [W] in 0.135 s? The mass of the motorist’s head is 5.40 kg. Assume that the force acting on the head is the same magnitude as the force on the torso. 12. What will be the change in momentum of a shoulder-launched rocket that experiences a thrust of 2.67 kN [W] for 0.204 s? Extensions 13. Experienced curlers know how to safely stop a moving stone. What do they do and why? 14. Research one safety device used in sports that applies the concept of varying Fnetave and t for a given impulse to prevent injury. Explain how the variables that affect impulse are changed by using this device. Begin your search at www.pearsoned.ca/school/physicssource. 8. What will be the magnitude of the impulse generated by a slapshot when an average net force of magnitude 520 N is applied to a puck for 0.012 s? e TEST To check your understanding of impulse, follow the eTest links at www.pearsoned.ca/school/ physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 467 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 468 info BIT The horns of a bighorn ram can account for more than 10% of its mass, which is about 125 kg. Rams collide at about 9 m/s, and average about 5 collisions per hour. Mating contests between any two rams may last for more than 24 h in total. 9.3 Collisions in One Dimension During mating season each fall, adult bighorn rams compete for supremacy in an interesting contest. Two rams will face each other, rear up, and then charge, leaping into the air to butt heads with tremendous force (Figure 9.30). Without being consciously aware of it, each ram attempts to achieve maximum momentum before the collision, because herd structure is determined by the outcome of the contest. Often, rams will repeat the head-butting interaction until a clear winner is determined. While most other mammals would be permanently injured by the force experienced during such a collision, the skull and brain structure of bighorn sheep enables them to emerge relatively undamaged from such interactions. In the previous section, many situations involved an object experiencing a change in momentum, or impulse, because of a collision with another object. When two objects such as bighorn rams collide, what relationship exists among the momenta of the objects both before and after collision? In order to answer this question, first consider one-dimensional collisions involving spheres in 9-4 QuickLab. Figure 9.30 By lunging toward each other, these bighorn rams will eventually collide head-on. During the collision, each ram will be provided with an impulse. 468 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 469 9-4 QuickLab 9-4 QuickLab Observing Collinear Collisions Problem What happens when spheres collide in one dimension? before direction of motion after Materials one set of four identical ball bearings or marbles (set A) a second set of four identical ball bearings or marbles of double the mass (set B) a third set of four identical ball bearings or marbles of half the mass (set C) 1-m length of an I-beam curtain rod or two metre-sticks with smooth edges masking tape Procedure 1 Lay the curtain rod flat on a bench to provide a horizontal track for the spheres. Tape the ends of the rod securely. If you are using metre-sticks, tape them 5 mm apart to form a uniform straight horizontal track. 2 Using set A, place three of the spheres tightly together at the centre of the track. 3 Predict what will happen when one sphere of set A moves along the track and collides with the three stationary spheres. Figure 9.31 4 Test your prediction. Ensure that the spheres remain on the track after collision. Record your observations using diagrams similar to Figure 9.31. 5 Repeat steps 2 to 4, but this time use set B, spheres of greater mass. 6 Repeat steps 2 to 4, but this time use set C, spheres of lesser mass. 7 Repeat steps 2 to 4 using different numbers of stationary spheres. The stationary spheres should all be the same mass, but the moving sphere should be of a different mass in some of the trials. Questions 1. Describe the motion of the spheres in steps 4 to 6. 2. Explain what happened when (a) a sphere of lesser mass collided with a number of spheres of greater mass, and (b) a sphere of greater mass collided with a number of spheres of lesser mass. In 9-4 QuickLab, for each set of spheres A to C, when one sphere hit a row of three stationary ones from the same set, the last sphere in the row moved outward at about the same speed as the incoming sphere. But when one sphere from set A hit a row of spheres from set B, the last sphere in the row moved outward at a much slower speed than the incoming sphere, and the incoming sphere may even have rebounded. When one sphere from set A hit a row of spheres from set C, the last sphere in the row moved outward at a greater speed than the incoming sphere, and the incoming sphere continued moving forward. To analyze these observations, it is important to first understand what a collision is. A collision is an interaction between two objects in which a force acts on each object for a period of time. In other words, the collision provides an impulse to each object. e MATH Explore how the masses of two colliding objects affect their velocities just after collision. Follow the eMath links at www.pearsoned.ca/school/ physicssource. collision: an interaction between two objects where each receives an impulse Chapter 9 The momentum of an isolated system of interacting objects is conserved. 469 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 470 system: two or more objects that interact with each other Systems of Objects in Collisions Each trial in 9-4 QuickLab involved two or more spheres colliding with each other. A group of two or more objects that interact is called a system. You encountered the concept of a system in Unit III in the context of energy. For each system in 9-4 QuickLab, the total mass remained constant because the mass of each sphere did not change as a result of the interaction. However, friction was an external force that acted on the system. For example, in steps 4 to 6 of 9-4 QuickLab, you likely observed that the speed of the sphere moving outward was a little less than the speed of the incoming sphere. In real life, a system of colliding objects is provided with two impulses: one due to external friction and the other due to the actual collision (Figure 9.32). External friction acts before, during, and after collision. The second impulse is only present during the actual collision. Since the actual collision time is very short, the impulse due to external friction during the collision is relatively small. before during after viA A 0 viB B Ff on A Ff on B 0 vfA A vfB B Ff on A Ff on B FB on A A B FA on B Ff on A tc ts Figure 9.32 External friction acts throughout the entire time interval of the interaction ts. But the action-reaction forces due to the objects only exist when the objects actually collide, and these forces only act for time interval tc. If you apply the form of Newton’s second law that relates net force to momentum to analyze the motion of a system of objects, you get net)sys (F p sys where p t sys is the momentum of the system The momentum of a system is defined as the sum of the momenta of all the objects in the system. So if objects A, B, and C form a system, the momentum of the system is p sys p A p B p C In the context of momentum, when the mass of a system is constant and no external net force acts on the system, the system is isolated. So 0. In 9-5 Inquiry Lab, a nearly isolated system of objects is net)sys (F involved in a one-dimensional collision. Find a quantitative relationship for the momentum of such a system in terms of momenta before and after collision. 470 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 471 9-5 Inquiry Lab 9-5 Inquiry Lab Relating Momentum Just Befo |
re and Just After a Collision Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question How does the momentum of a system consisting of two objects compare just before and just after a collision? Hypothesis State a hypothesis relating the momentum of a system immediately before and immediately after collision, where objects combine after impact. Remember to write an “if/then” statement. Variables Read the procedure carefully and identify the manipulated variables, the responding variables, and the controlled variables. Materials and Equipment one of these set-ups: air track, dynamics carts, Fletcher’s trolley, bead table or air table with linear guides colliding objects for the set-up chosen: gliders, carts, discs, blocks, etc. objects of different mass fastening material (Velcro™ strips, tape, Plasticine™, magnets, etc.) balance timing device (stopwatch, spark-timer, ticker-tape timer, electronic speed-timing device, or time-lapse camera) metre-sticks Procedure 1 Copy Tables 9.1 and 9.2 on page 472 into your notebook. 2 Set up the equipment in such a way that friction is minimized and the two colliding objects travel in the same straight line. 5 Set up the timing device to measure the velocity of object 1 just before and just after collision. Object 2 will be stationary before collision. The velocities of both objects will be the same after collision because they will stick together. 6 Send object 1 at a moderate speed on a collision course with object 2. Ensure that both objects will stick together and that the timing device is working properly. Make adjustments if needed. 7 Send object 1 at a moderate speed on a collision course with stationary object 2, recording the relevant observations and the masses as trial 1. 8 Send object 1 at a different speed on a collision course with stationary object 2, recording the relevant observations and the masses as trial 2. 9 Change the mass of one of the objects and again send object 1 at a moderate speed on a collision course with stationary object 2, recording the relevant observations and the masses as trial 3. 10 If you can simultaneously measure the speed of two objects, run trials where both objects are in motion before the collision. Do one trial in which they begin moving toward each other and stick together upon impact, and another trial where they move apart after impact. If you remove the fastening material, you will have to remeasure the masses of the objects. Include the direction of motion for both objects before and after collision. 11 If you did not do step 10, do two more trials, changing the mass of one of the objects each time. Include the direction of motion for both objects before and after collision. 3 Attach some fastening material to the colliding objects, so that the two objects remain together after impact. e LAB 4 Measure and record the masses of the two objects. If necessary, change the mass of one object so that the two objects have significantly different masses. For a probeware activity, go to www.pearsoned.ca/school/physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 471 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 472 Analysis 1. Determine the velocities for each colliding object in each trial, and record them in your data table. Show your calculations. 2. For each trial, calculate the momentum of each object just before and just after collision. Show your calculations. Record the values in your data table. 3. Calculate the momentum of the system just before and just after collision for each trial. Show your calculations. Record the values in your data table. 4. Calculate the difference between the momentum of the system just before and just after collision. Show your calculations. Record the values in your data table. 5. What is the relationship between the momentum of the system just before and just after collision? Does this relationship agree with your hypothesis? 6. What effect did friction have on your results? Explain. 7. Check your results with other groups. Account for any discrepancies. Table 9.1 Mass and Velocity Before and After for Object 1 Before and After for Object 2 Initial Velocity v 1i (m/s) Final Velocity v 1f (m/s) Mass m2 (g) Initial Velocity v 2i (m/s) Trial Mass m1 (g) 1 2 3 4 5 Table 9.2 Momentum Final Velocity v 2f (m/s) Change in Momentum of System sys g m s p Before and After for Object 1 Before and After for Object 2 Before and After for System Initial Final Momentum Momentum of System sysi g m p s of System sysf g m p s Initial Trial Momentum 1i g p m s Final Momentum 1f g p m s Initial Momentum 2i g p m s Final Momentum 2f g p m s 1 2 3 4 5 472 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 473 Momentum Is Conserved in One-dimensional Collisions In 9-5 Inquiry Lab, you discovered that, in one-dimensional collisions, the momentum of a system immediately before collision is about the same as the momentum of the system immediately after collision. If the external force of friction acting on the system is negligible, the momentum of the system is constant. This result is true no matter how many objects are in the system, how many of those objects collide, how massive the objects are, or how fast they are moving. The general form of Newton’s second law for a system is net)sys (F p sys t In an isolated system, the external net force on the system is zero, net)sys (F 0. So p sys 0 t p sys to be zero, the change in momentum of the system In order for t must be zero. p sysf sys p p sysi p sysi 0 0 p sysf In other words, p sys constant. This is a statement of the law of conservation of momentum. In Unit III, you encountered another conservation law, that is, in an isolated system the total energy of the system is conserved. Conservation laws always have one quantity that remains unchanged. In the law of conservation of momentum, it is momentum that remains unchanged. law of conservation of momentum: momentum of an isolated system is constant When no external net force acts on a system, the momentum of the system remains constant. p sysi p net)sys sysf where (F 0 Concept Check Why did cannons on 16th- to 19th-century warships need a rope around the back, tying them to the side of the ship (Figure 9.33)? Figure 9.33 Chapter 9 The momentum of an isolated system of interacting objects is conserved. 473 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 474 Freaction FB on A A B Faction FA on B Figure 9.34 The action-reaction forces when two objects collide e SIM Learn how the momentum of a system just before and just after a one- dimensional collision are related. Vary the ratio of the mass of two pucks. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Writing the Conservation of Momentum in Terms of Mass and Velocity Suppose a system consists of two objects, A and B. If the system is isolated, 0. Consider the internal forces of the system. At collision time, (F net)sys object A exerts a force on object B and object B exerts a force on object A (Figure 9.34). From Newton’s third law, these action-reaction forces are related by the equation F A on B F B on A Objects A and B interact for the same time interval t. If you multiply both sides of the equation by t, you get an equation in terms of impulse: F A on B t F B on A t Since impulse is equivalent to a change in momentum, the equation can be rewritten in terms of the momenta of each object: B B p p p Bi p Bi p 0 A 0 p Af p Bf A p p Bf p Ai p Af p Ai If the mass of each object is constant during the interaction, the equation can be written in terms of m and v: mBv Bi mAv Af mBv Bf mAv Ai This equation is the law of conservation of momentum in terms of the momenta of objects A and B. So if two bighorn rams head-butt each other, the sum of the momenta of both rams is constant during the collision, even though the momentum of each ram changes. The law of conservation of momentum has no known exceptions, and holds even when particles are travelling close to the speed of light, or when the mass of the colliding particles is very small, as in the case of electrons. In real life, when objects collide, external friction acts on nearly all systems and the instantaneous forces acting on each object are usually not known (Figure 9.35). Often, the details of the interaction are also unknown. However, you do not require such information to apply the law of conservation of momentum. Instead, it is the mass and instantaneous velocity of the objects immediately before and immediately after collision that are important, so that the effects of external friction are minimal, and do not significantly affect the outcome. Figure 9.35 During a vehicle collision, many forces cause a change in the velocity and shape of each vehicle. 474 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 475 Conservation of Momentum Applied to Rockets In Unit II, the motion of a rocket was explained using Newton’s third law. However, the conservation of momentum can be used to explain why a rocket can accelerate even in a vacuum. When the engines of a rocket burn fuel, the escaping exhaust gas has mass and considerable speed. When a rocket is in outer space, external friction is negligible. So the rocket-exhaust gas system is an isolated system. For a two-object system, the equation for the conservation of momentum is gas gas rocket rocket p p p 0 p where, during time interval t, p rocket is the change in momentum of the rocket including any unspent fuel and p gas is the change in momentum of the fuel that is expelled in the form of exhaust gas. It is the change in momentum of the exhaust gas that enables a rocket to accelerate (Figure 9.36). In the case of a very large rocket, such as a Saturn V, the magnitude of p would be very large (Figure 9.37). gas procket change in momentum of rocket pgas change in momentum of exhaust gas Figure 9.37 From the law |
of conservation of momentum, the magnitude of p gas is equal to the magnitude of p rocket. That is why a rocket can accelerate on Earth or in outer space. Figure 9.36 With a height of about 112 m, the Saturn V rocket was the largest and most powerful rocket ever built. info BIT None of the 32 Saturn rockets that were launched ever failed. Altogether 15 Saturn V rockets were built. Three Saturn V rockets are on display, one at each of these locations: the Johnson Space Center, the Kennedy Space Center, and the Alabama Space and Rocket Center. Of these three, only the rocket at the Johnson Space Center is made up entirely of former flight-ready, although mismatched, parts. info BIT Design of the Saturn V began in the 1950s with the intent to send astronauts to the Moon. In the early 1970s, this type of rocket was used to launch the Skylab space station. The rocket engines in the first stage burned a combination of kerosene and liquid oxygen, producing a total thrust of magnitude 3.34 107 N. The rocket engines in the second and third stages burned a combination of liquid hydrogen and liquid oxygen. The magnitude of the total thrust produced by the second-stage engines was 5.56 106 N, and the third-stage engine produced 1.11 106 N of thrust. Concept Check (a) Refer to the second infoBIT on this page. Why is less thrust needed by the second-stage engines of a rocket? (b) Why is even less thrust needed by the third-stage engine? In Example 9.5 on the next page, the conservation of momentum is applied to a system of objects that are initially stationary. This type of interaction is called an explosion. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 475 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 476 Example 9.5 A 75-kg hunter in a stationary kayak throws a 0.72-kg harpoon at 12 m/s [right]. The mass of the kayak is 10 kg. What will be the velocity of the kayak and hunter immediately after the harpoon is released? Given mp v pi 75 kg 0 m/s mk v ki 10 kg 0 m/s mh v hi v hf 0.72 kg 0 m/s 12 m/s [right] before left right after ? vTf 12 m/s vhf Figure 9.38 Practice Problems 1. A 110-kg astronaut and a 4000-kg spacecraft are attached by a tethering cable. Both masses are motionless relative to an observer a slight distance away from the spacecraft. The astronaut wants to return to the spacecraft, so he pulls on the cable until his velocity changes to 0.80 m/s [toward the spacecraft] relative to the observer. What will be the change in velocity of the spacecraft? 2. A student is standing on a stationary 2.3-kg skateboard. If the student jumps at a velocity of 0.37 m/s [forward], the velocity of the skateboard becomes 8.9 m/s [backward]. What is the mass of the student? Answers 1. 0.022 m/s [toward the astronaut] 2. 55 kg Required final velocity of hunter and kayak Analysis and Solution Choose the kayak, hunter, and harpoon as an isolated system. The hunter and kayak move together as a unit after the harpoon is released. So find the total mass of the hunter and kayak. mT mp mk 75 kg 10 kg 85 kg The hunter, kayak, and harpoon each have an initial velocity of zero. So the system has an initial momentum of zero. p sysi 0 Apply the law of conservation of momentum. sysf p p sysi p p p hf Tf sysi mhv 0 mTv hf Tf m hv m T 0.72 kg 85 kg v Tf hf (12 m/s) 0.10 m/s 0.10 m/s [left] v Tf Paraphrase and Verify The velocity of the kayak and hunter will be 0.10 m/s [left] immediately after the harpoon is released. Since the harpoon is thrown right, from Newton’s third law, you would expect the hunter and kayak to move left after the throw. So the answer is reasonable. 476 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:43 PM Page 477 In Example 9.6, a dart is fired at a stationary block sitting on a glider. This situation involves two objects (dart and block) that join together and move as a unit after interaction. This type of interaction is called a hit-and-stick interaction. Example 9.6 A wooden block attached to a glider has a combined mass of 0.200 kg. Both the block and glider are at rest on a frictionless air track. A dart gun shoots a 0.012-kg dart into the block. The velocity of the block-dart system after collision is 0.78 m/s [right]. What was the velocity of the dart just before it hit the block? Given mb v bi 0.200 kg 0 m/s before left ? vdi 0.012 kg 0.78 m/s [right] f md v right after vf 0.78 m/s Figure 9.39 Required initial velocity of dart (v di) Analysis and Solution Choose the block, glider, and dart as an isolated system. The dart, block, and glider move together as a unit after collision. The block-glider unit has an initial velocity of zero. So its initial momentum is zero. p bi 0 Apply the law of conservation of momentum. sysf p p sysf md)v (mb mdv mb m d f f 0.200 kg 0.012 kg 0.012 kg (0.78 m/s) (0.78 m/s) 0.212 kg 0.012 kg 14 m/s 14 m/s [right] p sysi p p di bi 0 mdv di v di v di Paraphrase Practice Problems 1. A student on a skateboard, with a combined mass of 78.2 kg, is moving east at 1.60 m/s. As he goes by, the student skilfully scoops his 6.4-kg backpack from the bench where he had left it. What will be the velocity of the student immediately after the pickup? 2. A 1050-kg car at an intersection has a velocity of 2.65 m/s [N]. The car hits the rear of a stationary truck, and their bumpers lock together. The velocity of the cartruck system immediately after collision is 0.78 m/s [N]. What is the mass of the truck? Answers 1. 1.5 m/s [E] 2. 2.5 103 kg The dart had a velocity of 14 m/s [right] just before it hit the block. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 477 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 478 Example 9.7 involves a basketball player, initially moving with some velocity, colliding with a stationary player. After the interaction, both players move in different directions. Example 9.7 A basketball player and her wheelchair (player A) have a combined mass of 58 kg. She moves at 0.60 m/s [E] and pushes off a stationary player (player B) while jockeying for a position near the basket. Player A ends up moving at 0.20 m/s [W]. The combined mass of player B and her wheelchair is 85 kg. What will be player B’s velocity immediately after the interaction? Given mA v Ai v Af 58 kg 0.60 m/s [E] 0.20 m/s [W] before 0.60 m/s vAi mB v Bi 85 kg 0 m/s W E after 0.20 m/s vAf ? vBf Figure 9.40 Player A Player B Player A Player B Practice Problems 1. A 0.25-kg volleyball is flying west at 2.0 m/s when it strikes a stationary 0.58-kg basketball dead centre. The volleyball rebounds east at 0.79 m/s. What will be the velocity of the basketball immediately after impact? 2. A 9500-kg rail flatcar moving forward at 0.70 m/s strikes a stationary 18 000-kg boxcar, causing it to move forward at 0.42 m/s. What will be the velocity of the flatcar immediately after collision if they fail to connect? Answers 1. 1.2 m/s [W] 2. 0.096 m/s [backward] Required final velocity of player B (v Bf) Analysis and Solution Choose players A and B as an isolated system. Player B has an initial velocity of zero. So her initial momentum is zero. p Bi 0 Apply the law of conservation of momentum. p Ai mAv Ai p sysi p Bi sysf p p Af 0 mAv Af m (v v A Bf m B 58 kg 85 kg p Bf mBv Bf v Ai Af) [0.60 m/s (0.20 m/s)] (0.60 m/s 0.20 m/s) 5 8 8 5 0.55 m/s 0.55 m/s [E] v Bf Paraphrase Player B’s velocity is 0.55 m/s [E] just after collision. 478 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 479 In Example 9.8, two football players in motion collide with each other. After the interaction, the players bounce apart. Example 9.8 A 110-kg Stampeders football fullback moving east at 1.80 m/s on a snowy playing field is struck by a 140-kg Eskimos defensive lineman moving west at 1.50 m/s. The fullback is bounced west at 0.250 m/s. What will be the velocity of the Eskimos defensive lineman just after impact? Given mS v Si v Sf 110 kg 1.80 m/s [E] 0.250 m/s [W] before vSi 1.80 m/s W m E v Ei 140 kg 1.50 m/s [W] E 1.50 m/s vEi after vSf 0.250 m/s ? vEf Figure 9.41 Required final velocity of Eskimos lineman (v Ef) Analysis and Solution Choose the fullback and lineman as an isolated system. Apply the law of conservation of momentum. p sysi p p Ei Si mEv Ei mEv Ef mSv Si v Ef v Ef Sf Si sysf v v Ei mSv Sf m Sv m E p Ef mEv Ef mEv Ei p p Sf mSv Sf mSv Si m S v m E m S(v Sf) v Si Ei m E 110 kg [(1.80 m/s) 140 kg (0.250 m/s)] (1.50 m/s) (1.80 m/s 0.250 m/s) 110 140 1.50 m/s 0.111 m/s 0.111 m/s [E] Practice Problems 1. A 72-kg snowboarder gliding at 1.6 m/s [E] bounces west at 0.84 m/s immediately after colliding with an 87-kg skier travelling at 1.4 m/s [W]. What will be the velocity of the skier just after impact? 2. A 125-kg bighorn ram butts heads with a younger 122-kg ram during mating season. The older ram is rushing north at 8.50 m/s immediately before collision, and bounces back at 0.11 m/s [S]. If the younger ram moves at 0.22 m/s [N] immediately after collision, what was its velocity just before impact? Answers 1. 0.62 m/s [E] 2. 8.6 m/s [S] Paraphrase The velocity of the Eskimos defensive lineman immediately after impact is 0.111 m/s [E]. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 479 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 480 PHYSICS INSIGHT The law of conservation of energy states that the total energy of an isolated system remains constant. The energy may change into several different forms. This law has no known exceptions. e SIM Predict the speed of two pucks just after a onedimensional collision using momentum and energy concepts. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Elastic and Inelastic Collisions in One Dimension In Examples 9.3 to 9.8, some of the collisions involved hard objects, such as the golf club hitting the golf ball. Other collisions, such as the block and dart, involved a dart that became embedded in a softer material (a block of wood). In all these collisions, it wa |
s possible to choose an isolated system so that the total momentum of the system was conserved. When objects collide, they sometimes deform, make a sound, give off light, or heat up a little at the moment of impact. Any of these observations indicate that the kinetic energy of the system before collision is not the same as after collision. However, the total energy of the system is constant. Concept Check (a) Is it possible for an object to have energy and no momentum? Explain, using an example. (b) Is it possible for an object to have momentum and no energy? Explain, using an example. Elastic Collisions Suppose you hit a stationary pool ball dead centre with another pool ball so that the collision is collinear and the balls move without spinning immediately after impact. What will be the resulting motion of both balls (Figure 9.42)? The ball that was initially moving will become stationary upon impact, while the other ball will start moving in the same direction as the incoming ball. If you measure the speed of both balls just before and just after collision, you will find that the speed of the incoming 1 mv2, 2 ball is almost the same as that of the outgoing ball. Since Ek the final kinetic energy of the system is almost the same as the initial kinetic energy of the system. Figure 9.42 Many collisions take place during a game of pool. What evidence suggests that momentum is conserved during the collision shown in the photo? What evidence suggests that energy is conserved? 480 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 481 If the initial kinetic energy of a system is equal to the final kinetic energy of the system after collision, the collision is elastic. In an elastic collision, the total kinetic energy of the system is conserved. elastic collision: a collision Ekf in which Eki Eki Ekf Most macroscopic interactions in the real world involve some of the initial kinetic energy of the system being converted to sound, light, or deformation (Figure 9.43). When deformation occurs, some of the initial kinetic energy of the system is converted to heat because friction acts on objects in almost all situations. These factors make it difficult to achieve an elastic collision. Even if two colliding objects are hard and do not appear to deform, energy is still lost in the form of sound, light, and/or heat due to friction. Usually, the measured speed of an object after interaction is a little less than the predicted speed, which indicates that the collision is inelastic. Example 9.9 demonstrates how to determine if the collision between a billiard ball and a snooker ball is elastic. Project LINK How will you apply the concepts of conservation of momentum and conservation of energy to the design of the water balloon protection? info BIT A steel sphere will bounce as high on a steel anvil as a rubber ball will on concrete. However, when a steel sphere is dropped on linoleum or hardwood, even more kinetic energy is lost and the sphere hardly bounces at all. The kinetic energy of the sphere is converted to sound, heat, and the deformation of the floor surface. To try this, use flooring samples. Do not try this on floors at home or at school. Figure 9.43 Is the collision shown in this photo elastic? What evidence do you have to support your answer? Chapter 9 The momentum of an isolated system of interacting objects is conserved. 481 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 482 Example 9.9 Practice Problems 1. A 45.9-g golf ball is stationary on the green when a 185-g golf club face travelling horizontally at 1.24 m/s [E] strikes it. After impact, the club face continues moving at 0.760 m/s [E] while the ball moves at 1.94 m/s [E]. Assume that the club face is vertical at the moment of impact so that the ball does not spin. Determine if the collision is elastic. 2. An argon atom with a mass of 6.63 1026 kg travels at 17 m/s [right] and strikes another identical argon atom dead centre travelling at 20 m/s [left]. The first atom rebounds at 20 m/s [left], while the second atom moves at 17 m/s [right]. Determine if the collision is elastic. Answers 1. inelastic 2. elastic A 0.160-kg billiard ball travelling at 0.500 m/s [N] strikes a stationary 0.180-kg snooker ball and rebounds at 0.0230 m/s [S]. The snooker ball moves off at 0.465 m/s [N]. Ignore possible rotational effects. Determine if the collision is elastic. 0.160 kg 0.180 kg 0.500 m/s [N] Given mb ms v bi v 0 m/s si v bf v 0.465 m/s [N] sf 0.0230 m/s [S] before after N S vsf 0.465 m/s Required determine if the collision is elastic snooker ball vbi 0.500 m/s snooker ball billiard ball vbf 0.0230 m/s Figure 9.44 billiard ball Analysis and Solution Choose the billiard ball and the snooker ball as an isolated system. Calculate the total initial kinetic energy and the total final kinetic energy of the system. Eki mb(vbi)2 1 1 ms(vsi)2 2 2 1 (0.160 kg)(0.500 m/s)2 0 2 Ekf 0.0200 kgm2/s2 0.0200 J mb(vbf)2 1 1 ms(vsf)2 2 2 1 (0.160 kg)(0.0230 m/s)2 2 1 (0.180 kg)(0.465 m/s)2 2 0.0195 kgm2/s2 0.0195 J Since Eki Ekf, the collision is inelastic. Paraphrase The collision between the billiard ball and the snooker ball is inelastic. 482 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 483 inelastic collision: a collision in which Eki Ekf e WEB Research examples of elastic and inelastic one- dimensional collisions. Then analyze how the momentum and energy change in those collisions. Begin your search at www.pearsoned.ca/ school/physicssource. Inelastic Collisions In 9-2 QuickLab on page 455, after the putty ball collided with a hard surface, the putty ball became stationary and had no kinetic energy. Upon impact, the putty ball deformed and the kinetic energy of the putty ball was converted mostly to thermal energy. Although the total energy of the system was conserved, the total initial kinetic energy of the system was not equal to the total final kinetic energy of the system after collision. This type of collision is inelastic. In an inelastic collision, the total kinetic energy of the system is not conserved. Eki Ekf One type of inelastic collision occurs when two objects stick together after colliding. However, this type of interaction does not necessarily mean that the final kinetic energy of the system is zero. For example, consider a ballistic pendulum, a type of pendulum used to determine the speed of bullets before electronic timing devices were invented (Figure 9.45). suspension wire ceiling pistol v bullet block height of swing Figure 9.45 When a bullet is fired into the block, both the block and bullet move together as a unit after impact. The pendulum consists of a stationary block of wood suspended from the ceiling by light ropes or cables. When a bullet is fired at the block, the bullet becomes embedded in the wood upon impact. The kinetic energy of the bullet is converted to sound, thermal energy, deformation of the wood and bullet, and the kinetic energy of the pendulum-bullet system. The initial momentum of the bullet causes the pendulum to move upon impact, but since the pendulum is suspended by cables, it swings upward just after the bullet becomes embedded in the block. As the pendulum-bullet system swings upward, its kinetic energy is converted to gravitational potential energy. Example 9.10 involves a ballistic pendulum. By using the conservation of energy, it is possible to determine the speed of the pendulum-bullet system immediately after impact. By applying the conservation of momentum to the collision, it is possible to determine the initial speed of the bullet. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 483 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 484 Example 9.10 A 0.0149-kg bullet from a pistol strikes a 2.0000-kg ballistic pendulum. Upon impact, the pendulum swings forward and rises to a height of 0.219 m. What was the velocity of the bullet immediately before impact? backward before after forward vbi ? h Given mb 0.0149 kg mp 2.0000 kg h 0.219 m Required initial velocity of bullet (v bi ) Ek 0 Ep 0 vf ? Ek 0 Ep (mb mp)gh Figure 9.46 Analysis and Solution Choose the pendulum and the bullet as an isolated system. Since the pendulum is stationary before impact, its initial velocity is zero. So its initial momentum is zero. p pi 0 Immediately after collision, the bullet and pendulum move together as a unit. The kinetic energy of the pendulum-bullet system just after impact is converted to gravitational potential energy. Ek Ep Practice Problems 1. A 2.59-g bullet strikes a stationary 1.00-kg ballistic pendulum, causing the pendulum to swing up to 5.20 cm from its initial position. What was the speed of the bullet immediately before impact? 2. A 7.75-g bullet travels at 351 m/s before striking a stationary 2.5-kg ballistic pendulum. How high will the pendulum swing? Answers 1. 391 m/s 2. 6.0 cm Apply the law of conservation of energy to find the speed of the pendulum-bullet system just after impact. 1 (mb 2 Ep Ek mp) (vf)2 (mb mp) g(h) (vf)2 2g(h) vf v f (0.219 m) 2g(h) 29.81 m s2 2.073 m/s 2.073 m/s [forward] Apply the law of conservation of momentum to find the initial velocity of the bullet. p bi mbv bi p sysi p pi sysf p p sysf mp)v 0 (mb mpv mb m b v bi f f 0.0149 kg 2.0000 kg 0.0149 kg (2.073 m/s) 484 Unit V Momentum and Impulse v bi (2.073 m/s) 2.0149 kg 0.0149 kg 280 m/s 280 m/s [forward] 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 485 Paraphrase The initial velocity of the bullet immediately before impact was 280 m/s [forward]. Example 9.11 demonstrates how to determine if the collision in Example 9.10 is elastic or inelastic by comparing the kinetic energy of the system just before and just after collision. Example 9.11 Determine if the collision in Example 9.10 is elastic or inelastic. Given mb mp 0.0149 kg 2.0000 kg v bi v f 280 m/s [forward] from Example 9.10 2.073 m/s [forward] from Example 9.10 Required initial and final kinetic energies (Eki and Ekf) to find |
if the collision is elastic Analysis and Solution Choose the pendulum and the bullet as an isolated system. Calculate the total initial kinetic energy and the total final kinetic energy of the system. Ekf Eki mb(vbi)2 1 1 mp(vpi)2 2 2 1 (0.0149 kg)(280 m/s)2 0 2 585 kgm2/s2 585 J mp)(vf)2 1 (mb 2 1 (0.0149 kg 2.0000 kg)(2.073 m/s)2 2 4.33 kgm2/s2 4.33 J Since Eki Ekf, the collision is inelastic. Paraphrase and Verify Since the kinetic energy of the system just before impact is much greater than the kinetic energy of the system just after impact, the collision is inelastic. This result makes sense since the bullet became embedded in the pendulum upon impact. Practice Problems 1. In Example 9.6 on page 477, how much kinetic energy is lost immediately after the interaction? 2. (a) Determine if the interaction in Example 9.8 on page 479 is elastic. (b) What percent of kinetic energy is lost? Answers 1. 1.1 J 2. (a) inelastic (b) 98.7% Chapter 9 The momentum of an isolated system of interacting objects is conserved. 485 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 486 9.3 Check and Reflect 9.3 Check and Reflect Knowledge 1. In your own words, state the law of conservation of momentum. 2. (a) In the context of momentum, what is an isolated system? (b) Why is it necessary to choose an isolated system when solving a momentum problem? 3. Explain the difference between an elastic and an inelastic collision. Include an example of each type of collision in your answer. 8. A 60.0-kg student on a 4.2-kg skateboard is travelling south at 1.35 m/s. A friend throws a 0.585-kg basketball to him with a velocity of 12.6 m/s [N]. What will be the velocity of the student and skateboard immediately after he catches the ball? 9. A hockey forward with a mass of 95 kg skates in front of the net at 2.3 m/s [E]. He is met by a 104-kg defenceman skating at 1.2 m/s [W]. What will be the velocity of the resulting tangle of players if they stay together immediately after impact? 4. What evidence suggests that a collision is 10. A 75.6-kg volleyball player leaps toward the (a) elastic? (b) inelastic? Applications 5. Give two examples, other than those in the text, of possible collinear collisions between two identical masses. Include a sketch of each situation showing the velocity of each object immediately before and immediately after collision. 6. A student is sitting in a chair with nearly frictionless rollers. Her homework bag is in an identical chair right beside her. The chair and bag have a combined mass of 20 kg. The student and her chair have a combined mass of 65 kg. If she pushes her homework bag away from her at 0.060 m/s relative to the floor, what will be the student’s velocity immediately after the interaction? 7. At liftoff, a space shuttle has a mass of 2.04 106 kg. The rocket engines expel 3.7 103 kg of exhaust gas during the first second of liftoff, giving the rocket a velocity of 5.7 m/s [up]. At what velocity is the exhaust gas leaving the rocket engines? Ignore the change in mass due to the fuel being consumed. The exhaust gas needed to counteract the force of gravity has already been accounted for and should not be part of this calculation. net to block the ball. At the top of his leap, he has a horizontal velocity of 1.18 m/s [right], and blocks a 0.275-kg volleyball moving at 12.5 m/s [left]. The volleyball rebounds at 6.85 m/s [right]. (a) What will be the horizontal velocity of the player immediately after the block? (b) Determine if the collision is elastic. 11. A 220-kg bumper car (A) going north at 0.565 m/s hits another bumper car (B) and rebounds at 0.482 m/s [S]. Bumper car B was initially travelling south at 0.447 m/s, and after collision moved north at 0.395 m/s. (a) What is the mass of bumper car B? (b) Determine if the collision is elastic. Extension 12. Summarize the concepts and ideas associated with one-dimensional collisions using a graphic organizer of your choice. See Student References 4: Using Graphic Organizers on pp. 869–871 for examples of different graphic organizers. Make sure that the concepts and ideas are clearly presented and appropriately linked. e TEST To check your understanding of the conservation of momentum and one-dimensional collisions, follow the eTest links at www.pearsoned.ca/school/physicssource. 486 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 487 9.4 Collisions in Two Dimensions Many interactions in the universe involve collisions. Comets, asteroids, and meteors sometimes collide with celestial bodies. Molecules and atoms are constantly colliding during chemical reactions throughout the universe: in stars, in Earth’s atmosphere, and even within your body. An interesting collision in recent history occurred on June 30, 1908, at Tunguska, Siberia, between a cosmic object and Earth (Figure 9.47). Eyewitnesses reported seeing a giant fireball that moved rapidly across the sky and eventually collided with the ground. Upon impact, a tremendous explosion occurred producing an atmospheric shock wave that circled Earth twice. About 2000 km2 of forest were levelled and thousands of trees were burned. In fact, there was so much fine dust in the atmosphere that people in London, England, could read a newspaper at night just from the scattered light. info BIT Scientists speculate that the cosmic object that hit Tunguska was about 100 m across and had a mass of about 1 106 t. The estimated speed of the object was about 30 km/s, which is 1.1 105 km/h. After the collision at Tunguska, a large number of diamonds were found scattered all over the impact site. So the cosmic object contained diamonds as well as other materials. Figure 9.47 The levelled trees and charred remnants of a forest at Tunguska, Siberia, after a cosmic object collided with Earth in 1908. Although the chance that a similar collision with Earth during your lifetime may seem remote, such collisions have happened throughout Earth’s history. In real life, most collisions occur in three dimensions. Only in certain situations, such as those you studied in section 9.3, does the motion of the interacting objects lie along a straight line. In this section, you will examine collisions that occur in two dimensions. These interactions occur when objects in a plane collide off centre. In 9-1 QuickLab on page 447, you found that when two coins collide off centre, the resulting path of each coin is in a different direction from its initial path. You may have noticed that certain soccer or hockey players seem to be at the right place at the right time whenever there is a rebound from the goalie. How do these players know where to position themselves so that they can score on the rebound? Find out by doing 9-6 Inquiry Lab. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 487 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 488 9-6 Inquiry Lab 9-6 Inquiry Lab Analyzing Collisions in Two Dimensions Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question How does the momentum of a two-body system in the x and y directions compare just before and just after a collision? Hypothesis State a hypothesis relating the momentum of a system in each direction immediately before and immediately after collision. Remember to write an “if/then” statement. Variables Read the procedure and identify the controlled, manipulated, and responding variables in the experiment. Materials and Equipment air table or bead table pucks spark-timer or camera set-up to measure velocities rulers or metre-sticks protractors Procedure 1 Copy Tables 9.3 and 9.4 on page 489 into your notebook. 2 Label the pucks as “puck 1” and “puck 2” respectively. Measure the mass of each puck and record it in Table 9.3. 3 Set up the apparatus so that puck 2 is at rest near the centre of the table. 4 Have each person in your group do one trial. Each time, send puck 1 aimed at the left side of puck 2, recording the paths of both pucks. Make sure the recording tracks of both pucks can be used to accurately measure their velocities before and after collision. 5 Have each person in your group measure and analyze one trial. Help each other as needed to ensure the measurements and calculations for each trial are accurate. 6 Find a suitable point on the recorded tracks to be the impact location. 7 On the path of puck 1 before collision, choose an interval where the speed is constant. Choose the positive x-axis to be in the initial direction of puck 1. 8 Using either the spark dots, the physical centre of the puck, or the leading or trailing edge of the puck, measure the distance and the time interval. Record those values in Table 9.3. 9 On the path of each puck after collision, choose an interval where the speed is constant. Measure the distance, direction of motion relative to the positive x-axis, and time interval. Record those values in Table 9.3. Analysis 1. Calculate the initial velocity and initial momentum of puck 1. Record the values in Table 9.4. 2. Calculate the velocity of puck 1 after collision. Resolve the velocity into x and y components. Record the values in Table 9.4. 3. Use the results of question 2 to calculate the x and y components of the final momentum of puck 1. Record the values in Table 9.4. 4. Repeat questions 2 and 3 but this time use the data for puck 2. Explain why the y component of the momentum of puck 2 is negative. 5. Record the calculated values from each member of your group as a different trial in Table 9.4. 6. For each trial, state the relationship between the initial momentum of the system in the x direction and the final momentum of the system in the x direction. Remember to consider measurement errors. Write this result as a mathematical statement. 7. The initial momentum of the system in the y direction was zero. For each trial, what was the final momentum of the system in the y direction? Remember to consider measurement errors. Write thi |
s result as a mathematical statement. 8. Compare your answers to questions 6 and 7 with other groups. Does this relationship agree with your hypothesis? Account for any discrepancies. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. 488 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 489 Table 9.3 Mass, Distance, Time Elapsed, and Angle Before and After for Puck 1 After for Puck 2 Trial Mass m1 (g) Initial Distance d1i (m) Initial Time Elapsed t1i (s) Final Distance d1f (m) Final Time Elapsed t1f (s) Final Angle () 1f Mass m2 (g) Final Distance d2f (m) Final Time Elapsed t2f (s) Final Angle () 2f 1 2 3 4 5 Table 9.4 Velocity and Momentum Before and After for Puck 1 After for Puck 2 Initial x Final x Initial x Velocity Momentum Velocity Final y Final x Velocity Momentum Momentum Velocity Final y Final x (m/s) p1ix gm v1fx v1ix (m/s) v1fy (m/s) p1fx gm p1fy gm v2fx (m/s) v2fy s s Final x Final y Final y Velocity Momentum Momentum p2fy gm s (m/s) p2fx gm s s Trial 1 2 3 4 5 Momentum Is Conserved in Two-dimensional Collisions In 9-6 Inquiry Lab, you found that along each direction, x and y, the momentum of the system before collision is about the same as the momentum of the system immediately after collision. In other words, momentum is conserved in two-dimensional collisions. This result agrees with what you saw in 9-5 Inquiry Lab, where only one-dimensional collisions were examined. As in one-dimensional collisions, the law of conservation of momentum is valid only when no external net force acts on the system. In two dimensions, the motion of each object in the system must be analyzed in terms of two perpendicular axes. To do this, you can either use a vector addition diagram drawn to scale or vector components. The law of conservation of momentum can be stated using components in the x and y directions. In two-dimensional collisions where no external net force acts on the system, the momentum of the system in both the x and y directions remains constant. psysix psysfx and psysiy psysfy where (F net)sys 0 e SIM Apply the law of conservation of momentum to twodimensional collisions. Go to www.pearsoned.ca/school/ physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 489 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 490 info BIT In championship curling, rebound angles and conservation of momentum are crucial for placing stones in counting position behind guards. Just nudging a stone several centimetres can make all the difference. Concept Check (a) Will the magnitude of the momentum of an object always increase if a non-zero net force acts on it? Explain, using an example. (b) How can the momentum of an object change but its speed remain the same? Explain, using an example. Example 9.12 involves a curling stone colliding off centre with an identical stone that is at rest. The momentum of each stone is analyzed in two perpendicular directions. Example 9.12 A 19.6-kg curling stone (A) moving at 1.20 m/s [N] strikes another identical stationary stone (B) off centre, and moves off with a velocity of 1.17 m/s [12.0° E of N]. What will be the velocity of stone B after the collision? Ignore frictional and rotational effects. Given mA v Bi 19.6 kg 0 m/s before N mB v Af 19.6 kg 1.17 m/s [12.0 E of N] v Ai 1.20 m/s [N] after N vAf 1.17 m/s vBf ? 12.0° E W Figure 9.48 S N W vBi 0 m/s E vAi 1.20 m/s S Required final velocity of stone B (v Bf) Analysis and Solution Choose both curling stones as an isolated system. Stone B has an initial velocity of zero. So its initial momentum is zero. p Bi 0 Resolve all velocities into east and north components (Figure 9.49). vAf Vector v Ai v Bi v Af East component North component 0 1.20 m/s 0 (1.17 m/s)(sin 12.0) 0 (1.17 m/s)(cos 12.0) 12.0° E Figure 9.49 490 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/28/08 9:26 AM Page 491 Apply the law of conservation of momentum to the system in the east and north directions. E direction psysiE pBiE pBfE pAiE pBfE pBiE psysfE pAfE pAiE mAvAiE 0 0 (19.6 kg)(1.17 m/s)(sin 12.0) 4.768 kgm/s pAfE mBvBiE mAvAfE The negative sign indicates the vector is directed west. N direction psysiN pBiN pBfN pAiN pBfN pBiN psysfN pAfN pAfN pAiN mAvAiN mAvAfN mBvBiN (19.6 kg)(1.20 m/s) 0 (19.6 kg)(1.17 m/s)(cos 12.0) 1.089 kgm/s Draw a vector diagram of the components of the final momentum of stone B and find the magnitude of the resultant p Pythagorean theorem. Bf using the N 1.089 kg m/s W Figure 9.50 p Bf θ 4.768 kg m/s E pBf (4.768 kgm/s)2 (1.089 kgm/s)2 4.8906 kgm/s Use the magnitude of the momentum and the mass of stone B to find its final speed. mBvBf pBf mB pBf vBf vBf 4.8906 kgm/s 19.6 kg 0.250 m/s Use the tangent function to find the direction of the final momentum. tan pBfN pBfW tan–1 12.9 1.089 kgm/s 4.768 kgm/s The final velocity will be in the same direction as the final momentum. Paraphrase The velocity of stone B after the collision is 0.250 m/s [12.9° N of W]. Practice Problems 1. A 97.0-kg hockey centre stops momentarily in front of the net. He is checked from the side by a 104-kg defenceman skating at 1.82 m/s [E], and bounces at 0.940 m/s [18.5 S of E]. What is the velocity of the defenceman immediately after the check? 2. A 1200-kg car, attempting to run a red light, strikes a stationary 1350-kg vehicle waiting to make a turn. Skid marks show that after the collision, the 1350-kg vehicle moved at 8.30 m/s [55.2 E of N], and the other vehicle at 12.8 m/s [36.8 W of N]. What was the velocity of the 1200-kg vehicle just before collision? Note this type of calculation is part of many vehicle collision investigations where charges may be pending. Answers 1. 1.03 m/s [15.7 N of E] 2. 15.6 m/s [N] Chapter 9 The momentum of an isolated system of interacting objects is conserved. 491 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 492 centre of mass: point where the total mass of an object can be assumed to be concentrated info BIT When an object is symmetric and has uniform density, its centre of mass is in the same location as the physical centre of the object. Example 9.13 involves a football tackle with two players. Each player has an initial velocity, but when they collide, both players move together as a unit. To analyze the motion, the centre of mass of the combination of both players must be used. The centre of mass is a point that serves as an average location of the total mass of an object or system. Depending on the distribution of mass, the centre of mass may be located even outside the object. Generally, momentum calculations are made using the centre of mass of an object. No matter where any external forces are acting on an object, whether the object is rotating or not, or whether the object is deformable or rigid, the translational motion of the object can be easily analyzed in terms of its centre of mass. Example 9.13 A 90-kg quarterback moving at 7.0 m/s [270] is tackled by a 110-kg linebacker running at 8.0 m/s [0]. What will be the velocity of the centre of mass of the combination of the two players immediately after impact? Practice Problems 1. A 2000-kg car travelling at 20.0 m/s [90.0] is struck at an intersection by a 2500-kg pickup truck travelling at 14.0 m/s [180]. If the vehicles stick together upon impact, what will be the velocity of the centre of mass of the cartruck combination immediately after the collision? 2. A 100-kg hockey centre is moving at 1.50 m/s [W] in front of the net. He is checked by a 108-kg defenceman skating at 4.20 m/s [S]. Both players move off together after collision. What will be the velocity of the centre of mass of the combination of the two players immediately after the check? Answers 1. 11.8 m/s [131] 2. 2.30 m/s [71.7 S of W] Given mq v qi 90 kg 7.0 m/s [270] before 8.0 m/s vli y 110 kg 8.0 m/s [0] ml v li after y x 7.0 m/s vqi x vf ? Figure 9.51 Required final velocity of centre of mass of the two players (v f) Analysis and Solution Choose the quarterback and the linebacker as an isolated system. The linebacker tackled the quarterback. So both players have the same final velocity. Resolve all velocities into x and y components. Vector v qi v li x component y component 0 7.0 m/s 8.0 m/s 0 Apply the law of conservation of momentum to the system in the x and y directions. 492 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/28/08 9:30 AM Page 493 x direction pqix psysix plix psysfx plix psysfx psysfx pqix mqvqix mlvlix 0 (110 kg)(8.0 m/s) 880 kgm/s y direction psysiy pliy psysfy pqiy pliy psysfy psysfy pqiy mqvqiy mlvliy (90 kg)(7.0 m/s) 0 630 kgm/s Draw a vector diagram of the components of the final momentum of the two players and find the magnitude of the resultant p Pythagorean theorem. sysf using the y 324º 880 kgm/s θ x 630 kgm/s psysf Figure 9.52 psysf (880kgm/s)2 (630 kgm/s)2 1082 kgm/s Use the magnitude of the momentum and combined masses of the two football players to find their final speed. psysf vf (mq ml)vf psysf ml) (mq 1082 kgm/s (90 kg 110 kg) = 5.4 m/s Use the tangent function to find the direction of the final momentum. tan psysfy psysfx tan1 630 kgm/s 880 kgm/s = 35.6° The final velocity will be in the same direction as the final momentum. From Figure 9.52, is below the positive x-axis. So the direction of v measured counterclockwise from the positive x-axis is 360 35.6 324.4. f v f 5.4 m/s [324] Paraphrase The final velocity of both players is 5.4 m/s [324]. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 493 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 494 Example 9.14 deals with a fireworks bundle that is initially stationary. After it explodes, three fragments (A, B, and C) fly in different directions in a plane. To demonstrate an alternative method of solving collision problems, a vector addition diagram is used to determine the momentum of fragment C. This quantity is then used to calculate its final velocity. Example 9.14 A 0.60-kg fireworks bundle is at |
rest just before it explodes into three fragments. A 0.20-kg fragment (A) flies at 14.6 m/s [W], and a 0.18-kg fragment (B) moves at 19.2 m/s [S]. What is the velocity of the third fragment (C) just after the explosion? Assume that no mass is lost, and that the motion of the fragments lies in a plane. Practice Problems 1. A 0.058-kg firecracker that is at rest explodes into three fragments. A 0.018-kg fragment moves at 2.40 m/s [N] while a 0.021-kg fragment moves at 1.60 m/s [E]. What will be the velocity of the third fragment? Assume that no mass is lost, and that the motion of the fragments lies in a plane. 2. A 65.2-kg student on a 2.50-kg skateboard moves at 0.40 m/s [W]. He jumps off the skateboard with a velocity of 0.38 m/s [30.0 S of W]. What will be the velocity of the skateboard immediately after he jumps? Ignore friction between the skateboard and the ground. Answers 1. 2.9 m/s [52 S of W] 2. 5.4 m/s [66 N of W] Given mT v i 0.60 kg 0 m/s before W N S E vi 0 m/s mA v Af 0.20 kg 14.6 m/s [W] after mB v Bf N 0.18 kg 19.2 m/s [S] vAf 14.6 m/s W vCf ? E vBf 19.2 m/s Figure 9.53 S Required final velocity of fragment C (v Cf ) Analysis and Solution Choose fragments A, B, and C as an isolated system. Since no mass is lost, find the mass of fragment C. mT mB) 0.60 kg (0.20 kg 0.18 kg) 0.22 kg (mA mC The original firework has an initial velocity of zero. So the system has an initial momentum of zero. p sysi 0 pAf The momentum of each fragment is in the same direction as its velocity. Calculate the momentum of fragments A and B. mBvBf (0.18 kg)(19.2 m/s) 3.46 kgm/s 3.46 kgm/s [S] mAvAf (0.20 kg)(14.6 m/s) 2.92 kgm/s 2.92 kgm/s [W] pBf p Af p Bf 494 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 495 Apply the law of conservation of momentum to the system. p p sysi 0 p sysf p Bf p Cf Af Use a vector addition diagram to determine the momentum of fragment C. N 1.00 kgm/s pAf 2.92 kgm/s pBf 3.46 kgm/s pCf ? θ E Figure 9.54 From Figure 9.54, careful measurements give pCf 50 N of E. 4.53 kgm/s and pCf vCf Divide the momentum of fragment C by its mass to find the velocity. mCvCf pC f m C 4.53 kgm s 0.22 kg 21 m/s 21 m/s [50 N of E] v Cf Paraphrase The velocity of the third fragment just after the explosion is 21 m/s [50 N of E]. Elastic and Inelastic Collisions in Two Dimensions As with one-dimensional collisions, collisions in two dimensions may be either elastic or inelastic. The condition for an elastic two-dimensional Ekf. collision is the same as for an elastic one-dimensional collision, Eki To determine if a collision is elastic, the kinetic energy values before and after collision must be compared. The kinetic energy of an object only depends on the magnitude of the velocity vector. So it does not matter if the velocity vector has only an x component, only a y component, or both x and y components. If you can determine the magnitude of the velocity vector, it is possible to calculate the kinetic energy. An example of an inelastic collision occurs when two objects join together and move as a unit immediately after impact. If two objects bounce apart after impact, the collision may be either elastic or inelastic, depending on the initial and final kinetic energy of the system. Usually, if one or both colliding objects deform upon impact, the collision is inelastic. e SIM and vfy Predict vfx for an object just after a twodimensional collision using momentum and energy concepts. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 495 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 496 e WEB Research some design improvements in running shoes. Use momentum and collision concepts to explain how these features affect athletes. Write a brief report of your findings, including diagrams where appropriate. Begin your search at www.pearsoned.ca/ school/physicssource. In track sports, the material used for the track has a profound effect on the elasticity of the collision between a runner’s foot and the running surface (Figure 9.55). If a track is made of a very hard material such as concrete, it experiences very little deformation when a runner’s foot comes in contact with it. The collision is more elastic than if the track were made of a more compressible material such as cork. So less kinetic energy of the runner is converted to other forms of energy upon impact. However, running on harder tracks results in a decreased interaction time and an increase in the net force acting on each foot, which could result in more injuries to joints, bones, and tendons. But a track that is extremely compressible is not desirable either, because it slows runners down. With all the pressure to achieve faster times in Olympic and world competitions, researchers and engineers continue to search for the optimum balance between resilience and safety in track construction. On the other hand, some collisions in sporting events present a very low risk of injury to contestants. Example 9.15 involves determining if the collision between two curling stones is elastic. Figure 9.55 Canadian runner Diane Cummins (far right) competing in the 2003 World Championships. The material of a running surface affects the interaction time and the net force acting on a runner’s feet. How would the net force change if the track were made of a soft material that deforms easily? Example 9.15 Determine if the collision in Example 9.12 on pages 490 and 491 is elastic. If it is not, what percent of the kinetic energy is retained? Given mA v Bi v Bf 19.6 kg 0 m/s 0.2495 m/s [77.1 W of N] mB v Af 19.6 kg 1.17 m/s [12.0 E of N] v Ai 1.20 m/s [N] Required determine if the collision is elastic Analysis and Solution Choose the two curling stones as an isolated system. Calculate the total initial kinetic energy and the total final kinetic energy of the system. 496 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 497 Eki 1 1 mA(vAi)2 mB(vBi)2 2 2 1 (19.6 kg)(1.20 m/s)2 0 2 14.11 kgm2/s2 14.11 J Since Eki Find the percent of Ek retained. Ekf, the collision is inelastic. E k 100% % Ek retained f E k i 0 3 . 4 1 100% 1 J 1 1 . 4 99.4% J Paraphrase The collision is inelastic, and 99.4% of the kinetic energy is retained. (Collisions like this, where very little kinetic energy is lost, may be called “near elastic collisions.”) Conservation Laws and the Discovery of Subatomic Particles Based on the results of experiments, scientists have gained great confidence in the laws of conservation of momentum and of conservation of energy, and have predicted that there are no known exceptions. This confidence has enabled scientists to make discoveries about the existence of particles within atoms as well. You will learn more about subatomic particles in Units VII and VIII. Ekf 1 1 mA(vAf)2 mB(vBf)2 2 2 1 1 (19.6 kg)(0.2495 m/s)2 (19.6 kg)(1.17 m/s)2 2 2 14.03 kgm2/s2 14.03 J Practice Problems 1. A 0.168-kg hockey puck flying at 45.0 m/s [252] is trapped in the pads of an 82.0-kg goalie moving at 0.200 m/s [0]. The velocity of the centre of mass of the goalie, pads, and puck immediately after collision is 0.192 m/s [333]. Was the collision elastic? If not, calculate the percent of total kinetic energy retained. 2. A 19.0-kg curling stone collides with another identical stationary stone. Immediately after collision, the first stone moves at 0.663 m/s. The second stone, which was stationary, moves at 1.31 m/s. If the collision was elastic, what would have been the speed of the first stone just before collision? Answers 1. inelastic, 0.882% 2. 1.47 m/s In 1930, German scientists Walther Bothe and Wilhelm Becker produced a very penetrating ray of unknown particles when they bombarded the element beryllium with alpha particles (Figure 9.56). An alpha particle is two protons and two neutrons bound together to form a stable particle. In 1932, British scientist James Chadwick (1891–1974) directed rays of these unknown particles at a thin paraffin strip and found that protons were emitted from the paraffin. He analyzed the speeds and angles of the emitted protons and, by using the conservation of momentum, he showed that the protons were being hit by particles of approximately the same mass. In other related experiments, Chadwick was able to determine the mass of these unknown particles very accurately using the conservation of momentum. Earlier experiments had shown that the unknown particles were neutral because they were unaffected by electric or magnetic fields. You will learn about electric and magnetic fields in Unit VI. Chadwick had attempted for several years to find evidence of a suggested neutral particle that was believed to be located in the nucleus of an atom. The discovery of these neutral particles, now called neutrons, resulted in Chadwick winning the Nobel Prize for Physics in 1935. alpha particle 2 neutron beryllium Figure 9.56 The experiment of Bothe and Becker using beryllium paved the way for Chadwick who later discovered the existence of neutrons by creating an experiment where he could detect them. e SIM Practise solving problems involving two-dimensional collisions. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 497 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 498 electron neutron pn 0 kgm/s proton pe pp Figure 9.57 If a neutron is initially stationary, p 0. If the neutron becomes transformed into a proton and an electron moving in the same direction, the momentum of the system is no longer zero. sysi Scientists later found that the neutron, when isolated, soon became transformed into a proton and an electron. Sometimes the electron and proton were both ejected in the same direction, which seemed to contradict the law of conservation of momentum (Figure 9.57). Furthermore, other experiments showed that the total energy of the neutron before transforma |
tion was greater than the total energy of both the proton and electron. It seemed as if the law of conservation of energy was not valid either. Austrian physicist Wolfgang Pauli (1900–1958) insisted that the conservation laws of momentum and of energy were still valid, and in 1930, he proposed that an extremely tiny neutral particle produced during the transformation must be moving in the opposite direction at an incredibly high speed. This new particle accounted for the missing momentum and missing energy (Figure 9.58). unknown particle pν neutron pn 0 kgm/s pe electron Many other scientists accepted Pauli’s explanation because they were convinced that the conservation laws of momentum and of energy were valid. For 25 years, they held their belief in the existence of this tiny particle, later called a neutrino, with no other evidence. Then in 1956, the existence of neutrinos was finally confirmed experimentally, further strengthening the universal validity of conservation laws. proton pp Figure 9.58 The existence of another particle accounted for the missing momentum and missing energy observed when a neutron transforms itself into a proton and an electron. THEN, NOW, AND FUTURE Neutrino Research in Canada Canada is a world leader in neutrino research. The SNO project (Sudbury Neutrino Observatory) is a special facility that allows scientists to gather data about these extremely tiny particles that are difficult to detect. The observatory is located in INCO’s Creighton Mine near Sudbury, Ontario, 2 km below Earth’s surface. Bedrock above the mine shields the facility from cosmic rays that might interfere with the observation of neutrinos. The experimental apparatus consists of 1000 t of heavy water encased in an acrylic vessel shaped like a 12-m diameter boiling flask (Figure 9.59). The vessel is surrounded by an array of about 1000 photo detectors, all immersed in a 10-storey chamber of purified water. When a neutrino collides with a heavy water molecule, a tiny burst of light is emitted, which the photo detectors pick up. Despite all that equipment, scientists only detect an average of about 10 neutrinos a day. So experiments acrylic vessel with heavy water vessel of purified water photo detectors Figure 9.59 The Sudbury Neutrino Observatory is a collaborative effort among 130 scientists from Canada, the U.S., and the U.K. must run for a long time in order to collect enough useful data. Scientists are interested in neutrinos originating from the Sun and other distant parts of the universe. At first, it seemed that the Sun was not emitting as many neutrinos as expected. Scientists thought they would have to modify their theories about the reactions taking place within the core of the Sun. Tripling the sensitivity of the detection process, by adding 2 t of salt to the heavy water, showed that 2 of the neutrinos from the 3 Sun were being transformed into different types of neutrinos as they travelled. This discovery has important implications about the basic properties of a neutrino, including its mass. It now appears that scientists’ theories about the reactions within the core of the Sun are very accurate. Continued research at this facility will help answer fundamental questions about matter and the universe. Questions 1. Why was Sudbury chosen as the site for this type of observatory? 2. Explain why, at first, it appeared that the Sun was not emitting the expected number of neutrinos. 3. If a neutrino has a very small mass and travels very fast, why doesn’t it run out of energy? 498 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 499 9.4 Check and Reflect 9.4 Check and Reflect Knowledge 1. How is a two-dimensional collision different from a one-dimensional collision? Explain, using examples. 2. In your own words, state the law of conservation of momentum for twodimensional collisions. Show how the law relates to x and y components by using an example. 3. In your own words, define the centre of mass of an object. 4. Explain why scientists accepted the existence of the neutrino for so long when there was no direct evidence for it. Applications 5. A cue ball travelling at 0.785 m/s [270] strikes a stationary five-ball, causing it to move at 0.601 m/s [230]. The cue ball and the five-ball each have a mass of 160 g. What will be the velocity of the cue ball immediately after impact? Ignore frictional and rotational effects. 6. A stationary 230-kg bumper car in a carnival is struck off centre from behind by a 255-kg bumper car moving at 0.843 m/s [W]. The more massive car bounces off at 0.627 m/s [42.0 S of W]. What will be the velocity of the other bumper car immediately after collision? 7. A 0.25-kg synthetic rubber ball bounces to a height of 46 cm when dropped from a height of 50 cm. Determine if this collision is elastic. If not, how much kinetic energy is lost? 8. A football halfback carrying the ball, with a combined mass of 95 kg, leaps toward the goal line at 4.8 m/s [S]. In the air at the goal line, he collides with a 115-kg linebacker travelling at 4.1 m/s [N]. If the players move together after impact, will the ball cross the goal line? 9. A 0.160-kg pool ball moving at 0.563 m/s [67.0 S of W] strikes a 0.180-kg snooker ball moving at 0.274 m/s [39.0 S of E]. The pool ball glances off at 0.499 m/s [23.0 S of E]. What will be the velocity of the snooker ball immediately after collision? 10. A 4.00-kg cannon ball is flying at 18.5 m/s [0] when it explodes into two fragments. One 2.37-kg fragment (A) goes off at 19.7 m/s [325]. What will be the velocity of the second fragment (B) immediately after the explosion? Assume that no mass is lost during the explosion, and that the motion of the fragments lies in the xy plane. 11. A 0.952-kg baseball bat moving at 35.2 m/s [0] strikes a 0.145-kg baseball moving at 40.8 m/s [180]. The baseball rebounds at 37.6 m/s [64.2]. What will be the velocity of the centre of mass of the bat immediately after collision if the batter exerts no force on the bat during and after the instant of impact? Extensions 12. Some running shoe designs contain springs. Research these types of shoes and the controversy surrounding them. How do momentum and impulse apply to these shoes? Write a brief report of your findings. Include diagrams where appropriate. Begin your search at www.pearsoned.ca/school/physicssource. 13. Italian physicist Enrico Fermi gave the name “neutrino” to the elusive particle that scientists were at first unable to detect. Research Fermi’s contributions to physics. Write a brief report of your findings. Begin your search at www.pearsoned.ca/school/physicssource. e TEST To check your understanding of two-dimensional collisions, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 9 The momentum of an isolated system of interacting objects is conserved. 499 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 500 CHAPTER 9 SUMMARY Key Terms and Concepts momentum impulse collision system law of conservation of momentum elastic collision inelastic collision centre of mass Key Equations Momentum: Impulse: p mv F net F netave t p or F net t p or F t mv t mv netave Conservation of momentum: p sysi p sysf Elastic collisions: Eki Ekf Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. Momentum has units of is defined as the of and applies to collisions applies throughout the universe because it is 2 types equation where where equation equation where a change in momentum is equivalent to an equation Figure 9.60 500 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/28/08 9:33 AM Page 501 UNIT V PROJECT An Impulsive Water Balloon Scenario Imagine you are part of a creative engineering design team commissioned to design and build an amusement ride that will make the West Edmonton Mall 14-storey “Space Shot” obsolete. The ride must allow patrons to experience the thrill of free fall in a safe environment. Accelerations have to remain within appropriate limits during a required change from vertical to horizontal motion, and as the people are brought to rest. A model must be constructed for sales demonstrations using a water balloon that experiences a minimum 2.4-m drop, where the impulse changes the magnitude and direction of the momentum while maintaining the integrity of the balloon. The water balloon must begin with a vertical drop equivalent to eight storeys. Then for the equivalent height of six storeys, the balloon must change direction and come to a stop horizontally. Planning Form a design team of three to five members. Plan and assign roles so that each team member has at least one major task. Roles may include researcher, engineer to perform mathematical calculations, creative designer, construction engineer, materials acquisition officer, and writer, among others. One person may need to perform several roles in turn. Ensure that all team members help along the way. Prepare a time schedule for each task, and for group planning and reporting sessions. Materials • small balloon filled with water • plastic zip-closing bag to contain the water balloon • cardboard and/or wooden frame for apparatus • vehicle or container for balloon • cushioning material • braking device • art materials CAUTION: Test your design in an appropriate area. Make sure no one is in the way during the drop. Assessing Results After completing the project, assess its success based on a rubric designed in class* that considers research strategies experiment and construction techniques clarity and thoroughness of the written report effectiveness of the team’s presentation quality and fairness of the teamwork Procedure 1 Research the range of acceptable accelerations that most people can tolerate. 2 Calculate the maximum speed obtained when an object is dropped from an eight-storey building (equivalent to 24.6 m). 3 Calculate the impulse necessary to chang |
e the direction of motion of a 75.0-kg person from vertical to horizontal in the remaining height of six storeys (equivalent to 18.4 m). The person must come to a stop at the end. Assume that the motion follows the arc of a circle. 4 Determine the time required so that the change in the direction of motion and stopping the person meets the maximum acceptable acceleration in step 1. 5 Include your calculations in a report that shows your design and method of changing the motion. 6 Build a working model and test it. Make modifications as necessary to keep the water balloon intact for a fall. Present the project to your teacher and the class. Thinking Further 1. Explain why eight storeys was used in the calculation in step 2, instead of 14 storeys. 2. What other amusement rides has your team thought of while working on this project? What would make each of these rides thrilling and appealing? 3. In what ways could your ideas have a practical use, such as getting people off a high oil derrick or out of a high-rise building quickly and safely? 4. What conditions would cause a person to be an unacceptable candidate for your ride? Write out a list of rules or requirements that would need to be posted. *Note: Your instructor will assess the project using a similar assessment rubric. Unit V Momentum and Impulse 501 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 502 UNIT V SUMMARY Unit Concepts and Skills: Quick Reference Concepts CHAPTER 9 Momentum Newton’s second law Impulse Effects of varying the net force and time interval for a given impulse Summary Resources and Skill Building The momentum of an isolated system of interacting objects is conserved. 9.1 Momentum Is Mass Times Velocity Momentum is the product of the mass of an object and its velocity. Momentum is a vector quantity measured in kilogram-metres per second (kgm/s). Newton’s second law states that the net force on an object is equal to the rate of change of its momentum. 9.2 Impulse Is Equivalent to a Change in Momentum The impulse provided to an object is defined as the product of the net force (or average net force if F the interaction time. Impulse is equivalent to the change in momentum of the object. constant) acting on the object during an interaction and net The magnitude of the net force during an interaction and the interaction time determine whether or not injuries or damage to an object occurs. Examples 9.1 & 9.2 9-2 QuickLab Figures 9.12 & 9.13 9-3 Design a Lab Example 9.3 Example 9.4 Net force-time graph Impulse can be determined by calculating the area under a net force-time graph. Conservation of momentum in one dimension Elastic collisions Inelastic collisions Conservation of momentum in two dimensions Elastic and inelastic collisions 9.3 Collisions in One Dimension Momentum is conserved when objects in an isolated system interact in one dimension. A system is the group of objects that interact with each other, and it is isolated if no external net force acts on these objects. 9-4 QuickLab 9-5 Inquiry Lab Examples 9.5–9.8, 9.10 Elastic collisions are collisions in which a system of objects has the same initial and final kinetic energy. So both the momentum and kinetic energy of the system are conserved. Example 9.9 Inelastic collisions are collisions in which a system of objects has different initial and final kinetic energy values. Example 9.11 9.4 Collisions in Two Dimensions Momentum is conserved when objects in an isolated system interact in two dimensions. An isolated system has no external net force acting on it. Elastic collisions in two dimensions satisfy the same conditions as one-dimensional elastic collisions, that is, Eki Ekf . 9-6 Inquiry Lab Examples 9.12–9.14 Example 9.15 502 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 503 UNIT V REVIEW Vocabulary 1. Using your own words, define these terms, concepts, principles, or laws. momentum impulse one-dimensional collisions conservation of momentum conservation of energy elastic collisions inelastic collisions two-dimensional collisions centre of mass Knowledge CHAPTER 9 2. Compare and contrast momentum and impulse. 3. Explain the relationship between the units in which momentum and impulse are measured. 4. A student calculated the answer to a problem and got 40 kgm/s [W]. Which quantities could the student have calculated? 5. In your own words, restate Newton’s second law in terms of momentum. 6. What difference does it make that momentum is a vector quantity rather than a scalar quantity? 7. Compare and contrast a net force and an average net force acting on an object during an interaction. 8. Statistics show that less massive vehicles tend to have fewer accidents than more massive vehicles. However, the survival rate for accidents in more massive vehicles is much greater than for less massive ones. How could momentum be used to explain these findings? 9. Using the concept of impulse, explain how the shocks on a high-end mountain bike reduce the chance of strain injuries to the rider. 10. State the quantities, including units, you would need to measure to determine the momentum of an object. 11. State the quantities that are conserved in oneand two-dimensional collisions. Give an example of each type of collision. 12. How do internal forces affect the momentum of a system? 13. What instructions would you give a young gymnast so that she avoids injury when landing on a hard surface? 14. Will the magnitude of the momentum of an object always increase if a net force acts on it? Explain, using an example. 15. What quantity do you get when p is divided by mass? 16. For a given impulse, what is the effect of (a) increasing the time interval? (b) decreasing the net force during interaction? 17. For each situation, explain how you would effectively provide the required impulse. • to catch a water balloon tossed from some distance • to design a hiking boot for back-country hiking on rough ground • to shoot an arrow with maximum velocity using a bow • for an athlete to win the gold medal in the javelin event with the longest throw • for a car to accelerate on an icy road 18. Why does a hunter always press the butt of a shotgun tight against the shoulder before firing? 19. Describe a method to find the components of a momentum vector. 20. Explain why the conservation of momentum and the conservation of energy are universal laws. 21. Why does the law of conservation of momentum require an isolated system? 22. Suppose a problem involves a two-dimensional collision between two objects, and the initial momentum of one object is unknown. Explain how to solve this problem using (a) a vector addition diagram drawn to scale (b) vector components 23. Explain, in terms of momentum, why a rocket does not need an atmosphere to push against when it accelerates. 24. If a firecracker explodes into two fragments of unequal mass, which fragment will have the greater speed? Why? Unit V Momentum and Impulse 503 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 504 38. Draw a momentum vector diagram to represent a 575-g basketball flying at 12.4 m/s [26.0 S of E]. 39. Calculate the momentum of a 1250-kg car travelling south at 14.8 m/s. 40. A bowling ball has a momentum of 28 kgm/s [E]. If its speed is 4.5 m/s, what is the mass of the ball? 41. A curling stone has a momentum of 32 kgm/s [W]. What would be the momentum if the mass of the stone is decreased to 7 of its original mass 8 and its speed is increased to 4 of its original 3 speed? 42. A soccer ball has a momentum of 2.8 kgm/s [W]. What would be the momentum if its mass decreased to 3 of its original mass and its 4 speed increased to 9 of its original speed? 8 43. The graph below shows the magnitude of the net force as a function of interaction time for a volleyball being blocked. The velocity of the ball changes from 18 m/s [N] to 11 m/s [S]. (a) Using the graph, calculate the magnitude of the impulse on the volleyball. (b) What is the mass of the ball? Magnitude of Net Force vs. Interaction Time for a Volleyball Block ) 5000 4000 3000 2000 1000 0 0.0 1.0 3.0 2.0 Time t (ms) 4.0 5.0 44. (a) Calculate the impulse on a soccer ball if a player heads the ball with an average net force of 120 N [210] for 0.0252 s. (b) If the mass of the soccer ball is 0.44 kg, calculate the change in velocity of the ball. 45. At a buffalo jump, a 900-kg bison is running at 6.0 m/s toward the drop-off ahead when it senses danger. What horizontal force must the bison exert to stop itself in 2.0 s? 25. When applying the conservation of momentum to a situation, why is it advisable to find the velocities of all objects in the system immediately after collision, instead of several seconds later? 26. Which physics quantities are conserved in a collision? 27. A curling stone hits another stationary stone off centre. Draw possible momentum vectors for each stone immediately before and immediately after collision, showing both the magnitude and direction of each vector. 28. A Superball™ of rubber-like plastic hits a wall perpendicularly and rebounds elastically. Explain how momentum is conserved. 29. What two subatomic particles were discovered using the conservation of momentum and the conservation of energy? 30. Explain how an inelastic collision does not violate the law of conservation of energy. 31. If a system is made up of only one object, show how the law of conservation of momentum can be used to derive Newton’s first law. 32. A Calgary company, Cerpro, is a world leader in ceramic armour plating for military protection. The ceramic structure of the plate transmits the kinetic energy of an armour-piercing bullet throughout the plate, reducing its penetrating power. Explain if this type of collision is elastic or inelastic. 33. A compact car and a heavy van travelling at approximately the same speed perpendicular to each other collide and stick together. Which vehicle will experience the greatest change in its direction of motion just after impact? Why? 34. Is it p |
ossible for the conservation of momentum to be valid if two objects move faster just before, than just after, collision? Explain, using an example. 35. Fighter pilots have reported that immediately after a burst of gunfire from their jet fighter, the speed of their aircraft decreased by 50–65 km/h. Explain the reason for this change in motion. 36. A cannon ball explodes into three fragments. One fragment goes north and another fragment goes east. Draw the approximate direction of the third fragment. What scientific law did you use to arrive at your answer? Applications 37. Calculate the momentum of a 1600-kg car travelling north at 8.5 m/s. 504 Unit V Momentum and Impulse 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 505 46. (a) What is the minimum impulse needed to 52. A 0.146-kg baseball pitched at 40 m/s is hit back give a 275-kg motorcycle and rider a velocity of 20.0 m/s [W] if the motorcycle is initially at rest? (b) If the wheels exert an average force of 710 N [E] to the road, what is the minimum time needed to reach a velocity of 20.0 m/s [W]? (c) Explain how the force directed east causes the motorcycle to accelerate westward. (d) Why is it necessary to specify a minimum impulse and a minimum time? 47. A 1.15-kg peregrine falcon flying at 15.4 m/s [W] captures a 0.423-kg pigeon flying at 4.68 m/s [S]. What will be the velocity of their centre of mass immediately after the interaction? 48. A 275-kg snowmobile carrying a 75-kg driver exerts a net backward force of 508 N on the snow for 15.0 s. (a) What impulse will the snow provide to the snowmobile and driver? (b) Calculate the change in velocity of the snowmobile. 49. The graph below shows the magnitude of the net force as a function of time for a 275-g volleyball being spiked. Assume the ball is motionless the instant before it is struck. (a) Using the graph, calculate the magnitude of the impulse on the volleyball. (b) What is the speed of the ball when it leaves the player’s hand Magnitude of Net Force vs. Interaction Time for a Volleyball Spike 2000 1500 1000 500 0 0.0 1.0 2.0 3.0 Time t (ms) 4.0 5.0 6.0 50. A Centaur rocket engine expels 520 kg of exhaust gas at 5.0 104 m/s in 0.40 s. What is the magnitude of the net force on the rocket that will be generated? 51. An elevator with passengers has a total mass of 1700 kg. What is the net force on the cable needed to give the elevator a velocity of 4.5 m/s [up] in 8.8 s if it is starting from rest? toward the pitcher at a speed of 45 m/s. (a) What is the impulse provided to the ball? (b) The bat is in contact with the ball for 8.0 ms. What is the average net force that the bat exerts on the ball? 53. An ice dancer and her 80-kg partner are both gliding at 2.5 m/s [225]. They push apart, giving the 45-kg dancer a velocity of 3.2 m/s [225]. What will be the velocity of her partner immediately after the interaction? 54. Two students at a barbecue party put on inflatable Sumo-wrestling outfits and take a run at each other. The 87.0-kg student (A) runs at 1.21 m/s [N] and the 73.9-kg student (B) runs at 1.51 m/s [S]. The students are knocked off their feet by the collision. Immediately after impact, student B rebounds at 1.03 m/s [N]. (a) Assuming the collision is completely elastic, calculate the speed of student A immediately after impact using energy considerations. (b) How different would your answer be if only conservation of momentum were used? Calculate to check. (c) How valid is your assumption in part (a)? 55. A cannon mounted on wheels has a mass of 1380 kg. It shoots a 5.45-kg projectile at 190 m/s [forward]. What will be the velocity of the cannon immediately after firing the projectile? 56. A 3650-kg space probe travelling at 1272 m/s [0.0] has a directional thruster rocket exerting a force of 1.80 104 N [90.0] for 15.6 s. What will be the newly adjusted velocity of the probe? 57. In a movie stunt, a 1.60-kg pistol is struck by a 15-g bullet travelling at 280 m/s [50.0]. If the bullet moves at 130 m/s [280] after the interaction, what will be the velocity of the pistol? Assume that no external force acts on the pistol. 58. A 52.5-kg snowboarder, travelling at 1.24 m/s [N] at the end of her run, jumps and kicks off her 4.06-kg snowboard. The snowboard leaves her at 2.63 m/s [62.5 W of N]. What is her velocity just after she kicks off the snowboard? 59. A 1.26-kg brown bocce ball travelling at 1.8 m/s [N] collides with a stationary 0.145-kg white ball, driving it off at 0.485 m/s [84.0 W of N]. (a) What will be the velocity of the brown ball immediately after impact? Ignore friction and rotational effects. (b) Determine if the collision is elastic. Unit V Momentum and Impulse 505 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 506 60. Two people with a combined mass of 128 kg are sliding downhill on a 2.0-kg toboggan at 1.9 m/s. A third person of mass 60 kg inadvertently stands in front and upon impact is swept along with the toboggan. If all three people remain on the toboggan after impact, what will be its velocity after impact? 61. An aerosol paint can is accidentally put in a fire pit. After the fire is lit, the can is heated and explodes into two fragments. A 0.0958-kg fragment (A) flies off at 8.46 m/s [E]. The other fragment (B) has a mass of 0.0627 kg. The 0.0562-kg of gas inside bursts out at 9.76 m/s [N]. What will be the velocity of fragment B immediately after the explosion? Assume that no mass is lost during the explosion, and that the motion of the fragments lies in a plane. 62. A 0.185-kg golf club head travelling horizontally at 28.5 m/s hits a 0.046-kg golf ball, driving it straight off at 45.7 m/s. (a) Suppose the golfer does not exert an external force on the golf club after initial contact with the ball. If the collision between the golf club and the ball is elastic, what will be the speed of the club head immediately after impact? (b) Show that the law of conservation of momentum is valid in this interaction. 63. A student on a skateboard is travelling at 4.84 m/s [0], carrying a 0.600-kg basketball. The combined mass of the student and skateboard is 50.2 kg. He throws the basketball to a friend at a velocity of 14.2 m/s [270]. What is the resulting velocity of the centre of mass of the student-skateboard combination immediately after the throw? Ignore frictional effects. 64. An oxygen molecule of mass 5.31 10–26 kg with a velocity of 4.30 m/s [0.0°] collides headon with a 7.31 10–26-kg carbon dioxide molecule which has a velocity of 3.64 m/s [180.0°]. After collision the oxygen molecule has a velocity of 4.898 m/s [180.0°]. (a) Calculate the velocity of the carbon dioxide molecule immediately after collision. (b) Determine by calculation whether or not the collision is elastic. 65. An isolated stationary neutron is transformed into a 9.11 1031-kg electron travelling at 4.35 105 m/s [E] and a 1.67 1027-kg proton travelling at 14.8 m/s [E]. What is the momentum of the neutrino that is released? 506 Unit V Momentum and Impulse 66. An 8.95-kg bowling ball moving at 3.62 m/s [N] hits a 0.856-kg bowling pin, sending it off at 3.50 m/s [58.6 E of N]. (a) What will be the velocity of the bowling ball immediately after collision? (b) Determine if the collision is elastic. 67. A wooden crate sitting in the back of a pickup truck travelling at 50.4 km/h [S] has a momentum of magnitude 560 kgm/s. (a) What is the mass of the crate? (b) What impulse would the driver have to apply with the brakes to stop the vehicle in 5.25 s at an amber traffic light? Use mT for the total mass of the truck. (c) If the coefficient of friction between the crate and the truck bed is 0.30, will the crate slide forward as the truck stops? Justify your answer with calculations. 68. A firecracker bursts into three fragments. An 8.5-g fragment (A) flies away at 25 m/s [S]. A 5.6-g fragment (B) goes east at 12 m/s. Calculate the velocity of the 6.7-g fragment (C). Assume that no mass is lost during the explosion, and that the motion of the fragments lies in a plane. 69. A spherical molecule with carbon atoms arranged like a geodesic dome is called a buckyball. A 60-atom buckyball (A) of mass 1.2 10–24 kg travelling at 0.92 m/s [E] collides with a 70-atom buckyball (B) of mass 1.4 10–24 kg with a velocity of 0.85 m/s [N] in a laboratory container. Buckyball (A) bounces away at a velocity of 1.24 m/s [65° N of E]. (a) Calculate the speed of buckyball (B) after the collision assuming that this is an elastic collision. (b) Use the conservation of momentum to find the direction of buckyball (B) after the collision. 70. A moose carcass on a sled is being pulled by a tow rope behind a hunter’s snowmobile on a horizontal snowy surface. The sled and moose have a combined mass of 650 kg and a momentum of 3.87 103 kgm/s [E]. (a) Calculate the velocity of the moose and sled. (b) The magnitude of the force of friction between the sled and the snow is 1400 N. As the hunter uniformly slows the snowmobile, what minimum length of time is needed for him to stop and keep the sled from running into the snowmobile (i.e., keep the same distance between the sled and the snowmobile)? 09-Phys20-Chap09.qxd 7/24/08 2:44 PM Page 507 71. A 940-kg car is travelling at 15 m/s [W] when it is struck by a 1680-kg van moving at 20 m/s [50.0 N of E]. If both vehicles join together after impact, what will be the velocity of their centre of mass immediately after impact? 78. Research the types of rockets being used in NASA’s current launchings. What is their thrust and time of firing? If possible, obtain data to calculate the impulse on the rocket. How much does the mass of a rocket change over time as it accelerates? 72. A 0.450-kg soccer ball is kicked parallel to the floor at 3.24 m/s [E]. It strikes a basketball sitting on a bench, driving it at 2.177 m/s [30.0 S of E]. The soccer ball goes off at 1.62 m/s [60.0 N of E]. What is the mass of the basketball? 73. A cue ball moving at 2.00 m/s [0.0] hits a stationary three-ball, sending it away at 1.58 m/s |
[36.0]. The cue ball and three-ball each have a mass of 0.160 kg. Calculate the velocity of the cue ball immediately after collision. Ignore friction and rotational effects. 74. A hunter claims to have shot a charging bear through the heart and “dropped him in his tracks.” To immediately stop the bear, the momentum of the bullet would have to be as great as the momentum of the charging bear. Suppose the hunter was shooting one of the largest hunting rifles ever sold, a 0.50 caliber Sharps rifle, which delivers a 2.27 102 kg bullet at 376 m/s. Evaluate the hunter’s claim by calculating the velocity of a 250-kg bear after impact if he was initially moving directly toward the hunter at a slow 0.675 m/s [S]. 75. An object explodes into three fragments (A, B, and C) of equal mass. What will be the approximate direction of fragment C if (a) both fragments A and B move north? (b) fragment A moves east and fragment B moves south? (c) fragment A moves [15.0] and fragment B moves [121]? Extensions 76. Research the physics principles behind the design of a Pelton wheel. Explain why it is more efficient than a standard water wheel. Begin your search at www.pearsoned.ca/school/physicssource. 77. A fireworks bundle is moving upward at 2.80 m/s when it bursts into three fragments. A 0.210-kg fragment (A) moves at 4.52 m/s [E]. A 0.195-kg fragment (B) flies at 4.63 m/s [N]. What will be the velocity of the third fragment (C) immediately after the explosion if its mass is 0.205 kg? Assume that no mass is lost during the explosion. 79. Two billiard balls collide off centre and move at right angles to each other after collision. In what directions did the impulsive forces involved in the collision act? Include a diagram in your answer. 80. A 2200-kg car travelling west is struck by a 2500-kg truck travelling north. The vehicles stick together upon impact and skid for 20 m [48.0 N of W]. The coefficient of friction for the tires on the road surface is 0.78. Both drivers claim to have been travelling at 90 km/h before the crash. Determine the truth of their statements. 81. Research the developments in running shoes that help prevent injuries. Interview running consultants, and consult sales literature and the Internet. How does overpronation or underpronation affect your body’s ability to soften the road shock on your knees and other joints? Write a brief report of your findings. Begin your search at www.pearsoned.ca/school/physicssource. 82. A 3.5-kg block of wood is at rest on a 1.75-m high fencepost. When a 12-g bullet is fired horizontally into the block, the block topples off the post and lands 1.25 m away. What was the speed of the bullet immediately before collision? Consolidate Your Understanding 83. Write a paragraph describing the differences between momentum and impulse. Include an example for each concept. 84. Write a paragraph describing how momentum and energy concepts can be used to analyze the motion of colliding objects. Include two examples: One is a one-dimensional collision and the other is a two-dimensional collision. Include appropriate diagrams. Think About It Review your answers to the Think About It questions on page 447. How would you answer each question now? e TEST To check your understanding of momentum and impulse, follow the eTest links at www.pearsoned.ca/school/physicssource. Unit V Momentum and Impulse 507 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 508 U N I T VI Forces Forces and Fields and Fields On huge metal domes, giant electrostatic charge generators can create voltages of 5 000 000 V, compared with 110 V in most of your household circuits. How are electrostatic charges produced? What is voltage? What happens when electric charges interact? e WEB The person in this photo is standing inside a Faraday cage. To find out how the Faraday cage protects her from the huge electrical discharges, follow the links at www.pearsoned.ca/school/physicssource. 508 Unit VI 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 509 Unit at a Glance C H A P T E R 1 0 Physics laws can explain the behaviour of electric charges. 10.1 Electrical Interactions 10.2 Coulomb’s Law C H A P T E R 1 1 Electric field theory describes electrical phenomena. 11.1 Forces and Fields 11.2 Electric Field Lines and Electric Potential 11.3 Electrical Interactions and the Law of Conservation of Energy C H A P T E R 1 2 Properties of electric and magnetic fields apply in nature and technology. 12.1 Magnetic Forces and Fields 12.2 Moving Charges and Magnetic Fields 12.3 Current-carrying Conductors and Magnetic Fields 12.4 Magnetic Fields, Moving Charges, and New and Old Technologies Unit Themes and Emphases • Energy and Matter • Nature of Science • Scientific Inquiry Focussing Questions While studying this unit, you will investigate how the science of electricity, magnetism, and electromagnetism evolved and its corresponding effect on technology. As you work through this unit, consider these questions. • How is the value of the elementary charge determined? • What is the relationship between electricity and magnetism? • How does magnetism assist in the understanding of fundamental particles? Unit Project Building a Model of a Direct Current Generator • By the time you complete this unit, you will have the knowledge and skills to build a model of a direct current generator. For this task, you will research wind power and design and build a model of an electric generator that uses wind energy. Unit VI Forces and Fields 509 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 510 Physics laws can explain the behaviour of electric charges. Figure 10.1 A thunderbird on a totem pole in Vancouver Abolt of lightning flashing across dark cloudy skies, followed a few moments later by the deafening sound of thunder, is still one of the most awe-inspiring physical events unleashed by nature. What is the cause of lightning? Why is it so dangerous? So powerful is this display that many early civilizations reasoned these events must be the actions of gods. To the Romans, lightning was the sign that Jove, the king of the gods, was angry at his enemies. In some First Nations traditions, lightning flashed from the eyes of the enormous thunderbird, while thunder boomed from the flapping of its huge wings (Figure 10.1). In this chapter, you will learn how relating lightning to simpler phenomena, such as the sparking observed as you stroke a cat, initially revealed the electrical nature of matter. Further studies of the nature of electric charges and the electrical interactions between them will enable you to understand laws that describe their behaviour. Finally, you will investigate the force acting on electric charges by studying the variables that determine this force and the law that describes how to calculate such forces. C H A P T E R 10 Key Concepts In this chapter, you will learn about: electric charge conservation of charge Coulomb’s law Learning Outcomes When you have completed this chapter, you will be able to: Knowledge explain electrical interactions using the law of conservation of charge explain electrical interactions in terms of the repulsion and attraction of charges compare conduction and induction explain the distribution of charge on the surfaces of conductors and insulators use Coulomb’s law to calculate the electric force on a point charge due to a second point charge explain the principles of Coulomb’s torsion balance experiment determine the magnitude and direction of the electric force on a point charge due to one or more stationary point charges in a plane compare, qualitatively and quantitatively, the inverse square relationship as it is expressed by Coulomb’s law and by Newton’s universal law of gravitation Science, Technology, and Society explain that concepts, models, and theories are often used in predicting, interpreting, and explaining observations explain that scientific knowledge may lead to new technologies and new technologies may lead to scientific discoveries 510 Unit VI 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 511 10-1 QuickLab 10-1 QuickLab Charging Objects Using a Van de Graaff Generator Problem What can demonstrations on the Van de Graaff generator reveal about the behaviour and interactions of electric charges? Materials and Equipment Van de Graaff generator and grounding rod small piece of animal fur (approximately 15 cm x 15 cm) 5 aluminium pie plates small foam-plastic cup with confetti soap bubble dispenser and soap CAUTION! Follow your teacher’s instructions to avoid getting an electric shock. Procedure 1 Copy Table 10.1 into your notebook. Make the table the full width of your page so you have room to write in your observations and explanations. Table 10.1 Observations and Explanations from Using a Van de Graaff Generator 2 Watch or perform each of the demonstrations in steps 3 to 9. 3 Place a piece of animal fur, with the fur side up, on the top of the charging sphere of the Van de Graaff generator. 4 Turn on the generator and let it run. 5 Record your observations in Table 10.1, making sure that your description is precise. 6 Ground the sphere with the grounding rod, and turn off the generator. 7 Repeat steps 3 to 6, replacing the animal fur with the aluminium pie plates (stacked upside down), and then the foam-plastic cup with confetti. 8 Turn on the Van de Graaff generator and let it run. 9 Dip the soap bubble dispenser into the soap and blow a stream of bubbles toward the charging sphere of the Van de Graaff generator. 10 Record your detailed observations in Table 10.1. 11 Ground the sphere with the grounding rod, and Demonstration Observation Explanation turn off the generator. Animal Fur Aluminium Pie Plates Foam-plastic Cup and Confetti Stream of Soap Bubbles Think About It Question 1. Using your knowledge of electricity, provide a possible explanation of the events that occurred during each demonstration. 1. How does the sphere of the Van de Graaff generator become charged? 2. Describe a situation during |
the demonstrations where the forces of interaction between the sphere of the generator and the various objects were: (a) attractive (b) repulsive 3. Why does touching the sphere with a grounding rod affect the charge on the sphere? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 10 Physics laws can explain the behaviour of electric charges. 511 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 512 10.1 Electrical Interactions Your world runs on electricity. The music you listen to, the movies you watch, the video games you play—all require electricity to run. Today, electricity is so familiar that you probably don’t even think about it when you turn on a light, pop a piece of bread into the toaster, or switch off the TV. Try to imagine a time before electricity was even named. People had noticed interesting effects in certain situations that seemed almost magical. The Greek philosopher Thales (624–546 BCE) recorded that when he rubbed amber (a hard fossilized form of tree resin), it could attract small pieces of straw or thread. This effect was called “electricity,” after the Greek word for amber, “elektron.” The ancient Greeks observed two important properties of electricity: • Charged objects could either attract or repel each other. These two types of interactions suggested that there must be two different types of charge. • Repulsion occurred when two similarly charged objects were placed near each other, and attraction occurred when two oppositely charged objects were placed near each other. These observations can be summarized as the law of charges: Like charges repel and unlike charges attract. M I N D S O N Electrical Attraction Rub an ebonite rod with fur and hold the rod close to a fine stream of water from a faucet. Then rub a glass rod with silk and hold this rod close to a fine stream of water. Observe what happens in each case. Using your knowledge of charging objects, explain why the ebonite rod or the glass rod affects the water. There was little progress in understanding the nature of electricity until the 1600s, when the English scientist William Gilbert (1544–1603) performed extensive investigations. In De Magnete, his book on magnetism, Gilbert compared the effects of electricity and magnetism. He concluded that: 1. Objects only exhibit electrical effects when recently rubbed; magnetic objects do not need to be rubbed. 2. Electrified objects can attract small pieces of many types of objects; magnetic objects can attract only a few types of objects. 3. Electrified objects attract objects toward one central region; magnetic objects appear to have two poles. Although Gilbert was able to describe certain effects of electricity, he still did not know the origins of electric charges. In the 1700s, the American scientist and inventor Benjamin Franklin (1706–1790) attempted to prove that lightning in the sky was the same electricity as the spark observed when you reach for a metal info BIT The plastic in contact lenses contains etafilcon, which is a molecule that attracts molecules in human tears. This electrostatic attraction holds a contact lens on the eye. info BIT Gilbert was also a medical doctor and held the prestigious position of personal physician of Queen Elizabeth I of England. 512 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 513 door handle after shuffling across a carpet. He performed his famous kite experiment to explore whether lightning was a form of electricity (Figure 10.2). Luckily, he did not get killed, and he succeeded in drawing electricity from the clouds. He observed that lightning behaves the same way that electricity produced in the laboratory does. Through further investigations, he identified and named the two different types of electric charges as positive and negative charges. It soon became apparent that electricity is in all substances. This idea caught the imagination of many different people. Scientists studied electricity’s intriguing effects (Figure 10.3), and entrepreneurs exploited it. Magicians and carnivals featured the “mysterious” effects of electricity. Figure 10.2 This figure is an artist’s representation of Benjamin Franklin’s famous kite experiment. info BIT Some historians think that Franklin may not have actually performed his kite experiment. They suspect that Franklin sent a description of this dangerous experiment to the Royal Society in London, England as a joke, because this British academy had largely ignored his earlier work. In 1753, the Royal Society awarded Franklin the prestigious Copley Medal for his electrical research. Figure 10.3 Boys were sometimes used in experiments such as this one in the early 1700s. The boy was suspended over the floor and electrostatically charged. His positive electric charge would attract pieces of paper. Studies to determine the nature of electricity continued. These studies were the beginning of the science of electrostatics, which is the study of electric charges at rest. It involves electric charges, the forces acting on them, and their behaviour in substances. electrostatics: the study of electric charges at rest The Modern Theory of Electrostatics Today’s theory of electrostatics and the nature of electric charges is based on the models of the atom that Ernest Rutherford (1871–1937) and Niels Bohr (1885–1962) proposed in the early 1900s. In their theories, an atom is composed of two types of charges: positively charged protons in a nucleus surrounded by negatively charged electrons. In nature, atoms have equal numbers of electrons and protons so that each atom is electrically neutral. Just as some materials are good thermal conductors or insulators, there are also good conductors and insulators of electric charges. Electrical conductivity depends on how tightly the electrons are bound to the nucleus of the atom. Some materials have electrons that are tightly bound to the nucleus and are not free to travel within the substance. These materials are called insulators. Materials that have electrons in the outermost regions of the atom that are free to travel are called conductors. insulator: material in which the electrons are tightly bound to the nucleus and are not free to move within the substance conductor: material in which electrons in the outermost regions of the atom are free to move Chapter 10 Physics laws can explain the behaviour of electric charges. 513 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 514 Figure 10.4 shows some examples of good conductors and insulators. Note that metals are usually good conductors. It is also interesting to note that a good conductor, such as silver, can have a conductivity 1023 times greater than that of a good insulator, such as rubber. Figure 10.4 Relative electrical conductivity of some materials Relative magnitude of conductivity 108 107 103 109 1010 1012 1015 Material silver copper aluminium iron mercury carbon germanium silicon wood glass rubber Conductors F O I L (computer chips) Semiconductors (transistors) Insulators Semiconductors Materials that lie in the middle, between good conductors and good insulators, are called semiconductors. Because of their nature, they are good conductors in certain situations, and good insulators in other situations. Selenium, for example, is an insulator in the dark, but in the presence of light, it becomes a good conductor. Because of this property, selenium is very useful in the operation of photocopiers (Figure 10.5). original copy face down movable light lens A mirror positively charged paper A negatively charged toner brush selenium-coated drum heater assembly mirror positively charged Figure 10.5 Photocopiers use the semiconductor selenium in the copying process. 514 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 515 The selenium-coated drum in the photocopier is initially given a positive charge and kept in the dark to retain the charge. When a flash of light shines on a document to be copied, an image of the document is transferred to the drum. Where the document is light-coloured, the selenium is illuminated, causing it to be conductive. Electrons flow into the conductive portions of the selenium coating, leaving them uncharged. The page remains light-coloured or white. Where the document is darkcoloured, the selenium remains non-conductive, and the positive charge remains. Negatively charged “toner” powder is sprinkled on the drum and attaches to the positively charged portions of the drum. When a sheet of paper is passed over the drum, the toner transfers to the paper and an image of the document is created. This toner image is then fused on the paper with heat, and the copying process is complete. Silicon and germanium are also semiconductors. They become conductors when atoms such as gallium or arsenic are added to them. This process is called “doping” with impurities. The field of solidstate electronics, which includes components such as transistors, diodes, and silicon chips, is based on this type of semiconductor. Superconductors Recall from earlier science studies that resistance is a measure of how difficult it is for electrons to flow through a material. Materials with a low electrical resistance are better conductors because very little energy is lost to heat in the conduction of electricity. Early attempts at conducting electricity efficiently used conducting materials with low electrical resistance, such as silver, copper, and gold. Researchers soon discovered that the electrical resistance of any material tends to decrease as its temperature decreases. Could the temperature of a material be lowered to the point that it loses all its resistive nature, creating the ideal conductor? This property of materials would have an enormous range of applications. Once a current is established in such a conductor, it s |
hould persist indefinitely with no energy loss. In the early 20th century, a class of materials called superconductors was developed. These conductors have no measurable resistance at very low temperatures. The Dutch physicist Heike Kammerlingh Onnes (1853–1926) discovered this effect in 1911 when he observed that solid mercury lost its electrical resistance when cooled to a temperature of 269 C. Although this discovery was significant, the usefulness of superconductors was limited because of the extremely low temperatures necessary for their operation. It was not until 1986 that materials were developed that were superconductors at much higher temperatures. These materials are ceramic alloys of rare earth elements, such as lanthanum and yttrium. As an example, one such alloy was made by grinding yttrium, barium, and copper oxide into a mixture and heating the mixture to form the alloy YBa2Cu3O7. This substance became a superconductor at 216 C. In 1987, another alloy was developed that displayed superconductivity at 175 C. More recent discoveries have reported copper oxide alloys that are superconductors at temperatures as high as 123 C. The ultimate goal is to develop superconductors that operate at room temperature, thus creating a whole new era of useful applications in technology. e WEB Find out what research is being done on superconductors today. How soon will you be seeing superconductors in use around the house? Write a brief summary of what you discover. To learn more about superconductors, follow the links at www.pearsoned.ca/school/ physicssource. Chapter 10 Physics laws can explain the behaviour of electric charges. 515 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 516 10-2 Inquiry Lab 10-2 Inquiry Lab Charging Objects Question How can objects become electrically charged? Materials and Equipment 2 white plastic polyethylene strips (or ebonite rods) fur (approximately 15 cm x 15 cm) 2 clear plastic acetate strips (or glass rods) silk (approximately 15 cm x 15 cm) electroscope silk thread 2 retort stands with clamps Procedure Part A: Charging by Friction 1 Copy Table 10.2 into your notebook. Table 10.2 Observations for Charging by Friction White Polyethylene Strip Rubbed with Fur Clear Acetate Strip Rubbed with Silk Hanging White Polyethylene Strip Hanging Clear Acetate Strip Hanging White Polyethylene Strip Hanging Clear Acetate Strip 2 Hang a white polyethylene strip from one retort stand and a clear acetate strip from another retort stand. 3 While holding the hanging white polyethylene strip in the middle, rub both ends of it with the fur. While holding the hanging clear acetate strip in the middle, rub both ends of it with the silk. 4 Rub the other white polyethylene strip with the fur. 5 Carefully bring this second polyethylene strip close to one end of the hanging white polyethylene strip. Do not allow the two plastic strips to touch each other. 6 Carefully bring the second polyethylene strip close to one end of the hanging clear acetate strip. Do not allow the two plastic strips to touch each other. 7 Observe what happens in each situation and record your observations in Table 10.2. 516 Unit VI Forces and Fields Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 8 Rub a clear acetate strip with the silk. 9 Carefully bring this strip close to the hanging clear acetate strip. Do not allow the two plastic strips to touch each other. Observe what happens and record your observations in Table 10.2. 10 Carefully bring this clear acetate strip close to one end of the hanging white polyethylene strip. Do not allow the two plastic strips to touch each other. Observe what happens and record your observations in Table 10.2. Part B: Charging by Conduction 11 Copy Table 10.3 into your notebook. Table 10.3 Observations for Charging by Conduction Electroscope Charged with the White Polyethylene Strip Rubbed with Fur Electroscope Charged with the Clear Acetate Strip Rubbed with Silk 12 Rub the unattached white polyethylene strip with the fur. Touch this white strip to the knob of the electroscope. 13 Carefully bring the electroscope close to one end of the hanging white polyethylene strip. Observe what happens to the leaves in the electroscope. Record your observations in Table 10.3. 14 Now bring the electroscope near one end of the hanging clear acetate strip. Observe what happens to the leaves in the electroscope. Record your observations in Table 10.3. 15 Ground the electroscope by touching the knob of the electroscope with your finger. 16 Rub a clear acetate strip with the silk and touch the strip to the knob of the grounded electroscope. 17 Repeat steps 13 and 14. 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 517 Part C: Charging by Induction 18 Copy Table 10.4 into your notebook. Table 10.4 Observations for Charging by Induction Grounded Electroscope Hanging White Polyethylene Strip Hanging Clear Acetate Strip 19 Bring an uncharged electroscope near one end of the hanging white polyethylene strip. Observe what happens to the leaves of the electroscope, and record your observations in Table 10.4. 20 Bring an uncharged electroscope near one end of the hanging clear acetate strip. Observe what happens to the leaves of the electroscope, and record your observations in Table 10.4. Analysis 1. What effect did you observe when two similarly charged white polyethylene strips were held near each other or when two similarly charged clear acetate strips were held near each other? 2. What effect did you observe when a charged white polyethylene strip was held near an oppositely charged hanging clear acetate strip or when a charged clear acetate strip was held near an oppositely charged hanging white polyethylene strip? 3. Based on your observations, what charge did the electroscope acquire when it was touched by the charged white polyethylene strip? when it was touched by the charged acetate strip? 4. What evidence shows a movement of charge in the electroscope when it is held near a charged object? 5. From your observations in Table 10.2, what general rule can you formulate about attraction and repulsion of charged objects? 6. From your observations in Table 10.3, what general rule can you formulate about the charge received by an object when it is touched by another charged object? 7. From your observations in Table 10.4, what general rule can you formulate about the charge received by an object when it is held near another charged object? 8. Does the electroscope acquire a net electrical charge during the process of charging by induction? Justify your answer. 9. What evidence is there from this investigation to prove that there are two types of electrical charges? 10. From the investigation, is there any evidence to prove which type of charge was developed on the white polyethylene strip and on the clear acetate strip? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Methods of Charging Objects According to the modern theory of electrostatics, objects can become charged through a transfer of electrons. Electron transfer can occur in three ways: by friction, by conduction, and by induction. Law of Conservation of Charge During any charging procedure, it is important to keep in mind that new charges are not being created. The charges existing in materials are merely being rearranged between the materials, as the law of conservation of charge states: The net charge of an isolated system is conserved. Net charge is the sum of all electric charge in the system. For example, if a system contains 3 C of charge and 5 C of charge, the system’s net coulomb (C): SI unit for charge, equivalent to the charge on 6.25 1018 electrons or protons Chapter 10 Physics laws can explain the behaviour of electric charges. 517 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 518 charge is 2 C. Suppose you have a system that initially consists of two electrically neutral objects, and there is a transfer of electrons from one object to the other. One object will lose electrons and become positively charged while the other object will gain these electrons and become equally negatively charged. However, the net charge of the system is still zero. Charges have not been created, they have only been rearranged. Charging Objects by Friction The most common method of charging objects is by rubbing or friction. You have probably had the unpleasant experience of receiving a shock when you touched a door handle after walking across a carpeted floor. Similarly, gently stroking a cat can result in the generation of small sparks, which are very uncomfortable for the cat. Charging by this method involves separating electrons from the atoms in one object through rubbing or friction, and then transferring and depositing these electrons to the atoms of another object. The object whose atoms lose electrons then possesses positively charged ions. The object whose atoms gain electrons possesses negatively charged ions. As shown in Figure 10.6, rubbing the ebonite rod with fur transferred some of the electrons in the fur to the rod. The fur becomes positively charged, and the rod becomes negatively charged. (a) (b) hold electrons tightly sulfur brass copper ebonite paraffin wax silk lead fur wool glass Figure 10.6 (a) A neutral ebonite rod and a neutral piece of fur have equal amounts of negative and positive charge. When the fur is rubbed against the rod, a transfer of electrons occurs. (b) After rubbing, the ebonite has gained electrons and has a net negative charge. The fur has lost electrons and has a net positive charge. Whether an object gains or loses electrons when rubbed by another object depends on how tightly the material holds onto its electrons. Figure 10.7 shows the electrostatic series, in which substances are listed according to how tightly they hold their electrons. Substances at the top have a strong hold on their electrons and do not lose electrons ea |
sily. Substances near the bottom have a weak hold on their electrons and lose them easily. hold electrons loosely Concept Check Figure 10.7 The electrostatic or triboelectric series 1. Using information from Figure 10.7, explain why ebonite acquires a greater charge when rubbed with fur rather than silk. 2. What type of charge does ebonite acquire when rubbed with fur? Charging objects by friction can also occur during collisions. The collisions of water vapour molecules in rain clouds, for example, cause 518 Unit VI Forces and Fields conduction: process of charging an object through the direct transfer of electrons when a charged object touches a neutral object 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 519 the separation and transfer of electrons. The result is that vapour molecules become positively or negatively charged, eventually resulting in lightning. You will learn more about lightning later in this chapter. This process of charging objects was also observed by the Voyageur spacecraft on its mission to Saturn. Colliding particles in the rings of Saturn create electrical discharges within the rings, similar to lightning on Earth. Charging Objects by Conduction Objects can become charged by the transfer of electrons from a charged object to an uncharged object by simply touching the objects together (Figure 10.8). This process is called charging by conduction. (a) (b) Figure 10.8 (a) During charging by conduction, electrons from a negatively charged metal conducting sphere transfer to a neutral metal conducting sphere, upon contact. (b) The neutral sphere gains electrons and is said to have been charged by conduction. The quantity of charge that transfers from one object to another depends on the size and shape of the two objects. If both objects are roughly the same size and shape, the charge transferred will be such that both objects are approximately equally charged (Figure 10.9(a)). If one sphere is larger than the other, then the larger sphere will receive more of the charge (Figure 10.9(b)). When the spheres are separated, the excess charges move to become equidistant from each other because of the forces of repulsion between like charges. Charging by conduction is similar to charging by friction because there is contact between two objects and some electrons transfer from one object to the other. (a) (b) Figure 10.9 Electrostatic repulsion of like charges forces excess charges within objects to redistribute so that the distances between charges are equal. (a) If two objects are the same size, the charges redistribute equally. (b) If the two objects are different sizes, the object with a larger surface area has more charges. Once the charge has transferred to another object, it will either be distributed over the surface of the object, if the object is a conductor, or remain on the surface at the point of contact, if the object is an insulator. Chapter 10 Physics laws can explain the behaviour of electric charges. 519 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 520 excess negative charge neutral Figure 10.10 A piece of paper appears to be attracted to a charged ebonite rod, even before they touch. induction: movement of charge caused by an external charged object Charging Objects by Induction If you bring a negatively charged ebonite rod slowly toward a small piece of uncharged paper, the rod will attract the piece of paper, as shown in Figure 10.10. In fact, the piece of paper would begin to jiggle and move toward the rod even before the rod touches it. This reaction is a result of the forces acting on electrostatic charges. You know that electrostatic attraction can occur only between oppositely charged objects, but how can a charged object attract a neutral or uncharged object? And why is there never a force of repulsion between a charged object and a neutral object? The answers to these questions are revealed in the third method of charging objects, which involves two processes: induction and charging by induction. Induction Induction is a process in which charges in a neutral object shift or migrate because of the presence of an external charged object. This temporary charge separation polarizes the neutral object. One side of the object becomes positively charged and the other side is equally negatively charged. Although the object now behaves as if it is charged, it is still electrically neutral. The charging object and the neutral object do not touch each other, so there is no actual transfer of charge. (a) (b) negatively charged negatively charged Figure 10.11 (a) A neutral metal sphere and a neutral piece of paper (b) The influence of the large negative charge of a rod causes charge migration within the conducting sphere, which polarizes the sphere. The influence of the rod causes charge shift within the atoms of the insulating paper. The atoms in the paper become polarized. Because of induction, the sides of the sphere and the atoms in the paper that are positively charged are closer to the negatively charged rod than their negatively charged sides are. The net result is attraction. The process of induction varies slightly, depending on whether the charging object is approaching a substance that is an insulator or a conductor. Figure 10.11(a) depicts a neutral metal sphere (conductor) and a neutral piece of paper (insulator). Figure 10.11(b) shows a negatively charged rod approaching each neutral object. The electrons in the two neutral objects are repelled by the negative charge of an ebonite rod. The metal sphere is a conductor, so the electrons can move easily through it to its other side. This process of charge migration causes the sphere to become polarized, where one side of the sphere is positive and the other side is negative. charge migration: movement of electrons in a neutral object where one side of the object becomes positive and the other side becomes negative 520 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 521 Since the paper is an insulator, its electrons cannot move easily through it, so they just shift slightly relative to the nuclei. This process of charge shift causes the atoms to become polarized, where one side of an atom becomes positive and the other side becomes negative. In both cases, the distance from the negatively charged rod to the positive end of the neutral object is less than the distance to the negative end of the object. Therefore, the attraction of the opposite charges is greater than repulsion of like charges, and the net force is attractive. Charge separation by induction, which results in polarization of objects, explains electrostatic situations such as the initial attraction of a neutral piece of paper to a negatively charged rod without contact, as you saw in Figure 10.10. Charging by Induction In the situation shown in Figure 10.11, the electrons in the metal sphere and the paper return to their original positions when the negatively charged rod is removed. The objects lose their polarity and remain electrically neutral. For conductors, like the metal sphere, it is possible to maintain a residual charge by adding a grounding step. Grounding involves touching or connecting a wire from the object to the ground, as shown in Figure 10.12(a). The grounding path is then removed while the source charge is still present. The grounding step allows the conductor to maintain a charge (Figure 10.12(b)). The complete process of charging by induction includes grounding. (a) (b) charge shift: movement of electrons in an atom where one side of an atom becomes positive and the other side becomes negative grounding: the process of transferring charge to and from Earth charging by induction: the process of polarizing an object by induction while grounding it Figure 10.12 (a) While the charged rod is held near the metal sphere, the sphere remains polarized by induction. Grounding the sphere removes excess charge. In this situation, the sphere appears to have excess electrons on one side, which are removed. The positive charges cannot move because they represent the fixed nuclei of atoms. (b) After the ground and charged rod are removed, the sphere retains a net positive charge because of the loss of electrons. It has been charged by induction. PHYSICS INSIGHT The symbol for ground is A grounded conductor that is polarized by the presence of a charged object will always have a net charge that is opposite to that of the charged object if the ground is removed before the charged object is removed. The conductor has been charged by induction. Chapter 10 Physics laws can explain the behaviour of electric charges. 521 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 522 Concept Check When you rub a balloon on your hair, you are charging it by friction. Explain why the balloon will then stick to a wall for a long time. How Lightning Gets Its Charge Many theories attempt to explain the formation of lightning. One theory relates the cause to the processes of evaporation and condensation of water in the clouds and different methods of charging objects. Under the right conditions, a churning cloud formation causes water vapour molecules to collide, resulting in a transfer of electrons between these molecules. The transfer of an electron from one water molecule to another leaves the molecules oppositely charged. plasma: highly ionized gas containing nearly equal numbers of free electrons and positive ions Cooling causes water vapour molecules to condense into water droplets. The atoms in these droplets hold onto electrons more readily than atoms in water vapour, and thus the droplets become negatively charged. Being heavier, these negatively charged water droplets accumulate at the bottom of the cloud, causing the bottom of the cloud to become negatively charged (Figure 10.13). The top of the cloud, containing the rising water vapour, becomes positively charged. The increasing polarization of the cloud ionizes the surrounding air, forming a conduct |
ive plasma. Excess electrons on the bottom of the cloud begin a zigzag journey through this plasma toward the ground at speeds of up to 120 km/s, creating a step leader. This is not the actual lightning strike. The presence of the large negative charge at the bottom of the cloud causes the separation of charges at that location on Earth’s surface. Earth’s surface at that spot becomes positively charged, and the area below the surface becomes negatively charged. Charge separation has polarized Earth’s surface. Air molecules near Earth’s surface become ionized and begin to drift upward. This rising positive charge is called a streamer. When the rising positive streamer meets the step leader from the clouds, at an altitude of about 100 m, a complete pathway is formed and the lightning begins. A transfer of negative charge in the form of a lightning strike from the cloud travels to Earth’s surface at speeds of up to 100 000 km/s (Figure 10.14). Figure 10.13 Lightning forms when the bottom of the cloud becomes negatively charged and Earth’s surface becomes positively charged. Figure 10.14 A streamer moving up from Earth’s surface meets a step leader coming down from the clouds and lightning lights up the sky. 522 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 523 10.1 Check and Reflect 10.1 Check and Reflect Knowledge 1. What is the science of electrostatics? 2. Describe a simple experiment that enabled early scientists to determine that there were two different types of charges. 3. In the 1600s, William Gilbert compared the effects of electricity and magnetism. (a) Describe two similarities between 7. Describe how you would charge the sphere in Figure 10.12 negatively by induction. 8. A negatively charged ebonite rod is brought near a neutral pith ball that is hanging by an insulating thread from a support. Describe what happens (a) before they touch (b) after they touch these effects. 9. Compare the distribution of charge (b) Describe two differences between these effects. 4. In the classification of substances by electrical conductivity, a substance may be a conductor, insulator, semiconductor, or superconductor. (a) What property of matter determines the electrical conductivity of a substance? (b) List the classifications given above in order of increasing electrical conductivity. (c) Give an example of a substance in each classification. (d) Describe the conditions when the semiconductor selenium becomes a conductor or an insulator. (a) on hanging aluminium and glass rods if both are touched at one end by a negatively charged ebonite rod (b) after a small negatively charged metal sphere momentarily touches a larger neutral metal sphere 10. A negatively charged ebonite rod is held near the knob of a neutral electroscope. (a) Explain what happens to the leaves of the electroscope. (b) Explain what happens to the leaves of the electroscope if the other side of the knob is now grounded while the rod is still in place. (c) Explain why removing the ground first and then the rod will leave a net charge on the electroscope. Applications Extensions 5. (a) An ebonite rod is rubbed with fur. 11. You are given an ebonite rod, fur, an How can the electrostatic series chart in Figure 10.7 on page 518 help you determine which object will become negatively charged? (b) Why is it better to rub an ebonite rod with fur rather than silk? 6. Describe how you could charge a glass sphere positively using the following methods: electroscope, and a sphere of unknown charge. Describe an experimental procedure that you could use to determine the charge on the sphere. 12. If a glass rod becomes positively charged when rubbed with silk, use the law of conservation of charge to explain why the silk must be negatively charged. (a) friction (b) conduction (c) induction e TEST To check your understanding of electrical interactions, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 10 Physics laws can explain the behaviour of electric charges. 523 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 524 info BIT Charles de Coulomb was given credit for investigations into the electrostatic forces acting on charged objects, but the actual discovery of the relationship between electrostatic forces and charged objects was made earlier by Henry Cavendish (1731–1810), an English scientist. However, he was so shy that he didn’t publish his results and thus was not credited with the discovery. PHYSICS INSIGHT Vertical bars around a vector symbol are an alternative method for representing the magnitude of a vector. This notation is used for the magnitude of force and field vectors from this chapter on. This notation avoids confusing the symbol for energy, E, with the magnitude of the electric field strength , which is introvector, E duced in the next chapter. 10.2 Coulomb’s Law In Chapter 4, you studied Newton’s law of universal gravitation and learned that any two objects in the universe exert a gravitational force g). The magnitude of this force of gravitational attraction on each other (F is directly proportional to the product of the two masses (m1 and m2): F g m1m2 and inversely proportional to the square of the distance between their centres (r): F g 1 r 2 These relationships can be summarized in the following equation: F g G m1m2 r 2 where G is the universal gravitational constant in newton-metres squared per kilogram squared. Charles de Coulomb suspected that the gravitational force that one mass exerts on another is similar to the electrostatic force that one charge exerts on another. To verify his hypothesis, he constructed an apparatus called a torsion balance to measure the forces of electrostatics. Although it could not be used to determine the quantity of charge on an object, Coulomb devised an ingenious method to vary the quantity of charge in a systematic manner. 10-3 Inquiry Lab 10-3 Inquiry Lab Investigating the Variables in Coulomb’s Law Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question Two charged objects exert electrostatic forces of magnitude Fe on each other. How does Fe depend on the charges carried by the objects and on the separation between the objects? Materials and Equipment Van de Graaff generator or ebonite rod and fur 3 small Styrofoam™ or pith spheres, about 1 cm in diameter, coated with aluminium or graphite paint Hypothesis State a hypothesis relating the electrostatic force and each of the variables. Remember to write an “if/then” statement. Variables Read the procedure for each part of the inquiry and identify the manipulated variable, the responding variable, and the controlled variables in each one. sewing needle about 65 cm of thread 3 drinking straws 2 retort stands and 2 clamps balance 2 rulers tape small mirror (about 10 cm long and 5 cm wide) marking pen 524 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 525 Procedure 1 Determine the mass of one sphere (sphere 1) and record this mass in kilograms in your notebook. 2 Using the sewing needle, draw the thread through the centre of this sphere and attach both ends of the thread, with tape, to both ends of a drinking straw so that the sphere is suspended in the centre with the thread forming a V pattern. Clamp this straw, horizontally, to a retort stand, as shown in Figure 10.15. 3 Carefully insert the second drinking straw into the second sphere (sphere 2). Then fasten this drinking straw, horizontally, to the clamp on the other retort stand. 7 Tape the mirror on the paper so that the image of sphere 1 aligns with the centre of the mirror. Tape a ruler over the mirror just below the image of sphere 1. Using the marking pen, mark the position of the centre of the image of sphere 1 on the mirror while sighting in such a way that the sphere lines up with its image. 8 Do part A of the activity, which begins below, followed by part B. mirror line indicating top position of thread attached to straw image of sphere 1 string sphere 1 sphere 3 4 Adjust the clamps so that both spheres are at the paper same height. sphere 2 5 Carefully insert the third drinking straw into the third sphere (sphere 3). (This sphere will be the grounding sphere that will be used to change the charges on spheres 1 and 2.) 6 Tape sheets of white paper to the wall. Place the retort stand with sphere 1 close to the wall, about 5 cm away, so that the centre of the sphere aligns with the centre of the paper. Using a ruler, draw a horizontal line on the paper indicating the top position of the string attached to the straw. Figure 10.15 Part A: The relationship between the quantity of charge on each object and the electrostatic force In this part of the lab, the distance between the spheres is held constant while the charges on the spheres are varied. 9 Copy Table 10.5 into your notebook, leaving out the numbers in parentheses in the first three columns. Use these numbers only if you are unable to measure the distances accurately. They are hypothetical values that you can use to complete the rest of the activity. Table 10.5 Data and Calculations for Part A Magnitude of Weight of Sphere 1 F g (103 N) F mg g (1.28) (1.28) (1.28) (1.28) (1.28) Vertical Height of Sphere 1 dy (m) Horizontal Distance to Centre of Sphere 1 from the Centre Mark dx (m) Product of Charges of Spheres 1 and 2 (q2) Magnitude of Force of Electrostatic Repulsion Acting on Spheres 1 and 2 (N) e F (0.300) (0.300) (0.300) (0.300) (0.300) (0.0500) (0.0250) (0.0130) (0.0062) (0.0030) q 1 q1 1 2 2 4 q 1 q1 1 4 8 2 1 q1 1 q 6 1 4 4 q1 1 1 q 8 4 2 3 1 q1 1 q 4 6 8 8 Chapter 10 Physics laws can explain the behaviour of electric charges. 525 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 526 10 Rub the ebonite rod with the fur or charge the Van de Graaff generator. 11 Carefully touch sphere 2 to the charged rod or generator to charge the sphere by conduction. Since it is nearly impossible to measure the q |
uantity of charge transferred to sphere 2, assume that the quantity of charge on the sphere is q. 15 To obtain more data, vary the charge on each sphere using sphere 3. Gently touch sphere 3 to sphere 1. Since the charge on each sphere should be shared equally, the new charge on sphere 1 is 1 q. The 4 q 1 q 1 charge product on spheres 1 and 2 is now 1 q 2. 8 2 4 Label the new position of sphere 1 as position 2. 12 Slide the stand holding charged sphere 2 toward 16 Remove sphere 3 to a safe distance and ground it by sphere 1 on the other stand and momentarily touch the two spheres together. The charge on each object can be assumed to be 1 q because, on contact, the charge 2 is equally divided between two similar spheres. q 1 q 1 Therefore, the charge product is 1 q 2. 4 2 2 13 Slide sphere 2, parallel to the wall, to the left until the centre of its image is 1.0 cm away from the centre mark on the mirror. 14 Mark the new position of the centre of the image of sphere 1 on the mirror. Label it position 1. gently touching it with your hand. 17 Repeat steps 14 to 16, keeping sphere 2 in position and alternately touching spheres 1 and 2 with sphere 3 to obtain three more readings. Label these positions 3, 4, and 5. 18 With a ruler, accurately measure the vertical distance from the centre of the image of sphere 1 to the top horizontal line. Record this vertical distance (dy ) in metres in the appropriate column of Table 10.5. 19 Measure the distance between the original centre mark on the mirror and the centre of the image of sphere 1 in its new position for each trial. Record this distance (dx) in metres in the appropriate column of Table 10.5. Part B: The relationship between the distance between two charges and the electrostatic force In this part of the lab, the charges on the spheres are fixed while the distances are varied. 20 Copy Table 10.6 into your notebook, leaving out the numbers in parentheses in the first three columns. Use these numbers only if you are unable to measure the distances accurately. They are hypothetical values that you can use to complete the rest of the activity. Table 10.6 Data and Calculations for Part B Magnitude of Weight of Sphere 1 F g (103 N) F mg g (1.28) (1.28) (1.28) (1.28) (1.28) Vertical Height of Sphere 1 dy (m) Horizontal Distance to Centre of Sphere 1 from the Centre Mark dx (m) Distance Between the Centres of Spheres 1 and 2 r (m) Magnitude of Force of Electrostatic Repulsion Acting on Spheres 1 and 2 (N) e F (0.300) (0.300) (0.300) (0.300) (0.300) (0.0500) (0.0467) (0.0438) (0.0409) (0.0383) (0.0600) (0.0617) (0.0638) (0.0659) (0.0683) 526 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 527 21 Rub the ebonite rod with the fur or charge the Van de 2. Construct a graph of the force of electrostatic Graaff generator. 22 Carefully touch sphere 2 to the rod or the generator to give it a charge. 23 Slide the stand holding charged sphere 2 toward sphere 1 on the other stand and momentarily touch the two spheres together. repulsion on the y-axis as a function of the charge product on the x-axis. 3. What does the shape of the graph indicate about the relationship between the force of electrostatic repulsion and the charge product? 4. Complete the calculations as indicated in Table 10.6 for 24 Slide sphere 2 to a position so that the centre of its Part B. image is 0.5 cm to the left of the original centre mark on the mirror. 25 Mark the new position of the image of sphere 1 on the mirror and label it position 1. 26 Repeat steps 24 and 25, sliding sphere 2 parallel to the wall so that its image positions are at 1.0 cm, 1.5 cm, 2.0 cm, and 2.5 cm away, to obtain four more readings. Label these positions 2, 3, 4, and 5. 27 With a ruler, accurately measure the vertical distance from the centre of the image of sphere 1 to the horizontal line representing the top position of the string. Record this vertical distance (dy) in metres in the appropriate column of Table 10.6. 28 Measure the distance from the original centre mark on the mirror to the centre of the image of sphere 1 in its new position for each trial. Record this distance (dx) in metres in the appropriate column of Table 10.6. 29 Measure the distance between the centres of the two spheres. For each trial, record this distance (r) in metres in the appropriate column of Table 10.6. Analysis 1. Although the force of electrostatic repulsion F e acting on two similarly charged spheres cannot be measured directly, it can be calculated as shown in Figure 10.16. Complete the calculations as indicated in Table 10.5 for Part A. 5. Construct a graph of the force of electrostatic repulsion on the y-axis as a function of the distance between the charges on the x-axis. 6. What does the shape of the graph indicate about the relationship between the force of electrostatic repulsion and the distance between the two charges? 7. From the investigation, identify the two variables that affect the force of electrostatic repulsion acting on two charges. 8. Using a variation statement, describe the relationship between these two variables and the force of electrostatic repulsion. 9. Does your investigation confirm your hypotheses about the relationship between the variables and the electrostatic force? Why or why not? 10. How does the relationship between the variables affecting the electrostatic force in this investigation compare with that of the variables affecting the force of gravitational attraction in Newton’s law of gravitation? 11. (a) What sources of error could have led to inaccuracy in the investigation? (b) What modifications to the investigation would you recommend? dy dx Fg dy Fe dx Fe Fg Figure 10.16(a) Use the concept of similar triangles: F dy dx dx dy e F g F e F g Figure 10.16(b) The shaded triangles are similar. Chapter 10 Physics laws can explain the behaviour of electric charges. 527 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 528 The Force of Electrostatic Repulsion or Attraction Coulomb correctly hypothesized that the two factors influencing the magnitude of the electrostatic force that one charge exerts on another were the magnitudes of the charges on each object and their separation distance. To experimentally derive the relationships between the two factors and the electrostatic force, Coulomb used a procedure similar to that used in the 10-3 Inquiry Lab but with a different apparatus—the torsion balance shown in Figure 10.17. To determine the force of electrostatic attraction or repulsion acting on two charged objects, a charged ball on a rod is brought near a charged object at one end of the arm of the torsion balance. Repulsion or attraction causes the ball on the arm to move, rotating the arm. As the arm rotates, a sensitive spring either tightens or loosens, causing the needle to move a proportional angle. This movement of the needle can be measured on a scale. The amount of movement is related to a measure of the force of electrical attraction or repulsion. Figure 10.17 Coulomb’s apparatus Determining Relative Charge e MATH In 10-3 Inquiry Lab, Investigating the Variables in Coulomb’s Law, you learned how separation and the magnitude of electric charges affect the electrostatic force. To graph the electrostatic force as a function of separation, and to analyze this relationship in more depth, visit www.pearsoned.ca/school/ physicssource. e SIM Explore the inverse square relationship through a simulation using a sphere of uniform charge density. Follow the eSim links at www.pearsoned.ca/ school/physicssource. Realizing that there was not yet any way of measuring charge, Coulomb devised a method of accurately determining the relative magnitude of a charge. He knew that if a charged object with a charge of q touches a similar uncharged object, then the charge would be shared equally so that each object would have a charge of 1 q. Using this 2 assumption, he was able to do his experiment. Investigating the relationship between the electrostatic force and the distance between the centres of the spheres, he first charged a sphere with a charge q and touched it momentarily to the sphere on the torsion balance. Each sphere would then have a similar and equal q and 1 charge of 1 q. Then, holding the first sphere a measured distance 2 2 from the sphere in the torsion balance, he was able to measure the electrostatic force acting on the two spheres by the movement of the needle on the calibrated scale. Changing the distance between the spheres and measuring the force each time, he demonstrated there was an inverse square relationship between the electrostatic force and the separation distance. This relationship can be expressed as F e 1 r 2 Investigating further the relationship between the magnitude of the force and the magnitudes of the charges, he was able to accurately vary the charges on each sphere. By charging one sphere with a charge q and touching it to the sphere on the balance, he knew that the charge would be shared equally. The two spheres would have charges of 1 q and 1 q 2 2 q)(1 each, and the charge product would be (1 q). By touching each 2 2 charged sphere alternately with a third neutral and similar sphere, he q)(1 q), then (1 q)(1 could vary the charge products as (1 q), and so on. 4 4 2 4 528 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 529 By varying the charges on both objects and measuring the electrostatic force acting on them, he demonstrated that the magnitude of the electrostatic force is proportional to the product of the two charges: F e q1q2 In 1785, using the results from his experimentation on charged objects, Charles de Coulomb summarized his conclusions about the electrostatic force. This force is also known as the Coulomb force. His summary of his conclusions is called Coulomb’s law. The magnitude of the force of electrostatic attraction or repulsion (F e • directly proportional to the product of the two charges q1 and q2: ) is: F e q1q2 • in |
versely proportional to the square of the distance between their centres r : F e 1 2 r If these are the only variables that determine the electrostatic force, then F e q1q2 r 2 The beautiful fact about Coulomb’s law and Newton’s law of gravitation is that they have exactly the same form even though they arise from different sets of operations and apply to completely different kinds of phenomena. The fact that they match so exactly is a fascinating aspect of nature. Although Coulomb was able to identify and determine the relationships of the variables that affect the electrostatic force acting on two charges, he was unable to calculate the actual force. To do so, the variation statement must be converted to an equation by determining a proportionality constant (k), whose value depends on the units of the charge, the distance, and the force. At the time, however, it was impossible to measure the exact quantity of charge on an object. The Magnitude of Charges The SI unit for electric charge is the coulomb (C). A bolt of lightning might transfer 1 C of charge to the ground, while rubbing an ebonite rod with fur typically separates a few microcoulombs (C). It is difficult to build up larger quantities of charge on small objects because of the tremendous repulsive forces between the like charges. As you will see in section 15.2, experiments at the beginning of the 20th century showed that an electron has a charge of about 1.60 1019 C. So, 1 C of negative charge corresponds to the charge on 6.25 1018 electrons, or 6.25 billion billion electrons. Similarly, the charge on a proton is about 1.60 1019 C. PHYSICS INSIGHT Newton’s law of gravitation is called an inverse square law because the gravitational force acting on any two masses is inversely proportional to the square of the distance between their centres. Chapter 10 Physics laws can explain the behaviour of electric charges. 529 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 530 info BIT The electrostatic force acting on two charges of 1 C each separated by 1 m is about 9 109 N. This electrostatic force is equal to the gravitational force that Earth exerts on a billion-kilogram object at sea level! Given these values, physicists were able to calculate the constant of proportionality for Coulomb’s law. With this constant, Coulomb’s law becomes: F e 1q q 2 k 2 r is the magnitude of the force of electrostatic attraction or where F e repulsion in newtons; q1 and q2 are the magnitudes of the two charges in coulombs; r is the distance between the centres of the charges in metres; k is the proportionality constant called Coulomb’s constant and is equal to 8.99 109 Nm2/C2. This electrostatic force is attractive if the two objects have opposite charges and repulsive if the two objects have like charges. This equation is used to determine electrostatic forces in many different types of problems involving charges and the electrostatic forces acting on them. Examples 10.1 and 10.2 show how to calculate the electrostatic force of attraction or repulsion acting on two charges in a one-dimensional situation. Example 10.1 Practice Problem 1. In a hydrogen atom, an electron is 5.29 1011 m from a proton. An electron has a charge of 1.60 1019 C, and the proton’s charge is 1.60 1019 C. Calculate the electrostatic force of attraction acting on the two charges. Answer 1. 8.22 108 N [attraction] A small metal sphere with a negative charge of 2.10 106 C is brought near an identical sphere with a positive charge of 1.50 106 C so that the distance between the centres of the two spheres is 3.30 cm (Figure 10.18). Calculate the magnitude and type (attraction or repulsion) of the force of one charge acting on another. Given q1 2.10 106 C 1.50 106 C q2 r 3.30 102 m 3.30 cm 2.10 106 C 1.50 106 C Required magnitude and type of the electrostatic force acting on the two charges (F e ) Figure 10.18 Analysis and Solution According to Newton’s third law, the electrostatic forces acting on the two spheres are the same in magnitude but opposite in direction. The magnitude of the electrostatic force is F e 1q q 2 k 2 r m (2.10 106 C)(1.50 106 C) 8.99 109 N C (3.30 102 m)2 2 2 26.0 N The magnitude calculation does not use the positive and negative signs for the charges. However, you can use these signs to determine whether the electrostatic force is attractive or repulsive. In this example, the charges have opposite signs, so the force is attractive. 530 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 531 Paraphrase The electrostatic force is one of attraction, with a magnitude of 26.0 N. In the next example, the two spheres touch and the charge is distributed between them. Example 10.2 The two spheres in Example 10.1 are momentarily brought together and then returned to their original separation distance. Determine the electrostatic force now exerted by one charge on the other. Given initial magnitude of the charges: 2.10 106 C q1 1.50 106 C q2 r 3.30 102 m 3.30 cm 2.10 106 C 1.50 106 C Figure 10.19 Practice Problem 1. A metal sphere with a negative charge of 3.00 C is placed 12.0 cm from another similar metal sphere with a positive charge of 2.00 C. The two spheres momentarily touch, then return to their original positions. Calculate the electrostatic force acting on the two metal spheres. Answer 1. 1.56 101 N [repulsion] Required magnitude and type of the electrostatic force acting on the e) two charges (F Analysis and Solution When a sphere with a negative charge of 2.10 106 C momentarily touches a sphere with a positive charge of 1.50 106 C, then 1.50 106 C of charge from the first sphere neutralizes the 1.50 106 C of charge on the second sphere. The remaining charge of 0.60 106 C from the first sphere then divides equally between the two identical spheres. Each sphere now has a charge of 3.0 107 C. The magnitude of the electrostatic force is now F e 1q q 2 k 2 r m (3.0 107 C)(3.0 107 C) 8.99 109 N 2 C (3.30 102 m)2 2 0.74 N Since both spheres have a negative charge, the electrostatic force is repulsive. Paraphrase The electrostatic force is one of repulsion, with a magnitude of 0.74 N. Chapter 10 Physics laws can explain the behaviour of electric charges. 531 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 532 Concept Check Compare gravitational forces and electrostatic forces by identifying two similarities and two differences between the two types of forces. Vector Analysis of Electrostatic Forces So far in this section, you have studied Coulomb’s law and applied the equation to calculate the magnitude of the electrostatic force that one charged particle exerts on another. However, many situations involve more than two charges. The rest of this section illustrates how to use Coulomb’s law to analyze the vector nature of electrostatic forces by determining the electrostatic forces of more than two charges in onedimensional and two-dimensional situations. Examples 10.3 and 10.4 illustrate how to apply Coulomb’s law to three or more collinear charges. Recall from unit I that collinear entities lie along the same straight line. Example 10.3 A small metal sphere (B) with a negative charge of 2.0 106 C is placed midway between two similar spheres (A and C) with positive charges of 1.5 106 C that are 3.0 cm apart (Figure 10.20). Use a vector diagram to find the net electrostatic force acting on sphere B. Analysis and Solution 1.5 cm 1.5 cm A 1.5 106 C B 2.0 106 C C 1.5 106 C Figure 10.20 Spheres A and C have equal charges and are the same distance from sphere B. As shown in Figure 10.21, the force vectors are equal in length and opposite in direction. FAB B FCB Figure 10.21 F net F F CB F AB CB F AB , so F net 0. Since the forces are equal in magnitude and opposite in direction, the net electrostatic force on charge B is 0. Practice Problems 1. Three small, hollow, metallic spheres hang from insulated threads as shown in the figure below. Draw a free-body diagram showing the electrostatic forces acting on sphere B. 1.0 cm 2.0 cm A 2.0 μC B 2.0 μC C 2.0 μC 2. For the figure in problem 1 above, draw a vector for the net electrostatic force on sphere B. Answers 1. FAB Fnet 2. B FCB 532 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 533 Example 10.4 A small metal sphere (B) with a negative charge of 2.10 106 C is placed midway between two similar spheres (A and C) 3.30 cm apart with positive charges of 1.00 106 C and 1.50 106 C, respectively, as shown in Figure 10.22. If the three charges are along the same line, calculate the net electrostatic force on the negative charge. Given qA qB qC rAC rAB 1.00 106 C 2.10 106 C 1.50 106 C 3.30 102 m 1 rAC rBC 2 3.30 cm 1.65 cm 1.00 106 C 2.10 106 C 1.50 106 C Figure 10.22 Required net) net electrostatic force on qB (F Analysis and Solution The charge on sphere B is negative and the charge on sphere A is positive, so the electrostatic force of qA on qB, AB, is an attractive force to the left. Similarly, the F CB, is an attractive force electrostatic force of qC on qB, F to the right (Figure 10.23). Consider right to be positive. FAB B FCB Figure 10.23 F The sum of these two force vectors is the net force on qB: F F net Applying F CB 1q q 2 gives k 2 r AB e F net 2 m (1.00 106 C)(2.10 106 C) 8.99 109 N C 3.30 102 m2 2 2 Practice Problems 1. A metal sphere with a charge of 2.50 109 C is 1.50 cm to the left of a second metal sphere with a charge of 1.50 109 C. A third metal sphere of 1.00 109 C is situated 2.00 cm to the right of the second charged sphere. If all three charges form a line, determine the net electrostatic force on the second sphere. 2. In the situation described above, if the first and third spheres remain at their original positions, where should the second sphere be situated so that the net electrostatic force on it would be zero? Answers 1. 1.16 104 N [to the left] 2. 2.14 102 m to the right of the 2.50 109 C charge [left] 2 m (1.50 106 C)(2.10 106 C) 8.99 109 N C 3.30 102 m2 2 2 [right] (69.34 N 104.0 N) [rig |
ht] 34.7 N [right] Paraphrase The net electrostatic force on charge B is 34.7 N to the right. In Examples 10.3 and 10.4, the forces act along the same line, so the calculations involve only a single dimension. Examples 10.5 and 10.6 demonstrate how to calculate net electrostatic forces in two dimensions. Chapter 10 Physics laws can explain the behaviour of electric charges. 533 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 534 Example 10.5 Practice Problems 1. A small metal sphere X with a negative charge of 2.50 C is 1.20 cm directly to the left of another similar sphere Y with a charge of 3.00 C. A third sphere Z with a charge of 4.00 C is 1.20 cm directly below sphere Y. The three spheres are at the vertices of a right triangle, with sphere Y at the right angle. Calculate the net electrostatic force on sphere Y, sketching diagrams as necessary. 2. Calculate the net electrostatic force on charge B shown in the figure below. 2.00 μC A 1.00 cm 90° 1.00 μC B Answers 1. 8.83 1014 N [122] 2. 2.54 102 N [225] 1.00 cm C 2.00 μC A small metal sphere A with a negative charge of 2.10 106 C is 2.00 102 m to the left of another similar sphere B with a positive charge of 1.50 106 C. A third sphere C with a positive charge of 1.80 106 C is situated 3.00 102 m directly below sphere B (Figure 10.24). Calculate the net electrostatic force on sphere B. 2.00 102 m A 2.10 106 C B 1.50 106 C Given qA qB qC rAB rBC 2.10 106 C 1.50 106 C 1.80 106 C 2.00 102 m 3.00 102 m Required net electrostatic force net) on sphere B (F Analysis and Solution The electrostatic force of qA on qB, F AB, is a force of attraction directed from charge B toward charge A (left). The electrostatic CB, is a force force of qC on qB, F of repulsion directly upward (Figure 10.25). 3.00 102 m C 1.80 106 C Figure 10.24 FCB B FAB Figure 10.25 Applying F e 1q q 2 gives k 2 r F AB 2.10 106 C)(1.50 106 C) 8.99 109 N C (2.00 102 m)2 2 2 70.80 N Similarly, F CB 1.50 106 C)(1.80 106 C) 8.99 109 N C (3.00 102 m)2 2 2 534 Unit VI Forces and Fields 26.97 N 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 535 Use trigonometry to find the net electrostatic force on charge B, as shown in Figure 10.26. Use the Pythagorean theorem to find the magnitude of the net force: F net (70.80 N)2 (26.97 N)2 75.8 N Determine the angle : 9 N 7 . 6 tan 2 7 N 0 8 . 0 20.9 Fnet FCB θ FAB Figure 10.26 The direction of the net force is [20.9 N of W] or [159]. Paraphrase The net electrostatic force on charge B is 75.8 N [20.9 N of W], or 75.8 N [159]. PHYSICS INSIGHT Recall that with the polar coordinates method, angles are measured counterclockwise from the positive x-axis of the coordinate system, which is given a value of 0. Practice Problems 1. Three metal spheres are situated in positions forming an equilateral triangle with sides of 1.20 cm, as shown below. X has a charge of 2.50 C; Y has a charge of 3.00 C; and Z has a charge of 4.00 C. Calculate the net electrostatic force on the Y charge. X 2.50 C 1.20 cm 1.20 cm Y 3.00 C 1.20 cm Z 4.00 C 2. Four charged spheres, with equal charges of 2.20 C, are situated in positions forming a rectangle, as shown in the figure below. Determine the net electrostatic force on the charge in the top right corner of the rectangle. 2.00 102 m A 2.10 106 C B 1.50 106 C rAC 3.00 102 m C 1.80 106 C Figure 10.27 Example 10.6 A small metal sphere A with a charge of 2.10 106 C is 2.00 102 m to the left of a second sphere B with a charge of 1.50 106 C. A third sphere C with a charge of 1.80 106 C is situated 3.00 102 m directly below sphere B. Calculate the net electrostatic force on sphere C. Given qA qB qC rAB rBC 2.10 106 C 1.50 106 C 1.80 106 C 2.00 102 m 3.00 102 m Required net electrostatic force net) on sphere C (F FAC Analysis and Solution The electrostatic force of qA on qC, AC, is an attractive force directed F from charge C toward charge A. The electrostatic force of qB on qC, BC, is a repulsive force directed F downward (Figure 10.28). C 2.20 C 40.0 cm 2.20 C 30.0 cm 30.0 cm 40.0 cm 2.20 C 2.20 C FBC Figure 10.28 Answers 1. 6.56 1014 N [142°] 2. 7.17 1011 N [55.0°] Chapter 10 Physics laws can explain the behaviour of electric charges. 535 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 536 Determine the distance between charges A and C by using the Pythagorean theorem (Figure 10.29): rAC (2.00 102 m)2 (3.00 102 m)2 3.606 102 m 3.61 102 m Applying F e 1q q 2 gives k 2 r 2.00 102 m B A 3.00 102 m θ 1 C F AC 2.10 106 C)(1.80 106 C) 8.99 109 N C (3.61 102 m)2 Figure 10.29 2 2 26.13 N Similarly, F BC 1.50 106 C)(1.80 106 C) 8.99 109 N C (3.00 102 m)2 2 2 26.97 N Use the component method to find the sum of the two force vectors. Use trigonometry to determine the angle 1 AC (Figure 10.29): for the direction of F tan 1 1 2 0 0 . 3 0 0 . 33.69 2 m 1 2 1 m 0 0 Then resolve F AC into x and y components, as shown in Figure 10.30: FACx FACy (26.13 N) (sin 33.69) 14.49 N (26.13 N) (cos 33.69) 21.74 N x 26.13 N y 33.7° C Figure 10.30 The electrostatic force of charge B on charge C has only a y component (see Figure 10.28). So, the x component of F BC is 0 N and the y component is 26.97 N. net. Now find the sum of the x and y components of F F net Fnetx F F BC AC FACx FBCx 14.49 N 0 N 14.49 N Fnety FBCy FACy 21.74 N (26.97 N) 5.23 N 536 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 537 Use trigonometry to determine the magnitude and direction of the net electrostatic force on charge C, as shown in Figure 10.31. Determine the magnitude of the net force using the Pythagorean theorem: (14.49 N)2 (5.23 N)2 15.4 N F net 5.23 N 14.49 N θ 2 Fnet C Figure 10.31 To determine the angle 2, use the tangent function: tan 2 2 3 N .2 5 1 N 9 .4 4 19.8 The direction of the net force is [19.8 S of W] or [200]. Paraphrase The net electrostatic force on charge C is 15.4 N [19.8 S of W] or 15.4 N [200]. THEN, NOW, AND FUTURE ESD Control Manager Since the early 1970s, electrostatic discharge (ESD) has evolved from an interesting, but generally harmless, phenomenon to one of the most rapidly expanding fields of research in science today. As electronic devices have become smaller and smaller, ESD has become a major cause of failure. Each year, billions of dollars’ worth of electronic devices and systems are destroyed or degraded by electrical stress caused by ESD. A dangerous property of ESD is its ability to cause fires in a flammable atmosphere. Property loss, injuries, and fatalities due to the accidental ignition of petrochemical vapours, dusts, and fuels by ESD are on the rise. ESD has been the proven ignition source in many fires. However, research into the firesparking nature of ESD is still in its infancy. Today, electronics manufacturers have ESD awareness and control programs, ESD control program managers, and, in some cases, entire departments dedicated to preventing the damaging effects of ESD. Ron Zezulka (Figure 10.32) is the chief technical officer of TB&S Consultants and has specialized in the science of ESD for over 30 years. He graduated from the Southern Alberta Institute of Technology with a diploma as a telecommunications technician and began his career as a failure analyst specializing in the for Alberta science of ESD Government Telephones. Ron completed many courses to become a control program manager. Because of the newness of the industry, there are no specific quali- Figure 10.32 Ron Zezulka fications for becoming a control program manager in the field of ESD. An ESD control program manager might have a technical diploma and related job experience, a Master’s degree, or a Ph.D. in physics. In 2001, Ron formed TB&S Consultants and has developed and delivered over 25 different training programs and management systems for the awareness and control of ESD in industry. He has written on the topic and lectured in industry, universities, and colleges on awareness and control of ESD. Static electricity is now tied to almost every aspect of the physical sciences. As technology advances, so does our need for a greater understanding of ESD phenomena. Questions 1. Describe two hazards associated with ESD. 2. How could ESD have damaging and harmful effects in your home? 3. How are ESD control program managers employed in industry? Chapter 10 Physics laws can explain the behaviour of electric charges. 537 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 538 10.2 Check and Reflect 10.2 Check and Reflect Knowledge 1. Identify the two factors that influence the force of electrostatic attraction or repulsion acting on two charges. Write a mathematical expression to describe the relationship. 2. Describe how the inverse square law, first proposed by Newton for gravitational forces, was applied to electrostatic forces by Coulomb. 3. (a) What is the SI unit for electric charge? (b) Compare the charge on an electron to that produced by rubbing an ebonite rod with fur. 4. Coulomb could not measure the amount of charge on his spheres, but he could vary the amount of charge on each sphere. Describe the procedure he used to do so. Applications 5. An electrostatic force of 10 N acts on two charged spheres, separated by a certain distance. What will be the new force in the following situations? 7. Two identical conducting spheres have charges of 5.00 105 C and 6.00 105 C and are in fixed positions, 2.00 m apart. (a) Calculate the electrostatic force acting on the two charges. (b) The spheres are touched together and returned to their original positions. Calculate the new electrostatic force acting on them. 8. Three charges are placed in a line, as shown in the diagram below. 2.00 m 3.00 m A 2.00 mC B 3.00 mC C 2.00 mC (a) What is the net electrostatic force on charge A? (b) What is the net electrostatic force on charge B? Extensions (a) The charge on one sphere is doubled. 9. A helium nucleus has a positive charge (b) The charge on both spheres is doubled. 6. (a) Why is it difficult to attain a large charge of 100 C on a small object? (b) During the rubbing process, an object |
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