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. 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 centrip
etal 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 perpendicular 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 Or
bits 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 motion. 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 and 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 Gmsatellite 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 close 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 CHAPTER 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 transferred. 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 in 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 2
W 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 the 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 Paraph
rase (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 F
app 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 isolated 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 pend
ulum 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) What 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 mid
point of each chosen interval. Starting 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 collision 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 settings. 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 while, 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 keep 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? (

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