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only describesmotion—it does not include any forces or masses that may affect rotation (these are part of dynamics). Recall the kinematics equation for linear motion: (constant a). As in linear kinematics, we assume a is constant, which means that angular acceleration The equation for the kinematics relationship between,, and tis is also a constant, because. is the initial angular velocity. Notice that the equation is identical to the linear version, except with angular analogs of where the linear variables. In fact, all of the linear kinematics equations have rotational analogs, which are given in Table 6.3. These equations can be used to solve rotational or linear kinematics problem in which a and are constant. Rotational Linear constant, a constant, a constant, a Table 6.3 Equations for Rotational Kinematics In these equations, and are initial values, is zero, and the average angular velocity and average velocity are 6.11 Access for free at openstax.org. FUN IN PHYSICS Storm Chasing 6.3 • Rotational Motion 215 Figure 6.10 Tornadoes descend from clouds in funnel-like shapes that spin violently. (Daphne Zaras, U.S. National Oceanic and Atmospheric Administration) Storm chasers tend to fall into one of three groups: Amateurs chasing tornadoes as a hobby, atmospheric scientists gathering data for research, weather watchers for news media, or scientists having fun under the guise of work. Storm chasing is a dangerous pastime because tornadoes can change course rapidly with little warning. Since storm chasers follow in the wake of the destruction left by tornadoes, changing flat tires due to debris left on the highway is common. The most active part of the world for tornadoes, called tornado alley, is in the central United States, between the Rocky Mountains and Appalachian Mountains. Tornadoes are perfect examples of rotational motion in action in nature. They come out of severe thunderstorms called supercells, which have a column of air rotating around a horizontal axis, usually about four miles across. The difference in wind speeds between the strong cold winds higher up in the atmosphere in the jet stream and weaker winds traveling north from the Gulf of Mexico causes the column of rotating air to shift so that it spins around a vertical axis, creating a tornado. Tornadoes produce wind speeds as high as 500 km/h (approximately 300 miles/h), particularly at the bottom where the funnel is narrowest because the
rate of rotation increases as the radius decreases. They blow houses away as if they were made of paper and have been known to pierce tree trunks with pieces of straw. GRASP CHECK What is the physics term for the eye of the storm? Why would winds be weaker at the eye of the tornado than at its outermost edge? a. The eye of the storm is the center of rotation. Winds are weaker at the eye of a tornado because tangential velocity is directly proportional to radius of curvature. b. The eye of the storm is the center of rotation. Winds are weaker at the eye of a tornado because tangential velocity is inversely proportional to radius of curvature. c. The eye of the storm is the center of rotation. Winds are weaker at the eye of a tornado because tangential velocity is directly proportional to the square of the radius of curvature. d. The eye of the storm is the center of rotation. Winds are weaker at the eye of a tornado because tangential velocity is inversely proportional to the square of the radius of curvature. Torque If you have ever spun a bike wheel or pushed a merry-go-round, you know that force is needed to change angular velocity. The farther the force is applied from the pivot point (or fulcrum), the greater the angular acceleration. For example, a door opens slowly if you push too close to its hinge, but opens easily if you push far from the hinges. Furthermore, we know that the more 216 Chapter 6 • Circular and Rotational Motion massive the door is, the more slowly it opens; this is because angular acceleration is inversely proportional to mass. These relationships are very similar to the relationships between force, mass, and acceleration from Newton’s second law of motion. Since we have already covered the angular versions of distance, velocity and time, you may wonder what the angular version of force is, and how it relates to linear force. The angular version of force is torque, which is the turning effectiveness of a force. See Figure 6.11. The equation for the magnitude of torque is where ris the magnitude of the lever arm, F is the magnitude of the linear force, and is the angle between the lever arm and the force. The lever arm is the vector from the point of rotation (pivot point or fulcrum) to the location where force is applied. Since the magnitude of the lever arm is a distance, its units are in meters, and torque has units of N�
��m. Torque is a vector quantity and has the same direction as the angular acceleration that it produces. Figure 6.11 A man pushes a merry-go-round at its edge and perpendicular to the lever arm to achieve maximum torque. Applying a stronger torque will produce a greater angular acceleration. For example, the harder the man pushes the merry-goround in Figure 6.11, the faster it accelerates. Furthermore, the more massive the merry-go-round is, the slower it accelerates for the same torque. If the man wants to maximize the effect of his force on the merry-go-round, he should push as far from the center as possible to get the largest lever arm and, therefore, the greatest torque and angular acceleration. Torque is also maximized when the force is applied perpendicular to the lever arm. Solving Rotational Kinematics and Torque Problems Just as linear forces can balance to produce zero net force and no linear acceleration, the same is true of rotational motion. When two torques of equal magnitude act in opposing directions, there is no net torque and no angular acceleration, as you can see in the following video. If zero net torque acts on a system spinning at a constant angular velocity, the system will continue to spin at the same angular velocity. WATCH PHYSICS Introduction to Torque This video (https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/v/introduction-totorque) defines torque in terms of moment arm (which is the same as lever arm). It also covers a problem with forces acting in opposing directions about a pivot point. (At this stage, you can ignore Sal’s references to work and mechanical advantage.) GRASP CHECK Click to view content (https://www.openstax.org/l/28torque) If the net torque acting on the ruler from the example was positive instead of zero, what would this say about the angular Access for free at openstax.org. 6.3 • Rotational Motion 217 acceleration? What would happen to the ruler over time? a. The ruler is in a state of rotational equilibrium so it will not rotate about its center of mass. Thus, the angular acceleration will be zero. b. The ruler is not in a state of rotational equilibrium so it will not rotate about its center of mass. Thus, the angular acceleration will
be zero. c. The ruler is not in a state of rotational equilibrium so it will rotate about its center of mass. Thus, the angular acceleration will be non-zero. d. The ruler is in a state of rotational equilibrium so it will rotate about its center of mass. Thus, the angular acceleration will be non-zero. Now let’s look at examples applying rotational kinematics to a fishing reel and the concept of torque to a merry-go-round. WORKED EXAMPLE Calculating the Time for a Fishing Reel to Stop Spinning A deep-sea fisherman uses a fishing rod with a reel of radius 4.50 cm. A big fish takes the bait and swims away from the boat, pulling the fishing line from his fishing reel. As the fishing line unwinds from the reel, the reel spins at an angular velocity of 220 rad/s. The fisherman applies a brake to the spinning reel, creating an angular acceleration of −300 rad/s2. How long does it take the reel to come to a stop? Strategy We are asked to find the time tfor the reel to come to a stop. The magnitude of the initial angular velocity is rad/s, rad/s2, and the magnitude of the final angular velocity. The signed magnitude of the angular acceleration is where the minus sign indicates that it acts in the direction opposite to the angular velocity. Looking at the rotational kinematic equations, we see all quantities but tare known in the equation problem., making it the easiest equation to use for this Solution The equation to use is. We solve the equation algebraically for t, and then insert the known values. 6.12 Discussion The time to stop the reel is fairly small because the acceleration is fairly large. Fishing lines sometimes snap because of the forces involved, and fishermen often let the fish swim for a while before applying brakes on the reel. A tired fish will be slower, requiring a smaller acceleration and therefore a smaller force. 218 Chapter 6 • Circular and Rotational Motion WORKED EXAMPLE Calculating the Torque on a Merry-Go-Round Consider the man pushing the playground merry-go-round in Figure 6.11. He exerts a force of 250 N at the edge of the merrygo-round and perpendicular to the radius, which is 1.50 m. How much torque does he produce? Assume that friction acting on the merry-go-round is negligible. Strategy To find the torque,
note that the applied force is perpendicular to the radius and that friction is negligible. Solution 6.13 Discussion The man maximizes the torque by applying force perpendicular to the lever arm, so that maximizes his torque by pushing at the outer edge of the merry-go-round, so that he gets the largest-possible lever arm. and. The man also Practice Problems 15. How much torque does a person produce if he applies a force away from the pivot point, perpendicularly to the lever arm? a. b. c. d. 16. An object’s angular velocity changes from 3 rad/s clockwise to 8 rad/s clockwise in 5 s. What is its angular acceleration? a. 0.6 rad/s2 1.6 rad/s2 b. 1 rad/s2 c. 5 rad/s2 d. Check Your Understanding 17. What is angular acceleration? a. Angular acceleration is the rate of change of the angular displacement. b. Angular acceleration is the rate of change of the angular velocity. c. Angular acceleration is the rate of change of the linear displacement. d. Angular acceleration is the rate of change of the linear velocity. 18. What is the equation for angular acceleration, α? Assume θis the angle, ωis the angular velocity, and tis time. a. b. c. d. 19. Which of the following best describes torque? It is the rotational equivalent of a force. It is the force that affects linear motion. It is the rotational equivalent of acceleration. It is the acceleration that affects linear motion. a. b. c. d. 20. What is the equation for torque? Access for free at openstax.org. a. b. c. d. 6.3 • Rotational Motion 219 220 Chapter 6 • Key Terms KEY TERMS angle of rotation the ratio of the arc length to the radius of time curvature of a circular path lever arm the distance between the point of rotation (pivot angular acceleration the rate of change of angular velocity point) and the location where force is applied with time radius of curvature the distance between the center of a angular velocity ( ) the rate of change in the angular circular path and the path position of an object following a circular path rotational motion the circular motion of an object about an arc length ( ) the distance traveled by an object along a axis of rotation circular path spin rotation about an axis that goes through the center of centrifugal force a
fictitious force that acts in the direction mass of the object opposite the centripetal acceleration tangential acceleration the acceleration in a direction centripetal acceleration the acceleration of an object moving in a circle, directed toward the center of the circle centripetal force any force causing uniform circular tangent to the circular path of motion and in the same direction or opposite direction as the tangential velocity tangential velocity the instantaneous linear velocity of an motion object in circular or rotational motion circular motion the motion of an object along a circular torque the effectiveness of a force to change the rotational path speed of an object kinematics of rotational motion the relationships between rotation angle, angular velocity, angular acceleration, and uniform circular motion the motion of an object in a circular path at constant speed that always points toward the center of rotation, perpendicular to the linear velocity, in the same direction as the net force, and in the direction opposite that of the radius vector. • The standard unit for centripetal acceleration is m/s2. • Centripetal force Fc is any net force causing uniform circular motion. 6.3 Rotational Motion • Kinematics is the description of motion. • The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, angular acceleration, and time. • Torque is the effectiveness of a force to change the rotational speed of an object. Torque is the rotational analog of force. • The lever arm is the distance between the point of rotation (pivot point) and the location where force is applied. • Torque is maximized by applying force perpendicular to the lever arm and at a point as far as possible from the pivot point or fulcrum. If torque is zero, angular acceleration is zero. SECTION SUMMARY 6.1 Angle of Rotation and Angular Velocity • Circular motion is motion in a circular path. • The angle of rotation is defined as the ratio of the arc length to the radius of curvature. • The arc length is the distance traveled along a circular path and ris the radius of curvature of the circular path. • The angle of rotation (rad), where • Angular velocity where a rotation is measured in units of radians revolution. is the rate of change of an angle, occurs in a time. • The units of angular velocity are radians per second (rad/s). • Tangential speed vand angular speed are related by, and tangential velocity has units of m/s. • The
direction of angular velocity is along the axis of rotation, toward (away) from you for clockwise (counterclockwise) motion. 6.2 Uniform Circular Motion • Centripetal acceleration ac is the acceleration experienced while in uniform circular motion. • Centripetal acceleration force is a center-seekingforce Access for free at openstax.org. KEY EQUATIONS 6.1 Angle of Rotation and Angular Velocity 6.3 Rotational Motion Chapter 6 • Key Equations 221 Angle of rotation Angular speed: Tangential speed: 6.2 Uniform Circular Motion Centripetal acceleration Centripetal force or,, Angular acceleration Rotational kinematic equations Tangential (linear) acceleration Torque,,, CHAPTER REVIEW Concept Items 6.2 Uniform Circular Motion 6.1 Angle of Rotation and Angular Velocity 4. What is the equation for centripetal acceleration in terms 1. One revolution is equal to how many radians? Degrees? a. b. c. d. 2. What is tangential velocity? a. Tangential velocity is the average linear velocity of an object in a circular motion. b. Tangential velocity is the instantaneous linear velocity of an object undergoing rotational motion. c. Tangential velocity is the average angular velocity of an object in a circular motion. d. Tangential velocity is the instantaneous angular velocity of an object in a circular motion. 3. What kind of motion is called spin? a. Spin is rotational motion of an object about an axis parallel to the axis of the object. of angular velocity and the radius? a. b. c. d. 5. How can you express centripetal force in terms of centripetal acceleration? a. b. c. d. 6. What is meant by the word centripetal? a. b. c. d. center-seeking center-avoiding central force central acceleration b. Spin is translational motion of an object about an 6.3 Rotational Motion axis parallel to the axis of the object. c. Spin is the rotational motion of an object about its center of mass. d. Spin is translational motion of an object about its own axis. 7. Conventionally, for which direction of rotation of an object is angular acceleration considered positive? a. b. c. d. the positive xdirection of the coordinate system the negative xdirection of the coordinate system the counterclockwise direction the clockwise direction 222 Chapter 6 • Chapter
Review 8. When you push a door closer to the hinges, why does it 9. When is angular acceleration negative? open more slowly? a. It opens slowly, because the lever arm is shorter so the torque is large. It opens slowly because the lever arm is longer so the torque is large. It opens slowly, because the lever arm is shorter so the torque is less. It opens slowly, because the lever arm is longer so the torque is less. b. c. d. Critical Thinking Items 6.1 Angle of Rotation and Angular Velocity 10. When the radius of the circular path of rotational motion increases, what happens to the arc length for a given angle of rotation? a. The arc length is directly proportional to the radius of the circular path, and it increases with the radius. b. The arc length is inversely proportional to the radius of the circular path, and it decreases with the radius. c. The arc length is directly proportional to the radius of the circular path, and it decreases with the radius. d. The arc length is inversely proportional to the radius of the circular path, and it increases with the radius. 11. Consider a CD spinning clockwise. What is the sum of the instantaneous velocities of two points on both ends of its diameter? a. b. c. d. 6.2 Uniform Circular Motion 12. What are the directions of the velocity and acceleration of an object in uniform circular motion? a. Velocity is tangential, and acceleration is radially outward. b. Velocity is tangential, and acceleration is radially inward. c. Velocity is radially outward, and acceleration is tangential. d. Velocity is radially inward, and acceleration is tangential. 13. Suppose you have an object tied to a rope and are rotating it over your head in uniform circular motion. If Access for free at openstax.org. displacement and is negative when a. Angular acceleration is the rate of change of the increases. b. Angular acceleration is the rate of change of the decreases. displacement and is negative when c. Angular acceleration is the rate of change of angular velocity and is negative when increases. d. Angular acceleration is the rate of change of angular velocity and is negative when decreases. you increase the length of the rope, would you have to apply more or less force to maintain the same speed? a. More force is required, because the force is inversely proportional to the radius of the circular orbit. b. More force is required because
the force is directly proportional to the radius of the circular orbit. c. Less force is required because the force is inversely proportional to the radius of the circular orbit. d. Less force is required because the force is directly proportional to the radius of the circular orbit. 6.3 Rotational Motion 14. Consider two spinning tops with different radii. Both have the same linear instantaneous velocities at their edges. Which top has a higher angular velocity? a. the top with the smaller radius because the radius of curvature is inversely proportional to the angular velocity the top with the smaller radius because the radius of curvature is directly proportional to the angular velocity the top with the larger radius because the radius of curvature is inversely proportional to the angular velocity b. c. d. The top with the larger radius because the radius of curvature is directly proportional to the angular velocity 15. A person tries to lift a stone by using a lever. If the lever arm is constant and the mass of the stone increases, what is true of the torque necessary to lift it? a. It increases, because the torque is directly proportional to the mass of the body. It increases because the torque is inversely proportional to the mass of the body. It decreases because the torque is directly proportional to the mass of the body. It decreases, because the torque is inversely proportional to the mass of the body. b. c. d. Chapter 6 • Test Prep 223 Problems d. 13, 333 N 6.1 Angle of Rotation and Angular Velocity 16. What is the angle of rotation (in degrees) between two hands of a clock, if the radius of the clock is the arc length separating the two hands is a. b. c. d. and? 17. A clock has radius of. The outermost point on its minute hand travels along the edge. What is its tangential speed? a. b. c. d. 6.2 Uniform Circular Motion 18. What is the centripetal force exerted on a 1,600 kg car that rounds a 100 m radius curve at 12 m/s? 192 N a. b. 1, 111 N c. 2, 300 N Performance Task 6.3 Rotational Motion 22. Design a lever arm capable of lifting a 0.5 kg object such as a stone. The force for lifting should be provided by TEST PREP Multiple Choice 6.1 Angle of Rotation and Angular Velocity 23. What is 1 radian approximately in degrees? 57.3
° 360° a. b. c. π° d. 2π° 24. If the following objects are spinning at the same angular velocities, the edge of which one would have the highest speed? a. Mini CD b. Regular CD c. Vinyl record 25. What are possible units for tangential velocity? a. b. 19. Find the frictional force between the tires and the road that allows a 1,000 kg car traveling at 30 m/s to round a 20 m radius curve. a. 22 N b. 667 N c. d. 45, 000 N 1, 500 N 6.3 Rotational Motion 20. An object’s angular acceleration is 36 rad/s2. If it were initially spinning with a velocity of 6.0 m/s, what would its angular velocity be after 5.0 s? 186 rad/s a. 190 rad/s2 b. c. −174 rad/s d. −174 rad/s2 21. When a fan is switched on, it undergoes an angular acceleration of 150 rad/s2. How long will it take to achieve its maximum angular velocity of 50 rad/s? a. −0.3 s b. 0.3 s 3.0 s c. placing coins on the other end of the lever. How many coins would you need? What happens if you shorten or lengthen the lever arm? What does this say about torque? in radians? c. 26. What is a. b. c. d. 27. For a given object, what happens to the arc length as the angle of rotation increases? a. The arc length is directly proportional to the angle of rotation, so it increases with the angle of rotation. b. The arc length is inversely proportional to the angle of rotation, so it decreases with the angle of rotation. c. The arc length is directly proportional to the angle of rotation, so it decreases with the angle of rotation. 224 Chapter 6 • Test Prep d. The arc length is inversely proportional to the angle of rotation, so it increases with the angle of rotation. 6.2 Uniform Circular Motion 28. Which of these quantities is constant in uniform circular motion? a. Speed b. Velocity c. Acceleration d. Displacement 29. Which of these quantities impact centripetal force? a. Mass and speed only b. Mass and radius only c. Speed and radius only d. Mass, speed,
and radius all impact centripetal force 30. An increase in the magnitude of which of these quantities causes a reduction in centripetal force? a. Mass b. Radius of curvature c. Speed 31. What happens to centripetal acceleration as the radius of curvature decreases and the speed is constant, and why? a. It increases, because the centripetal acceleration is inversely proportional to the radius of the curvature. It increases, because the centripetal acceleration is directly proportional to the radius of curvature. It decreases, because the centripetal acceleration is inversely proportional to the radius of the curvature. It decreases, because the centripetal acceleration is directly proportional to the radius of the curvature. b. c. d. 32. Why do we experience more sideways acceleration while driving around sharper curves? Short Answer 6.1 Angle of Rotation and Angular Velocity 37. What is the rotational analog of linear velocity? a. Angular displacement b. Angular velocity c. Angular acceleration d. Angular momentum 38. What is the rotational analog of distance? a. Rotational angle b. Torque c. Angular velocity d. Angular momentum Access for free at openstax.org. a. Centripetal acceleration is inversely proportional to the radius of curvature, so it increases as the radius of curvature decreases. b. Centripetal acceleration is directly proportional to the radius of curvature, so it decreases as the radius of curvature decreases. c. Centripetal acceleration is directly proportional to the radius of curvature, so it decreases as the radius of curvature increases. d. Centripetal acceleration is directly proportional to the radius of curvature, so it increases as the radius of curvature increases. 6.3 Rotational Motion 33. Which of these quantities is not described by the kinematics of rotational motion? a. Rotation angle b. Angular acceleration c. Centripetal force d. Angular velocity 34. In the equation, what is F? a. Linear force b. Centripetal force c. Angular force 35. What happens when two torques act equally in opposite directions? a. Angular velocity is zero. b. Angular acceleration is zero. 36. What is the mathematical relationship between angular and linear accelerations? a. b. c. d. 39. What is the equation that relates the linear speed of a point on a rotating object with the object's angular
quantities? a. b. c. d. 40. As the angular velocity of an object increases, what happens to the linear velocity of a point on that object? It increases, because linear velocity is directly a. proportional to angular velocity. It increases, because linear velocity is inversely proportional to angular velocity. b. Chapter 6 • Test Prep 225 c. d. It decreases because linear velocity is directly proportional to angular velocity. It decreases because linear velocity is inversely proportional to angular velocity. a. b. c. d. 41. What is angular speed in terms of tangential speed and 47. What are the standard units for centripetal force? the radius? a. b. c. d. 42. Why are radians dimensionless? a. Radians are dimensionless, because they are defined as a ratio of distances. They are defined as the ratio of the arc length to the radius of the circle. b. Radians are dimensionless because they are defined as a ratio of distances. They are defined as the ratio of the area to the radius of the circle. c. Radians are dimensionless because they are defined as multiplication of distance. They are defined as the multiplication of the arc length to the radius of the circle. d. Radians are dimensionless because they are defined as multiplication of distance. They are defined as the multiplication of the area to the radius of the circle. 6.2 Uniform Circular Motion 43. What type of quantity is centripetal acceleration? a. Scalar quantity; centripetal acceleration has magnitude only but no direction b. Scalar quantity; centripetal acceleration has magnitude as well as direction c. Vector quantity; centripetal acceleration has magnitude only but no direction d. Vector quantity; centripetal acceleration has magnitude as well as direction 44. What are the standard units for centripetal acceleration? a. m/s b. c. d. 45. What is the angle formed between the vectors of tangential velocity and centripetal force? a. b. c. d. 46. What is the angle formed between the vectors of centripetal acceleration and centripetal force? a. m b. m/s c. m/s2 d. newtons 48. As the mass of an object in uniform circular motion increases, what happens to the centripetal force required to keep it moving at the same speed? a. It increases, because the centrip
etal force is directly proportional to the mass of the rotating body. It increases, because the centripetal force is inversely proportional to the mass of the rotating body. It decreases, because the centripetal force is directly proportional to the mass of the rotating body. It decreases, because the centripetal force is inversely proportional to the mass of the rotating body. b. c. d. 6.3 Rotational Motion 49. The relationships between which variables are described by the kinematics of rotational motion? a. The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, and angular acceleration. b. The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, angular acceleration, and angular momentum. c. The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, angular acceleration, and time. d. The kinematics of rotational motion describes the relationships between rotation angle, angular velocity, angular acceleration, torque, and time. 50. What is the kinematics relationship between,, and? a. b. c. d. 51. What kind of quantity is torque? a. Scalar b. Vector 226 Chapter 6 • Test Prep c. Dimensionless d. Fundamental quantity 52. If a linear force is applied to a lever arm farther away from the pivot point, what happens to the resultant torque? a. b. c. d. It decreases. It increases. It remains the same. It changes the direction. 53. How can the same force applied to a lever produce different torques? a. By applying the force at different points of the lever Extended Response 6.1 Angle of Rotation and Angular Velocity 54. Consider two pits on a CD, one close to the center and one close to the outer edge. When the CD makes one full rotation, which pit would have gone through a greater angle of rotation? Which one would have covered a greater arc length? a. The one close to the center would go through the greater angle of rotation. The one near the outer edge would trace a greater arc length. b. The one close to the center would go through the greater angle of rotation. The one near the center would trace a greater arc length. c. Both would go through the same angle of rotation. The one near the outer edge would trace a greater arc length. d. Both would go through the same angle of rotation. The one near the center would
trace a greater arc length. 55. Consider two pits on a CD, one close to the center and one close to the outer edge. For a given angular velocity of the CD, which pit has a higher angular velocity? Which has a higher tangential velocity? a. The point near the center would have the greater angular velocity and the point near the outer edge would have the higher linear velocity. arm along the length of the lever or by changing the angle between the lever arm and the applied force. b. By applying the force at the same point of the lever arm along the length of the lever or by changing the angle between the lever arm and the applied force. c. By applying the force at different points of the lever arm along the length of the lever or by maintaining the same angle between the lever arm and the applied force. d. By applying the force at the same point of the lever arm along the length of the lever or by maintaining the same angle between the lever arm and the applied force. the same? a. It increases because tangential velocity is directly proportional to the radius. It increases because tangential velocity is inversely proportional to the radius. It decreases because tangential velocity is directly proportional to the radius. It decreases because tangential velocity is inversely proportional to the radius. b. c. d. 6.2 Uniform Circular Motion 57. Is an object in uniform circular motion accelerating? Why or why not? a. Yes, because the velocity is not constant. b. No, because the velocity is not constant. c. Yes, because the velocity is constant. d. No, because the velocity is constant. 58. An object is in uniform circular motion. Suppose the centripetal force was removed. In which direction would the object now travel? a. b. In the direction of the centripetal force In the direction opposite to the direction of the centripetal force In the direction of the tangential velocity In the direction opposite to the direction of the tangential velocity c. d. b. The point near the edge would have the greater 59. An object undergoes uniform circular motion. If the angular velocity and the point near the center would have the higher linear velocity. c. Both have the same angular velocity and the point near the outer edge would have the higher linear velocity. d. Both have the same angular velocity and the point near the center would have the higher linear velocity. 56. What happens to tangential velocity as the radius of an object
increases provided the angular velocity remains Access for free at openstax.org. radius of curvature and mass of the object are constant, what is the centripetal force proportional to? a. b. c. d. 6.3 Rotational Motion 60. Why do tornadoes produce more wind speed at the Chapter 6 • Test Prep 227 bottom of the funnel? a. Wind speed is greater at the bottom because rate of rotation increases as the radius increases. b. The force should be applied perpendicularly to the lever arm as far as possible from the pivot point. c. The force should be applied parallel to the lever arm b. Wind speed is greater at the bottom because rate of as far as possible from the pivot point. rotation increases as the radius decreases. d. The force should be applied parallel to the lever arm c. Wind speed is greater at the bottom because rate of as close as possible from the pivot point. rotation decreases as the radius increases. d. Wind speed is greater at the bottom because rate of rotation decreases as the radius increases. 61. How can you maximize the torque applied to a given lever arm without applying more force? a. The force should be applied perpendicularly to the lever arm as close as possible from the pivot point. 62. When will an object continue spinning at the same angular velocity? a. When net torque acting on it is zero b. When net torque acting on it is non zero c. When angular acceleration is positive d. When angular acceleration is negative 228 Chapter 6 • Test Prep Access for free at openstax.org. CHAPTER 7 Newton's Law of Gravitation Figure 7.1 Johannes Kepler (left) showed how the planets move, and Isaac Newton (right) discovered that gravitational force caused them to move that way. ((left) unknown, Public Domain; (right) Sir Godfrey Kneller, Public Domain) Chapter Outline 7.1 Kepler's Laws of Planetary Motion 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity INTRODUCTION What do a falling apple and the orbit of the moon have in common? You will learn in this chapter that each is caused by gravitational force. The motion of all celestial objects, in fact, is determined by the gravitational force, which depends on their mass and separation. Johannes Kepler discovered three laws of planetary motion that all orbiting planets and moons follow. Years later, Isaac Newton found these laws useful in developing his law of universal gravitation. This law relates gravitational force to
the masses of objects and the distance between them. Many years later still, Albert Einstein showed there was a little more to the gravitation story when he published his theory of general relativity. 7.1 Kepler's Laws of Planetary Motion Section Learning Objectives By the end of this section, you will be able to do the following: • Explain Kepler’s three laws of planetary motion • Apply Kepler’s laws to calculate characteristics of orbits 230 Chapter 7 • Newton's Law of Gravitation Section Key Terms aphelion Copernican model eccentricity Kepler’s laws of planetary motion perihelion Ptolemaic model Concepts Related to Kepler’s Laws of Planetary Motion Examples of orbits abound. Hundreds of artificial satellites orbit Earth together with thousands of pieces of debris. The moon’s orbit around Earth has intrigued humans from time immemorial. The orbits of planets, asteroids, meteors, and comets around the sun are no less interesting. If we look farther, we see almost unimaginable numbers of stars, galaxies, and other celestial objects orbiting one another and interacting through gravity. All these motions are governed by gravitational force. The orbital motions of objects in our own solar system are simple enough to describe with a few fairly simple laws. The orbits of planets and moons satisfy the following two conditions: • The mass of the orbiting object, m, is small compared to the mass of the object it orbits, M. • The system is isolated from other massive objects. Based on the motion of the planets about the sun, Kepler devised a set of three classical laws, called Kepler’s laws of planetary motion, that describe the orbits of all bodies satisfying these two conditions: 1. The orbit of each planet around the sun is an ellipse with the sun at one focus. 2. Each planet moves so that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal times. 3. The ratio of the squares of the periods of any two planets about the sun is equal to the ratio of the cubes of their average distances from the sun. These descriptive laws are named for the German astronomer Johannes Kepler (1571–1630). He devised them after careful study (over some 20 years) of a large amount of meticulously recorded observations of planetary motion done by Tycho Brahe (1546–1601). Such careful collection and detailed recording of methods and data are hallmarks of good science. Data constitute the evidence from which new interpretations and meanings can be constructed. Let’
s look closer at each of these laws. Kepler’s First Law The orbit of each planet about the sun is an ellipse with the sun at one focus, as shown in Figure 7.2. The planet’s closest approach to the sun is called aphelion and its farthest distance from the sun is called perihelion. Access for free at openstax.org. 7.1 • Kepler's Laws of Planetary Motion 231 Figure 7.2 (a) An ellipse is a closed curve such that the sum of the distances from a point on the curve to the two foci (f1 and f2) is constant. (b) For any closed orbit, mfollows an elliptical path with Mat one focus. (c) The aphelion (ra) is the closest distance between the planet and the sun, while the perihelion (rp) is the farthest distance from the sun. If you know the aphelion (ra) and perihelion (rp) distances, then you can calculate the semi-major axis (a) and semi-minor axis (b). Figure 7.3 You can draw an ellipse as shown by putting a pin at each focus, and then placing a loop of string around a pen and the pins and tracing a line on the paper. 232 Chapter 7 • Newton's Law of Gravitation Kepler’s Second Law Each planet moves so that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal times, as shown in Figure 7.4. Figure 7.4 The shaded regions have equal areas. The time for mto go from A to B is the same as the time to go from C to D and from E to F. The mass mmoves fastest when it is closest to M. Kepler’s second law was originally devised for planets orbiting the sun, but it has broader validity. TIPS FOR SUCCESS Note that while, for historical reasons, Kepler’s laws are stated for planets orbiting the sun, they are actually valid for all bodies satisfying the two previously stated conditions. Kepler’s Third Law The ratio of the periods squared of any two planets around the sun is equal to the ratio of their average distances from the sun cubed. In equation form, this is where Tis the period (time for one orbit) and ris the average distance (also called orbital radius). This equation is valid only for comparing
two small masses orbiting a single large mass. Most importantly, this is only a descriptive equation; it gives no information about the cause of the equality. LINKS TO PHYSICS History: Ptolemy vs. Copernicus Before the discoveries of Kepler, Copernicus, Galileo, Newton, and others, the solar system was thought to revolve around Earth as shown in Figure 7.5 (a). This is called the Ptolemaic model, named for the Greek philosopher Ptolemy who lived in the second century AD. The Ptolemaic model is characterized by a list of facts for the motions of planets, with no explanation of cause and effect. There tended to be a different rule for each heavenly body and a general lack of simplicity. Figure 7.5 (b) represents the modern or Copernican model. In this model, a small set of rules and a single underlying force explain not only all planetary motion in the solar system, but also all other situations involving gravity. The breadth and simplicity of the laws of physics are compelling. Access for free at openstax.org. 7.1 • Kepler's Laws of Planetary Motion 233 Figure 7.5 (a) The Ptolemaic model of the universe has Earth at the center with the moon, the planets, the sun, and the stars revolving about it in complex circular paths. This geocentric (Earth-centered) model, which can be made progressively more accurate by adding more circles, is purely descriptive, containing no hints about the causes of these motions. (b) The Copernican heliocentric (sun-centered) model is a simpler and more accurate model. Nicolaus Copernicus (1473–1543) first had the idea that the planets circle the sun, in about 1514. It took him almost 20 years to work out the mathematical details for his model. He waited another 10 years or so to publish his work. It is thought he hesitated because he was afraid people would make fun of his theory. Actually, the reaction of many people was more one of fear and anger. Many people felt the Copernican model threatened their basic belief system. About 100 years later, the astronomer Galileo was put under house arrest for providing evidence that planets, including Earth, orbited the sun. In all, it took almost 300 years for everyone to admit that Copernicus had been right all along. GRASP CHECK Explain why Earth does actually appear to be the center
of the solar system. a. Earth appears to be the center of the solar system because Earth is at the center of the universe, and everything revolves around it in a circular orbit. b. Earth appears to be the center of the solar system because, in the reference frame of Earth, the sun, moon, and planets all appear to move across the sky as if they were circling Earth. c. Earth appears to be at the center of the solar system because Earth is at the center of the solar system and all the heavenly bodies revolve around it. d. Earth appears to be at the center of the solar system because Earth is located at one of the foci of the elliptical orbit of the sun, moon, and other planets. Virtual Physics Acceleration This simulation allows you to create your own solar system so that you can see how changing distances and masses determines the orbits of planets. Click Helpfor instructions. Click to view content (https://archive.cnx.org/specials/ee816dff-0b5f-4e6f-8250-f9fb9e39d716/my-solar-system/) GRASP CHECK When the central object is off center, how does the speed of the orbiting object vary? a. The orbiting object moves fastest when it is closest to the central object and slowest when it is farthest away. b. The orbiting object moves slowest when it is closest to the central object and fastest when it is farthest away. 234 Chapter 7 • Newton's Law of Gravitation c. The orbiting object moves with the same speed at every point on the circumference of the elliptical orbit. d. There is no relationship between the speed of the object and the location of the planet on the circumference of the orbit. Calculations Related to Kepler’s Laws of Planetary Motion Kepler’s First Law Refer back to Figure 7.2 (a). Notice which distances are constant. The foci are fixed, so distance of an ellipse states that the sum of the distances perimeter of triangle distances of objects in a system that includes one object orbiting another. is also constant. These two facts taken together mean that the must also be constant. Knowledge of these constants will help you determine positions and is a constant. The definition Kepler’s Second Law Refer back to Figure 7.4. The second law says that the segments have equal area and that it takes equal time to sweep through each segment. That is,
the time it takes to travel from A to B equals the time it takes to travel from C to D, and so forth. Velocity v equals distance ddivided by time t:, so distance divided by velocity is also a constant. For example, if we know the average velocity of Earth on June 21 and December 21, we can compare the distance Earth travels on those days.. Then, The degree of elongation of an elliptical orbit is called its eccentricity (e). Eccentricity is calculated by dividing the distance f from the center of an ellipse to one of the foci by half the long axis a. When, the ellipse is a circle. The area of an ellipse is given by sweeps out in a given period of time, you can calculate the fraction of the year that has elapsed., where bis half the short axis. If you know the axes of Earth’s orbit and the area Earth 7.1 WORKED EXAMPLE Kepler’s First Law At its closest approach, a moon comes within 200,000 km of the planet it orbits. At that point, the moon is 300,000 km from the other focus of its orbit, f2. The planet is focus f1 of the moon’s elliptical orbit. How far is the moon from the planet when it is 260,000 km from f2? Strategy Show and label the ellipse that is the orbit in your solution. Picture the triangle f1mf2 collapsed along the major axis and add up the lengths of the three sides. Find the length of the unknown side of the triangle when the moon is 260,000 km from f2. Solution Perimeter of Discussion The perimeter of triangle f1mf2 must be constant because the distance between the foci does not change and Kepler’s first law says the orbit is an ellipse. For any ellipse, the sum of the two sides of the triangle, which are f1mand mf2, is constant. WORKED EXAMPLE Kepler’s Second Law Figure 7.6 shows the major and minor axes of an ellipse. The semi-major and semi-minor axes are half of these, respectively. Access for free at openstax.org. 7.1 • Kepler's Laws of Planetary Motion 235 Figure 7.6 The major axis is the length of the ellipse, and the minor axis is the width of the ellipse. The semi
-major axis is half the major axis, and the semi-minor axis is half the minor axis. Earth’s orbit is slightly elliptical, with a semi-major axis of 152 million km and a semi-minor axis of 147 million km. If Earth’s period is 365.26 days, what area does an Earth-to-sun line sweep past in one day? Strategy Each day, Earth sweeps past an equal-sized area, so we divide the total area by the number of days in a year to find the area swept past in one day. For total area use a year (i.e., its period).. Calculate A, the area inside Earth’s orbit and divide by the number of days in Solution 7.2 The area swept out in one day is thus. Discussion The answer is based on Kepler’s law, which states that a line from a planet to the sun sweeps out equal areas in equal times. Kepler’s Third Law Kepler’s third law states that the ratio of the squares of the periods of any two planets (T1, T2) is equal to the ratio of the cubes of their average orbital distance from the sun (r1, r2). Mathematically, this is represented by From this equation, it follows that the ratio r3/T2 is the same for all planets in the solar system. Later we will see how the work of Newton leads to a value for this constant. WORKED EXAMPLE Kepler’s Third Law Given that the moon orbits Earth each 27.3 days and that it is an average distance of calculate the period of an artificial satellite orbiting at an average altitude of 1,500 km above Earth’s surface. Strategy The period, or time for one orbit, is related to the radius of the orbit by Kepler’s third law, given in mathematical form by from the center of Earth, 236 Chapter 7 • Newton's Law of Gravitation. Let us use the subscript 1 for the moon and the subscript 2 for the satellite. We are asked to find T2. The given information tells us that the orbital radius of the moon is, and that the period of the moon is. The height of the artificial satellite above Earth’s surface is given, so to get the distance r2 from the center of. Now all Earth we must add the height to the radius of Earth (6380 km). This gives quantities are known, so T2 can be found
. Solution To solve for T2, we cross-multiply and take the square root, yielding 7.3 Discussion This is a reasonable period for a satellite in a fairly low orbit. It is interesting that any satellite at this altitude will complete one orbit in the same amount of time. Practice Problems 1. A planet with no axial tilt is located in another solar system. It circles its sun in a very elliptical orbit so that the temperature varies greatly throughout the year. If the year there has 612 days and the inhabitants celebrate the coldest day on day 1 of their calendar, when is the warmest day? a. Day 1 b. Day 153 c. Day 306 d. Day 459 2. A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth’s rotation). The ratio for the moon is. Calculate the radius of the orbit of such a satellite. a. b. c. d. Check Your Understanding 3. Are Kepler’s laws purely descriptive, or do they contain causal information? a. Kepler’s laws are purely descriptive. b. Kepler’s laws are purely causal. c. Kepler’s laws are descriptive as well as causal. d. Kepler’s laws are neither descriptive nor causal. 4. True or false—According to Kepler’s laws of planetary motion, a satellite increases its speed as it approaches its parent body and decreases its speed as it moves away from the parent body. a. True b. False 5. Identify the locations of the foci of an elliptical orbit. a. One focus is the parent body, and the other is located at the opposite end of the ellipse, at the same distance from the center as the parent body. b. One focus is the parent body, and the other is located at the opposite end of the ellipse, at half the distance from the center as the parent body. Access for free at openstax.org. 7.2 • Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 237 c. One focus is the parent body and the other is located outside of the elliptical orbit, on the line on which is the semi- major axis of the ellipse. d. One focus is on the line containing
the semi-major axis of the ellipse, and the other is located anywhere on the elliptical orbit of the satellite. 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity Section Learning Objectives By the end of this section, you will be able to do the following: • Explain Newton’s law of universal gravitation and compare it to Einstein’s theory of general relativity • Perform calculations using Newton’s law of universal gravitation Section Key Terms Einstein’s theory of general relativity gravitational constant Newton’s universal law of gravitation Concepts Related to Newton’s Law of Universal Gravitation Sir Isaac Newton was the first scientist to precisely define the gravitational force, and to show that it could explain both falling bodies and astronomical motions. See Figure 7.7. But Newton was not the first to suspect that the same force caused both our weight and the motion of planets. His forerunner, Galileo Galilei, had contended that falling bodies and planetary motions had the same cause. Some of Newton’s contemporaries, such as Robert Hooke, Christopher Wren, and Edmund Halley, had also made some progress toward understanding gravitation. But Newton was the first to propose an exact mathematical form and to use that form to show that the motion of heavenly bodies should be conic sections—circles, ellipses, parabolas, and hyperbolas. This theoretical prediction was a major triumph. It had been known for some time that moons, planets, and comets follow such paths, but no one had been able to propose an explanation of the mechanism that caused them to follow these paths and not others. Figure 7.7 The popular legend that Newton suddenly discovered the law of universal gravitation when an apple fell from a tree and hit him on the head has an element of truth in it. A more probable account is that he was walking through an orchard and wondered why all the apples fell in the same direction with the same acceleration. Great importance is attached to it because Newton’s universal law of gravitation and his laws of motion answered very old questions about nature and gave tremendous support to the notion of underlying simplicity and unity in nature. Scientists still expect underlying simplicity to emerge from their ongoing inquiries into nature. The gravitational force is relatively simple. It is always attractive, and it depends only on the masses involved and the distance 238 Chapter 7 • Newton's Law of Gravitation between them. Expressed in modern language, Newton’
s universal law of gravitation states that every object in the universe attracts every other object with a force that is directed along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This attraction is illustrated by Figure 7.8. Figure 7.8 Gravitational attraction is along a line joining the centers of mass (CM) of the two bodies. The magnitude of the force on each body is the same, consistent with Newton’s third law (action-reaction). For two bodies having masses mand Mwith a distance rbetween their centers of mass, the equation for Newton’s universal law of gravitation is where F is the magnitude of the gravitational force and Gis a proportionality factor called the gravitational constant. Gis a universal constant, meaning that it is thought to be the same everywhere in the universe. It has been measured experimentally to be. If a person has a mass of 60.0 kg, what would be the force of gravitational attraction on him at Earth’s surface? Gis given above, Earth’s mass Mis 5.97 × 1024 kg, and the radius rof Earth is 6.38 × 106 m. Putting these values into Newton’s universal law of gravitation gives We can check this result with the relationship: You may remember that g, the acceleration due to gravity, is another important constant related to gravity. By substituting g for a in the equation for Newton’s second law of motion we get gravitation gives. Combining this with the equation for universal Cancelling the mass mon both sides of the equation and filling in the values for the gravitational constant and mass and radius of the Earth, gives the value of g,which may look familiar. This is a good point to recall the difference between mass and weight. Mass is the amount of matter in an object; weight is the Access for free at openstax.org. 7.2 • Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 239 force of attraction between the mass within two objects. Weight can change because gis different on every moon and planet. An object’s mass mdoes not change but its weight mg can. Virtual Physics Gravity and Orbits Move the sun, Earth, moon and space station in this simulation to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies. Turn off gravity to see what
would happen without it! Click to view content (https://archive.cnx.org/specials/a14085c8-96b8-4d04-bb5a-56d9ccbe6e69/gravity-and-orbits/) GRASP CHECK Why doesn’t the Moon travel in a smooth circle around the Sun? a. The Moon is not affected by the gravitational field of the Sun. b. The Moon is not affected by the gravitational field of the Earth. c. The Moon is affected by the gravitational fields of both the Earth and the Sun, which are always additive. d. The moon is affected by the gravitational fields of both the Earth and the Sun, which are sometimes additive and sometimes opposite. Snap Lab Take-Home Experiment: Falling Objects In this activity you will study the effects of mass and air resistance on the acceleration of falling objects. Make predictions (hypotheses) about the outcome of this experiment. Write them down to compare later with results. • Four sheets of -inch paper Procedure • Take four identical pieces of paper. ◦ Crumple one up into a small ball. ◦ Leave one uncrumpled. ◦ Take the other two and crumple them up together, so that they make a ball of exactly twice the mass of the other crumpled ball. ◦ Now compare which ball of paper lands first when dropped simultaneously from the same height. 1. Compare crumpled one-paper ball with crumpled two-paper ball. 2. Compare crumpled one-paper ball with uncrumpled paper. GRASP CHECK Why do some objects fall faster than others near the surface of the earth if all mass is attracted equally by the force of gravity? a. Some objects fall faster because of air resistance, which acts in the direction of the motion of the object and exerts more force on objects with less surface area. b. Some objects fall faster because of air resistance, which acts in the direction opposite the motion of the object and exerts more force on objects with less surface area. c. Some objects fall faster because of air resistance, which acts in the direction of motion of the object and exerts more force on objects with more surface area. d. Some objects fall faster because of air resistance, which acts in the direction opposite the motion of the object and exerts more force on objects with more surface area. It is possible to derive Kepler’s third law from Newton�
�s law of universal gravitation. Applying Newton’s second law of motion to 240 Chapter 7 • Newton's Law of Gravitation angular motion gives an expression for centripetal force, which can be equated to the expression for force in the universal gravitation equation. This expression can be manipulated to produce the equation for Kepler’s third law. We saw earlier that the expression r3/T2is a constant for satellites orbiting the same massive object. The derivation of Kepler’s third law from Newton’s law of universal gravitation and Newton’s second law of motion yields that constant: where Mis the mass of the central body about which the satellites orbit (for example, the sun in our solar system). The usefulness of this equation will be seen later. The universal gravitational constant Gis determined experimentally. This definition was first done accurately in 1798 by English scientist Henry Cavendish (1731–1810), more than 100 years after Newton published his universal law of gravitation. The measurement of Gis very basic and important because it determines the strength of one of the four forces in nature. Cavendish’s experiment was very difficult because he measured the tiny gravitational attraction between two ordinary-sized masses (tens of kilograms at most) by using an apparatus like that in Figure 7.9. Remarkably, his value for Gdiffers by less than 1% from the modern value. Figure 7.9 Cavendish used an apparatus like this to measure the gravitational attraction between two suspended spheres (m) and two spheres on a stand (M) by observing the amount of torsion (twisting) created in the fiber. The distance between the masses can be varied to check the dependence of the force on distance. Modern experiments of this type continue to explore gravity. Einstein’s Theory of General Relativity Einstein’s theory of general relativity explained some interesting properties of gravity not covered by Newton’s theory. Einstein based his theory on the postulate that acceleration and gravity have the same effect and cannot be distinguished from each other. He concluded that light must fall in both a gravitational field and in an accelerating reference frame. Figure 7.10 shows this effect (greatly exaggerated) in an accelerating elevator. In Figure 7.10(a), the elevator accelerates upward in zero gravity. In Figure 7.10(b), the room is not accelerating but is subject to gravity. The effect on light is the same: it “falls
” downward in both situations. The person in the elevator cannot tell whether the elevator is accelerating in zero gravity or is stationary and subject to gravity. Thus, gravity affects the path of light, even though we think of gravity as acting between masses, while photons are massless. Access for free at openstax.org. 7.2 • Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 241 Figure 7.10 (a) A beam of light emerges from a flashlight in an upward-accelerating elevator. Since the elevator moves up during the time the light takes to reach the wall, the beam strikes lower than it would if the elevator were not accelerated. (b) Gravity must have the same effect on light, since it is not possible to tell whether the elevator is accelerating upward or is stationary and acted upon by gravity. Einstein’s theory of general relativity got its first verification in 1919 when starlight passing near the sun was observed during a solar eclipse. (See Figure 7.11.) During an eclipse, the sky is darkened and we can briefly see stars. Those on a line of sight nearest the sun should have a shift in their apparent positions. Not only was this shift observed, but it agreed with Einstein’s predictions well within experimental uncertainties. This discovery created a scientific and public sensation. Einstein was now a folk hero as well as a very great scientist. The bending of light by matter is equivalent to a bending of space itself, with light following the curve. This is another radical change in our concept of space and time. It is also another connection that any particle with mass orenergy (e.g., massless photons) is affected by gravity. Figure 7.11 This schematic shows how light passing near a massive body like the sun is curved toward it. The light that reaches the Earth then seems to be coming from different locations than the known positions of the originating stars. Not only was this effect observed, but the amount of bending was precisely what Einstein predicted in his general theory of relativity. To summarize the two views of gravity, Newton envisioned gravity as a tug of war along the line connecting any two objects in the universe. In contrast, Einstein envisioned gravity as a bending of space-time by mass. 242 Chapter 7 • Newton's Law of Gravitation BOUNDLESS PHYSICS NASA gravity probe B NASA’s Gravity Probe B (GP-B) mission has confirmed two key predictions derived from Albert Einstein’s general theory of relativity.
The probe, shown in Figure 7.12 was launched in 2004. It carried four ultra-precise gyroscopes designed to measure two effects hypothesized by Einstein’s theory: • The geodetic effect, which is the warping of space and time by the gravitational field of a massive body (in this case, Earth) • The frame-dragging effect, which is the amount by which a spinning object pulls space and time with it as it rotates Figure 7.12 Artist concept of Gravity Probe B spacecraft in orbit around the Earth. (credit: NASA/MSFC) Both effects were measured with unprecedented precision. This was done by pointing the gyroscopes at a single star while orbiting Earth in a polar orbit. As predicted by relativity theory, the gyroscopes experienced very small, but measureable, changes in the direction of their spin caused by the pull of Earth’s gravity. The principle investigator suggested imagining Earth spinning in honey. As Earth rotates it drags space and time with it as it would a surrounding sea of honey. GRASP CHECK According to the general theory of relativity, a gravitational field bends light. What does this have to do with time and space? a. Gravity has no effect on the space-time continuum, and gravity only affects the motion of light. b. The space-time continuum is distorted by gravity, and gravity has no effect on the motion of light. c. Gravity has no effect on either the space-time continuum or on the motion of light. d. The space-time continuum is distorted by gravity, and gravity affects the motion of light. Calculations Based on Newton’s Law of Universal Gravitation TIPS FOR SUCCESS When performing calculations using the equations in this chapter, use units of kilograms for mass, meters for distances, newtons for force, and seconds for time. The mass of an object is constant, but its weight varies with the strength of the gravitational field. This means the value of g varies from place to place in the universe. The relationship between force, mass, and acceleration from the second law of motion can be written in terms of g. In this case, the force is the weight of the object, which is caused by the gravitational attraction of the planet or moon on which the object is located. We can use this expression to compare weights of an object on different moons and planets. Access for free at openstax.org. 7.2 • Newton's Law of Universal Gravitation and Einstein
's Theory of General Relativity 243 WATCH PHYSICS Mass and Weight Clarification This video shows the mathematical basis of the relationship between mass and weight. The distinction between mass and weight are clearly explained. The mathematical relationship between mass and weight are shown mathematically in terms of the equation for Newton’s law of universal gravitation and in terms of his second law of motion. Click to view content (https://www.khanacademy.org/embed_video?v=IuBoeDihLUc) GRASP CHECK Would you have the same mass on the moon as you do on Earth? Would you have the same weight? a. You would weigh more on the moon than on Earth because gravity on the moon is stronger than gravity on Earth. b. You would weigh less on the moon than on Earth because gravity on the moon is weaker than gravity on Earth. c. You would weigh less on the moon than on Earth because gravity on the moon is stronger than gravity on Earth. d. You would weigh more on the moon than on Earth because gravity on the moon is weaker than gravity on Earth. Two equations involving the gravitational constant, G, are often useful. The first is Newton’s equation,. Several of the values in this equation are either constants or easily obtainable. F is often the weight of an object on the surface of a large object with mass M, which is usually known. The mass of the smaller object, m, is often known, and Gis a universal constant with the same value anywhere in the universe. This equation can be used to solve problems involving an object on or orbiting Earth or other massive celestial object. Sometimes it is helpful to equate the right-hand side of the equation to mg and cancel the mon both sides. The equation is also useful for problems involving objects in orbit. Note that there is no need to know the mass of the object. Often, we know the radius ror the period Tand want to find the other. If these are both known, we can use the equation to calculate the mass of a planet or star. WATCH PHYSICS Mass and Weight Clarification This video demonstrates calculations involving Newton’s universal law of gravitation. Click to view content (https://www.khanacademy.org/embed_video?v=391txUI76gM) GRASP CHECK and. are both the acceleration due to gravity Identify the constants a. b. c. d. and
is acceleration due to gravity on Earth and is the gravitational constant and and are both the universal gravitational constant. is the universal gravitational constant. is the acceleration due to gravity on Earth. WORKED EXAMPLE Change in g The value of g on the planet Mars is 3.71 m/s2. If you have a mass of 60.0 kg on Earth, what would be your mass on Mars? What would be your weight on Mars? Strategy Weight equals acceleration due to gravity times mass: on Mars gMand weight on Mars WM.. An object’s mass is constant. Call acceleration due to gravity 244 Chapter 7 • Newton's Law of Gravitation Solution Mass on Mars would be the same, 60 kg. Discussion The value of g on any planet depends on the mass of the planet and the distance from its center. If the material below the surface varies from point to point, the value of g will also vary slightly. 7.4 WORKED EXAMPLE Earth’s g at the Moon Find the acceleration due to Earth’s gravity at the distance of the moon. Express the force of gravity in terms of g. Combine with the equation for universal gravitation. Solution Cancel mand substitute. 7.5 7.6 7.7 Discussion The value of g for the moon is 1.62 m/s2. Comparing this value to the answer, we see that Earth’s gravitational influence on an object on the moon’s surface would be insignificant. 7.8 Practice Problems 6. What is the mass of a person who weighs? a. b. c. d. 7. Calculate Earth’s mass given that the acceleration due to gravity at the North Pole is and the radius of the Earth is from pole to center. a. b. c. d. Check Your Understanding 8. Some of Newton’s predecessors and contemporaries also studied gravity and proposed theories. What important advance did Newton make in the study of gravity that the other scientists had failed to do? a. He gave an exact mathematical form for the theory. Access for free at openstax.org. 7.2 • Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 245 b. He added a correction term to a previously existing formula. c. Newton found the value of the universal gravitational constant. d. Newton showed that gravitational force is always attractive. 9. State the law of universal gravitation in words only. a. Gravitational force between two objects
is directly proportional to the sum of the squares of their masses and inversely proportional to the square of the distance between them. b. Gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. c. Gravitational force between two objects is directly proportional to the sum of the squares of their masses and inversely proportional to the distance between them. d. Gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the distance between them. 10. Newton’s law of universal gravitation explains the paths of what? a. A charged particle b. A ball rolling on a plane surface c. A planet moving around the sun d. A stone tied to a string and whirled at constant speed in a horizontal circle 246 Chapter 7 • Key Terms KEY TERMS aphelion closest distance between a planet and the sun (called apoapsis for other celestial bodies) Copernican model the model of the solar system where the sun is at the center of the solar system and all the planets orbit around it; this is also called the heliocentric model eccentricity a measure of the separation of the foci of an ellipse Johannes Kepler that describe the properties of all orbiting satellites Newton’s universal law of gravitation states that gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. perihelion farthest distance between a planet and the sun Einstein’s theory of general relativity the theory that (called periapsis for other celestial bodies) gravitational force results from the bending of spacetime by an object’s mass gravitational constant the proportionality constant in Newton’s law of universal gravitation Kepler’s laws of planetary motion three laws derived by SECTION SUMMARY 7.1 Kepler's Laws of Planetary Motion • All satellites follow elliptical orbits. • The line from the satellite to the parent body sweeps out equal areas in equal time. • The radius cubed divided by the period squared is a Ptolemaic model the model of the solar system where Earth is at the center of the solar system and the sun and all the planets orbit around it; this is also called the geocentric model 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity • Newton’s law of universal gravitation provides a mathematical basis for gravitational force and
Kepler’s laws of planetary motion. constant for all satellites orbiting the same parent body. • Einstein’s theory of general relativity shows that KEY EQUATIONS 7.1 Kepler's Laws of Planetary Motion Kepler’s third law eccentricity area of an ellipse semi-major axis of an ellipse semi-minor axis of an ellipse gravitational fields change the path of light and warp space and time. • An object’s mass is constant, but its weight changes when acceleration due to gravity, g, changes. 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity Newton’s second law of motion Newton’s universal law of gravitation acceleration due to gravity constant for satellites orbiting the same massive object CHAPTER REVIEW Concept Items 7.1 Kepler's Laws of Planetary Motion is different from other ellipses. a. The foci of a circle are at the same point and are located at the center of the circle. 1. A circle is a special case of an ellipse. Explain how a circle b. The foci of a circle are at the same point and are Access for free at openstax.org. located at the circumference of the circle. c. The foci of a circle are at the same point and are located outside of the circle. d. The foci of a circle are at the same point and are located anywhere on the diameter, except on its midpoint. 2. Comets have very elongated elliptical orbits with the sun at one focus. Using Kepler's Law, explain why a comet travels much faster near the sun than it does at the other end of the orbit. a. Because the satellite sweeps out equal areas in equal times b. Because the satellite sweeps out unequal areas in Chapter 7 • Chapter Review 247 illustration of this is any description of the feeling of constant velocity in a situation where no outside frame of reference is considered. c. Gravity and acceleration have the same effect and cannot be distinguished from each other. An acceptable illustration of this is any description of the feeling of acceleration in a situation where no outside frame of reference is considered. d. Gravity and acceleration have different effects and can be distinguished from each other. An acceptable illustration of this is any description of the feeling of acceleration in a situation where no outside frame of reference is considered. equal times 6. Titan, with a radius of, is the largest c. Because the satellite is at the other focus of the moon
of the planet Saturn. If the mass of Titan is ellipse d. Because the square of the period of the satellite is proportional to the cube of its average distance from the sun 3. True or False—A planet-satellite system must be isolated from other massive objects to follow Kepler’s laws of planetary motion. a. True b. False 4. Explain why the string, pins, and pencil method works for drawing an ellipse. a. The string, pins, and pencil method works because the length of the two sides of the triangle remains constant as you are drawing the ellipse. b. The string, pins, and pencil method works because the area of the triangle remains constant as you are drawing the ellipse. c. The string, pins, and pencil method works because the perimeter of the triangle remains constant as you are drawing the ellipse. d. The string, pins, and pencil method works because the volume of the triangle remains constant as you are drawing the ellipse. 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 5. Describe the postulate on which Einstein based the theory of general relativity and describe an everyday experience that illustrates this postulate. a. Gravity and velocity have the same effect and cannot be distinguished from each other. An acceptable illustration of this is any description of the feeling of constant velocity in a situation where no outside frame of reference is considered. b. Gravity and velocity have different effects and can be distinguished from each other. An acceptable, what is the acceleration due to gravity on the surface of this moon? a. b. c. d. 7. Saturn’s moon Titan has an orbital period of 15.9 days. If Saturn has a mass of 5.68×1023 kg, what is the average distance from Titan to the center of Saturn? 1.22×106 m a. b. 4.26×107 m 5.25×104 km c. d. 4.26×1010 km 8. Explain why doubling the mass of an object doubles its weight, but doubling its distance from the center of Earth reduces its weight fourfold. a. The weight is two times the gravitational force between the object and Earth. b. The weight is half the gravitational force between the object and Earth. c. The weight is equal to the gravitational force between the object and Earth, and the gravitational force is inversely proportional to the distance squared between the object and Earth. d
. The weight is directly proportional to the square of the gravitational force between the object and Earth. 9. Explain why a star on the other side of the Sun might appear to be in a location that is not its true location. It can be explained by using the concept of a. atmospheric refraction. It can be explained by using the concept of the special theory of relativity. It can be explained by using the concept of the general theory of relativity. It can be explained by using the concept of light d. b. c. 248 Chapter 7 • Chapter Review scattering in the atmosphere. 10. The Cavendish experiment marked a milestone in the study of gravity. Part A. What important value did the experiment determine? Part B. Why was this so difficult in terms of the masses used in the apparatus and the strength of the gravitational force? a. Part A. The experiment measured the acceleration due to gravity, g. Part B. Gravity is a very weak force but despite this limitation, Cavendish was able to measure the attraction between very massive objects. b. Part A. The experiment measured the gravitational Critical Thinking Items 7.1 Kepler's Laws of Planetary Motion 11. In the figure, the time it takes for the planet to go from A to B, C to D, and E to F is the same. constant, G. Part B. Gravity is a very weak force but, despite this limitation, Cavendish was able to measure the attraction between very massive objects. c. Part A. The experiment measured the acceleration due to gravity, g. Part B. Gravity is a very weak force but despite this limitation, Cavendish was able to measure the attraction between less massive objects. d. Part A. The experiment measured the gravitational constant, G. Part B. Gravity is a very weak force but despite this limitation, Cavendish was able to measure the attraction between less massive objects. a. Area X < Area Y; the speed is greater for area X. b. Area X > Area Y; the speed is greater for area Y. c. Area X = Area Y; the speed is greater for area X. d. Area X = Area Y; the speed is greater for area Y. 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 14. Rhea, with a radius of 7.63×105 m, is the second-largest moon of the planet Saturn. If the mass of Rhea is 2.31×1021 kg, what is the acceleration due
to gravity on the surface of this moon? a. 2.65×10−1 m/s b. 2.02×105 m/s c. 2.65×10−1 m/s2 d. 2.02×105 m/s2 15. Earth has a mass of 5.971×1024 kg and a radius of 6.371×106 m. Use the data to check the value of the gravitational constant. a. it matches the value of the gravitational constant G. it matches the value of the gravitational constant G. it matches the value of the b. c. gravitational constant G. Compare the areas A1, A2, and A3 in terms of size. a. A1 ≠ A2 ≠ A3 b. A1 = A2 = A3 c. A1 = A2 > A3 d. A1 > A2 = A3 12. A moon orbits a planet in an elliptical orbit. The foci of the ellipse are 50, 000 km apart. The closest approach of the moon to the planet is 400, 000 km. What is the length of the major axis of the orbit? a. 400, 000 km b. 450, 000, km c. 800, 000 km d. 850, 000 km 13. In this figure, if f1 represents the parent body, which set of statements holds true? Access for free at openstax.org. d. it matches the value of the gravitational constant G. 16. The orbit of the planet Mercury has a period of 88.0 days and an average radius of 5.791×1010 m. What is the mass of the sun? Problems 7.1 Kepler's Laws of Planetary Motion 17. The closest Earth comes to the sun is 1.47×108 km, and Earth’s farthest distance from the sun is 1.52×108 km. What is the area inside Earth’s orbit? a. 2.23×1016 km2 b. 6.79×1016 km2 7.02×1016 km2 c. 7.26×1016 km2 d. Performance Task 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 19. Design an experiment to test whether magnetic force is inversely proportional to the square of distance. Gravitational, magnetic, and electrical fields all act at a distance, but do they all follow the inverse square law? One difference in the forces related
to these fields is that gravity is only attractive, but the other two can repel as well. In general, the inverse square law says that force F equals a constant Cdivided by the distance between objects, d, squared: Incorporate these materials into your design:. TEST PREP Multiple Choice 7.1 Kepler's Laws of Planetary Motion 20. A planet of mass m circles a sun of mass M. Which distance changes throughout the planet’s orbit? a. b. c. d. 21. The focal point of the elliptical orbit of a moon is from the center of the orbit. If the, what is the length of the eccentricity of the orbit is semi-major axis? a. b. c. d. Chapter 7 • Test Prep 249 3.43×1019 kg a. 1.99×1030 kg b. c. 2.56×1029 kg 1.48×1040 kg d. 18. Earth is 1.496×108 km from the sun, and Neptune is 4.490×109 km from the sun. What best represents the number of Earth years it takes for Neptune to complete one orbit around the sun? 10 years a. 30 years b. 160 years c. d. 900 years • Two strong, permanent bar magnets • A spring scale that can measure small forces • A short ruler calibrated in millimeters Use the magnets to study the relationship between attractive force and distance. a. What will be the independent variable? b. What will be the dependent variable? c. How will you measure each of these variables? If you plot the independent variable versus the d. dependent variable and the inverse square law is upheld, will the plot be a straight line? Explain. e. Which plot would be a straight line if the inverse square law were upheld? 22. An artificial satellite orbits the Earth at a distance of 1.45×104 km from Earth’s center. The moon orbits the Earth at a distance of 3.84×105 km once every 27.3 days. How long does it take the satellite to orbit the Earth? a. 0.200 days b. 3.07 days c. 243 days 3721 days d. 23. Earth is 1.496×108 km from the sun, and Venus is 1.08×108 km from the sun. One day on Venus is 243 Earth days long. What best represents the number of Venusian days in a Venusian year? a. 0.78 days b
. 0.92 days 1.08 days c. 1.21 days d. 250 Chapter 7 • Test Prep 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 24. What did the Cavendish experiment measure? a. The mass of Earth b. The gravitational constant c. Acceleration due to gravity d. The eccentricity of Earth’s orbit 25. You have a mass of and you have just landed on one of the moons of Jupiter where you have a weight of, on. What is the acceleration due to gravity, the moon you are visiting? a. b. c. d. 26. A person is in an elevator that suddenly begins to descend. The person knows, intuitively, that the feeling of suddenly becoming lighter is because the elevator is accelerating downward. What other change would Short Answer 7.1 Kepler's Laws of Planetary Motion 27. Explain how the masses of a satellite and its parent body must compare in order to apply Kepler’s laws of planetary motion. a. The mass of the parent body must be much less than that of the satellite. b. The mass of the parent body must be much greater than that of the satellite. c. The mass of the parent body must be equal to the mass of the satellite. d. There is no specific relationship between the masses for applying Kepler’s laws of planetary motion. 28. Hyperion is a moon of the planet Saturn. Its orbit has an eccentricity of and a semi-major axis of. How far is the center of the orbit from the center of Saturn? a. b. c. d. 29. The orbits of satellites are elliptical. Define an ellipse. a. An ellipse is an open curve wherein the sum of the distance from the foci to any point on the curve is constant. b. An ellipse is a closed curve wherein the sum of the distance from the foci to any point on the curve is constant. Access for free at openstax.org. produce the same feeling? How does this demonstrate Einstein’s postulate on which he based the theory of general relativity? a. It would feel the same if the force of gravity suddenly became weaker. This illustrates Einstein’s postulates that gravity and acceleration are indistinguishable. It would feel the same if the force of gravity suddenly became stronger. This illustrates Einstein’s postulates that gravity and acceleration are indistinguishable. It would feel the same if the force
of gravity suddenly became weaker. This illustrates Einstein’s postulates that gravity and acceleration are distinguishable. It would feel the same if the force of gravity suddenly became stronger. This illustrates Einstein’s postulates that gravity and acceleration are distinguishable. b. c. d. c. An ellipse is an open curve wherein the distances from the two foci to any point on the curve are equal. d. An ellipse is a closed curve wherein the distances from the two foci to any point on the curve are equal. 30. Mars has two moons, Deimos and Phobos. The orbit of. The average radius of the and an average Deimos has a period of radius of. According to orbit of Phobos is Kepler’s third law of planetary motion, what is the period of Phobos? a. b. c. d. 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 31. Newton’s third law of motion says that, for every action force, there is a reaction force equal in magnitude but that acts in the opposite direction. Apply this law to gravitational forces acting between the Washington Monument and Earth. a. The monument is attracted to Earth with a force equal to its weight, and Earth is attracted to the monument with a force equal to Earth’s weight. The situation can be represented with two force vectors of unequal magnitude and pointing in the same direction. b. The monument is attracted to Earth with a force equal to its weight, and Earth is attracted to the monument with a force equal to Earth’s weight. The situation can be represented with two force vectors of unequal magnitude but pointing in opposite directions. c. The monument is attracted to Earth with a force equal to its weight, and Earth is attracted to the monument with an equal force. The situation can be represented with two force vectors of equal magnitude and pointing in the same direction. d. The monument is attracted to Earth with a force equal to its weight, and Earth is attracted to the monument with an equal force. The situation can be represented with two force vectors of equal magnitude but pointing in opposite directions. 32. True or false—Gravitational force is the attraction of the mass of one object to the mass of another. Light, either Extended Response 7.1 Kepler's Laws of Planetary Motion 35. The orbit of Halley’'s Comet has an eccentricity of 0.967 and stretches to the edge of
the solar system. Part A. Describe the shape of the comet’s orbit. Part B. Compare the distance traveled per day when it is near the sun to the distance traveled per day when it is at the edge of the solar system. Part C. Describe variations in the comet's speed as it completes an orbit. Explain the variations in terms of Kepler's second law of planetary motion. a. Part A. The orbit is circular, with the sun at the center. Part B. The comet travels much farther when it is near the sun than when it is at the edge of the solar system. Part C. The comet decelerates as it approaches the sun and accelerates as it leaves the sun. b. Part A. The orbit is circular, with the sun at the center. Part B. The comet travels much farther when it is near the sun than when it is at the edge of the solar system. Part C. The comet accelerates as it approaches the sun and decelerates as it leaves the sun. c. Part A. The orbit is very elongated, with the sun near one end. Part B. The comet travels much farther when it is near the sun than when it is at the edge of the solar system. Part C. The comet decelerates as it approaches the sun and accelerates as it moves away from the sun. 36. For convenience, astronomers often use astronomical units (AU) to measure distances within the solar system. One AU equals the average distance from Earth to the Chapter 7 • Test Prep 251 as a particle or a wave, has no rest mass. Despite this fact gravity bends a beam of light. a. True b. False 33. The average radius of Earth is. What is Earth’s mass? a. b. c. d. 34. What is the gravitational force between two apart? people sitting a. b. c. d. sun. Halley’s Comet returns once every 75.3 years. What is the average radius of the orbit of Halley’s Comet in AU? a. 0.002 AU b. 0.056 AU c. 17.8 AU d. 653 AU 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity 37. It took scientists a long time to arrive at the understanding of gravity as explained by Galileo and Newton. They were hindered by two ideas that seemed like common sense but were serious misconceptions. First was the fact that
heavier things fall faster than light things. Second, it was believed impossible that forces could act at a distance. Explain why these ideas persisted and why they prevented advances. a. Heavier things fall faster than light things if they have less surface area and greater mass density. In the Renaissance and before, forces that acted at a distance were considered impossible, so people were skeptical about scientific theories that invoked such forces. b. Heavier things fall faster than light things because they have greater surface area and less mass density. In the Renaissance and before, forces that act at a distance were considered impossible, so people were skeptical about scientific theories that invoked such forces. c. Heavier things fall faster than light things because they have less surface area and greater mass density. In the Renaissance and before, forces that 252 Chapter 7 • Test Prep act at a distance were considered impossible, so people were quick to accept scientific theories that invoked such forces. d. Heavier things fall faster than light things because they have larger surface area and less mass density. In the Renaissance and before, forces that act at a distance were considered impossible because of people’s faith in scientific theories. 38. The masses of Earth and the moon are 5.97×1024 kg and 7.35×1022 kg, respectively. The distance from Earth to the moon is 3.80×105 km. At what point between the Earth and the moon are the opposing gravitational forces equal? (Use subscripts e and m to represent Earth and moon.) a. b. c. d. 3.42×105 km from the center of Earth 3.80×105 km from the center of Earth 3.42×106 km from the center of Earth 3.10×107 km from the center of Earth Access for free at openstax.org. CHAPTER 8 Momentum Figure 8.1 NFC defensive backs Ronde Barber and Roy Williams along with linebacker Jeremiah Trotter gang tackle AFC running back LaDainian Tomlinson during the 2006 Pro Bowl in Hawaii. (United States Marine Corps) Chapter Outline 8.1 Linear Momentum, Force, and Impulse 8.2 Conservation of Momentum 8.3 Elastic and Inelastic Collisions We know from everyday use of the word momentumthat it is a tendency to continue on course in the same INTRODUCTION direction. Newscasters speak of sports teams or politicians gaining, losing, or maintaining the momentum to win. As we learned when studying about inertia, which is Newton's first law
of motion, every object or system has inertia—that is, a tendency for an object in motion to remain in motion or an object at rest to remain at rest. Mass is a useful variable that lets us quantify inertia. Momentum is mass in motion. Momentum is important because it is conserved in isolated systems; this fact is convenient for solving problems where objects collide. The magnitude of momentum grows with greater mass and/or speed. For example, look at the football players in the photograph (Figure 8.1). They collide and fall to the ground. During their collisions, momentum will play a large part. In this chapter, we will learn about momentum, the different types of collisions, and how to use momentum equations to solve collision problems. 254 Chapter 8 • Momentum 8.1 Linear Momentum, Force, and Impulse Section Learning Objectives By the end of this section, you will be able to do the following: • Describe momentum, what can change momentum, impulse, and the impulse-momentum theorem • Describe Newton’s second law in terms of momentum • Solve problems using the impulse-momentum theorem Section Key Terms change in momentum impulse impulse–momentum theorem linear momentum Momentum, Impulse, and the Impulse-Momentum Theorem Linear momentum is the product of a system’s mass and its velocity. In equation form, linear momentum p is You can see from the equation that momentum is directly proportional to the object’s mass (m) and velocity (v). Therefore, the greater an object’s mass or the greater its velocity, the greater its momentum. A large, fast-moving object has greater momentum than a smaller, slower object. Momentum is a vector and has the same direction as velocity v. Since mass is a scalar, when velocity is in a negative direction (i.e., opposite the direction of motion), the momentum will also be in a negative direction; and when velocity is in a positive direction, momentum will likewise be in a positive direction. The SI unit for momentum is kg m/s. Momentum is so important for understanding motion that it was called the quantity of motionby physicists such as Newton. Force influences momentum, and we can rearrange Newton’s second law of motion to show the relationship between force and momentum. Recall our study of Newton’s second law of motion (Fnet = ma). Newton actually stated his second law of motion in terms of
momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. The change in momentum is the difference between the final and initial values of momentum. In equation form, this law is where Fnet is the net external force, is the change in momentum, and is the change in time. We can solve for by rearranging the equation to be is known as impulse and this equation is known as the impulse-momentum theorem. From the equation, we see that the impulse equals the average net external force multiplied by the time this force acts. It is equal to the change in momentum. The effect of a force on an object depends on how long it acts, as well as the strength of the force.Impulse is a useful concept because it quantifies the effect of a force. A very large force acting for a short time can have a great effect on the momentum of an object, such as the force of a racket hitting a tennis ball. A small force could cause the same change in momentum, but it would have to act for a much longer time. Newton’s Second Law in Terms of Momentum When Newton’s second law is expressed in terms of momentum, it can be used for solving problems where mass varies, since. In the more traditional form of the law that you are used to working with, mass is assumed to be constant. In fact, this traditional form is a special case of the law, where mass is constant. is actually derived from the equation: Access for free at openstax.org. 8.1 • Linear Momentum, Force, and Impulse 255 For the sake of understanding the relationship between Newton’s second law in its two forms, let’s recreate the derivation of from by substituting the definitions of acceleration and momentum. The change in momentum is given by If the mass of the system is constant, then By substituting for, Newton’s second law of motion becomes for a constant mass. Because we can substitute to get the familiar equation when the mass of the system is constant. TIPS FOR SUCCESS We just showed how would not apply would be a moving rocket that burns enough fuel to significantly change the mass of the rocket. In this case, you would need to use Newton’s second law expressed in terms of momentum to account for the changing mass. applies only when the mass of the system is constant. An example of when this formula Snap Lab Hand Movement and Impulse
In this activity you will experiment with different types of hand motions to gain an intuitive understanding of the relationship between force, time, and impulse. • one ball • one tub filled with water Procedure: 1. Try catching a ball while givingwith the ball, pulling your hands toward your body. 2. Next, try catching a ball while keeping your hands still. 3. Hit water in a tub with your full palm. Your full palm represents a swimmer doing a belly flop. 4. After the water has settled, hit the water again by diving your hand with your fingers first into the water. Your diving hand represents a swimmer doing a dive. 5. Explain what happens in each case and why. GRASP CHECK What are some other examples of motions that impulse affects? 256 Chapter 8 • Momentum a. a football player colliding with another, or a car moving at a constant velocity b. a car moving at a constant velocity, or an object moving in the projectile motion c. a car moving at a constant velocity, or a racket hitting a ball d. a football player colliding with another, or a racket hitting a ball LINKS TO PHYSICS Engineering: Saving Lives Using the Concept of Impulse Cars during the past several decades have gotten much safer. Seat belts play a major role in automobile safety by preventing people from flying into the windshield in the event of a crash. Other safety features, such as airbags, are less visible or obvious, but are also effective at making auto crashes less deadly (see Figure 8.2). Many of these safety features make use of the concept of impulse from physics. Recall that impulse is the net force multiplied by the duration of time of the impact. This was expressed mathematically as. Figure 8.2 Vehicles have safety features like airbags and seat belts installed. Airbags allow the net force on the occupants in the car to act over a much longer time when there is a sudden stop. The momentum change is the same for an occupant whether an airbag is deployed or not. But the force that brings the occupant to a stop will be much less if it acts over a larger time. By rearranging the equation for impulse to solve for force see how increasing padded dashboard increases the time over which the force of impact acts, thereby reducing the force of impact. you can stays the same will decrease Fnet. This is another example of an inverse relationship. Similarly, a while Cars today have many plastic components. One advantage of plastics is their lighter weight, which
results in better gas mileage. Another advantage is that a car will crumple in a collision, especially in the event of a head-on collision. A longer collision time means the force on the occupants of the car will be less. Deaths during car races decreased dramatically when the rigid frames of racing cars were replaced with parts that could crumple or collapse in the event of an accident. GRASP CHECK You may have heard the advice to bend your knees when jumping. In this example, a friend dares you to jump off of a park bench onto the ground without bending your knees. You, of course, refuse. Explain to your friend why this would be a foolish thing. Show it using the impulse-momentum theorem. a. Bending your knees increases the time of the impact, thus decreasing the force. b. Bending your knees decreases the time of the impact, thus decreasing the force. c. Bending your knees increases the time of the impact, thus increasing the force. d. Bending your knees decreases the time of the impact, thus increasing the force. Access for free at openstax.org. 8.1 • Linear Momentum, Force, and Impulse 257 Solving Problems Using the Impulse-Momentum Theorem WORKED EXAMPLE Calculating Momentum: A Football Player and a Football (a) Calculate the momentum of a 110 kg football player running at 8 m/s. (b) Compare the player’s momentum with the momentum of a 0.410 kg football thrown hard at a speed of 25 m/s. Strategy No information is given about the direction of the football player or the football, so we can calculate only the magnitude of the momentum, p. (A symbol in italics represents magnitude.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum: Solution for (a) To find the player’s momentum, substitute the known values for the player’s mass and speed into the equation. Solution for (b) To find the ball’s momentum, substitute the known values for the ball’s mass and speed into the equation. The ratio of the player’s momentum to the ball’s momentum is Discussion Although the ball has greater velocity, the player has a much greater mass. Therefore, the momentum of the player is about 86 times greater than the momentum of the football. WORKED EXAMPLE Calculating Force: Venus Williams
’ Racquet During the 2007 French Open, Venus Williams (Figure 8.3) hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h). What was the average force exerted on the 0.057 kg tennis ball by Williams’ racquet? Assume that the ball’s speed just after impact was 58 m/s, the horizontal velocity before impact is negligible, and that the ball remained in contact with the racquet for 5 ms (milliseconds). Figure 8.3 Venus Williams playing in the 2013 US Open (Edwin Martinez, Flickr) Strategy Recall that Newton’s second law stated in terms of momentum is 258 Chapter 8 • Momentum As noted above, when mass is constant, the change in momentum is given by where vf is the final velocity and vi is the initial velocity. In this example, the velocity just after impact and the change in time are given, so after we solve for to find the force., we can use Solution To determine the change in momentum, substitute the values for mass and the initial and final velocities into the equation above. Now we can find the magnitude of the net external force using 8.1 8.2 Discussion This quantity was the average force exerted by Venus Williams’ racquet on the tennis ball during its brief impact. This problem could also be solved by first finding the acceleration and then using Fnet = ma, but we would have had to do one more step. In this case, using momentum was a shortcut. Practice Problems 1. What is the momentum of a bowling ball with mass and velocity? a. b. c. d. 2. What will be the change in momentum caused by a net force of acting on an object for seconds? a. b. c. d. Check Your Understanding 3. What is linear momentum? a. b. c. d. the sum of a system’s mass and its velocity the ratio of a system’s mass to its velocity the product of a system’s mass and its velocity the product of a system’s moment of inertia and its velocity 4. If an object’s mass is constant, what is its momentum proportional to? a. b. c. d. Its velocity Its weight Its displacement Its moment of inertia 5. What is the equation for Newton’s second law of motion, in terms of mass, velocity, and time, when the mass
of the system is Access for free at openstax.org. 8.2 • Conservation of Momentum 259 constant? a. b. c. d. 6. Give an example of a system whose mass is not constant. a. A spinning top b. A baseball flying through the air c. A rocket launched from Earth d. A block sliding on a frictionless inclined plane 8.2 Conservation of Momentum Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the law of conservation of momentum verbally and mathematically Section Key Terms angular momentum isolated system law of conservation of momentum Conservation of Momentum It is important we realize that momentum is conserved during collisions, explosions, and other events involving objects in motion. To say that a quantity is conserved means that it is constant throughout the event. In the case of conservation of momentum, the total momentum in the system remains the same before and after the collision. You may have noticed that momentum was notconserved in some of the examples previously presented in this chapter. where forces acting on the objects produced large changes in momentum. Why is this? The systems of interest considered in those problems were not inclusive enough. If the systems were expanded to include more objects, then momentum would in fact be conserved in those sample problems. It is always possible to find a larger system where momentum is conserved, even though momentum changes for individual objects within the system. For example, if a football player runs into the goalpost in the end zone, a force will cause him to bounce backward. His momentum is obviously greatly changed, and considering only the football player, we would find that momentum is not conserved. However, the system can be expanded to contain the entire Earth. Surprisingly, Earth also recoils—conserving momentum—because of the force applied to it through the goalpost. The effect on Earth is not noticeable because it is so much more massive than the player, but the effect is real. Next, consider what happens if the masses of two colliding objects are more similar than the masses of a football player and Earth—in the example shown in Figure 8.4 of one car bumping into another. Both cars are coasting in the same direction when the lead car, labeled m2, is bumped by the trailing car, labeled m1. The only unbalanced force on each car is the force of the collision, assuming that the effects due to friction are negligible. Car m1 slows down as a
result of the collision, losing some momentum, while car m2 speeds up and gains some momentum. If we choose the system to include both cars and assume that friction is negligible, then the momentum of the two-car system should remain constant. Now we will prove that the total momentum of the two-car system does in fact remain constant, and is therefore conserved. 260 Chapter 8 • Momentum Figure 8.4 Car of mass m1 moving with a velocity of v1 bumps into another car of mass m2 and velocity v2. As a result, the first car slows down to a velocity of v′1 and the second speeds up to a velocity of v′2. The momentum of each car is changed, but the total momentum ptot of the two cars is the same before and after the collision if you assume friction is negligible. Using the impulse-momentum theorem, the change in momentum of car 1 is given by where F1 is the force on car 1 due to car 2, and is the time the force acts, or the duration of the collision. Similarly, the change in momentum of car 2 is duration of the collision is the same for both cars. We know from Newton’s third law of motion that F2 = –F1, and so. where F2 is the force on car 2 due to car 1, and we assume the Therefore, the changes in momentum are equal and opposite, and. Because the changes in momentum add to zero, the total momentum of the two-car system is constant. That is, where p′1 and p′2 are the momenta of cars 1 and 2 after the collision. This result that momentum is conserved is true not only for this example involving the two cars, but for any system where the net external force is zero, which is known as an isolated system. The law of conservation of momentum states that for an isolated system with any number of objects in it, the total momentum is conserved. In equation form, the law of conservation of momentum for an isolated system is written as or where ptot is the total momentum, or the sum of the momenta of the individual objects in the system at a given time, and p′tot is the total momentum some time later. The conservation of momentum principle can be applied to systems as diverse as a comet striking the Earth or a gas containing huge numbers of atoms and molecules. Conservation of momentum appears to be violated only when the net external force is not zero. But another
larger system can always be considered in which momentum is conserved by simply including the source of the external force. For example, in the collision of two cars considered above, the two-car system conserves momentum while each one-car system does not. Access for free at openstax.org. 8.2 • Conservation of Momentum 261 TIPS FOR SUCCESS Momenta is the plural form of the word momentum. One object is said to have momentum, but two or more objects are said to have momenta. FUN IN PHYSICS Angular Momentum in Figure Skating So far we have covered linear momentum, which describes the inertia of objects traveling in a straight line. But we know that many objects in nature have a curved or circular path. Just as linear motion has linear momentum to describe its tendency to move forward, circular motion has the equivalent angular momentum to describe how rotational motion is carried forward. This is similar to how torque is analogous to force, angular acceleration is analogous to translational acceleration, and mr2 is analogous to mass or inertia. You may recall learning that the quantity mr2 is called the rotational inertia or moment of inertia of a point mass mat a distance rfrom the center of rotation. We already know the equation for linear momentum, p = mv. Since angular momentum is analogous to linear momentum, the moment of inertia (I) is analogous to mass, and angular velocity is analogous to linear velocity, it makes sense that angular momentum (L) is defined as Angular momentum is conserved when the net external torque ( external force is zero. ) is zero, just as linear momentum is conserved when the net Figure skaters take advantage of the conservation of angular momentum, likely without even realizing it. In Figure 8.5, a figure skater is executing a spin. The net torque on her is very close to zero, because there is relatively little friction between her skates and the ice, and because the friction is exerted very close to the pivot point. Both F and rare small, and so is negligibly small. Figure 8.5 (a) An ice skater is spinning on the tip of her skate with her arms extended. In the next image, (b), her rate of spin increases greatly when she pulls in her arms. Consequently, she can spin for quite some time. She can do something else, too. She can increase her rate of spin by pulling her arms and legs in. Why does pulling her arms and legs in increase her rate of
spin? The answer is that her angular momentum is constant, so that L = L′. Expressing this equation in terms of the moment of inertia, where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because I′ is smaller, the angular velocity must increase to keep the angular momentum constant. This allows her to spin much faster without exerting any extra torque. A video (http://openstax.org/l/28figureskater) is also available that shows a real figure skater executing a spin. It discusses the physics of spins in figure skating. 262 Chapter 8 • Momentum GRASP CHECK Based on the equation L = Iω, how would you expect the moment of inertia of an object to affect angular momentum? How would angular velocity affect angular momentum? a. Large moment of inertia implies large angular momentum, and large angular velocity implies large angular momentum. b. Large moment of inertia implies small angular momentum, and large angular velocity implies small angular momentum. c. Large moment of inertia implies large angular momentum, and large angular velocity implies small angular momentum. d. Large moment of inertia implies small angular momentum, and large angular velocity implies large angular momentum. Check Your Understanding 7. When is momentum said to be conserved? a. When momentum is changing during an event b. When momentum is increasing during an event c. When momentum is decreasing during an event d. When momentum is constant throughout an event 8. A ball is hit by a racket and its momentum changes. How is momentum conserved in this case? a. Momentum of the system can never be conserved in this case. b. Momentum of the system is conserved if the momentum of the racket is not considered. c. Momentum of the system is conserved if the momentum of the racket is also considered. d. Momentum of the system is conserved if the momenta of the racket and the player are also considered. 9. State the law of conservation of momentum. a. Momentum is conserved for an isolated system with any number of objects in it. b. Momentum is conserved for an isolated system with an even number of objects in it. c. Momentum is conserved for an interacting system with any number of objects in it. d. Momentum is conserved for an interacting system with an even number of objects in it. 8.3 Elastic and Inelastic Collisions Section Learning Objectives By the end of this
section, you will be able to do the following: • Distinguish between elastic and inelastic collisions • Solve collision problems by applying the law of conservation of momentum Section Key Terms elastic collision inelastic collision point masses recoil Elastic and Inelastic Collisions When objects collide, they can either stick together or bounce off one another, remaining separate. In this section, we’ll cover these two different types of collisions, first in one dimension and then in two dimensions. In an elastic collision, the objects separate after impact and don’t lose any of their kinetic energy. Kinetic energy is the energy of motion and is covered in detail elsewhere. The law of conservation of momentum is very useful here, and it can be used whenever the net external force on a system is zero. Figure 8.6 shows an elastic collision where momentum is conserved. Access for free at openstax.org. 8.3 • Elastic and Inelastic Collisions 263 Figure 8.6 The diagram shows a one-dimensional elastic collision between two objects. An animation of an elastic collision between balls can be seen by watching this video (http://openstax.org/l/28elasticball). It replicates the elastic collisions between balls of varying masses. Perfectly elastic collisions can happen only with subatomic particles. Everyday observable examples of perfectly elastic collisions don’t exist—some kinetic energy is always lost, as it is converted into heat transfer due to friction. However, collisions between everyday objects are almost perfectly elastic when they occur with objects and surfaces that are nearly frictionless, such as with two steel blocks on ice. Now, to solve problems involving one-dimensional elastic collisions between two objects, we can use the equation for conservation of momentum. First, the equation for conservation of momentum for two objects in a one-dimensional collision is Substituting the definition of momentum p = mv for each initial and final momentum, we get where the primes (') indicate values after the collision; In some texts, you may see ifor initial (before collision) and ffor final (after collision). The equation assumes that the mass of each object does not change during the collision. WATCH PHYSICS Momentum: Ice Skater Throws a Ball This video covers an elastic collision problem in which we find the recoilvelocityof an ice skater who throws a ball straight forward. To clarify, Sal is using the equation Click to view content (https://www.khanacademy.
org/embed_video?v=vPkkCOlGND4). GRASP CHECK The resultant vector of the addition of vectors respectively. Which of the following is true? a. b. and is. The magnitudes of,, and are,, and, 264 Chapter 8 • Momentum c. d. Now, let us turn to the second type of collision. An inelastic collision is one in which objects stick together after impact, and kinetic energy is notconserved. This lack of conservation means that the forces between colliding objects may convert kinetic energy to other forms of energy, such as potential energy or thermal energy. The concepts of energy are discussed more thoroughly elsewhere. For inelastic collisions, kinetic energy may be lost in the form of heat. Figure 8.7 shows an example of an inelastic collision. Two objects that have equal masses head toward each other at equal speeds and then stick together. The two objects come to rest after sticking together, conserving momentum but not kinetic energy after they collide. Some of the energy of motion gets converted to thermal energy, or heat. Figure 8.7 A one-dimensional inelastic collision between two objects. Momentum is conserved, but kinetic energy is not conserved. (a) Two objects of equal mass initially head directly toward each other at the same speed. (b) The objects stick together, creating a perfectly inelastic collision. In the case shown in this figure, the combined objects stop; This is not true for all inelastic collisions. Since the two objects stick together after colliding, they move together at the same speed. This lets us simplify the conservation of momentum equation from to for inelastic collisions, where v′ is the final velocity for both objects as they are stuck together, either in motion or at rest. WATCH PHYSICS Introduction to Momentum This video reviews the definitions of momentum and impulse. It also covers an example of using conservation of momentum to solve a problem involving an inelastic collision between a car with constant velocity and a stationary truck. Note that Sal accidentally gives the unit for impulse as Joules; it is actually N s or k gm/s. Click to view content (https://www.khanacademy.org/embed_video?v=XFhntPxow0U) GRASP CHECK How would the final velocity of the car-plus-truck system change if the truck had some initial velocity moving in
the same direction as the car? What if the truck were moving in the opposite direction of the car initially? Why? a. If the truck was initially moving in the same direction as the car, the final velocity would be greater. If the truck was initially moving in the opposite direction of the car, the final velocity would be smaller. If the truck was initially moving in the same direction as the car, the final velocity would be smaller. If the truck was initially moving in the opposite direction of the car, the final velocity would be greater. b. c. The direction in which the truck was initially moving would not matter. If the truck was initially moving in either Access for free at openstax.org. direction, the final velocity would be smaller. d. The direction in which the truck was initially moving would not matter. If the truck was initially moving in either direction, the final velocity would be greater. 8.3 • Elastic and Inelastic Collisions 265 Snap Lab Ice Cubes and Elastic Collisions In this activity, you will observe an elastic collision by sliding an ice cube into another ice cube on a smooth surface, so that a negligible amount of energy is converted to heat. • Several ice cubes (The ice must be in the form of cubes.) • A smooth surface Procedure 1. Find a few ice cubes that are about the same size and a smooth kitchen tabletop or a table with a glass top. 2. Place the ice cubes on the surface several centimeters away from each other. 3. Flick one ice cube toward a stationary ice cube and observe the path and velocities of the ice cubes after the collision. Try to avoid edge-on collisions and collisions with rotating ice cubes. 4. Explain the speeds and directions of the ice cubes using momentum. GRASP CHECK Was the collision elastic or inelastic? a. perfectly elastic b. perfectly inelastic c. Nearly perfect elastic d. Nearly perfect inelastic TIPS FOR SUCCESS Here’s a trick for remembering which collisions are elastic and which are inelastic: Elastic is a bouncy material, so when objects bounceoff one another in the collision and separate, it is an elastic collision. When they don’t, the collision is inelastic. Solving Collision Problems The Khan Academy videos referenced in this section show examples of elastic and inelastic collisions in one dimension. In onedimensional collisions, the incoming and outgoing velocities are all along the same line. But what
about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and just as we did with twodimensional forces, we will solve these problems by first choosing a coordinate system and separating the motion into its xand y components. One complication with two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass each other, they will spin in circles. We will not consider such rotation until later, and so for now, we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses—that is, structureless particles that cannot rotate or spin. We start by assuming that Fnet = 0, so that momentum p is conserved. The simplest collision is one in which one of the particles is initially at rest. The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in Figure 8.8. Because momentum is conserved, the components of momentum along the x- and y-axes, displayed as px and py, will also be conserved. With the chosen coordinate system, pyis initially zero and pxis the momentum of the incoming particle. 266 Chapter 8 • Momentum Figure 8.8 A two-dimensional collision with the coordinate system chosen so that m2 is initially at rest and v1 is parallel to the x-axis. Now, we will take the conservation of momentum equation, p1 + p2 = p′1 + p′2 and break it into its xand ycomponents. Along the x-axis, the equation for conservation of momentum is In terms of masses and velocities, this equation is But because particle 2 is initially at rest, this equation becomes 8.3 8.4 The components of the velocities along the x-axis have the form v cos θ. Because particle 1 initially moves along the x-axis, we find v1x= v1. Conservation of momentum along the x-axis gives the equation where and are as shown in Figure 8.8. Along the y-axis, the equation for conservation of momentum is or But v1yis zero, because particle 1 initially moves along the x-axis. Because particle 2 is initially at rest, v2yis also zero. The equation for conservation of momentum along the y-axis becomes 8.5 8.6
8.7 The components of the velocities along the y-axis have the form v sin. Therefore, conservation of momentum along the y-axis gives the following equation: Virtual Physics Collision Lab In this simulation, you will investigate collisions on an air hockey table. Place checkmarks next to the momentum vectors Access for free at openstax.org. 8.3 • Elastic and Inelastic Collisions 267 and momenta diagram options. Experiment with changing the masses of the balls and the initial speed of ball 1. How does this affect the momentum of each ball? What about the total momentum? Next, experiment with changing the elasticity of the collision. You will notice that collisions have varying degrees of elasticity, ranging from perfectly elastic to perfectly inelastic. Click to view content (https://archive.cnx.org/specials/2c7acb3c-2fbd-11e5-b2d9-e7f92291703c/collision-lab/) GRASP CHECK If you wanted to maximize the velocity of ball 2 after impact, how would you change the settings for the masses of the balls, the initial speed of ball 1, and the elasticity setting? Why? Hint—Placing a checkmark next to the velocity vectors and removing the momentum vectors will help you visualize the velocity of ball 2, and pressing the More Data button will let you take readings. a. Maximize the mass of ball 1 and initial speed of ball 1; minimize the mass of ball 2; and set elasticity to 50 percent. b. Maximize the mass of ball 2 and initial speed of ball 1; minimize the mass of ball 1; and set elasticity to 100 percent. c. Maximize the mass of ball 1 and initial speed of ball 1; minimize the mass of ball 2; and set elasticity to 100 percent. d. Maximize the mass of ball 2 and initial speed of ball 1; minimize the mass of ball 1; and set elasticity to 50 percent. WORKED EXAMPLE Calculating Velocity: Inelastic Collision of a Puck and a Goalie Find the recoil velocity of a 70 kg ice hockey goalie who catches a 0.150-kg hockey puck slapped at him at a velocity of 35 m/s. Assume that the goalie is at rest before catching the puck, and friction between the ice and the puck-goalie system is negligible (see Figure 8.9). Figure 8.
9 An ice hockey goalie catches a hockey puck and recoils backward in an inelastic collision. Strategy Momentum is conserved because the net external force on the puck-goalie system is zero. Therefore, we can use conservation of momentum to find the final velocity of the puck and goalie system. Note that the initial velocity of the goalie is zero and that the final velocity of the puck and goalie are the same. Solution For an inelastic collision, conservation of momentum is where v′ is the velocity of both the goalie and the puck after impact. Because the goalie is initially at rest, we know v2 = 0. This simplifies the equation to 8.8 8.9 Solving for v′ yields 268 Chapter 8 • Momentum Entering known values in this equation, we get 8.10 8.11 Discussion This recoil velocity is small and in the same direction as the puck’s original velocity. WORKED EXAMPLE Calculating Final Velocity: Elastic Collision of Two Carts Two hard, steel carts collide head-on and then ricochet off each other in opposite directions on a frictionless surface (see Figure 8.10). Cart 1 has a mass of 0.350 kg and an initial velocity of 2 m/s. Cart 2 has a mass of 0.500 kg and an initial velocity of −0.500 m/s. After the collision, cart 1 recoils with a velocity of −4 m/s. What is the final velocity of cart 2? Figure 8.10 Two carts collide with each other in an elastic collision. Strategy Since the track is frictionless, Fnet = 0 and we can use conservation of momentum to find the final velocity of cart 2. Solution As before, the equation for conservation of momentum for a one-dimensional elastic collision in a two-object system is The only unknown in this equation is v′2. Solving for v′2 and substituting known values into the previous equation yields 8.12 8.13 Discussion The final velocity of cart 2 is large and positive, meaning that it is moving to the right after the collision. Access for free at openstax.org. 8.3 • Elastic and Inelastic Collisions 269 WORKED EXAMPLE Calculating Final Velocity in a Two-Dimensional Collision Suppose the following experiment is performed (Figure 8.11). An object of mass 0.250 kg (m1) is slid on a frictionless surface into a dark room, where it strikes an
initially stationary object of mass 0.400 kg (m2). The 0.250 kg object emerges from the room at an angle of 45º with its incoming direction. The speed of the 0.250 kg object is originally 2 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity (v′2 and ) of the 0.400 kg object after the collision. Figure 8.11 The incoming object of mass m1 is scattered by an initially stationary object. Only the stationary object’s mass m2 is known. By measuring the angle and speed at which the object of mass m1 emerges from the room, it is possible to calculate the magnitude and direction of the initially stationary object’s velocity after the collision. Strategy Momentum is conserved because the surface is frictionless. We chose the coordinate system so that the initial velocity is parallel to the x-axis, and conservation of momentum along the x- and y-axes applies. Everything is known in these equations except v′2 and θ2, which we need to find. We can find two unknowns because we have two independent equations—the equations describing the conservation of momentum in the xand ydirections. Solution First, we’ll solve both conservation of momentum equations ( ) for v′2 sin. For conservation of momentum along x-axis, let’s substitute sin comes from rearranging the definition of the trigonometric identity tan = sin /cos for cos /tan Solving for v′2 sin yields For conservation of momentum along y-axis, solving for v′2 sin yields and so that terms may cancel out later on. This. This gives us 8.14 8.15 8.16 270 Chapter 8 • Momentum Since both equations equal v′2 sin, we can set them equal to one another, yielding Solving this equation for tan, we get Entering known values into the previous equation gives Therefore, Since angles are defined as positive in the counterclockwise direction, m2 is scattered to the right. We’ll use the conservation of momentum along the y-axis equation to solve for v′2. Entering known values into this equation gives Therefore, 8.17 8.18 8.19 8.20 8.21 8.22 8.23 Discussion Either equation for the x- or y-axis could have been used to solve for v′2, but the equation for the y-axis is easier
because it has fewer terms. Practice Problems 10. In an elastic collision, an object with momentum collides with another object moving to the right that has a. After the collision, both objects are still moving to the right, but the first object’s momentum. What is the final momentum of the second object? momentum changes to a. b. c. d. 11. In an elastic collision, an object with momentum 25 kg ⋅ m/s collides with another that has a momentum 35 kg ⋅ m/s. The first object’s momentum changes to 10 kg ⋅ m/s. What is the final momentum of the second object? 10 kg ⋅ m/s a. b. 20 kg ⋅ m/s 35 kg ⋅ m/s c. 50 kg ⋅ m/s d. Check Your Understanding 12. What is an elastic collision? a. An elastic collision is one in which the objects after impact are deformed permanently. b. An elastic collision is one in which the objects after impact lose some of their internal kinetic energy. Access for free at openstax.org. 8.3 • Elastic and Inelastic Collisions 271 c. An elastic collision is one in which the objects after impact do not lose any of their internal kinetic energy. d. An elastic collision is one in which the objects after impact become stuck together and move with a common velocity. 13. Are perfectly elastic collisions possible? a. Perfectly elastic collisions are not possible. b. Perfectly elastic collisions are possible only with subatomic particles. c. Perfectly elastic collisions are possible only when the objects stick together after impact. d. Perfectly elastic collisions are possible if the objects and surfaces are nearly frictionless. 14. What is the equation for conservation of momentum for two objects in a one-dimensional collision? a. p1 + p1′ = p2 + p2′ b. p1 + p2 = p1′ + p2′ c. p1 − p2 = p1′ − p2′ d. p1 + p2 + p1′ + p2′ = 0 272 Chapter 8 • Key Terms KEY TERMS angular momentum the product of the moment of inertia and angular velocity change in momentum the difference between the final and initial values of momentum; the mass times the change in velocity elastic collision collision in which objects separate after after impact and kinetic energy is not conserved isolated system system in which the
net external force is zero law of conservation of momentum when the net external force is zero, the total momentum of the system is conserved or constant impact and kinetic energy is conserved linear momentum the product of a system's mass and impulse average net external force multiplied by the time the force acts; equal to the change in momentum impulse–momentum theorem the impulse, or change in momentum, is the product of the net external force and the time over which the force acts inelastic collision collision in which objects stick together SECTION SUMMARY 8.1 Linear Momentum, Force, and Impulse • Linear momentum, often referenced as momentumfor short, is defined as the product of a system’s mass multiplied by its velocity, p = mv. • The SI unit for momentum is kg m/s. • Newton’s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes,. • Impulse is the average net external force multiplied by the time this force acts, and impulse equals the change in momentum,. • Forces are usually not constant over a period of time, so we use the average of the force over the time it acts. 8.2 Conservation of Momentum velocity point masses spin structureless particles that cannot rotate or recoil backward movement of an object caused by the transfer of momentum from another object in a collision In an isolated system, the net external force is zero. • • Conservation of momentum applies only when the net external force is zero, within the defined system. 8.3 Elastic and Inelastic Collisions • If objects separate after impact, the collision is elastic; If they stick together, the collision is inelastic. • Kinetic energy is conserved in an elastic collision, but not in an inelastic collision. • The approach to two-dimensional collisions is to choose a convenient coordinate system and break the motion into components along perpendicular axes. Choose a coordinate system with the x-axis parallel to the velocity of the incoming particle. • Two-dimensional collisions of point masses, where mass 2 is initially at rest, conserve momentum along the initial direction of mass 1, or the x-axis, and along the direction perpendicular to the initial direction, or the y-axis. • The law of conservation of momentum is written ptot = • Point masses are structureless particles that cannot constant or ptot = p′tot (isolated system), where ptot is
the initial total momentum and p′tot is the total momentum some time later. spin. KEY EQUATIONS 8.1 Linear Momentum, Force, and Impulse Newton’s second law in terms of momentum impulse impulse–momentum theorem linear momentum Access for free at openstax.org. 8.2 Conservation of Momentum law of conservation of momentum ptot = constant, or ptot = p′tot Chapter 8 • Chapter Review 273 conservation of momentum along x-axis for 2D collisions conservation of momentum along y-axis for 2D collisions conservation of momentum for two objects p1 + p2 = constant, or p1 + p2 = p′1 + p′2 angular momentum L = I 8.3 Elastic and Inelastic Collisions conservation of momentum in an elastic collision conservation of momentum in an inelastic collision CHAPTER REVIEW Concept Items 8.2 Conservation of Momentum 8.1 Linear Momentum, Force, and Impulse 5. What is angular momentum? 1. What is impulse? a. Change in velocity b. Change in momentum c. Rate of change of velocity d. Rate of change of momentum 2. In which equation of Newton’s second law is mass assumed to be constant? a. b. c. d. 3. What is the SI unit of momentum? a. b. c. d. 4. What is the equation for linear momentum? a. b. c. d. a. The sum of moment of inertia and angular velocity b. The ratio of moment of inertia to angular velocity c. The product of moment of inertia and angular velocity d. Half the product of moment of inertia and square of angular velocity 6. What is an isolated system? a. A system in which the net internal force is zero b. A system in which the net external force is zero c. A system in which the net internal force is a nonzero constant d. A system in which the net external force is a nonzero constant 8.3 Elastic and Inelastic Collisions 7. In the equation p1 + p2 = p'1 + p'2 for the collision of two objects, what is the assumption made regarding the friction acting on the objects? a. Friction is zero. b. Friction is nearly zero. c. Friction acts constantly. d. Friction before and after the impact remains the same. 8. What is an inelastic collision? 274 Chapter 8 • Chapter Review a. when
objects stick together after impact, and their c. when objects stick together after impact, and always internal energy is not conserved come to rest instantaneously after collision b. when objects stick together after impact, and their d. when objects stick together after impact, and their internal energy is conserved internal energy increases Critical Thinking Items 8.1 Linear Momentum, Force, and Impulse 9. Consider two objects of the same mass. If a force of acts on the first for a duration of and on the, which of the following other for a duration of statements is true? a. The first object will acquire more momentum. b. The second object will acquire more momentum. c. Both objects will acquire the same momentum. d. Neither object will experience a change in momentum. 10. Cars these days have parts that can crumple or collapse in the event of an accident. How does this help protect the passengers? a. It reduces injury to the passengers by increasing the time of impact. It reduces injury to the passengers by decreasing the time of impact. It reduces injury to the passengers by increasing the change in momentum. It reduces injury to the passengers by decreasing the change in momentum. b. c. d. 11. How much force would be needed to cause a 17 kg ⋅ m/s change in the momentum of an object, if the force acted for 5 seconds? 3.4 N a. b. 12 N c. 22 N d. 85 N 8.2 Conservation of Momentum 12. A billiards ball rolling on the table has momentum p1. It hits another stationary ball, which then starts rolling. Considering friction to be negligible, what will happen to the momentum of the first ball? Problems 8.1 Linear Momentum, Force, and Impulse is applied to an object for, and it, what could be the mass 16. If a force of changes its velocity by of the object? a. b. Access for free at openstax.org. a. b. c. d. It will decrease. It will increase. It will become zero. It will remain the same. 13. A ball rolling on the floor with momentum p1 collides with a stationary ball and sets it in motion. The momentum of the first ball becomes p'1, and that of the second becomes p'2. Compare the magnitudes of p1 and p'2. a. Momenta p1 and p'2 are the same in magnitude. b. The sum of
the magnitudes of p1 and p'2 is zero. c. The magnitude of p1 is greater than that of p'2. d. The magnitude of p'2 is greater than that of p1. 14. Two cars are moving in the same direction. One car with momentum p1 collides with another, which has momentum p2. Their momenta become p'1 and p'2 respectively. Considering frictional losses, compare (p'1 + p'2 ) with (p1 + p2). a. The value of (p'1 + p'2 ) is zero. b. The values of (p1 + p2) and (p'1 + p'2 ) are equal. c. The value of (p1 + p2) will be greater than (p'1 + p'2 ). d. The value of (p'1 + p'2 ) will be greater than (p1 + p2). 8.3 Elastic and Inelastic Collisions 15. Two people, who have the same mass, throw two different objects at the same velocity. If the first object is heavier than the second, compare the velocities gained by the two people as a result of recoil. a. The first person will gain more velocity as a result of recoil. b. The second person will gain more velocity as a result of recoil. c. Both people will gain the same velocity as a result of recoil. d. The velocity of both people will be zero as a result of recoil. c. d. 17. For how long should a force of 130 N be applied to an object of mass 50 kg to change its speed from 20 m/s to 60 m/s? a. 0.031 s b. 0.065 s 15.4 s c. Chapter 8 • Test Prep 275 d. 40 s d. 50.0 m/s 8.3 Elastic and Inelastic Collisions 18. If a man with mass 70 kg, standing still, throws an object with mass 5 kg at 50 m/s, what will be the recoil velocity of the man, assuming he is standing on a frictionless surface? a. −3.6 m/s b. 0 m/s c. 3.6 m/s 19. Find the recoil velocity of a ice hockey goalie who hockey puck slapped at him at a. Assume that the goalie is at rest catches a velocity of before catching the puck, and
friction between the ice and the puck-goalie system is negligible. a. b. c. d. Performance Task 8.3 Elastic and Inelastic Collisions 20. You will need the following: • balls of different weights • a ruler or wooden strip • some books • a paper cup Make an inclined plane by resting one end of a ruler on a stack of books. Place a paper cup on the other end. Roll TEST PREP Multiple Choice 8.1 Linear Momentum, Force, and Impulse 21. What kind of quantity is momentum? a. Scalar b. Vector 22. When does the net force on an object increase? a. When Δp decreases b. When Δtincreases c. When Δtdecreases 23. In the equation Δp = m(vf − vi), which quantity is considered to be constant? a. Initial velocity b. Final velocity c. Mass d. Momentum 24. For how long should a force of be applied to change the momentum of an object by a. b. c. d. 8.2 Conservation of Momentum 25. In the equation L = Iω, what is I? a ball from the top of the ruler so that it hits the paper cup. Measure the displacement of the paper cup due to the collision. Now use increasingly heavier balls for this activity and see how that affects the displacement of the cup. Plot a graph of mass vs. displacement. Now repeat the same activity, but this time, instead of using different balls, change the incline of the ruler by varying the height of the stack of books. This will give you different velocities of the ball. See how this affects the displacement of the paper cup. a. Linear momentum b. Angular momentum c. Torque d. Moment of inertia 26. Give an example of an isolated system. a. A cyclist moving along a rough road b. A figure skater gliding in a straight line on an ice rink c. A baseball player hitting a home run d. A man drawing water from a well 8.3 Elastic and Inelastic Collisions 27. In which type of collision is kinetic energy conserved? a. Elastic b. Inelastic 28. In physics, what are structureless particles that cannot? rotate or spin called? a. Elastic particles b. Point masses c. Rigid masses 29. Two objects having equal masses and velocities collide with each other and come to a rest. What type of a collision is this
and why? a. Elastic collision, because internal kinetic energy is conserved 276 Chapter 8 • Test Prep b. Inelastic collision, because internal kinetic energy is not conserved c. Elastic collision, because internal kinetic energy is d. not conserved Inelastic collision, because internal kinetic energy is conserved 30. Two objects having equal masses and velocities collide with each other and come to a rest. Is momentum conserved in this case? a. Yes b. No Short Answer 8.1 Linear Momentum, Force, and Impulse 31. If an object’s velocity is constant, what is its momentum proportional to? a. b. c. d. Its shape Its mass Its length Its breadth 32. If both mass and velocity of an object are constant, what can you tell about its impulse? a. b. c. d. Its impulse would be constant. Its impulse would be zero. Its impulse would be increasing. Its impulse would be decreasing. 33. When the momentum of an object increases with respect to time, what is true of the net force acting on it? a. It is zero, because the net force is equal to the rate of change of the momentum. It is zero, because the net force is equal to the product of the momentum and the time interval. It is nonzero, because the net force is equal to the rate of change of the momentum. It is nonzero, because the net force is equal to the product of the momentum and the time interval. b. c. d. 34. How can you express impulse in terms of mass and velocity when neither of those are constant? a. b. c. d. 35. How can you express impulse in terms of mass and initial and final velocities? a. b. c. d. 36. Why do we use average force while solving momentum problems? How is net force related to the momentum of the object? a. Forces are usually constant over a period of time, Access for free at openstax.org. and net force acting on the object is equal to the rate of change of the momentum. b. Forces are usually not constant over a period of time, and net force acting on the object is equal to the product of the momentum and the time interval. c. Forces are usually constant over a period of time, and net force acting on the object is equal to the product of the momentum and the time interval. d. Forces are usually not constant over a period
of time, and net force acting on the object is equal to the rate of change of the momentum. 8.2 Conservation of Momentum 37. Under what condition(s) is the angular momentum of a system conserved? a. When net torque is zero b. When net torque is not zero c. When moment of inertia is constant d. When both moment of inertia and angular momentum are constant 38. If the moment of inertia of an isolated system increases, what happens to its angular velocity? a. b. c. d. It increases. It decreases. It stays constant. It becomes zero. 39. If both the moment of inertia and the angular velocity of a system increase, what must be true of the force acting on the system? a. Force is zero. b. Force is not zero. c. Force is constant. d. Force is decreasing. 8.3 Elastic and Inelastic Collisions 40. Two objects collide with each other and come to a rest. How can you use the equation of conservation of momentum to describe this situation? a. m1v1 + m2v2 = 0 b. m1v1 − m2v2 = 0 c. m1v1 + m2v2 = m1v1′ d. m1v1 + m2v2 = m1v2 41. What is the difference between momentum and impulse? a. Momentum is the sum of mass and velocity. Impulse is the change in momentum. b. Momentum is the sum of mass and velocity. Impulse is the rate of change in momentum. c. Momentum is the product of mass and velocity. Impulse is the change in momentum. d. Momentum is the product of mass and velocity. Impulse is the rate of change in momentum. 42. What is the equation for conservation of momentum along the x-axis for 2D collisions in terms of mass and velocity, where one of the particles is initially at rest? Chapter 8 • Test Prep 277 a. m1v1 = m1v1′cos θ1 b. m1v1 = m1v1′cos θ1 + m2v2′cos θ2 c. m1v1 = m1v1′cos θ1 − m2v2′cos θ2 d. m1v1 = m1v1′sin θ1 + m2v2′sin θ2 43. What is the
equation for conservation of momentum along the y-axis for 2D collisions in terms of mass and velocity, where one of the particles is initially at rest? a. 0 = m1v1′sin θ1 b. 0 = m1v1′sin θ1 + m2v2′sin θ2 c. 0 = m1v1′sin θ1 − m2v2′sin θ2 d. 0 = m1v1′cos θ1 + m2v2′cos θ2 Extended Response 8.2 Conservation of Momentum 8.1 Linear Momentum, Force, and Impulse 47. Why does a figure skater spin faster if he pulls his arms 44. Can a lighter object have more momentum than a heavier one? How? a. No, because momentum is independent of the velocity of the object. b. No, because momentum is independent of the mass of the object. c. Yes, if the lighter object’s velocity is considerably high. d. Yes, if the lighter object’s velocity is considerably low. 45. Why does it hurt less when you fall on a softer surface? a. The softer surface increases the duration of the impact, thereby reducing the effect of the force. b. The softer surface decreases the duration of the impact, thereby reducing the effect of the force. c. The softer surface increases the duration of the and legs in? a. Due to an increase in moment of inertia b. Due to an increase in angular momentum c. Due to conservation of linear momentum d. Due to conservation of angular momentum 8.3 Elastic and Inelastic Collisions 48. A driver sees another car approaching him from behind. He fears it is going to collide with his car. Should he speed up or slow down in order to reduce damage? a. He should speed up. b. He should slow down. c. He should speed up and then slow down just before the collision. d. He should slow down and then speed up just before the collision. impact, thereby increasing the effect of the force. 49. What approach would you use to solve problems d. The softer surface decreases the duration of the impact, thereby increasing the effect of the force. involving 2D collisions? a. Break the momenta into components and then 46. Can we use the equation when the mass is constant? a. No, because the given equation is applicable for the variable mass only. b. No, because the
given equation is not applicable for the constant mass. c. Yes, and the resultant equation is F = mv d. Yes, and the resultant equation is F = ma choose a coordinate system. b. Choose a coordinate system and then break the momenta into components. c. Find the total momenta in the x and y directions, and then equate them to solve for the unknown. d. Find the sum of the momenta in the x and y directions, and then equate it to zero to solve for the unknown. 278 Chapter 8 • Test Prep Access for free at openstax.org. CHAPTER 9 Work, Energy, and Simple Machines Figure 9.1 People on a roller coaster experience thrills caused by changes in types of energy. (Jonrev, Wikimedia Commons) Chapter Outline 9.1 Work, Power, and the Work–Energy Theorem 9.2 Mechanical Energy and Conservation of Energy 9.3 Simple Machines Roller coasters have provided thrills for daring riders around the world since the nineteenth century. INTRODUCTION Inventors of roller coasters used simple physics to build the earliest examples using railroad tracks on mountainsides and old mines. Modern roller coaster designers use the same basic laws of physics to create the latest amusement park favorites. Physics principles are used to engineer the machines that do the work to lift a roller coaster car up its first big incline before it is set loose to roll. Engineers also have to understand the changes in the car’s energy that keep it speeding over hills, through twists, turns, and even loops. What exactly is energy? How can changes in force, energy, and simple machines move objects like roller coaster cars? How can machines help us do work? In this chapter, you will discover the answer to this question and many more, as you learn about 280 Chapter 9 • Work, Energy, and Simple Machines work, energy, and simple machines. 9.1 Work, Power, and the Work–Energy Theorem Section Learning Objectives By the end of this section, you will be able to do the following: • Describe and apply the work–energy theorem • Describe and calculate work and power Section Key Terms energy gravitational potential energy joule kinetic energy mechanical energy potential energy power watt work work–energy theorem The Work–Energy Theorem In physics, the term work has a very specific definition. Work is application of force, the direction that the force is applied. Work, W, is described by the equation, to move an object over
a distance, d, in Some things that we typically consider to be work are not work in the scientific sense of the term. Let’s consider a few examples. Think about why each of the following statements is true. • Homework is notwork. • Lifting a rock upwards off the ground iswork. • Carrying a rock in a straight path across the lawn at a constant speed is notwork. The first two examples are fairly simple. Homework is not work because objects are not being moved over a distance. Lifting a rock up off the ground is work because the rock is moving in the direction that force is applied. The last example is less obvious. Recall from the laws of motion that force is notrequired to move an object at constant velocity. Therefore, while some force may be applied to keep the rock up off the ground, no net force is applied to keep the rock moving forward at constant velocity. Work and energy are closely related. When you do work to move an object, you change the object’s energy. You (or an object) also expend energy to do work. In fact, energy can be defined as the ability to do work. Energy can take a variety of different forms, and one form of energy can transform to another. In this chapter we will be concerned with mechanical energy, which comes in two forms: kinetic energy and potential energy. • Kinetic energy is also called energy of motion. A moving object has kinetic energy. • Potential energy, sometimes called stored energy, comes in several forms. Gravitational potential energy is the stored energy an object has as a result of its position above Earth’s surface (or another object in space). A roller coaster car at the top of a hill has gravitational potential energy. Let’s examine how doing work on an object changes the object’s energy. If we apply force to lift a rock off the ground, we increase the rock’s potential energy, PE. If we drop the rock, the force of gravity increases the rock’s kinetic energy as the rock moves downward until it hits the ground. The force we exert to lift the rock is equal to its weight, w, which is equal to its mass, m, multiplied by acceleration due to gravity, g. The work we do on the rock equals the force we exert multiplied by the distance, d, that we lift the rock. The work we do on the rock also equals the rock’s gain in gravitational potential
energy, PEe. Kinetic energy depends on the mass of an object and its velocity, v. Access for free at openstax.org. 9.1 • Work, Power, and the Work–Energy Theorem 281 When we drop the rock the force of gravity causes the rock to fall, giving the rock kinetic energy. When work done on an object increases only its kinetic energy, then the net work equals the change in the value of the quantity. This is a statement of the work–energy theorem, which is expressed mathematically as The subscripts 2 and 1 indicate the final and initial velocity, respectively. This theorem was proposed and successfully tested by James Joule, shown in Figure 9.2. Does the name Joule sound familiar? The joule (J) is the metric unit of measurement for both work and energy. The measurement of work and energy with the same unit reinforces the idea that work and energy are related and can be converted into one another. 1.0 J = 1.0 N∙m, the units of force multiplied by distance. 1.0 N = 1.0 k∙m/s2, so 1.0 J = 1.0 k∙m2/s2. Analyzing the units of the term (1/2)mv2 will produce the same units for joules. Figure 9.2 The joule is named after physicist James Joule (1818–1889). (C. H. Jeens, Wikimedia Commons) WATCH PHYSICS Work and Energy This video explains the work energy theorem and discusses how work done on an object increases the object’s KE. Click to view content (https://www.khanacademy.org/embed_video?v=2WS1sG9fhOk) GRASP CHECK True or false—The energy increase of an object acted on only by a gravitational force is equal to the product of the object's weight and the distance the object falls. a. True b. False Calculations Involving Work and Power In applications that involve work, we are often interested in how fast the work is done. For example, in roller coaster design, the amount of time it takes to lift a roller coaster car to the top of the first hill is an important consideration. Taking a half hour on the ascent will surely irritate riders and decrease ticket sales. Let’s take a look at how to calculate the time it takes to do
work. Recall that a rate can be used to describe a quantity, such as work, over a period of time. Power is the rate at which work is done. In this case, rate means per unit of time. Power is calculated by dividing the work done by the time it took to do the work. Let’s consider an example that can help illustrate the differences among work, force, and power. Suppose the woman in Figure 9.3 lifting the TV with a pulley gets the TV to the fourth floor in two minutes, and the man carrying the TV up the stairs takes five 282 Chapter 9 • Work, Energy, and Simple Machines minutes to arrive at the same place. They have done the same amount of work mass over the same vertical distance, which requires the same amount of upward force. However, the woman using the pulley has generated more power. This is because she did the work in a shorter amount of time, so the denominator of the power formula, t, is smaller. (For simplicity’s sake, we will leave aside for now the fact that the man climbing the stairs has also done work on himself.) on the TV, because they have moved the same Figure 9.3 No matter how you move a TV to the fourth floor, the amount of work performed and the potential energy gain are the same. Power can be expressed in units of watts (W). This unit can be used to measure power related to any form of energy or work. You have most likely heard the term used in relation to electrical devices, especially light bulbs. Multiplying power by time gives the amount of energy. Electricity is sold in kilowatt-hours because that equals the amount of electrical energy consumed. The watt unit was named after James Watt (1736–1819) (see Figure 9.4). He was a Scottish engineer and inventor who discovered how to coax more power out of steam engines. Figure 9.4 Is James Watt thinking about watts? (Carl Frederik von Breda, Wikimedia Commons) Access for free at openstax.org. 9.1 • Work, Power, and the Work–Energy Theorem 283 LINKS TO PHYSICS Watt’s Steam Engine James Watt did not invent the steam engine, but by the time he was finished tinkering with it, it was more useful. The first steam engines were not only inefficient, they only produced a back and forth, or reciprocal, motion. This was natural because pistons move in and out
as the pressure in the chamber changes. This limitation was okay for simple tasks like pumping water or mashing potatoes, but did not work so well for moving a train. Watt was able build a steam engine that converted reciprocal motion to circular motion. With that one innovation, the industrial revolution was off and running. The world would never be the same. One of Watt's steam engines is shown in Figure 9.5. The video that follows the figure explains the importance of the steam engine in the industrial revolution. Figure 9.5 A late version of the Watt steam engine. (Nehemiah Hawkins, Wikimedia Commons) WATCH PHYSICS Watt's Role in the Industrial Revolution This video demonstrates how the watts that resulted from Watt's inventions helped make the industrial revolution possible and allowed England to enter a new historical era. Click to view content (https://www.youtube.com/embed/zhL5DCizj5c) GRASP CHECK Which form of mechanical energy does the steam engine generate? a. Potential energy b. Kinetic energy c. Nuclear energy d. Solar energy Before proceeding, be sure you understand the distinctions among force, work, energy, and power. Force exerted on an object over a distance does work. Work can increase energy, and energy can do work. Power is the rate at which work is done. WORKED EXAMPLE Applying the Work–Energy Theorem An ice skater with a mass of 50 kg is gliding across the ice at a speed of 8 m/s when her friend comes up from behind and gives her a push, causing her speed to increase to 12 m/s. How much work did the friend do on the skater? 284 Chapter 9 • Work, Energy, and Simple Machines Strategy The work–energy theorem can be applied to the problem. Write the equation for the theorem and simplify it if possible. Solution Identify the variables. m= 50 kg, Substitute. 9.1 9.2 Discussion Work done on an object or system increases its energy. In this case, the increase is to the skater’s kinetic energy. It follows that the increase in energy must be the difference in KE before and after the push. TIPS FOR SUCCESS This problem illustrates a general technique for approaching problems that require you to apply formulas: Identify the unknown and the known variables, express the unknown variables in terms of the known variables, and then enter all the known values. Practice Problems 1. How much work is done when a weightl
ifter lifts a barbell from the floor to a height of? a. b. c. d. 2. Identify which of the following actions generates more power. Show your work. • • carrying a carrying a TV to the second floor in or watermelon to the second floor in? a. Carrying a TV generates more power than carrying a watermelon to the same height because power is defined as work done times the time interval. b. Carrying a TV generates more power than carrying a watermelon to the same height because power is defined as the ratio of work done to the time interval. c. Carrying a watermelon generates more power than carrying a TV to the same height because power is defined as work done times the time interval. d. Carrying a watermelon generates more power than carrying a TV to the same height because power is defined as the ratio of work done and the time interval. Check Your Understanding 3. Identify two properties that are expressed in units of joules. a. work and force b. energy and weight c. work and energy d. weight and force Access for free at openstax.org. 9.2 • Mechanical Energy and Conservation of Energy 285 4. When a coconut falls from a tree, work Wis done on it as it falls to the beach. This work is described by the equation 9.3 Identify the quantities F, d, m, v1, and v2 in this event. a. Fis the force of gravity, which is equal to the weight of the coconut, dis the distance the nut falls, mis the mass of the earth, v1 is the initial velocity, and v2 is the velocity with which it hits the beach. b. Fis the force of gravity, which is equal to the weight of the coconut, dis the distance the nut falls, mis the mass of the coconut, v1 is the initial velocity, and v2 is the velocity with which it hits the beach. c. Fis the force of gravity, which is equal to the weight of the coconut, dis the distance the nut falls, mis the mass of the earth, v1 is the velocity with which it hits the beach, and v2 is the initial velocity. d. Fis the force of gravity, which is equal to the weight of the coconut, dis the distance the nut falls, mis the mass of the coconut, v1 is the velocity with which it hits the beach, and v2 is the initial velocity. 9.
2 Mechanical Energy and Conservation of Energy Section Learning Objectives By the end of this section, you will be able to do the following: • Explain the law of conservation of energy in terms of kinetic and potential energy • Perform calculations related to kinetic and potential energy. Apply the law of conservation of energy Section Key Terms law of conservation of energy Mechanical Energy and Conservation of Energy We saw earlier that mechanical energy can be either potential or kinetic. In this section we will see how energy is transformed from one of these forms to the other. We will also see that, in a closed system, the sum of these forms of energy remains constant. Quite a bit of potential energy is gained by a roller coaster car and its passengers when they are raised to the top of the first hill. Remember that the potentialpart of the term means that energy has been stored and can be used at another time. You will see that this stored energy can either be used to do work or can be transformed into kinetic energy. For example, when an object that has gravitational potential energy falls, its energy is converted to kinetic energy. Remember that both work and energy are expressed in joules. Refer back to. The amount of work required to raise the TV from point A to point B is equal to the amount of gravitational potential energy the TV gains from its height above the ground. This is generally true for any object raised above the ground. If all the work done on an object is used to raise the object above the ground, the amount work equals the object’s gain in gravitational potential energy. However, note that because of the work done by friction, these energy–work transformations are never perfect. Friction causes the loss of some useful energy. In the discussions to follow, we will use the approximation that transformations are frictionless. Now, let’s look at the roller coaster in Figure 9.6. Work was done on the roller coaster to get it to the top of the first rise; at this point, the roller coaster has gravitational potential energy. It is moving slowly, so it also has a small amount of kinetic energy. As the car descends the first slope, its PEis converted to KE. At the low point much of the original PEhas been transformed to KE, and speed is at a maximum. As the car moves up the next slope, some of the KEis transformed back into PEand the car slows down. 286 Chapter 9 • Work, Energy, and Simple Machines Figure 9.6 During this roller coaster ride,
there are conversions between potential and kinetic energy. Virtual Physics Energy Skate Park Basics This simulation shows how kinetic and potential energy are related, in a scenario similar to the roller coaster. Observe the changes in KEand PEby clicking on the bar graph boxes. Also try the three differently shaped skate parks. Drag the skater to the track to start the animation. Click to view content (http://phet.colorado.edu/sims/html/energy-skate-park-basics/latest/energy-skate-parkbasics_en.html) GRASP CHECK This simulation (http://phet.colorado.edu/en/simulation/energy-skate-park-basics (http://phet.colorado.edu/en/ simulation/energy-skate-park-basics) ) shows how kinetic and potential energy are related, in a scenario similar to the roller coaster. Observe the changes in KE and PE by clicking on the bar graph boxes. Also try the three differently shaped skate parks. Drag the skater to the track to start the animation. The bar graphs show how KE and PE are transformed back and forth. Which statement best explains what happens to the mechanical energy of the system as speed is increasing? a. The mechanical energy of the system increases, provided there is no loss of energy due to friction. The energy would transform to kinetic energy when the speed is increasing. b. The mechanical energy of the system remains constant provided there is no loss of energy due to friction. The energy would transform to kinetic energy when the speed is increasing. c. The mechanical energy of the system increases provided there is no loss of energy due to friction. The energy would transform to potential energy when the speed is increasing. d. The mechanical energy of the system remains constant provided there is no loss of energy due to friction. The energy would transform to potential energy when the speed is increasing. On an actual roller coaster, there are many ups and downs, and each of these is accompanied by transitions between kinetic and potential energy. Assume that no energy is lost to friction. At any point in the ride, the total mechanical energy is the same, and it is equal to the energy the car had at the top of the first rise. This is a result of the law of conservation of energy, which says that, in a closed system, total energy is conserved—that is, it is constant. Using subscripts 1 and 2 to represent initial
and final energy, this law is expressed as Either side equals the total mechanical energy. The phrase in aclosed systemmeans we are assuming no energy is lost to the surroundings due to friction and air resistance. If we are making calculations on dense falling objects, this is a good assumption. For the roller coaster, this assumption introduces some inaccuracy to the calculation. Access for free at openstax.org. 9.2 • Mechanical Energy and Conservation of Energy 287 Calculations involving Mechanical Energy and Conservation of Energy TIPS FOR SUCCESS When calculating work or energy, use units of meters for distance, newtons for force, kilograms for mass, and seconds for time. This will assure that the result is expressed in joules. WATCH PHYSICS Conservation of Energy This video discusses conversion of PEto KEand conservation of energy. The scenario is very similar to the roller coaster and the skate park. It is also a good explanation of the energy changes studied in the snap lab. Click to view content (https://www.khanacademy.org/embed_video?v=kw_4Loo1HR4) GRASP CHECK Did you expect the speed at the bottom of the slope to be the same as when the object fell straight down? Which statement best explains why this is not exactly the case in real-life situations? a. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is large amount of friction. In real life, much of the mechanical energy is lost as heat caused by friction. b. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is small amount of friction. In real life, much of the mechanical energy is lost as heat caused by friction. c. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is large amount of friction. In real life, no mechanical energy is lost due to conservation of the mechanical energy. d. The speed was the same in the scenario in the animation because the object was sliding on the ice, where there is small amount of friction. In real life, no mechanical energy is lost due to conservation of the mechanical energy. WORKED EXAMPLE Applying the Law of Conservation of Energy A 10 kg rock falls from a 20 m cliff. What is the kinetic and potential energy when the rock has fallen 10 m? Strategy Choose the equation. 9.
4 9.5 9.6 9.7 List the knowns. m= 10 kg, v1 = 0, g = 9.80 h1 = 20 m, h2 = 10 m Identify the unknowns. KE2 and PE2 Substitute the known values into the equation and solve for the unknown variables. 288 Chapter 9 • Work, Energy, and Simple Machines Solution 9.8 9.9 Discussion Alternatively, conservation of energy equation could be solved for v2 and KE2 could be calculated. Note that mcould also be eliminated. TIPS FOR SUCCESS Note that we can solve many problems involving conversion between KEand PEwithout knowing the mass of the object in question. This is because kinetic and potential energy are both proportional to the mass of the object. In a situation where KE= PE, we know that mgh= (1/2)mv2. Dividing both sides by mand rearranging, we have the relationship 2gh= v2. Practice Problems 5. A child slides down a playground slide. If the slide is 3 m high and the child weighs 300 N, how much potential energy does ) the child have at the top of the slide? (Round gto a. 0 J 100 J b. c. 300 J d. 900 J 6. A 0.2 kg apple on an apple tree has a potential energy of 10 J. It falls to the ground, converting all of its PE to kinetic energy. What is the velocity of the apple just before it hits the ground? a. 0 m/s b. 2 m/s 10 m/s c. 50 m/s d. Snap Lab Converting Potential Energy to Kinetic Energy In this activity, you will calculate the potential energy of an object and predict the object’s speed when all that potential energy has been converted to kinetic energy. You will then check your prediction. You will be dropping objects from a height. Be sure to stay a safe distance from the edge. Don’t lean over the railing too far. Make sure that you do not drop objects into an area where people or vehicles pass by. Make sure that dropping objects will not cause damage. You will need the following: Materials for each pair of students: • Four marbles (or similar small, dense objects) • Stopwatch Materials for class: • Metric measuring tape long enough to measure the chosen height • A scale Instructions Access for free at openstax.org. 9.2 • Mechanical Energy and
Conservation of Energy 289 Procedure 1. Work with a partner. Find and record the mass of four small, dense objects per group. 2. Choose a location where the objects can be safely dropped from a height of at least 15 meters. A bridge over water with a safe pedestrian walkway will work well. 3. Measure the distance the object will fall. 4. Calculate the potential energy of the object before you drop it using PE= mgh= (9.80)mh. 5. Predict the kinetic energy and velocity of the object when it lands using PE= KEand so, 6. One partner drops the object while the other measures the time it takes to fall. 7. Take turns being the dropper and the timer until you have made four measurements. 8. Average your drop multiplied by and calculate the velocity of the object when it landed using v = at= gt= (9.80)t. 9. Compare your results to your prediction. GRASP CHECK Galileo’s experiments proved that, contrary to popular belief, heavy objects do not fall faster than light objects. How do the equations you used support this fact? a. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy of the system, the mass term gets cancelled and the velocity is independent of the mass. In real life, the variation in the velocity of the different objects is observed because of the non-zero air resistance. b. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy of the system, the mass term does not get cancelled and the velocity is dependent on the mass. In real life, the variation in the velocity of the different objects is observed because of the non-zero air resistance. c. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy the system, the mass term gets cancelled and the velocity is independent of the mass. In real life, the variation in the velocity of the different objects is observed because of zero air resistance. d. Heavy objects do not fall faster than the light objects because while conserving the mechanical energy of the system, the mass term does not get cancelled and the velocity is dependent on the mass. In real life, the variation in the velocity of the different objects is observed because of zero air resistance. Check Your Understanding 7. Describe the transformation between forms of mechanical energy that is happening to a falling skydiver before his parachute opens. a. Kinetic
energy is being transformed into potential energy. b. Potential energy is being transformed into kinetic energy. c. Work is being transformed into kinetic energy. d. Kinetic energy is being transformed into work. 8. True or false—If a rock is thrown into the air, the increase in the height would increase the rock’s kinetic energy, and then the increase in the velocity as it falls to the ground would increase its potential energy. a. True b. False 9. Identify equivalent terms for stored energyand energy of motion. a. Stored energy is potential energy, and energy of motion is kinetic energy. b. Energy of motion is potential energy, and stored energy is kinetic energy. c. Stored energy is the potential as well as the kinetic energy of the system. d. Energy of motion is the potential as well as the kinetic energy of the system. 290 Chapter 9 • Work, Energy, and Simple Machines 9.3 Simple Machines Section Learning Objectives By the end of this section, you will be able to do the following: • Describe simple and complex machines • Calculate mechanical advantage and efficiency of simple and complex machines Section Key Terms complex machine efficiency output ideal mechanical advantage inclined plane input work lever mechanical advantage output work pulley screw simple machine wedge wheel and axle Simple Machines Simple machines make work easier, but they do not decrease the amount of work you have to do. Why can’t simple machines change the amount of work that you do? Recall that in closed systems the total amount of energy is conserved. A machine cannot increase the amount of energy you put into it. So, why is a simple machine useful? Although it cannot change the amount of work you do, a simple machine can change the amount of force you must apply to an object, and the distance over which you apply the force. In most cases, a simple machine is used to reduce the amount of force you must exert to do work. The down side is that you must exert the force over a greater distance, because the product of force and distance, fd, (which equals work) does not change. Let’s examine how this works in practice. In Figure 9.7(a), the worker uses a type of lever to exert a small force over a large distance, while the pry bar pulls up on the nail with a large force over a small distance. Figure 9.7(b) shows the how a lever works mathematically. The effort force, applied at Fe, lifts the load
(the resistance force) which is pushing down at Fr. The triangular pivot is called the fulcrum; the part of the lever between the fulcrum and Feis the effort arm, Le; and the part to the left is the resistance arm, Lr. The mechanical advantage is a number that tells us how many times a simple machine multiplies the effort force. The ideal mechanical advantage, IMA, is the mechanical advantage of a perfect machine with no loss of useful work caused by friction between moving parts. The equation for IMAis shown in Figure 9.7(b). Figure 9.7 (a) A pry bar is a type of lever. (b) The ideal mechanical advantage equals the length of the effort arm divided by the length of the resistance arm of a lever. In general, the IMA= the resistance force, Fr, divided by the effort force, Fe. IMAalso equals the distance over which the effort is applied, de, divided by the distance the load travels, dr. Getting back to conservation of energy, for any simple machine, the work put into the machine, Wi, equals the work the machine puts out, Wo. Combining this with the information in the paragraphs above, we can write Access for free at openstax.org. 9.3 • Simple Machines 291 The equations show how a simple machine can output the same amount of work while reducing the amount of effort force by increasing the distance over which the effort force is applied. WATCH PHYSICS Introduction to Mechanical Advantage This video shows how to calculate the IMAof a lever by three different methods: (1) from effort force and resistance force; (2) from the lengths of the lever arms, and; (3) from the distance over which the force is applied and the distance the load moves. Click to view content (https://www.youtube.com/embed/pfzJ-z5Ij48) GRASP CHECK Two children of different weights are riding a seesaw. How do they position themselves with respect to the pivot point (the fulcrum) so that they are balanced? a. The heavier child sits closer to the fulcrum. b. The heavier child sits farther from the fulcrum. c. Both children sit at equal distance from the fulcrum. d. Since both have different weights, they will never be in balance. Some levers exert a large force to a short effort arm. This results in a smaller force acting over
a greater distance at the end of the resistance arm. Examples of this type of lever are baseball bats, hammers, and golf clubs. In another type of lever, the fulcrum is at the end of the lever and the load is in the middle, as in the design of a wheelbarrow. The simple machine shown in Figure 9.8 is called a wheel and axle. It is actually a form of lever. The difference is that the effort arm can rotate in a complete circle around the fulcrum, which is the center of the axle. Force applied to the outside of the wheel causes a greater force to be applied to the rope that is wrapped around the axle. As shown in the figure, the ideal mechanical advantage is calculated by dividing the radius of the wheel by the radius of the axle. Any crank-operated device is an example of a wheel and axle. Figure 9.8 Force applied to a wheel exerts a force on its axle. An inclined plane and a wedge are two forms of the same simple machine. A wedge is simply two inclined planes back to back. Figure 9.9 shows the simple formulas for calculating the IMAs of these machines. All sloping, paved surfaces for walking or driving are inclined planes. Knives and axe heads are examples of wedges. 292 Chapter 9 • Work, Energy, and Simple Machines Figure 9.9 An inclined plane is shown on the left, and a wedge is shown on the right. The screw shown in Figure 9.10 is actually a lever attached to a circular inclined plane. Wood screws (of course) are also examples of screws. The lever part of these screws is a screw driver. In the formula for IMA, the distance between screw threads is called pitchand has the symbol P. Figure 9.10 The screw shown here is used to lift very heavy objects, like the corner of a car or a house a short distance. Figure 9.11 shows three different pulley systems. Of all simple machines, mechanical advantage is easiest to calculate for pulleys. Simply count the number of ropes supporting the load. That is the IMA. Once again we have to exert force over a longer distance to multiply force. To raise a load 1 meter with a pulley system you have to pull Nmeters of rope. Pulley systems are often used to raise flags and window blinds and are part of the mechanism of construction cranes. Figure 9.11 Three pulley systems are shown here. Access for free at openstax.
org. 9.3 • Simple Machines 293 WATCH PHYSICS Mechanical Advantage of Inclined Planes and Pulleys The first part of this video shows how to calculate the IMAof pulley systems. The last part shows how to calculate the IMAof an inclined plane. Click to view content (https://www.khanacademy.org/embed_video?v=vSsK7Rfa3yA) GRASP CHECK How could you use a pulley system to lift a light load to great height? a. Reduce the radius of the pulley. b. Increase the number of pulleys. c. Decrease the number of ropes supporting the load. Increase the number of ropes supporting the load. d. A complex machine is a combination of two or more simple machines. The wire cutters in Figure 9.12 combine two levers and two wedges. Bicycles include wheel and axles, levers, screws, and pulleys. Cars and other vehicles are combinations of many machines. Figure 9.12 Wire cutters are a common complex machine. Calculating Mechanical Advantage and Efficiency of Simple Machines In general, the IMA= the resistance force, Fr, divided by the effort force, Fe. IMAalso equals the distance over which the effort is applied, de, divided by the distance the load travels, dr. Refer back to the discussions of each simple machine for the specific equations for the IMAfor each type of machine. No simple or complex machines have the actual mechanical advantages calculated by the IMAequations. In real life, some of the applied work always ends up as wasted heat due to friction between moving parts. Both the input work (Wi) and output work (Wo) are the result of a force, F, acting over a distance, d. The efficiency output of a machine is simply the output work divided by the input work, and is usually multiplied by 100 so that it is expressed as a percent. Look back at the pictures of the simple machines and think about which would have the highest efficiency. Efficiency is related to friction, and friction depends on the smoothness of surfaces and on the area of the surfaces in contact. How would lubrication affect the efficiency of a simple machine? WORKED EXAMPLE Efficiency of a Lever The input force of 11 N acting on the effort arm of a lever moves 0.4 m, which lifts a 40 N weight resting on the resistance arm a 294 Chapter 9 • Work, Energy, and
Simple Machines distance of 0.1 m. What is the efficiency of the machine? Strategy State the equation for efficiency of a simple machine, are the product Fd. Solution = (11)(0.4) = 4.4 J and = (40)(0.1) = 4.0 J, then and calculate Woand Wi. Both work values Discussion Efficiency in real machines will always be less than 100 percent because of work that is converted to unavailable heat by friction and air resistance. Woand Wican always be calculated as a force multiplied by a distance, although these quantities are not always as obvious as they are in the case of a lever. Practice Problems 10. What is the IMA of an inclined plane that is long and high? a. b. c. d. 11. If a pulley system can lift a 200N load with an effort force of 52 N and has an efficiency of almost 100 percent, how many ropes are supporting the load? 1 rope is required because the actual mechanical advantage is 0.26. a. b. 1 rope is required because the actual mechanical advantage is 3.80. c. 4 ropes are required because the actual mechanical advantage is 0.26. d. 4 ropes are required because the actual mechanical advantage is 3.80. Check Your Understanding 12. True or false—The efficiency of a simple machine is always less than 100 percent because some small fraction of the input work is always converted to heat energy due to friction. a. True b. False 13. The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1 cm and the IMA of the machine is 6, what is the radius of the handle? A. 0.08 cm B. 0.17 cm C. 3.0 cm D. 6.0 cm Access for free at openstax.org. Chapter 9 • Key Terms 295 KEY TERMS complex machine a machine that combines two or more output work output force multiplied by the distance over simple machines which it acts efficiency output work divided by input work energy the ability to do work gravitational potential energy energy acquired by doing potential energy stored energy power pulley a simple machine consisting of a rope that passes the rate at which work is done work against gravity ideal mechanical advantage the mechanical advantage of an idealized machine that loses no energy to friction inclined plane a simple machine consisting of a slope input work effort force multiplied by the distance over which it is applied
over one or more grooved wheels screw a simple machine consisting of a spiral inclined plane simple machine a machine that makes work easier by changing the amount or direction of force required to move an object joule the metric unit for work and energy; equal to 1 watt the metric unit of power; equivalent to joules per newton meter (N∙m) second kinetic energy energy of motion law of conservation of energy states that energy is neither wedge a simple machine consisting of two back-to-back inclined planes created nor destroyed wheel and axle a simple machine consisting of a rod fixed lever a simple machine consisting of a rigid arm that pivots to the center of a wheel on a fulcrum mechanical advantage the number of times the input force is multiplied mechanical energy kinetic or potential energy work force multiplied by distance work–energy theorem states that the net work done on a system equals the change in kinetic energy SECTION SUMMARY 9.1 Work, Power, and the Work–Energy Theorem • Doing work on a system or object changes its energy. • The work–energy theorem states that an amount of work that changes the velocity of an object is equal to the change in kinetic energy of that object.The work–energy theorem states that an amount of work that changes the velocity of an object is equal to the change in kinetic energy of that object. • Power is the rate at which work is done. 9.2 Mechanical Energy and Conservation of Energy • Mechanical energy may be either kinetic (energy of KEY EQUATIONS 9.1 Work, Power, and the Work–Energy Theorem equation for work force work equivalencies motion) or potential (stored energy). • Doing work on an object or system changes its energy. • Total energy in a closed, isolated system is constant. 9.3 Simple Machines • The six types of simple machines make work easier by changing the fdterm so that force is reduced at the expense of increased distance. • The ratio of output force to input force is a machine’s mechanical advantage. • Combinations of two or more simple machines are called complex machines. • The ratio of output work to input work is a machine’s efficiency. kinetic energy work–energy theorem power 296 Chapter 9 • Chapter Review 9.2 Mechanical Energy and Conservation of Energy conservation of energy 9.3 Simple Machines ideal mechanical advantage (general) ideal mechanical advantage (lever) ideal mechanical advantage (wheel and axle) CHAPTER REVIEW Concept Items 9.1 Work, Power, and the Work–Energy
Theorem 1. Is it possible for the sum of kinetic energy and potential energy of an object to change without work having been done on the object? Explain. a. No, because the work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in KE requires a change in velocity. It is assumed that mass is constant. b. No, because the work-energy theorem states that work done on an object is equal to the sum of kinetic energy, and the change in KE requires a change in displacement. It is assumed that mass is constant. c. Yes, because the work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in KE requires a change in velocity. It is assumed that mass is constant. d. Yes, because the work-energy theorem states that work done on an object is equal to the sum of kinetic energy, and the change in KE requires a change in displacement. It is assumed that mass is constant. 2. Define work for one-dimensional motion. a. Work is defined as the ratio of the force over the distance. b. Work is defined as the sum of the force and the distance. c. Work is defined as the square of the force over the Access for free at openstax.org. ideal mechanical advantage (inclined plane) ideal mechanical advantage (wedge) ideal mechanical advantage (pulley) ideal mechanical advantage (screw) input work output work efficiency output distance. d. Work is defined as the product of the force and the distance. 3. A book with a mass of 0.30 kg falls 2 m from a shelf to the floor. This event is described by the work–energy theorem: Explain why this is enough information to calculate the speed with which the book hits the floor. a. The mass of the book, m, and distance, d, are stated. Fis the weight of the book mg. v1 is the initial velocity and v2 is the final velocity. The final velocity is the only unknown quantity. b. The mass of the book, m, and distance, d, are stated. Fis the weight of the book mg. v1 is the final velocity and v2 is the initial velocity. The final velocity is the only unknown quantity. c. The mass of the book, m, and distance, d, are stated. Fis the weight of the book mg. v1 is the initial
velocity and v2 is the final velocity. The final velocity and the initial velocities are the only unknown quantities. d. The mass of the book, m, and distance, d, are stated. Fis the weight of the book mg. v1 is the final velocity and v2 is the initial velocity. The final velocity and the initial velocities are the only unknown quantities. Chapter 9 • Chapter Review 297 9.2 Mechanical Energy and Conservation of Energy 4. Describe the changes in KE and PE of a person jumping up and down on a trampoline. a. While going up, the person’s KE would change to PE. While coming down, the person’s PE would change to KE. b. While going up, the person’s PE would change to KE. While coming down, the person’s KE would change to PE. c. While going up, the person’s KE would not change, but while coming down, the person’s PE would change to KE. d. While going up, the person’s PE would change to KE, but while coming down, the person’s KE would not change. 6. The starting line of a cross country foot race is at the bottom of a hill. Which form(s) of mechanical energy of the runners will change when the starting gun is fired? a. Kinetic energy only b. Potential energy only c. Both kinetic and potential energy d. Neither kinetic nor potential energy 9.3 Simple Machines 7. How does a simple machine make work easier? a. b. c. d. It reduces the input force and the output force. It reduces the input force and increases the output force. It increases the input force and reduces the output force. It increases the input force and the output force. 5. You know the height from which an object is dropped. 8. Which type of simple machine is a knife? Which equation could you use to calculate the velocity as the object hits the ground? a. b. c. d. a. A ramp b. A wedge c. A pulley d. A screw Critical Thinking Items 9.1 Work, Power, and the Work–Energy Theorem 9. Which activity requires a person to exert force on an object that causes the object to move but does not change the kinetic or potential energy of the object? a. Moving an object to a greater height with acceleration b. Moving an object to a greater height without
acceleration c. Carrying an object with acceleration at the same height 9.2 Mechanical Energy and Conservation of Energy 11. True or false—A cyclist coasts down one hill and up another hill until she comes to a stop. The point at which the bicycle stops is lower than the point at which it started coasting because part of the original potential energy has been converted to a quantity of heat and this makes the tires of the bicycle warm. a. True b. False 9.3 Simple Machines d. Carrying an object without acceleration at the same 12. We think of levers being used to decrease effort force. height 10. Which statement explains how it is possible to carry books to school without changing the kinetic or potential energy of the books or doing any work? a. By moving the book without acceleration and keeping the height of the book constant b. By moving the book with acceleration and keeping the height of the book constant c. By moving the book without acceleration and changing the height of the book d. By moving the book with acceleration and changing the height of the book Which of the following describes a lever that requires a large effort force which causes a smaller force to act over a large distance and explains how it works? a. Anything that is swung by a handle, such as a hammer or racket. Force is applied near the fulcrum over a short distance, which makes the other end move rapidly over a long distance. b. Anything that is swung by a handle, such as a hammer or racket. Force is applied far from the fulcrum over a large distance, which makes the other end move rapidly over a long distance. c. A lever used to lift a heavy stone. Force is applied near the fulcrum over a short distance, which 298 Chapter 9 • Chapter Review makes the other end lift a heavy object easily. d. A lever used to lift a heavy stone. Force is applied far from the fulcrum over a large distance, which makes the other end lift a heavy object easily 13. A baseball bat is a lever. Which of the following explains how a baseball bat differs from a lever like a pry bar? a. In a baseball bat, effort force is smaller and is applied over a large distance, while the resistance force is smaller and is applied over a long distance. Problems 9.1 Work, Power, and the Work–Energy Theorem 14. A baseball player exerts a force of on a ball for a as he throws it. If the ball has a mass,
what is its velocity as it leaves his hand? distance of of a. b. c. d. 15. A boy pushes his little sister on a sled. The sled accelerates from 0 to 3.2 m/s. If the combined mass of his sister and the sled is 40.0 kg and 18 W of power were generated, how long did the boy push the sled? a. 205 s b. 128 s c. 23 s 11 s d. 9.2 Mechanical Energy and Conservation of Energy 16. What is the kinetic energy of a bullet traveling? at a velocity of a. b. c. d. 17. A marble rolling across a flat, hard surface at rolls and no energy is up a ramp. Assuming that lost to friction, what will be the vertical height of the marble when it comes to a stop before rolling back down? Ignore effects due to the rotational kinetic energy. a. b. c. d. 18. The potential energy stored in a compressed spring is Access for free at openstax.org. b. c. d. In a baseball bat, effort force is smaller and is applied over a large distance, while the resistance force is smaller and is applied over a short distance. In a baseball bat, effort force is larger and is applied over a short distance, while the resistance force is smaller and is applied over a long distance. In a baseball bat, effort force is larger and is applied over a short distance, while the resistance force is smaller and is applied over a short distance., where kis the force constant and xis the distance the spring is compressed from the equilibrium position. Four experimental setups described below can be used to determine the force constant of a spring. Which one(s) require measurement of the fewest number of variables to determine k? Assume the acceleration due to gravity is known. I. An object is propelled vertically by a compressed spring. II. An object is propelled horizontally on a frictionless surface by a compressed spring. III. An object is statically suspended from a spring. IV. An object suspended from a spring is set into oscillatory motion. a. b. c. d. I only III only I and II only III and IV only 9.3 Simple Machines 19. A man is using a wedge to split a block of wood by hitting the wedge with a hammer. This drives the wedge into the wood creating a crack in the wood. When he hits the wedge with a force of 400 N it travels 4 cm into
the wood. This caused the wedge to exert a force of 1,400 N sideways increasing the width of the crack by 1 cm. What is the efficiency of the wedge? a. 0.875 percent b. 0.14 c. 0.751 d. 87.5 percent 20. An electrician grips the handles of a wire cutter, like the one shown, 10 cm from the pivot and places a wire between the jaws 2 cm from the pivot. If the cutter blades are 2 cm wide and 0.3 cm thick, what is the overall IMA of this complex machine? Performance Task 9.3 Simple Machines 21. Conservation of Energy and Energy Transfer; Cause and Effect; and S&EP, Planning and Carrying Out Investigations Plan an investigation to measure the mechanical advantage of simple machines and compare to the IMA of the machine. Also measure the efficiency of each machine studied. Design an investigation to make these measurements for these simple machines: lever, inclined plane, wheel and axle and a pulley system. In addition to these machines, include a spring scale, a tape measure, and a weight with a loop on top that can be attached to the hook on the spring scale. A spring scale is shown in the image. Chapter 9 • Test Prep 299 a. b. c. d. 1.34 1.53 33.3 33.5 LEVER: Beginning with the lever, explain how you would measure input force, output force, effort arm, and resistance arm. Also explain how you would find the distance the load travels and the distance over which the effort force is applied. Explain how you would use this data to determine IMAand efficiency. INCLINED PLANE: Make measurements to determine IMAand efficiency of an inclined plane. Explain how you would use the data to calculate these values. Which property do you already know? Note that there are no effort and resistance arm measurements, but there are height and length measurements. WHEEL AND AXLE: Again, you will need two force measurements and four distance measurements. Explain how you would use these to calculate IMAand efficiency. SCREW: You will need two force measurements, two distance traveled measurements, and two length measurements. You may describe a screw like the one shown in Figure 9.10 or you could use a screw and screw driver. (Measurements would be easier for the former). Explain how you would use these to calculate IMAand efficiency. PULLEY SYSTEM: Explain how you would determine the IMAand
efficiency of the four-pulley system shown in Figure 9.11. Why do you only need two distance measurements for this machine? Design a table that compares the efficiency of the five simple machines. Make predictions as to the most and least efficient machines. A spring scale measures weight, not mass. TEST PREP Multiple Choice 9.1 Work, Power, and the Work–Energy Theorem 22. Which expression represents power? a. b. c. d. 23. The work–energy theorem states that the change in the kinetic energy of an object is equal to what? a. The work done on the object b. The force applied to the object c. The loss of the object’s potential energy d. The object’s total mechanical energy minus its kinetic energy 24. A runner at the start of a race generates 250 W of power as he accelerates to 5 m/s. If the runner has a mass of 60 kg, how long did it take him to reach that speed? a. 0.33 s b. 0.83 s c. d. 1.2 s 3.0 s 300 Chapter 9 • Test Prep 25. A car’s engine generates 100,000 W of power as it exerts a force of 10,000 N. How long does it take the car to travel 100 m? a. 0.001 s b. 0.01 s 10 s c. 1,000 s d. 9.2 Mechanical Energy and Conservation of Energy 26. Why is this expression for kinetic energy incorrect? is missing.. a. The constant b. The term should not be squared. c. The expression should be divided by. d. The energy lost to friction has not been subtracted. 27. What is the kinetic energy of a object moving at? a. b. c. d. 28. Which statement best describes the PE-KE transformations for a javelin, starting from the instant the javelin leaves the thrower's hand until it hits the ground. a. Initial PE is transformed to KE until the javelin reaches the high point of its arc. On the way back down, KE is transformed into PE. At every point in the flight, mechanical energy is being transformed into heat energy. Initial KE is transformed to PE until the javelin reaches the high point of its arc. On the way back down, PE is transformed into KE. At every point in the flight, mechanical energy is being transformed into heat energy. Initial
PE is transformed to KE until the javelin reaches the high point of its arc. On the way back down, there is no transformation of mechanical energy. At every point in the flight, mechanical energy is being transformed into heat energy. Initial KE is transformed to PE until the javelin reaches the high point of its arc. On the way back down, there is no transformation of mechanical energy. At every point in the flight, mechanical energy is being transformed into heat energy. b. c. d. 29. At the beginning of a roller coaster ride, the roller coaster car has an initial energy mostly in the form of PE. Which statement explains why the fastest speeds of the car will be at the lowest points in the ride? a. At the bottom of the slope kinetic energy is at its Access for free at openstax.org. maximum value and potential energy is at its minimum value. b. At the bottom of the slope potential energy is at its maximum value and kinetic energy is at its minimum value. c. At the bottom of the slope both kinetic and potential energy reach their maximum values d. At the bottom of the slope both kinetic and potential energy reach their minimum values. 9.3 Simple Machines 30. A large radius divided by a small radius is the expression used to calculate the IMA of what? a. A screw b. A pulley c. A wheel and axle d. An inclined plane. 31. What is the IMA of a wedge that is long and thick? a. b. c. d. 32. Which statement correctly describes the simple machines, like the crank in the image, that make up an Archimedes screw and the forces it applies? a. The crank is a wedge in which the IMA is the length of the tube divided by the radius of the tube. The applied force is the effort force and the weight of the water is the resistance force. b. The crank is an inclined plane in which the IMA is the length of the tube divided by the radius of the tube. The applied force is the effort force and the weight of the water is the resistance force. c. The crank is a wheel and axle. The effort force of the crank becomes the resistance force of the screw. d. The crank is a wheel and axle. The resistance force of the crank becomes the effort force of the screw. 33. Refer to the pulley system on right in the image. Assume this pulley system is an ideal machine. How hard
would you have to pull on the rope to lift a 120 N Chapter 9 • Test Prep 301 load? How many meters of rope would you have to pull out of the system to lift the load 1 m? a. 480 N 4 m b. 480 N c. d. m 30 N 4 m 30 N m Short Answer 9.1 Work, Power, and the Work–Energy Theorem 34. Describe two ways in which doing work on an object can increase its mechanical energy. a. Raising an object to a higher elevation does work as it increases its PE; increasing the speed of an object does work as it increases its KE. b. Raising an object to a higher elevation does work as it increases its KE; increasing the speed of an object does work as it increases its PE. c. Raising an object to a higher elevation does work as it increases its PE; decreasing the speed of an object does work as it increases its KE. d. Raising an object to a higher elevation does work as it increases its KE; decreasing the speed of an object does work as it increases its PE. 35. True or false—While riding a bicycle up a gentle hill, it is fairly easy to increase your potential energy, but to increase your kinetic energy would make you feel exhausted. a. True b. False 36. Which statement best explains why running on a track with constant speed at 3 m/s is not work, but climbing a mountain at 1 m/s is work? a. At constant speed, change in the kinetic energy is zero but climbing a mountain produces change in the potential energy. b. At constant speed, change in the potential energy is zero, but climbing a mountain produces change in the kinetic energy. c. At constant speed, change in the kinetic energy is finite, but climbing a mountain produces no change in the potential energy. d. At constant speed, change in the potential energy is finite, but climbing a mountain produces no change in the kinetic energy. 37. You start at the top of a hill on a bicycle and coast to the bottom without applying the brakes. By the time you reach the bottom of the hill, work has been done on you and your bicycle, according to the equation: If is the mass of you and your bike, what are a. and? is your speed at the top of the hill, and is your speed at the bottom. b. c. d. is your speed at the bottom of the hill, and is your speed at the
top. is your displacement at the top of the hill, and is your displacement at the bottom. is your displacement at the bottom of the hill, and is your displacement at the top. 9.2 Mechanical Energy and Conservation of Energy 38. True or false—The formula for gravitational potential energy can be used to explain why joules, J, are equivalent to kg × mg2 / s2. Show your work. a. True b. False 39. Which statement best explains why accelerating a car from a. Because kinetic energy is directly proportional to quadruples its kinetic energy? to the square of the velocity. b. Because kinetic energy is inversely proportional to the square of the velocity. c. Because kinetic energy is directly proportional to the fourth power of the velocity. d. Because kinetic energy is inversely proportional to 302 Chapter 9 • Test Prep the fourth power of the velocity. plane. 40. A coin falling through a vacuum loses no energy to friction, and yet, after it hits the ground, it has lost all its potential and kinetic energy. Which statement best explains why the law of conservation of energy is still valid in this case? a. When the coin hits the ground, the ground gains potential energy that quickly changes to thermal energy. b. When the coin hits the ground, the ground gains kinetic energy that quickly changes to thermal energy. c. When the coin hits the ground, the ground gains thermal energy that quickly changes to kinetic energy. d. When the coin hits the ground, the ground gains thermal energy that quickly changes to potential energy. 41. True or false—A marble rolls down a slope from height h1 and up another slope to height h2, where (h2 < h1). The difference mg(h1 – h2) is equal to the heat lost due to the friction. a. True b. False 9.3 Simple Machines 42. Why would you expect the lever shown in the top image to have a greater efficiency than the inclined plane shown in the bottom image? b. The effort arm is shorter in case of the inclined plane. c. The area of contact is greater in case of the inclined plane. 43. Why is the wheel on a wheelbarrow not a simple machine in the same sense as the simple machine in the image? a. The wheel on the wheelbarrow has no fulcrum. b. The center of the axle is not the fulcrum for the wheels of a wheelbarrow. c. The wheelbarrow differs in
the way in which load is attached to the axle. d. The wheelbarrow has less resistance force than a wheel and axle design. 44. A worker pulls down on one end of the rope of a pulley system with a force of 75 N to raise a hay bale tied to the other end of the rope. If she pulls the rope down 2.0 m and the bale raises 1.0 m, what else would you have to know to calculate the efficiency of the pulley system? a. b. c. d. the weight of the worker the weight of the hay bale the radius of the pulley the height of the pulley from ground 45. True or false—A boy pushed a box with a weight of 300 N up a ramp. He said that, because the ramp was 1.0 m high and 3.0 m long, he must have been pushing with force of exactly 100 N. a. True b. False a. The resistance arm is shorter in case of the inclined Extended Response 9.1 Work, Power, and the Work–Energy Theorem 46. Work can be negative as well as positive because an object or system can do work on its surroundings as well as have work done on it. Which of the following statements describes: a situation in which an object does work on its surroundings by decreasing its velocity and a situation in which an object can do work on its surroundings by decreasing its altitude? a. A gasoline engine burns less fuel at a slower speed. Solar cells capture sunlight to generate electricity. Access for free at openstax.org. Chapter 9 • Test Prep 303 b. A hybrid car charges its batteries as it decelerates. Falling water turns a turbine to generate electricity. c. Airplane flaps use air resistance to slow down for landing. Rising steam turns a turbine to generate electricity. d. An electric train requires less electrical energy as it decelerates. A parachute captures air to slow a skydiver’s fall. 47. A boy is pulling a girl in a child’s wagon at a constant speed. He begins to pull harder, which increases the speed of the wagon. Which of the following describes two ways you could calculate the change in energy of the wagon and girl if you had all the information you needed? a. Calculate work done from the force and the velocity. Calculate work done from the change in the potential energy of the system. b. Calculate work done from the force and the displacement.
Calculate work done from the change in the potential energy of the system. c. Calculate work done from the force and the velocity. Calculate work done from the change in the kinetic energy of the system. be 12 m/s. Due to the lack of air friction, there would be complete transformation of the potential energy into the kinetic energy as the rock hits the moon’s surface. d. The velocity of the rock as it hits the ground would be 12 m/s. Due to the lack of air friction, there would be incomplete transformation of the potential energy into the kinetic energy as the rock hits the moon’s surface. 49. A boulder rolls from the top of a mountain, travels across a valley below, and rolls part way up the ridge on the opposite side. Describe all the energy transformations taking place during these events and identify when they happen. a. As the boulder rolls down the mountainside, KE is converted to PE. As the boulder rolls up the opposite slope, PE is converted to KE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction. b. As the boulder rolls down the mountainside, KE is converted to PE. As the boulder rolls up the opposite slope, KE is converted to PE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction. d. Calculate work done from the force and the c. As the boulder rolls down the mountainside, PE is displacement. Calculate work done from the change in the kinetic energy of the system. 9.2 Mechanical Energy and Conservation of Energy 48. Acceleration due to gravity on the moon is 1.6 m/s2 or about 16% of the value of gon Earth. If an astronaut on the moon threw a moon rock to a height of 7.8 m, what would be its velocity as it struck the moon’s surface? How would the fact that the moon has no atmosphere affect the velocity of the falling moon rock? Explain your answer. a. The velocity of the rock as it hits the ground would be 5.0 m/s. Due to the lack of air friction, there would be complete transformation of the potential energy into the kinetic energy as the rock hits the moon’s surface. b. The velocity of the rock as it hits the ground would be 5.0 m/s. Due to the lack of air friction, there would
be incomplete transformation of the potential energy into the kinetic energy as the rock hits the moon’s surface. c. The velocity of the rock as it hits the ground would converted to KE. As the boulder rolls up the opposite slope, PE is converted to KE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction. d. As the boulder rolls down the mountainside, PE is converted to KE. As the boulder rolls up the opposite slope, KE is converted to PE. The boulder rolls only partway up the ridge because some of the PE has been converted to thermal energy due to friction. 9.3 Simple Machines 50. To dig a hole, one holds the handles together and thrusts the blades of a posthole digger, like the one in the image, into the ground. Next, the handles are pulled apart, which squeezes the dirt between them, making it possible to remove the dirt from the hole. This complex machine is composed of two pairs of two different simple machines. Identify and describe the parts that are simple machines and explain how you would find the IMA of each type of simple machine. a. Each handle and its attached blade is a lever with the 304 Chapter 9 • Test Prep fulcrum at the hinge. Each blade is a wedge. The IMA of a lever would be the length of the handle divided by the length of the blade. The IMA of the wedges would be the length of the blade divided by its width. b. Each handle and its attached to blade is a lever with the fulcrum at the end. Each blade is a wedge. The IMA of a lever would be the length of the handle divided by the length of the blade. The IMA of the wedges would be the length of the blade divided by its width. c. Each handle and its attached blade is a lever with the fulcrum at the hinge. Each blade is a wedge. The IMA of a lever would be the length of the handle multiplied by the length of the blade. The IMA of the wedges would be the length of the blade multiplied by its width. d. Each handle and its attached blade is a lever with the fulcrum at the end. Each blade is a wedge. The IMA of a lever would be the length of the handle multiplied by the length of the blade. The IMA of the wedges would be the length of the blade multiplied by its
width. 51. A wooden crate is pulled up a ramp that is 1.0 m high and 6.0 m long. The crate is attached to a rope that is wound around an axle with a radius of 0.020 m. The axle is turned by a 0.20 m long handle. What is the overall IMA of the complex machine? A. 6 B. 10 16 C. D. 60 Access for free at openstax.org. CHAPTER 10 Special Relativity Figure 10.1 Special relativity explains why travel to other star systems, such as these in the Orion Nebula, is unlikely using our current level of technology. (s58y, Flickr) Chapter Outline 10.1 Postulates of Special Relativity 10.2 Consequences of Special Relativity Have you ever dreamed of traveling to other planets in faraway star systems? The trip might seem possible by INTRODUCTION traveling fast enough, but you will read in this chapter why it is not. In 1905, Albert Einstein developed the theory of special relativity. Einstein developed the theory to help explain inconsistencies between the equations describing electromagnetism and Newtonian mechanics, and to explain why the ether did not exist. This theory explains the limit on an object’s speed among other implications. Relativity is the study of how different observers moving with respect to one another measure the same events. Galileo and Newton developed the first correct version of classical relativity. Einstein developed the modern theory of relativity. Modern relativity is divided into two parts. Special relativity deals with observers moving at constant velocity. General relativity deals with observers moving at constant acceleration. Einstein’s theories of relativity made revolutionary predictions. Most importantly, his predictions have been verified by experiments. In this chapter, you learn how experiments and puzzling contradictions in existing theories led to the development of the theory of special relativity. You will also learn the simple postulates on which the theory was based; a postulate is a statement that is assumed to be true for the purposes of reasoning in a scientific or mathematic argument. 10.1 Postulates of Special Relativity Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the experiments and scientific problems that led Albert Einstein to develop the special theory of relativity • Understand the postulates on which the special theory of relativity was based 306 Chapter 10 • Special Relativity Section Key Terms ether frame of reference inertial reference frame general relativity postulate relativity simultaneity special relativity Scientific Experiments and Problems Relativity is
not new. Way back around the year 1600, Galileo explained that motion is relative. Wherever you happen to be, it seems like you are at a fixed point and that everything moves with respect to you. Everyone else feels the same way. Motion is always measured with respect to a fixed point. This is called establishing a frame of reference. But the choice of the point is arbitrary, and all frames of reference are equally valid. A passenger in a moving car is not moving with respect to the driver, but they are both moving from the point of view of a person on the sidewalk waiting for a bus. They are moving even faster as seen by a person in a car coming toward them. It is all relative. TIPS FOR SUCCESS A frame of reference is not a complicated concept. It is just something you decide is a fixed point or group of connected points. It is completely up to you. For example, when you look up at celestial objects in the sky, you choose the earth as your frame of reference, and the sun, moon, etc., seem to move across the sky. Light is involved in the discussion of relativity because theories related to electromagnetism are inconsistent with Galileo’s and Newton’s explanation of relativity. The true nature of light was a hot topic of discussion and controversy in the late 19th century. At the time, it was not generally believed that light could travel across empty space. It was known to travel as waves, and all other types of energy that propagated as waves needed to travel though a material medium. It was believed that space was filled with an invisible medium that light waves traveled through. This imaginary (as it turned out) material was called the ether (also spelled aether). It was thought that everything moved through this mysterious fluid. In other words, ether was the one fixed frame of reference. The Michelson–Morley experiment proved it was not. In 1887, Albert Michelson and Edward Morley designed the interferometer shown in Figure 10.2 to measure the speed of Earth through the ether. A light beam is split into two perpendicular paths and then recombined. Recombining the waves produces an inference pattern, with a bright fringe at the locations where the two waves arrive in phase; that is, with the crests of both waves arriving together and the troughs arriving together. A dark fringe appears where the crest of one wave coincides with a trough of the other, so that the two cancel. If Earth is traveling through
the ether as it orbits the sun, the peaks in one arm would take longer than in the other to reach the same location. The places where the two waves arrive in phase would change, and the interference pattern would shift. But, using the interferometer, there was no shift seen! This result led to two conclusions: that there is no ether and that the speed of light is the same regardless of the relative motion of source and observer. The Michelson–Morley investigation has been called the most famous failed experiment in history. Access for free at openstax.org. 10.1 • Postulates of Special Relativity 307 Figure 10.2 This is a diagram of the instrument used in the Michelson–Morley experiment. To see what Michelson and Morley expected to find when they measured the speed of light in two directions, watch this animation (http://openstax.org/l/28MMexperiment). In the video, two people swimming in a lake are represented as an analogy to light beams leaving Earth as it moves through the ether (if there were any ether). The swimmers swim away from and back to a platform that is moving through the water. The swimmers swim in different directions with respect to the motion of the platform. Even though they swim equal distances at the same speed, the motion of the platform causes them to arrive at different times. Einstein’s Postulates The results described above left physicists with some puzzling and unsettling questions such as, why doesn’t light emitted by a fast-moving object travel faster than light from a street lamp? A radical new theory was needed, and Albert Einstein, shown in Figure 10.3, was about to become everyone’s favorite genius. Einstein began with two simple postulates based on the two things we have discussed so far in this chapter. 1. The laws of physics are the same in all inertial reference frames. 2. The speed of light is the same in all inertial reference frames and is not affected by the speed of its source. Figure 10.3 Albert Einstein (1879–1955) developed modern relativity and also made fundamental contributions to the foundations of quantum mechanics. (The Library of Congress) The speed of light is given the symbol cand is equal to exactly 299,792,458 m/s. This is the speed of light in vacuum; that is, in the absence of air. For most purposes, we round this number off to The term inertial reference frame simply 308 Chapter
10 • Special Relativity refers to a frame of reference where all objects follow Newton’s first law of motion: Objects at rest remain at rest, and objects in motion remain in motion at a constant velocity in a straight line, unless acted upon by an external force. The inside of a car moving along a road at constant velocity and the inside of a stationary house are inertial reference frames. WATCH PHYSICS The Speed of Light This lecture on light summarizes the most important facts about the speed of light. If you are interested, you can watch the whole video, but the parts relevant to this chapter are found between 3:25 and 5:10, which you find by running your cursor along the bottom of the video. Click to view content (https://www.youtube.com/embed/rLNM8zI4Q_M) GRASP CHECK An airliner traveling at 200 m/s emits light from the front of the plane. Which statement describes the speed of the light? a. b. c. d. It travels at a speed of c+ 200 m/s. It travels at a speed of c– 200 m/s. It travels at a speed c, like all light. It travels at a speed slightly less than c. Snap Lab Measure the Speed of Light In this experiment, you will measure the speed of light using a microwave oven and a slice of bread. The waves generated by a microwave oven are not part of the visible spectrum, but they are still electromagnetic radiation, so they travel at the speed of light. If we know the wavelength, λ, and frequency, f, of a wave, we can calculate its speed, v, using the equation v = λf. You can measure the wavelength. You will find the frequency on a label on the back of a microwave oven. The wave in a microwave is a standing wave with areas of high and low intensity. The high intensity sections are one-half wavelength apart. • High temperature: Very hot temperatures are encountered in this lab. These can cause burns. • a microwave oven • one slice of plain white bread • a centimeter ruler • a calculator 1. Work with a partner. 2. Turn off the revolving feature of the microwave oven or remove the wheels under the microwave dish that make it turn. It is important that the dish does not turn. 3. Place the slice of bread on the dish, set the microwave on high, close the door, run the microwave for about 15 seconds. 4
. A row of brown or black marks should appear on the bread. Stop the microwave as soon as they appear. Measure the distance between two adjacent burn marks and multiply the result by 2. This is the wavelength. 5. The frequency of the waves is written on the back of the microwave. Look for something like “2,450 MHz.” Hz is the unit hertz, which means per second. The M represents mega, which stands for million, so multiply the number by 106. 6. Express the wavelength in meters and multiply it times the frequency. If you did everything correctly, you will get a number very close to the speed of light. Do not eat the bread. It is a general laboratory safety rule never to eat anything in the lab. GRASP CHECK How does your measured value of the speed of light compare to the accepted value (% error)? Access for free at openstax.org. 10.1 • Postulates of Special Relativity 309 a. The measured value of speed will be equal to c. b. The measured value of speed will be slightly less than c. c. The measured value of speed will be slightly greater than c. d. The measured value of speed will depend on the frequency of the microwave. Einstein’s postulates were carefully chosen, and they both seemed very likely to be true. Einstein proceeded despite realizing that these two ideas taken together and applied to extreme conditions led to results that contradict Newtonian mechanics. He just took the ball and ran with it. In the traditional view, velocities are additive. If you are running at 3 m/s and you throw a ball forward at a speed of 10 m/s, the ball should have a net speed of 13 m/s. However, according to relativity theory, the speed of a moving light source is not added to the speed of the emitted light. In addition, Einstein’s theory shows that if you were moving forward relative to Earth at nearly c(the speed of light) and could throw a ball forward at c, an observer at rest on the earth would not see the ball moving at nearly twice the speed of light. The observer would see it moving at a speed that is still less than c. This result conforms to both of Einstein’s postulates: The speed of light has a fixed maximum and neither reference frame is privileged. Consider how we measure elapsed time. If we use a stopwatch, for example, how do we know when to start and stop
the watch? One method is to use the arrival of light from the event, such as observing a light turn green to start a drag race. The timing will be more accurate if some sort of electronic detection is used, avoiding human reaction times and other complications. Now suppose we use this method to measure the time interval between two flashes of light produced by flash lamps on a moving train. (See Figure 10.4) 310 Chapter 10 • Special Relativity Figure 10.4 Light arriving to observer A as seen by two different observers. A woman (observer A) is seated in the center of a rail car, with two flash lamps at opposite sides equidistant from her. Multiple light rays that are emitted from the flash lamps move towards observer A, as shown with arrows. A velocity vector arrow for the rail car is shown towards the right. A man (observer B) standing on the platform is facing the woman and also observes the flashes of light. Observer A moves with the lamps on the rail car as the rail car moves towards the right of observer B. Observer B receives the light flashes simultaneously, and sees the bulbs as both having flashed at the same time. However, he sees observer A receive the flash from the right first. Because the pulse from the right reaches her first, in her frame of reference she sees the bulbs as not having flashed simultaneously. Here, a relative velocity between observers affects whether two events at well-separated locations are observed to be simultaneous. Simultaneity, or whether different events occur at the same instant, depends on the frame of reference of the observer. Remember that velocity equals distance divided by time, so t = d/v. If velocity appears to be different, then duration of time appears to be different. This illustrates the power of clear thinking. We might have guessed incorrectly that, if light is emitted simultaneously, then two observers halfway between the sources would see the flashes simultaneously. But careful analysis shows this not to be the case. Einstein was brilliant at this type of thought experiment (in German, Gedankenexperiment). He very carefully considered how an observation is made and disregarded what might seem obvious. The validity of thought experiments, of course, is determined by actual observation. The genius of Einstein is evidenced by the fact that experiments have repeatedly confirmed his theory of relativity. No experiments after that of Michelson and Morley were able to detect any ether medium. We will describe later how experiments also confirmed other predictions of special relativity, such as the distance between two objects and

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