Solutions to Chapter 6 Exercises
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- Chapter 8 Problem Solutions 1. Since the bicycle moves 2 m with each turn of the wheel, and the wheel turns once each second, the linear speed of the bicycle is 2 m/s
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Rotational MotionConceptual Physics Instructor’s Manual, 10th Edition Solutions to Chapter 8 Exercises 1. Tape moves faster when r is greater, in accord with v = r So the reel with the most-filled reel corresponds to the tape with the greatest linear speed v. 2. The tangential speeds are equal, for they have the same speed as the belt. The smaller wheel rotates twice as fast because for the same tangential speed, and r half, must be twice. v(big wheel) = r; v(small wheel) = (r/2 2 ). 3. Large diameter tires mean you travel farther with each revolution of the tire. So you’ll be moving faster than your speedometer indicates. (A speedometer actually measures the RPM of the wheels and displays this as mi/h or km/h. The conversion from RPM to the mi/h or km/h reading assumes the wheels are a certain size.) Oversize wheels give too low a reading because they really travel farther per revolution than the speedometer indicates, and undersize wheels give too high a reading because the wheels do not go as far per revolution. 4. Sue’s tires have a greater rotational speed for they have to turn more times to cover the same distance. 5. The amount of taper is related to the amount of curve the railroad tracks take. On a curve where the outermost track is say 10% longer than the inner track, the wide part of the wheel will also have to be at least 10% wider than the narrow part. If it’s less than this, the outer wheel will rely on the rim to stay on the track, and scraping will occur as the train makes the curve. The “sharper” the curve, the more the taper needs to be on the wheels. 6. For the same twisting speed the greater distance r means a much greater speed v. 7. The CM is directly above the bird’s foot. 8. Two conditions are necessary for mechanical equilibrium, F = 0 and = 0. 9. Rotational inertia and torque are most predominantly illustrated here, and the conservation of angular momentum also plays a role. The long distance to the front wheels increases the rotational inertia of the vehicle relative to the back wheels and also increases the lever arm of the front wheels without appreciably adding to the vehicle’s weight. As the back wheels are driven clockwise, the chassis tends to rotate counterclockwise (conservation of angular momentum) and thereby lift the front wheels off the ground. The greater rotational inertia and the increased clockwise torque of the more distant front wheels counter this effect. 10. Before leaving the cliff, front and back wheels provide the support base to support the car’s weight. The car’s center of mass is well within this support base. But when the car drives off the cliff, the front wheels are the first to leave the surface. This reduces the support base to the region between the rear wheels, so the car tips forward. In terms of torques, before driving off the cliff, the torques are balanced about the center of mass produced by the support forces at front and back wheels. But when the support force of the front wheels is absent, torque due to the support force of the rear wheels rotates the car forward about its center of mass making it nose forward as shown. 
11. Friction by the road on the tires produces a torque about the car’s CM. When the car accelerates forward, the friction force points forward and rotates the car upward. When braking, the direction of friction is rearward, and the torque rotates the car in the opposite direction so the rear end rotates upward (and the nose downward). 12. The bowling ball wins. A solid sphere of any mass and size beats both a solid cylinder and a hollow ball of any mass and size. That’s because a solid sphere has less rotational inertia per mass than the other shapes. A solid sphere has the bulk of its mass nearer the rotational axis that extends through its center of mass, whereas a cylinder or hollow ball has more of its mass farther from the axis. The object with the least rotational inertia per mass is the “least lazy” and will win races. 13. The ball to reach the bottom first is the one with the least rotational inertia compared with its mass—that’s the hollow basketball. 14. If you roll them down an incline, the solid ball will roll faster. (The hollow ball has more rotational inertia compared with its weight.) 15. Don’t say the same, for the water slides inside the can while the ice is made to roll along with the can. When the water inside slides, it contributes weight rather than rotational inertia to the can. So the can of water will roll faster. (It will even beat a hollow can.) 16. Lightweight tires have less rotational inertia, and easier to get up to speed. 17. Advise the youngster to use wheels with the least rotational inertia—lightweight solid ones without spokes. 18. The lever arm is the same whether a person stands, sits, or hangs from the end of the seesaw, and certainly the person’s weight is the same. So the net torque is the same also. 19. In the horizontal position the lever arm equals the length of the sprocket arm, but in the vertical position, the lever arm is zero because the line of action of forces passes right through the axis of rotation. (With cycling cleats, a cyclist pedals in a circle, which means they push their feet over the top of the spoke and pull around the bottom and even pull up on the recovery. This allows torque to be applied over a greater portion of the revolution.) 20. No, for by definition, a torque requires both force and a lever arm. 21. No, because there is zero lever arm about the CM. Zero lever arm means zero torque. 22. In all three cases the spool moves to the right. In (a) there is a torque about the point of contact with the table that rotates the spool clockwise, so the spool rolls to the right. In (b) the pull’s line of action extends through (not about) the point of table contact, yielding no lever arm and therefore no torque; but with a force component to the right; hence the spool slides to the right without rolling. In (c) the torque produces clockwise rotation so the spool rolls to the right. 
Friction between the ball and the lane provides a torque, which rotates the ball. 24. A rocking bus partially rotates about its center of mass, which is near its middle. The farther one sits from the center of mass, the greater is the up and down motion—as on a seesaw. Likewise for motion of a ship in choppy water or an airplane in turbulent air. 25. With your legs straight out, your CG is farther away and you exert more torque sitting up. So sit-ups are more difficult with legs straight out. 26. The long drooping pole lowers the CG of the balanced system—the tightrope walker and the pole. The rotational inertia of the pole contributes to the stability of the system also. 27. The wobbly motion of a star is an indication that it is revolving about a center of mass that is not at its geometric center, implying that there is some other mass nearby to pull the center of mass away from the star’s center. This is the way in which astronomers have discovered that planets exist around stars other than our own. 28. You bend forward when carrying a heavy load on your back to shift the CG of you and your load above the area bounded by your feet—otherwise you topple backward. 29. Two buckets are easier because you may stand upright while carrying a bucket in each hand. With two buckets, the CG will be in the center of the support base provided by your feet, so there is no need to lean. (The same can be accomplished by carrying a single bucket on your head.) 30. The weight of the boy is counterbalanced by the weight of the board, which can be considered to be concentrated at its CG on the opposite side of the fulcrum. He is in balance when his weight multiplied by his distance from the fulcrum is equal to the weight of the entire board multiplied by the distance between the fulcrum and the midpoint (CG) of the board. (How do the relative weight of boy and board relate to the relative lever arms?)  31. The CG of a ball is not above a point of support when the ball is on an incline. The weight of the ball therefore acts at some distance from the point of support which behaves like a fulcrum. A torque is produced and the ball rotates. This is why a ball rolls down a hill. 32. The top brick would overhang 3/4 of a brick length as shown. This is best explained by considering the top brick and moving downward; i.e., the CG of the top brick is at its midpoint; the CG of the top two bricks is midway between their combined length. Inspection will show that this is 1/4 of a brick length, the overhang of the middle brick. (Interestingly, with a few more bricks, the overhang can be greater than a brick length, and with a limitless number of bricks, the overhang can be made as large as you like.)  33. The Earth’s atmosphere is a nearly spherical shell, which like a basketball, has its center of mass at its center, i.e., at the center of the Earth. 34. It is dangerous to pull open the upper drawers of a fully-loaded file cabinet that is not secured to the floor because the CG of the cabinet can easily be shifted beyond the support base of the cabinet. When this happens, the torque that is produced causes the cabinet to topple over. 35. An object is stable when its PE must be raised in order to tip it over, or equivalently, when its PE must be increased before it can topple. By inspection, the first cylinder undergoes the least change in PE compared to its weight in tipping. This is because of its narrow base.  36. The CG of Truck 1 is not above its support base; the CGs of Trucks 2 and 3 are above their support bases. Therefore, only Truck 1 will tip. 37. The track will remain in equilibrium as the balls roll outward. This is because the CG of the system remains over the fulcrum. For example, suppose the billiard ball has twice the mass of the golf ball. By conservation of momentum, the twice-as-massive ball will roll outward at half the speed of the lighter ball, and at any time be half as far from the starting point as the lighter ball. So there is no CG change in the system of the two balls. We can see also that the torques produced by the weights of the balls multiplied by their relative distances from the fulcrum are equal at all points—because at any time the less massive ball has a correspondingly larger lever arm.
45. Centripetal force will be adequate when only the radial component of the normal force equals the mass of the car times the square of its speed divided by an appropriate radial distance from the center of the curve. 46. In accord with Newton’s first law, at every moment her tendency is to move in a straight-line path. But the floor intercepts this path and a pair of forces occur; the floor pressing against her feet and her feet pressing against the floor—Newton’s third law. The push by the floor on her feet provides the centripetal force that keeps her moving in a circle with the habitat. She sense this as an artificial gravity. 47.  48. The resultant is a centripetal force.  49. (a) Except for the vertical force of friction, no other vertical force except the weight of the motorcycle + rider exists. Since there is no change of motion in the vertical direction, the force of friction must be equal and opposite to the weight of motorcycle + rider. (b) The horizontal vector indeed represents the normal force. Since it is the only force acting in the radial direction, horizontally, it is also the centripetal force. So it’s both. 50.  51. The rotational inertia of you and the rotating turntable is least when you are at the rotational axis. As you crawl outward, the rotational inertia of the system increases (like the masses held outward in Figure 8.53). In accord with the conservation of angular momentum, as you crawl toward the outer rim, you increase the rotational inertia of the spinning system and decrease the angular speed. You can also see that if you don’t slip as you crawl out, you exert a friction force on the turntable opposite to its direction of rotation, thereby slowing it down. 52. Soil that washed down the river is being deposited at a greater distance from the Earth’s rotational axis. Just as the man on the turntable slows down when one of the masses is extended, the Earth slows down in its rotational motion, extending the length of the day. The amount of slowing, of course, is exceedingly small. (Interestingly, the construction of many dams in the Northern Hemisphere has the opposite effect; shortening our days!) 53. In accord with the conservation of angular momentum, as the radial distance of mass increases, the angular speed decreases. The mass of material used to construct skyscrapers is lifted, slightly increasing the radial distance from the Earth’s spin axis. This would tend to slightly decrease the Earth’s rate of rotation, which in turn tends to make the days a bit longer. The opposite effect occurs for falling leaves, as their radial distance from the Earth’s axis decreases. As a practical matter, these effects are entirely negligible! 54. Rotational inertia would increase. By angular momentum conservation, the rotation of the Earth would decrease (as a skater spins slower with arms outstretched), making a longer day. 55. In accord with the conservation of angular momentum, if mass moves closer to the axis of rotation, rotational speed increases. So the day would be ever so slightly shorter. 56. In accord with the conservation of angular momentum, if mass moves farther from the axis of rotation, rotational speed decreases. So the Earth would slow in its daily rotation. 57. The angular momentum of the wheel-train system will not change in the absence of an external torque. So when the train moves clockwise, the wheel moves counterclockwise with an equal and opposite angular momentum. When the train stops, the wheel stops. When the train backs up, the wheel moves clockwise. If masses of the train and wheel are equal, they will move with equal speeds (since the mass of the wheel is as far from the axis as the mass of the train—equal masses at equal radial distances having equal rotational inertias). If the train is more massive than the wheel, the wheel will “recoil” with more speed than the train, and vice versa. (This is a favorite demonstration of Paul Robinson, whose children David and Kristen are shown in the photo.) 58. Without the small rotor on its tail, the helicopter and the main rotor would rotate in opposite directions. The small rotor provides a torque to offset the rotational motion that the helicopter would otherwise have. 59. Gravitational force acting on every particle by every other particle causes the cloud to condense. The decreased radius of the cloud is then accompanied by an increased angular speed because of angular momentum conservation. The increased speed results in many stars being thrown out into a dish-like shape. 60. In accord with Newton’s first law, moving things tend to travel in straight lines. Surface regions of a rotating planet tend to fly off tangentially, especially at the equator where tangential speed is greatest. More predominantly, the surface is also pulled by gravity toward the center of the planet. Gravity wins, but bulging occurs at the equator because the tendency to fly off is greater there. Hence a rotating planet has a greater diameter at the equator than along the polar axis. Chapter 8 Problem Solutions 1. Since the bicycle moves 2 m with each turn of the wheel, and the wheel turns once each second, the linear speed of the bicycle is 2 m/s. 2. The linear speed, more correctly, tangential speed distance/time, will be the circumference of the Ferris wheel divided by the time for one revolution, 30 s. From geometry, the circumference = 2πr = 2(3.14)(10 m) = 62.8 m. So the linear speed v = (62.8 m)/(30 s) = 2.1 m/s. 3. The center of mass of the two weights is where a fulcrum would balance both—where the torques about the fulcrum would balance to zero. Call the distance (lever arm) from the 1-kg weight to the fulcrum x. Then the distance (lever arm) from the fulcrum to the 3-kg weight is (100 – x). Equating torques: 1x = (100 – x)3 x = 300 – 3x x = 75. So the center of mass of the system is just below the 75-cm mark. Then the three-times-as-massive weight is one-third as far from the fulcrum. 4. 7500 N at the left and 2500 N at the right. Since the distances from vehicle to supporting ends take the ratio 3/1, the support forces will have the inverse ratio; 1/3. That is, the left side will support 3 times as much as the farther right side. (Since the bridge is not set into rotation by the shared weight of the vehicle, the torques supplied by these forces must be equal. That is, Fd = fD, where the torque at the left is F times distance d from the vehicle, and the torque at the right is f times longer distance D from the vehicle. We consider torques about the CG of the vehicle, although torques about any point will produce the same result.) Note that F + f = 10,000 N, so f = (10,000 N - F), and distance D = 3d. Then Fd = (10,000 - F)3d, where F = 7500 N. This means f = 2500 N. 
5. The mass of the stick is 1 kg. (This is a “freebie”; see Check Yourself question and answer in the chapter!) 6. (a) Torque = force lever arm = (80 N)(0.25 m) = 20 N m. (b) Force = 200 N. Then (200 N)(0.10 m) = 20 N m. (c) Yes. These answers assume that you are pushing perpendicular to the wrench handle. Otherwise, you would need to exert more force to get the same torque. 7. Centripetal force (and “weight” and “g” in the rotating habitat) is directly proportional to radial distance from the hub. At half the radial distance, the g force will be half that at his feet. The man will literally be “light-headed.” (Gravitational variations of greater than 10% head-to-toe are uncomfortable for most people.) 8. This follows the preceding answer. If one’s feet at full radius R undergo a centripetal acceleration of g, then the fractions of g closer to the hub will be in direct proportion to the fractions of R. At a distance from the hub of R/2, we get g/2, at 3R/4 from the hub, we get 3g/4, and so forth. Some thought will show that the fraction by which we decrease R is matched by an equal decrease of g. So for a decrease of 1/100 for g, we must move 1/100 closer to the hub. This means the hub would have to be 1.01 times as far from the feet as from the head. The radius of the habitat would have to be 100 times a person’s height. 9. The artist will rotate 3 times per second. By the conservation of angular momentum, the artist will increase rotation rate by 3. That is Ibefore = Iafter Ibefore = [(I/3)I(3)]after 10. From data listed on the inside back cover, by ratio; = = where speed V of Earth about Sun is its circumference divided by the time for one orbit, 2πR/12 months, and speed v of Moon about the Earth is 2πr/1 month. Mass M of Earth is 6 1024 kg, and mass m of Moon is 7.4 1022 kg. Earth-Sun distance R is 1.5 1011 m, and Earth-Moon distance r is 3.8 108 m. Then ≈ ≈ 106. So the Earth has a million times as much angular momentum about the Sun as the Moon has about the Earth. Directory: media media -> Applications to Drill media -> Unicef voices of Youth Chat Theme: Children and aids nigeria and Zimbabwe 13 October 2006 Background media -> Tsunami Terror Alert: Voices of Youth media -> Biblical Eschatology Presentation by: D. Paul Beck May 4, 2016 Ground Rules media -> Guide to completing the collection using the Omnibus system media -> The milk carton kids media -> Events Date and Location media -> The Gilded Age: The First Generation of Historians by H. Wayne Morgan University of Oklahoma, April 18, 1997 media -> Analysis of Law in the United Kingdom pertaining to Cross-Border Disaster Relief Prepared by: For the 30 June 2010 Foreword media -> Cuba fieldcourse 2010 Download 0.89 Mb. Share with your friends:
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