Circular Motion Textbook Solutions Multiple Choice: 5. (d). 8. (d). 9. (b). 10. (d). Concept Questions 7. There is insufficient centripetal force (provided by friction and adhesive forces) on the water drops, so the water drops fly out along a tangent, and the clothes get dry. 8. Yes . It can be travelinG in a curved path at a constant 100 km/h, but it would be accelerating because the direction (but not the maGnitude) of its velocity vector would be changing. 9. The floats of the little mass will move in the direction of the accelertion, inward . It works the same way as the accelerometer in FiG. 4.26. No , it does not make a difference since the centripetal acceleration is always inward 10. All points on the Earth have the same anGular speed, so the centripetal acceleration is proportional to the perpendicular distance from the Earth’s spin axis. (a) Thus the greatest centripetal acceleration is at the equator . (b) The least centripetal acceleration is at either geographic pole . (c) Since both locations are the same distance from the Earth’s spin axis, their centripetal accelerations are the same . The gravity of the Earth provides the centripetal force. feeling thrown outward 11. Due to its inertia, your body has a tendency to keep movinG Straight line motion forward alonG a straiGht line (Newton’s first law), and the car makes (Newton’s first law) a turn by the centripetal force between the tires and the road. So we feel as if we were being “thrown outward.” 12. Centripetal force is required for a car to maintain its circular path. When a car is on a banked turn, the horizontal component of the normal force on the car is pointing toward the center of the circular path. This component will enable the car to negotiate the turn even when there is no friction. Exercises: 1 1 −5 21. (a) f = 12 500 rev/min, so T = f = 12 500 rev/min = 8.000 × 10 min = 4.80 × 10−3 s . 1 −4 −3 (b) T = 9 500 rev/min = 1.053 × 10 min = 6.32 × 10 s . v2 (33.33 m/s)2 30. 120 km/h = 33.33 m/s. a = = = 1.11m / s2 . c r (1.00 × 103 m) v2 2 31. ac = r , F v = ac r = (1.2 m/s )(1.5 m) = 1.3 m/s . 2 2 ac 9.80 m / s 32. ac = rω , F ω = = = r (35 000 m) 0.053 rad/s or about 730 rev/day . 33. a = v2/r r = v2/a = (50 m/s)2/[4(9.80 m/s2)] = 64 m 34. (a) No , it cannot be exactly horizontal. There must be somethinG T θ upward (a component of the tension) to balance the downward gravitational Fc force. mg d 2π r 2π (1.50 m) (b) v = t t = 1.20 s = 7.85 m/s . v2 (0.250 kG)(7.85 m/s)2 (c) Fc = mac = m r = 1.50 m = 10.3 N . 2 −1 ⎛⎞mg −1 ⎡⎤(0.250 kg)(9.80 m/s ) 35. θ = sin ⎜⎟= sin ⎢⎥= 11.3° . ⎝⎠T ⎣⎦12.5 N 36. The answer is No . v2 (23.06 m/s)2 83.0 km/h = 23.06 m/s, a = = = 1.33m / s22 1.25m / s . c r 0.400 × 103 m > 37. (a) The weight is supplying the centripetal force to maintain itself in the circular motion. Or it “falls” alonG a circle now. v2 2 (b) mg = Fc = m r , F v = gr = (9.80 m/s )(1.0 m) = 3.1 m/s . 40. (a) The normal force is Greater at the bottom , because at that position, N mg the normal force has to provide the upward centripetal force, in addition to O supporting the weight of the pilot. N (b) 700 km/h = 194 m/s. At the bottom, the centripetal force is provided by the difference mg v2 − − N mg. So Fc = N mg = m r , v2 (194 m/s)2 N = mg + m = mg + m = mg + m(18.8 m/s2) = mg + 1.9mg = r 2.0 × 103 m 2.9mg . (b) At the top, the centripetal force is provided by N + mg. v2 (194 m/s)2 N = m − mg = m − mg = m(18.8 m/s2) − mg = 1.93mg − mg = r 2.0 × 103 m 0.93mg . 42. To clear the Gully, the truck must travel a minimum horizontal distance of 10.0 m + 4.25 m = 14.25 m. First use kinematics to calculate the velocity of the truck, which is a horizontal projectile. 2 y 2(−2.96 m) The time of fliGht is t = − = − = 0.777 s (from y = v t g 9.80 m/s2 o 1 2 – 2 gt ). 14.25 m v2 (18.3 m/s)2 So vx = 0.777 s = 18.3 m/s. Therefore ac = r = 333 m = 1.01 m/s2 . .