06_InstSolManual_PDF

06_InstSolManual_PDF - CIRCULAR MOTION AND GRAVITATION 6...

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6-1 C IRCULAR M OTION AND G RAVITATION 6 Answers to Multiple-Choice Problems 1. D 2. B 3. C 4. A 5. C 6. C 7. C 8. A 9. D 10. B 11. B 12. C 13. A 14. A 15. C Solutions to Problems 6.1. Set Up: The path shows the relative size of the radius of curvature R of each part of the track. A straight line is an arc of a circle of inFnite radius, so Solve: At A ,I n e a c h c a s e , i s d i r e c t e d t o w a r d t h e c e n t e r o f t h e c i r c u l a r p a t h . The net force is proportional to and is shown in ±igure 6.1. Figure 6.1 Refect: The road must exert a greater friction force on the car when it makes tighter turns that have smaller radii of curvature. 6.2. Set Up: Since the stone travels in a circular path, its acceleration is directed toward the center of the circle. The only horizontal force on the stone is the tension of the string. Set the tension in the string equal to its maximum value. Solve: (a) The free-body diagram for the stone is given in ±igure 6.2. In the diagram the stone is at a point to the right of the center of the path. a rad 5 v 2 / R , B A C F net, B F net, A 5 0 F net, C a S rad a S rad a rad, B , a rad, C . a rad 5 0. a rad 5 v 2 R . R A S ` . R B . R C .
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Figure 6.2 (b) gives 6.3. Set Up: Each hand travels in a circle of radius 0.750 m and has mass and weight 6.4 N. The period for each hand is Let be toward the center of the circular path. (a) The free-body diagram for one hand is given in Figure 6.3. is the force exerted on the hand by the wrist. This force has both horizontal and vertical components. Figure 6.3 (b) gives (c) The horizontal force from the wrist is 12 times the weight of the hand. Refect: The wrist must also exert a vertical force on the hand equal to its weight. 6.4. Set Up: Since the car travels in an arc of a circle, it has acceleration directed toward the center of the arc. The only horizontal force on the car is the static friction force exerted by the roadway. To calculate the minimum coefficient of friction that is required, set the static friction force equal to its maximum value, Friction is static friction because the car is not sliding in the radial direction. Solve: (a) The free-body diagram for the car is given in Figure 6.4. The diagram assumes the center of the curve is to the left of the car. Figure 6.4 y n mg x f s 5m s n a rad f s s n . a rad 5 v 2 / R , F w 5 77 N 6.4 N 5 12. F x 5 ma rad 5 1 0.65 kg 21 118 m / s 2 2 5 77 N g F x 5 ma x a rad 5 4 p 2 R T 2 5 4 p 2 1 0.750 m 2 1 0.50 s 2 2 5 118 m / s 2 w x y F F x F y a rad F S 1 x T 5 1 1.0 s 2 / 1 2.0 2 5 0.50 s. 1 0.0125 52 kg 2 5 0.65 kg v 5 Å TR m 5 Å 1 60.0 N 0.90 m 2 0.80 kg 5 8.22 m / s. T 5 m v 2 R . g F x 5 ma x y x mg T n a rad 6-2 Chapter 6
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(b) gives gives and 6.5. Set Up: The person moves in a circle of radius The acceleration of the person is directed horizontally to the left in the f gure in the problem. The time for one revolution is the period Solve: (a) The free-body diagram is given in Figure 6.5. is the force applied to the seat by the rod.
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This note was uploaded on 06/10/2010 for the course PHY 185363 taught by Professor R during the Summer '09 term at Florida College.

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06_InstSolManual_PDF - CIRCULAR MOTION AND GRAVITATION 6...

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