06_InstSolManual_PDF_Part17

# 06_InstSolManual_PDF_Part17 - arc is of a revolution and in...

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6.49. Set Up: The person moves in a horizontal circle of radius Set the static friction force equal to its maximum value, The person has an acceleration directed toward the center of the circle. The period is and the person has speed Solve: (a) The free-body diagram is given in Figure 6.49. The diagram is for when the wall is to the right of her, so the center of the cylinder is to the left. Figure 6.49 (b) gives gives so Combining these two equations gives and (c) The mass m of the person divides out of the equation for the answer to (b) does not depend on the mass of the person. Reflect: The greater the rotation rate the larger the normal force exerted by the wall and the larger the friction force. Therefore, for smaller the rotation rate must be larger. 6.50. Set Up: Use coordinates where is upward. A
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Unformatted text preview: arc is of a revolution and in one swing the weights travel a distance of Solve: (a) The free-body diagram is given in Figure 6.50. F is the force exerted by the hand. Figure 6.50 (b) and gives and F 5 m 1 g 1 a rad 2 5 1 8.0 kg 21 9.80 m / s 2 1 1.77 m / s 2 2 5 93 N F 2 w 5 ma rad g F y 5 ma y a rad 5 v 2 R 5 1.77 m / s 2 v 5 1 12 1 2 p r 2 1 3 s 5 1 12 1 2 p 21 0.72 m 2 1 3 s 5 1.13 m / s F y w x a rad 1 12 1 2 p r 2 . 30 1 12 30 1 y m s m s ; m s 5 gr v 2 5 1 9.80 m / s 2 21 2.5 m 2 1 9.41 m / s 2 2 5 0.28. m s m v 2 r 5 mg m s n 5 mg . f s 5 mg , g F y 5 ma y n 5 m v 2 r . g F x 5 ma x y n f s 5 m s n mg x a rad v 5 2 p r T 5 9.41 m / s. T 5 1 0.60 rev / s 5 1.67 s a rad 5 v 2 r , f s 5 m s n . r 5 2.5 m. Circular Motion and Gravitation 6-17...
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## This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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