problem13_62

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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13.62: a) The normal force on the cowboy must always be upward if he is not holding on. He leaves the saddle when the normal force goes to zero (that is, when he is no longer in contact with the saddle, and the contact force vanishes). At this point the cowboy is in free fall, and so his acceleration is g - ; this must have been the acceleration just before he left contact with the saddle, and so this is also the saddle’s acceleration. b) m. 110 . 0 )) Hz 50 . 1 ( 2 ) s m 80 . 9 ( ) 2 ( 2 2 2 = + = + = π f π a x c) The cowboy’s speed will be the saddle’s speed, s. m 11 . 2 ) 2 ( 2 2 = - = x A πf v d) Taking 0 = t at the time when the cowboy leaves, the position of the saddle as a function of time is given by Eq. (13.13), with ; cos 2 A ω g - = φ this is checked by setting 0 = t and finding that
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Unformatted text preview: . 2 2 ω a ω g x-= = The cowboy’s position is . ) 2 ( 2 c t g t v x x-+ = Finding the time at which the cowboy and the saddle are again in contact involves a transcendental equation which must be solved numerically; specifically, rad), 11 . 1 s) rad 42 . 9 (( cos m) 25 . ( ) s m 90 . 4 ( s) m 11 . 2 ( m) 110 . ( 2 2-=-+ t t t which has as its least non-zero solution s. 538 . = t e) The speed of the saddle is , s m 72 . 1 ) ( sin s) m 36 . 2 ( = +-t ω and the cowboy’s speed is (2.11 ) s m 80 . 9 ( ) s m 2-s, m 16 . 3 s) 538 . (-= × giving a relative speed of s m 87 . 4 (extra figures were kept in the intermediate calculations)....
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