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posted by  bria1 on 11/2/2009 4:05:27 PM  |  status: Closed  |  Earned Karma: 50

Pretty tough question, I cannot get the right answer from...

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N/A N/A N/A N/A 11/3/2009 at 3:00:00 PM
Question Details:
A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.00 m/s on a horizontal section of a track, as shown in Figure P10.56. It rolls around the inside of a vertical circular loop 90.0 cm in diameter and finally leaves the track at a point 23.0 cm below the horizontal section.


Figure P10.56

(a) Find the speed of the ball at the top of the loop.
1 m/s
(b) Find its speed as it leaves the track.
2 m/s
(c) Suppose that static friction between ball and track were negligible, so that the ball slid instead of rolling. Would its speed then be higher, lower, or the same speed at the top of the loop? Explain.
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posted by john_r on 11/3/2009 3:20:12 PM  |  status: Live
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Response Details:

   for the given problem consider the diagram as shown

                                                                   

(a)
   we apply the energy conservation for the system of the ball and the earth between the horizontal section and the top of
    the loop
   (1 / 2) m v22 + (1 / 2) I ω22 + m g y2 = (1 / 2) m v12 + (1 / 2) I ω12
   (1 / 2) m v22 + (1 / 2)[(2 / 3) m r2] (v2 / r)2 + m g y2 = (1 / 2) m v12 + (1 / 2)[(2 / 3) m r2] (v1 / r)2
   (5 / 6) v22 + g y2 = (5 / 6) m v12
   v2 = √[v12 - (6 / 5) g y2]
       = ...... m / s
   the centripetal acceleration is
   v22 / r = ........ m / s2
   if this is greater than g the ball must be in contact with the track, pushing downward on it
(b)
   (1 / 2) m v32 + (1 / 2) [(2 / 3) m r2] (v3 / r)2 + m g y3 = (1 / 2) m v12 + (1 / 2)[(2 / 3) m r2] (v1 / r)2
   v3 = √[v12 - (6 / 5) g y3]
       = ...... m / s
(c)
   (1 / 2) m v22 + m g y2 = (1 / 2) m v12  
   v2 = √(v12 - 2 g y2)
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