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Circular rotational KE motionl.

  1. Oct 2, 2005 #1
    A tennis ball is set rolling without slipping on a horizontal section of a track. It rolls into a vertical loop and goes up the loop.

    I was just wondering: Suppose that the static friction betweeen the ball and the track were negligible, so that the ball slid instead of rolling, would its speed be higher or lower or the same as that of the loop?

    For me, my explanation: rotational KE = 0 (because the ball isnt rolling, but slipping) and linear KE is the same, the initial total energy would be lesser than the original question, the speed of the ball at the top should be lesser then.

    But from another pt of argument, if static friction is negligible, it means that the pt of the ball which is in contact with the ground at any pt is 0. As such, rotational KE remains the same at the bottom of the track and at the top of the loop. Rotational KE (initial) = rotational KE (Final). Thus we need not consider the rotational KE in the Conservation of energy in the loop. Thus, we only need to consider translational KE and the potential energy at the start and at the top, ignoring rotational KE.

    Is this the right way? Or is there another explanation for the speed at the top of the loop?
     
  2. jcsd
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