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Rotational Motion of a bowling ball

  1. Mar 30, 2016 #1
    1. The problem statement, all variables and given/known data
    A bowling ball is released with speed V and no rotational kinetic energy. After a period of sliding and rotating, the ball enters pure rotational motion. The coefficient of friction between the ball and the ground while sliding is Uk.
    a. Show that the rotational acceleration of the ball during the initial period of diluting is alpha=5gUk/2R, where g is the acceleration due to gravity and r is the radius of the ball (a solid sphere).
    b. Show that when sliding finishes and rolling begins, the speed of the center of mass is Vc=5V/7

    2. Relevant equations
    Rotational Kinetic Energy= 1/2IW^2
    Force due to friction= Ukmg
    Kinetic Energy= 1/2mv^2

    3. The attempt at a solution
    I tried this problem from the angle of energy conservation but it quickly gets complicated. I tried to work from:
    Potential energy of the ball + energy lost due to friction + Rotational energy = Total Kinetic, and then subbing in I and solving for V, plugging V into w=v/R and then l x V = alpha. Any help is appreciated.
  2. jcsd
  3. Mar 30, 2016 #2
    Have you tried using the torque equation?
    Torque = I * alpha (Hint: use the parallel axis theorem and use the frictional force as the force producing the torque).
    Conservation of energy can be used to get the final rotational speed, but
    a very concise solution can be obtained using conservation of momentum.
    You wrote alpha=5gUk/2R for the angular acceleration.
    Are you sure the denominator should be 2 R and not 7 R.
  4. Mar 30, 2016 #3
    My mistake - 2 R should be correct because the parallel axis theorem would not apply
    while the ball is slipping, but it can be used when the ball reaches its final speed.
  5. Mar 31, 2016 #4


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    To clarify, the difficulty with applying the parallel axis theorem while the ball is slipping is that the instantaneous centre of rotation is moving. It starts off infinitely below ground (i.e., not rotating) and finishes at ground level.
    That's conservation of angular momentum, right?
  6. Mar 31, 2016 #5
    Okay, so if I start with the equations for Torque this is where it takes me:

    alpha=T/I = (F x R)/(2/5)MR^2 = Mguk x R/ (2/5)Mr^2 = 5guk/2R

    This seems right to me because, as you said, the friction is the external force causing the torque. Thank you!
  7. Mar 31, 2016 #6


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    Looks right.
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