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Motion involving both translation and rotation

  1. Nov 13, 2006 #1
    Hello everyone. Thanks for all the help today. I know I have asked a lot of questions. This is my last one today.

    A spherical billard ball is sliding with a speed v_o. It has a radius R, mass M, and there is a friction coefficient mu. Determine the distance the ball travels before it begins to roll smoothly on the surface.

    The correct answer is supposed to be 12v_o^2/(49*mu*g).

    I know that the kinetic energy K = (1/2)MV_cm^2 (1 + B)

    K = (1/2)MV_cm^2 ( 1 + 2/5) - mu*Mg*x
    K = (1/2)Mv_cm^2 ( 7/5) -mu*Mg*x

    K = (7/10)Mv_cm^2 -mu*Mg*x
    x = (7/10)Mv_cm^2 /(mu*mg)

    However, this is not the right answer. Where did I go wrong? Is it possible to solve this with energy concept? I don't know how to do it with Newton's Law. I don't know how to get velocity out of the problem.
  2. jcsd
  3. Nov 13, 2006 #2


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    It would probably be a bit messy to do this with just work/energy. The work done by friction as the ball slides on the table is not the force times the distance the ball moves. If the ball is sliding without rolling, the work done is friction times the distance the center of the ball moves. If the ball is rolling, the work done by friction is zero. As the ball begins to rotate while still sliding, the work done per unit distance the ball moves gradually diminishes.

    It is easier to look at the force of friction causing two things to happen simultaneously. First, it declelerates the center of mass of the ball by Newton's 2nd law: F = ma. Second, it applies a torque to the ball that causes it to rotate: torque = Fr = Iα. Using the connection between v and ω when the ball is rolling you can determine when slipping stops and pure rolling begins
    Last edited: Nov 13, 2006
  4. Nov 13, 2006 #3
    I don't really understand what do.

    From what you told me,

    The net force in the horizontal direction is ma, with friction being mu*mg
    The net torque is I*(alpha), and I know that I for a sphere is (2/5)MR^2
    alpha is a/R

    so I = (2/5)*MA*R

    F = mu*Mg

    The answer has a 12 and 49, but I don't know where they're coming from. I know d = (1/2)at^2, but I don't know what to do with it.
  5. Nov 13, 2006 #4


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    It would have been better if I had said what I meant. I made a bit of a change to the previous post. The ball will decelerate while slipping so its velocity will decrease. Its angular velocity will increase while it is slipping. At some time, the angular velocity and the liner velocity will have the right relationship for pure rolling. From that point on the velocity and angular velocity will remain proportional, both decreaing only slightly because of rolling friction. You have the moment of inertia of the sphere and the force acting. Write the angular velocity as a function of time and write the linear velocity as a function of time. Find the time when the two are in the right ratio and then solve for the distance moved.
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