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Billard Ball Rolling problem (tricky)

  1. Feb 6, 2007 #1
    1. The problem statement, all variables and given/known data
    A billard ball of radius 'a' is intially spinning about a horizontal axis with angular speed w and with zero forward speed. If mu is the coefficent of sliding friction between the ball and the table, find the distance the ball travels before slipping ceases to occur.

    2. Relevant equations
    Radius of gyration= 2/5*a^2, through the center of mass.


    I*w=F(sub p)*r, where I is the moment of inertia, and F(sub p) is the fictional force parallel to the plane.

    3. The attempt at a solution
    So far, I have been able to determine that the intial velocity is equal to a*w(sub intial). But this doesn't do me much good.

    I am more or less having a hard time trying to set up a force equation to solve.

    I was thinking that the equation of motion would look something like this

    m*a= (I*w)/a - f, where f (lower case) is the force due to friction).

    But I don't think this is right.

    Any ideas?
  2. jcsd
  3. Feb 6, 2007 #2
    I would get two functions, one based on the acceleration of the CM of the ball--that is--frictional force over the mass. The other would be based acceleration of the tip of the ball (the part in contact with the surface). Find out how long it takes for the speeds to become equal and then it's easy.
  4. Feb 6, 2007 #3
    Ok I tried your suggestion and well: I'll just show you:

    [tex] Let F_p = \mu_k * F_n [/tex] since by rolling only, all motion intially should be slipping.

    Noting this:

    And using newton's second law

    m * x(double dot)_cm = [tex] \mu_k* mg[/tex] Since, the center of mass is perpendicular to the ground.

    Next I set up an equation for the motion of a point on the ball, noting that the motion would be uniformly circular:

    [tex] a*w^2 = [/tex] x(double dotted)_point

    Integrating and setting the equations equal to each other and taking into account that the intial rotational velocity (the intial condition for the first equation) I get:

    [tex] t = w/( \mu_k * g - a * w^2) [/tex]

    Then noting that the motion of the ball on the flat, horizontal, plane should follow the kinematics solution:

    I find that the distance traveled is equal to:

    [tex] ( g * \mu_k *w^2 )/ ( 2 * ( g * \mu_k - a * w^2 )^2 ) [/tex]

    But I have a feeling I made a mistake, as this just seems too easy (and to me that is a bad thing).

    Any help
  5. Feb 7, 2007 #4
    I could have really easily screwed this up but here is what I did. first I said that the acceleration of the center of mass is (mu)g so the speed is (mu)g(t). then I said that the angular acceleration of the ball is T/I=(mu)mga/(2/5)ma^2.
    I then got a speed funcion of the tip which is wa-(5mu(g))/2)t
    setting the two velocity funtions equal to each other, I got an answer which was different from yours, but I could have screwed up.
  6. Feb 7, 2007 #5
    I think yours is right; I talked a couple people stuck on the same problem and they manged to work it out to the same answer as you.

    I am turning in the problem in 2-days or so, so I will try to get the solution to it up here in case anyone else is curious (also in case someone wants to help point out where I made my mistake).


    What is "T" in your equation...is it torque?
    Last edited: Feb 7, 2007
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