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A ball sliding on a ring

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data
    A thin ring of mass M and radius R rotates around its vertical axis. A small ball of mass m can slide, without constraint nor friction, on the ring. If the angular speed of the ring when the ball is at the top is [\tex]\omega_0[\tex], what is the angular speed when the ball is at [tex]\theta=\pi/4[/tex].
    th_probleme.png


    2. Relevant equations
    This is what I can't find. The problem seems easy enough, provided the right path of solution.


    3. The attempt at a solution

    I tried working with the angular momentum conservation law. That gives:

    [tex] L = I\omega[/tex]
    where
    [tex]I= MR^2[/tex]

    The contribution of the ball to the inertia is null, because it is on the axis of rotation at that instant. Hence, we have

    [tex] \omega_0 = \frac{L}{MR^2}[/tex]

    However, if we include a dependance on [tex]\theta[/tex], we obtain

    [tex] L = (I_{ring} + I_{ball})\omega [/tex]
    [tex] L = (MR^2 + mR^2\sin^2\theta)\omega[/tex]

    Now, that seems bad to me, because (1) it's a bad application of the parallel axes theorem because I can't write the inertia of the ball relative to the center of mass since I do not have its radius and (2) it does give an answer as a fraction of the initial angular speed.

    Any ideas?
     
    Last edited: Sep 13, 2010
  2. jcsd
  3. Sep 13, 2010 #2
    Since there is no net external torque on the system, angular momentum is conserved
    apply conservation of angular momentum.
    initially L= {(MR^2)/2}w1 (applying perpendicular axis theorem)
    finally, L = {MR^2 + mR^2sin^2(theta)}w2
    substitute theta=45
    and you get w2= M/(M+m) w1
     
  4. Sep 13, 2010 #3
    Thank you. I just arrived at the same solution.
     
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