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valandil
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Homework Statement
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].
Homework Equations
This is what I can't find. The problem seems easy enough, provided the right path of solution.
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 dependence 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?
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