# A ball sliding on a ring

1. Sep 13, 2010

### valandil

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 $$\theta=\pi/4$$.

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:

$$L = I\omega$$
where
$$I= MR^2$$

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

$$\omega_0 = \frac{L}{MR^2}$$

However, if we include a dependance on $$\theta$$, we obtain

$$L = (I_{ring} + I_{ball})\omega$$
$$L = (MR^2 + mR^2\sin^2\theta)\omega$$

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. Sep 13, 2010

### zorro

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

3. Sep 13, 2010

### valandil

Thank you. I just arrived at the same solution.