# Spinning Sphere

## Homework Statement

A solid sphere of mass M and radius R rotates freely in space with an angular velocity w about a fixed diameter. A particle of mass m, initially at one pole, movies with a constant velocity v along a great circle of the sphere. Show that when the particle has reached the other pole, the rotation of the sphere will have been retarded by an angle
$$\alpha = \omega T ( 1 - \sqrt{\frac{2M}{2M+5m}})$$

## The Attempt at a Solution

So I have a picture of a rotating sphere with a mass on it. I know that $$I_{total} = 2/5 M R^2 + mr^2$$ where r is the vector from the initial pole to the position of the particle on the sphere. I believe I also need to take into account the polar angle phi, and the azimuthal angle theta.

I know that angular momentum is conserved. So dL/dt = 0
I also know that L = Iw

I was thinking of doing the following: d/dt(Iw) = 0 = d/dt(I) w + I d/dt(w)
so far that hasn't worked, but I think it may be because I haven't been able to come up with a decent guess for the angular velocity w. My best guess so far is w = v/(r*sin(phi)), which is clearly incorrect.

Anyone have any hints?

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So I think I have narrowed it down to:

$$L = (2/5 M R^2 + m r^2) (d \phi / dt)$$
$$\phi = \int_0^T \frac{L dt}{2/5MR^2 + mr^2}$$

Let
$$\omega = 5/2 L / (MR^2)$$
Unfortunately, I can't seem to find a relationship between r(t) and t? In other words, how do I integrate that? I can't seem to recall anything that would work? I think I am getting close, I just can't get this darn integral to work out.

I know v is constant, but I can't seem to make anything out of that work.
if v is constant then dv/dt = 0.
I was thinking that $$v = r \theta$$ where theta is the azimuthal angle. But this implies that $$d^2( \theta)dt^2 = 0$$ which means that $$d/dt (\theta ) = constant$$ but that means that my expression for v doesn't work, so I am going to guess that it is incorrect.