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# Dynamics - A solid sphere of mass M and radius R?

A solid sphere of mass M and radius R rotates freely in space with angular velocity ω about a fixed diameter.

A particle of mass m , initially at one pole , moves 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 na angle

ß = ω T ( 1 - √[2M/(2M+5m)] )

Where T is the total time required for the particle to move from one pole to the other

Thanks for the help.

### 2 Answers

- RLv 51 decade agoFavorite Answer
Interesting problem! I'm guessing the the particle is in contact with the sphere, and its velocity is relative to the sphere, "south" for example. Angular momentum is transferred from the sphere to the particle and back. During the time it was slowed, it covered less angle, hence beta is the angle it is "behind" another sphere that did not have a particle move.

During a short time interval dt, the particle has covered a distance pi R t/T, and so has covered a latitude angle measured from the "north pole" of pi t/T, and is at a distance r = R sin (pi t/T) from the axis of rotation. It then has moment of inertia I1=m r^2. So at that moment, the total moment of inertia is I(total) = I0 + I1 = 2/5 * MR^2 + mr^2, and the angular velocity (I'll use w for omega, pardon my font ignorance) is

w(t) = w (2/5 MR^2)/(2/5MR^2 + mr^2)

and the sphere covers an angle of w(t) dt in that same time. The total angle covered by a sphere without a particle is wT, and the total angle covered by our sphere is Integral from 0 to T of w(t)dt. Subtract those to obtain beta.

Plug everything in and you have an integral in terms of constants and t.

I hope that's enough. Good luck!

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