Yeah, I'm totally going to fail my defense. I don't remember how to do anything. Gah. Anyway, the answer I get is the same as for a physical pendulum. But I'm probably wrong.Quoting lovejuice (view post)
Say the line between the center of mass and the point of contact with the ground has a length L and makes an angle x with the vertical. Then the torque T about the point of contact is -MgLsin(x), where M is the mass of the half-sphere and g is the acceleration due to gravity. Since x is small, L is approximately R-r, and sin(x) is approximately x, so T=-Mg(R-r)x. I'm not sure what axis of rotation your moment of inertia is defined with respect to, so I'll introduce a new one, I. Then the equation of motion Ix''=T becomes x''=-Mg(R-r)x/I, where a prime indicates a time-derivative. This is the equation for a simple harmonic oscillator with frequency sqrt(Mg(R-r)/I).
In order to fully solve the problem in terms of given quantities, I'd need to relate I and M to J. If J is defined with respect to an axis through the center of mass and parallel to the actual axis of rotation, then I=J+M(R-r)^2, according to the parallel-axis theorem. Relating M to J seems to require evaluating an integral, which I can't be bothered with.
Did you get the same answer as me? There's something weird about the way the pivot point is constantly moving, so I'm not sure that I set up the problem correctly.