What is the period of a ball rolling in a circle?

Question

Imagine a uniform spherical ball of mass $m$ and radius $r$ rolling without slipping in a spherical bowl of radius $R$ with $r<<R$.

There are two special cases of small amplitude oscillations: a ball that rolls directly back and forth at the bottom of a bowl, as well as a ball that rolls in a circle around the bottom of the bowl. In the case of a ball rolling back and forth, I would treat it as approximating simple harmonic motion, and find the period that way.

Q: How would I find the period of a ball rolling in a circle around the bottom of the bowl?

Specifically, I'm not sure how the rolling aspect affects the calculation and makes the question unique from a block sliding frictionlessly around in a circle.

Answer 1

A block sliding without friction would behave itself in exactly the same way as a pendulum, the normal force being provided by the bowl rather than a string. Its period can be calculated from energy conservation, without linearization: $$\frac{m}{2}\left(\frac{dx}{dt}\right)^2+V(x)=E,$$ so that $$\frac{dt}{dx}=\sqrt{\frac{m}{2}}\frac{1}{\sqrt{E-V(x)}},\\ \frac{T}{2}= \sqrt{\frac{m}{2}}\int_{-a}^{a}\frac{dx}{\sqrt{E-V(x)}}, $$ where the amplitude and the net energy are related as $E=V(a)$.

As I have already mentioned, the potential energy is the same as for a pendulum: $$V(x) = mgh = mg(R - R\cos\theta) = mg(R - \sqrt{R^2-x^2}),$$ where $x$ is the horizontal displacement (naturally, one might facilitate the calculation by using the angular coordinate).

The above will not produce two distinct periods, although the period of oscillations is amplitude dependent. Yet, I do agree on the empirical grounds that two different regimes exist. However their existence is likely to deal with the very things neglected in the question: slipping and the size of the ball $r\ll R$ - it involves rotational energy.