Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 4 / 11A / b / iii / Solution

For nonzero $\omega$, the equilibrium points satisfy $\sin 2\theta =0$. The equilibrium point $\theta =0$ is stable: it puts the bead at the point farthest from the rotation axis. Writing $\theta =\eta$ small in the tangential ordinary differential equation givesThus the small oscillation is simple harmonic motion, with angular frequency $|\omega |$ and ordinary frequency $|\omega |/(2\pi )$.

The endpoints $\theta =\pm \pi /2$ represent the same rim point $O$. A perturbation there instead satisfies $\ddot{\eta }=+{\omega }^2\eta$ and is unstable. Equivalently, the centrifugal potential along the hoop is $-\frac{1}{2}{\omega }^2|\mathbf{r}{|}^2=-2a^2{\omega }^2{\cos }^2\theta$, whose minimum is at zero and maximum at $O$. If $\omega =0$, every bead position with zero relative velocity is an equilibrium point, but perturbing that velocity produces unrestored drift; there is no nonzero restoring frequency.