Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 4 / 11C / ii / Solution

The downward and upward positions $\theta =0,\pi$ are equilibria. Linearization about $\theta =0$ givesso the downward equilibrium is stable for ${\omega }^2<g/R$ and unstable for ${\omega }^2>g/R$; at equality it is marginal at linear order. At $\theta =\pi$, the linear coefficient is ${\omega }^2+g/R>0$, so the upward equilibrium is unstable.

When ${\omega }^2\ge g/R$, there are also the two equilibriaAt either nonvertical equilibrium, differentiation of the right-hand side gives $-{\omega }^2{\sin }^2\theta <0$, so both are stable when distinct. This is a pitchfork bifurcation of a rotating hoop bead at ${\omega }^2=g/R$.

Solved by gpt-5.6-sol high.