 Solution

An equilibrium point has $\dot{\varphi }=0$ and $\boxed{J^{\prime }({\varphi }_0)=0}$. Since $K({\varphi }_0)>0$, the linearized displacement $\psi =\varphi -{\varphi }_0$ satisfiesThus $J^{\prime \prime }({\varphi }_0)>0$ gives a stable equilibrium, while $J^{\prime \prime }({\varphi }_0)<0$ gives an unstable equilibrium point. More generally, a strict local minimum of $J$ is sufficient for stability by the conserved first integral in b. When $J^{\prime \prime }({\varphi }_0)=0$, the second-derivative test alone is inconclusive, and the local shape of $J$ must be examined.

For the specified ellipse, put $D=1+e\cos \varphi >0$. Direct differentiation givesConsequently, with $\eta =g/{\omega }^2$ and $\omega \ne 0$,Two equilibria always occur at $\boxed{\varphi =0,\pi (\mathrm{mod}2\pi )}$. Any further equilibria must satisfyTo decide exactly when this is possible, view the equation as $\eta =(c+e)/(1+ec)$ with $c=\cos \varphi$. Its derivative $(1-e^2)/(1+ec{)}^2$ is positive on $[-1,1]$, and it takes endpoint values $-1$ and $1$. Since $g>0$ implies $\eta >0$, the complete list is:

These are positions of the bead at rest relative to the rotating hoop. No stability classification of the particular ellipse is needed.