= Solution
{c}
An <equilibrium point> has $\dot\phi=0$ and $\boxed{J'(\phi_0)=0}$. Since $K(\phi_0)>0$, the linearized displacement $\psi=\phi-\phi_0$ satisfies
$$
K(\phi_0)\ddot\psi+J''(\phi_0)\psi=0.
$$
Thus $J''(\phi_0)>0$ gives a <stable equilibrium>, while $J''(\phi_0)<0$ gives an unstable <equilibrium point>. More generally, a <strict local minimum> of $J$ is sufficient for stability by the conserved first <integral> in <6/b>. \b[When $J''(\phi_0)=0$, the second-<derivative> test alone is inconclusive], and the local shape of $J$ must be examined.
For the specified <ellipse>, put $D=1+e\cos\phi>0$. Direct <differentiation> gives
$$
f'=\frac{\cos\phi+e}{D^2},\qquad q'=-\frac{\sin\phi}{D^2}.
$$
Consequently, with $\eta=g/\omega^2$ and $\omega\ne0$,
$$
J'(\phi)=\frac{\omega^2\sin\phi}{D^3}\left[\eta(1+e\cos\phi)-(\cos\phi+e)\right].
$$
Two <equilibria> always occur at $\boxed{\phi=0,\pi\pmod{2\pi}}$. Any further <equilibria> must satisfy
$$
\boxed{\cos\phi=\frac{\eta-e}{1-e\eta}.}
$$
To decide exactly when this is possible, view the <equation> as $\eta=(c+e)/(1+ec)$ with $c=\cos\phi$. 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:
* If $0<\eta<1$, there are \b[four distinct <equilibria> modulo $2\pi$]: $0$, $\pi$, and $\displaystyle\pm\arccos\frac{\eta-e}{1-e\eta}$. For $0<\eta<e$ the additional positive <angle> lies in $(\pi/2,\pi)$; at $\eta=e$ the additional pair is $\pm\pi/2$; for $e<\eta<1$ it lies in $(0,\pi/2)$.
* If $\eta=1$, the additional pair merges with $\phi=0$, leaving \b[two distinct <equilibria>], $0$ and $\pi$.
* If $\eta>1$, there are \b[only the two <equilibria>] $0$ and $\pi$. In particular, $\eta=1/e$ does not introduce a missing case: the undivided bracket is then the positive constant $1/e-e$, so there is no extra root.
These are positions of the bead at rest relative to the rotating hoop. No stability classification of the particular <ellipse> is needed.
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