= Solution
In the <rotating-hoop bead> problem, use a signed angle in the hoop plane from the downward vertical. The gravitational <potential energy> is $-mga\cos\theta$, the rotating-frame <speed> is $a\dot\theta$, and the distance from the rotation axis is $a\sin\theta$. The <centrifugal potential> gives
$$
E=\frac{ma^2}{2}\dot\theta^2-mga\cos\theta-\frac{m\omega^2a^2}{2}\sin^2\theta.
$$
Gravity is stationary in these rotating coordinates, and the frictionless constraint does no work. Projecting the rotating equation along the hoop, or differentiating its effective potential in the one-dimensional <Euler-Lagrange equation>, gives
$$
\boxed{\ddot\theta+\left(\frac ga-\omega^2\cos\theta\right)\sin\theta=0.}
$$
If $\omega^2>g/a$, the two off-axis positions are
$$
\boxed{\theta=\pm\theta_0,\qquad\theta_0=\arccos\left(\frac{g}{a\omega^2}\right).}
$$
Their unsigned angle to the downward vertical is the same; they lie on opposite sides of the hoop. The second <derivative> of the effective potential there is $m\omega^2a^2\sin^2\theta_0>0$. Equivalently, writing $\theta=\theta_0+\delta$ gives to first order
$$
\ddot\delta+\omega^2\sin^2\theta_0\,\delta=0.
$$
The same holds at $-\theta_0$. Thus \b[both equilibria are stable], with small-oscillation angular frequency $|\omega|\sin\theta_0$.
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