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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 givesGravity 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, givesIf ${\omega }^2>g/a$, the two off-axis positions areTheir 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 $m{\omega }^2{a}^2{\sin }^2{\theta }_0>0$. Equivalently, writing $\theta ={\theta }_0+\delta$ gives to first orderThe same holds at $-{\theta }_0$. Thus both equilibria are stable, with small-oscillation angular frequency $|\omega |\sin {\theta }_0$.