Lower bound of stable equilibrium for mass on rotating ring

Question

Consider a point mass $m$ constrained to move without friction along a ring of radius $R$. The ring rotates with angular frequency $\omega$ about the $z$-axis which runs through a diameter of the ring, with the origin at the ring's center. There is a uniform gravitational field with acceleration $g$ in the negative $z$-direction.

The Lagrangian with a generalized polar coordinate $\theta$ is

$$L = \frac{1}{2}mR^2(\dot{\theta}^2+\omega^2 \sin^2\theta)-mgR\cos\theta$$

and the equation of motion

$$\ddot{\theta} - \sin\theta\left(\frac{g}{R}+\omega^2\cos\theta\right) = 0$$

The point of stable equilibrium is $\theta=\theta_0$ for

$$\cos\theta_0 = \frac{-g}{R\omega^2} \geq -1$$

What does the lower bound $g=R\omega^2 \Rightarrow \theta_0=\pi$ correspond to physically?

Answer 1

$\theta =\pi$ is always the equilibrium point in this problem. If $g> R\omega^2$, then this equilibrium point is stable and there are no other stable equilibrium points. If $g <R\omega^2$, then $\theta = \pi$ is an unstable equilibrium point. So, the physical picture looks like this. If the frequency $\omega$ is not large enough, then the point mass is in the lowest position, otherwise the centrifugal force forces the mass to occupy a higher position. The value $\omega = \sqrt{g/R}$ corresponds to the loss of stability of the equilibrium point $\theta = \pi$.