# Lower bound of stable equilibrium for mass on rotating ring

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?

$$\theta =\pi$$ 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 $$g , then $$\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$$.