A small sleeve of mass $m$ can slide freely along a smooth, rigid wire bent into a semicircle of rad

Question

A small sleeve of mass $m$ can slide freely along a smooth, rigid wire bent into a semicircle of radius $R$. The semicircular wire rotates with a constant angular velocity $\omega$ about its vertical diameter. What is the angle $	heta$ (measured from the downward vertical axis) corresponding to the stable equilibrium position of the sleeve relative to the rotating wire, assuming $\omega > \sqrt{\frac{g}{R}}$?

Professor PC Thomas Classes · Answer

In the rotating reference frame of the wire, the sleeve experiences gravity downward ($mg$), a normal reaction from the wire, and a horizontal centrifugal force directed away from the axis of rotation: $F_{cf} = m \omega^2 r = m \omega^2 R \sin	heta$. For equilibrium along the tangential direction of the wire, the tangential components of gravity and centrifugal force must balance: $mg \sin	heta = (m \omega^2 R \sin	heta) \cos	heta$. For $\sin	heta 
eq 0$ (away from the bottom vertex), dividing by $m \sin	heta$ yields $\cos	heta = \frac{g}{\omega^2 R}$. This non-zero equilibrium position exists and is stable whenever $\omega > \sqrt{\frac{g}{R}}$.