Question

A sleeve $A$ can slide freely along a smooth rod bent in the shape of a half-circle of radius $R$ (shown in figure above). The system is set in rotation with a constant angular velocity $\omega$ about a vertical axis OO'. Find the angle $\theta$ corresponding to the steady position of the sleeve.

Answer 1

There are two forces on the sleeve, the weight $F_1$ and the centrifugal force $F_2$. We resolve both forces into tangential and normal component then the net downward tangential force on the sleeve is,
$mg\sin \theta (1-\frac{{\omega }^{2}R}{g}\cos \theta )$
This vanishes for $\theta =0$ and for
$\theta ={\theta }_0={\cos }^{-1}\left(\frac{g}{{\omega }^{2}R}\right)$, which is real if ${\omega }^{2}R>g$. If ${\omega }^{2}R<g$, then $1-\frac{{\omega }^{2}R}{g}\cos \theta$ is always positive for small values of $\theta$ and hence the net tangential force near $\theta =0$ opposes any displacement away from it. $\theta =0$ is then stable.
If ${\omega }^{2}R>g$, $1-\frac{{\omega }^{2}R}{g}\cos \theta$ is negative for small $\theta$ near $\theta =0$ and $\theta =0$ is then unstable.
However $\theta ={\theta }_0$ is stable because the force tends to bring the sleeve near the equilibrium position $\theta ={\theta }_0$.
If ${\omega }^{2}R=g$, the two positions coincide and becomes a stable equilibrium point.