Q. A smooth wire of length $2\pi r$ is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed $\omega$ about the vertical diameter $AB$, as shown in figure, the bead is at rest with respect to the circular ring at position $P$ as shown. Then the value of $\omega^2$ is equal to :
Solution:
$N \, sin \theta = m \frac{r}{2} \omega^2 \, \, \, \, \, \, \, \, \, \, \, \, ...(1)$
$N \, cos \theta = mg \, \, \, \, \, \, \, \, \, \, \, \, ...(2)$
$tan \theta \, \, = \, \, \frac{r \omega^2}{2g}$
$\frac{r}{2 \frac{\sqrt{3}r}{2}} \, \, = \frac{r \omega^2}{2g}$
$\omega^2 = \frac{2g}{\sqrt{3}r}$
