Question

A smooth wire is bent into a vertical circle of radius $a$. A bead $P$ can slide smoothly on the wire. The circle is rotated about vertical diameter $AB$ as axis with a constant speed as shown in figure. The bead $P$ is at rest w.r.t the wire in the position shown. Then ${\omega }^2$ is equal to:

Answer 1

Here the bead $P$ is rest with respect to the circular ring.

$N\sin \theta =m\left(\frac{a}{2}\right)w^{2}N\cos \theta =mgDividing\tan \theta =\frac{a{w}^2}{2g}{w}^2=\frac{2g}{a}\times \frac{\frac{a}{2}}{\frac{\sqrt{3}a}{2}}=\frac{2g}{\sqrt{3}a}$