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:
Answer: (C) $\dfrac{2g}{r\sqrt3}$
Let OP make angle $\theta$ with the downward vertical; P is at distance $r\sin\theta = \dfrac r2$ from the axis, so $\theta = 30^\circ$.
The normal force acts along PO. Its vertical component balances gravity and its horizontal component supplies the centripetal force:
$$N\cos\theta = mg,\qquad N\sin\theta = m\omega^2r\sin\theta$$
$$\omega^2 = \frac{g}{r\cos\theta} = \frac{2g}{\sqrt3\,r}$$
Solution by Sreeraj P, M.Sc Physics