Question 173 of 280

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:

A$2g/r$B$\dfrac{\sqrt{3}g}{2r}$C$2g/\left(r\sqrt{3}\right)$D$\left(g\sqrt{3}\right)/r$

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Answer: (C) $2g/\left(r\sqrt{3}\right)$

From free body diagram of bead,
$Ncos30^{\circ}=mg$ .... (1)
$Nsin30^{\circ}=m\dfrac{r}{2}{\omega }^{2}$ ..... (2)
By dividing equation (2) by equation (1)
$tan30^{\circ}=\dfrac{r{\omega }^{2}}{2g}$
${\omega }^{2}=\dfrac{2g}{r\sqrt{3}}$