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 $\omega$ 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:

• A. $\frac{2g}{a}$
• B. $\frac{2g}{a\sqrt{3}}$
• C. $\frac{q\sqrt{3}}{a}$
• D. $\frac{2a}{g\sqrt{3}}$

Answer 1

The correct option is B $\frac{2g}{a\sqrt{3}}$

$Ncos\theta =m{w}^2\left(\frac{a}{2}\right)Nsin\theta =mgtan\theta =\frac{2g}{w^{2}a}{w}^2=\frac{2g}{atan\theta }astan\theta =\sqrt{3}{w}^2=\frac{2g}{a\sqrt{3}}$