Question

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 $\text{AB}$, 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. $\frac{\sqrt{3g}}{2r}$
• B. $\frac{2g}{\left(r\sqrt{3}\right)}$
• C. $\frac{\left(g\sqrt{3}\right)}{r}$
• D. $\frac{2g}{r}$

Answer 1

The correct option is B $\frac{2g}{\left(r\sqrt{3}\right)}$


Now, $\sin \theta =\frac{r/2}{r}=\frac{1}{2}\Rightarrow \theta ={30}^{\circ }$

From the free body diagram, we can write,

$N\sin \theta =m{\omega }^2\left(\frac{r}{2}\right)$
And, $N\cos \theta =mg$

$\Rightarrow \tan \theta =\frac{{\omega }^{2}r}{2g}$

$\tan {30}^{\circ }=\frac{{\omega }^{2}r}{2g}$

$\frac{1}{\sqrt{3}}=\frac{{\omega }^{2}r}{2g}$

$\therefore {\omega }^2=\frac{2g}{r\sqrt{3}}$

Hence, $\left(B\right)$ is the correct answer.