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 circlular frame is rotating with angular speed $\omega$ about the vertical diameter $AB$, the bead is at rest with respect to the circular frame at the position shown in the figure. Then, the value of ${\omega }^2$ is equal to:

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

Answer 1

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

Given the perimeter of circular wire $=2\pi r$, hence radius $=r$

From FBD, $N$ is the normal reaction of the wire loop on the bead, acting towards centre $\text{C}$.


Since the bead is at rest w.r.t the rotating wire frame, its angular speed of rotation will be the same as that of the wire frame.
Then, component $N\cos \theta$ balances the weight of the bead, because bead is in equilibrium in vertical direction.
$\Rightarrow N\cos \theta =mg...\left(i\right)$

Component of normal reaction $N\sin \theta$ provides necessary centripetal force for the bead:
$\Rightarrow N\sin \theta =F_c=m\times {\omega }^2\times \left(\frac{r}{2}\right)...\left(ii\right)$

Dividing Eqs. $\left(ii\right)$ and $\left(i\right)$ we get:

$\tan \theta =\frac{{\omega }^{2}r}{2g}...\left(iii\right)$

Now, from the geometry of the figure:


$\tan \theta =\frac{\frac{r}{2}}{\sqrt{r^2-{\left(\frac{r}{2}\right)}^2}}=\frac{1}{\sqrt{3}}...\left(iv\right)$
Put value of $\tan \theta$ in Eq. ($iii$):
$\frac{1}{\sqrt{3}}=\frac{{\omega }^{2}r}{2g}$
$\therefore {\omega }^2=\frac{2g}{\sqrt{3}r}$