Question

A small ring $P$ is threaded on a smooth wire bent in the form of a circle of radius $a$ and center $O$. The wire is rotating with constant angular speed $\omega$ about a vertical diameter $xy$, while the ring remains at rest relative to the wire at a distance $\frac{a}{2}$ from $xy$. Then ${\omega }^2$ is equal to

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

Answer 1

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


From FBD:
$\sum F_y=0\Rightarrow N\cos \theta =mg$... (i)
$\sum F_x=0\Rightarrow N\sin \theta =\frac{m{\omega }^{2}a}{2}$ ... (ii)
Dividing Eq. (ii) $÷$ (i), $\tan \theta =\frac{{\omega }^{2}a}{2g}\Rightarrow {\omega }^2=\frac{2g\tan \theta }{a}$
Now, $\sin \theta =\frac{a/2}{a}=\frac{1}{2}$ or $\theta ={30}^{\circ }$
$\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}$

$(\because \sin \theta =\frac{\text{opp}}{\text{hyp}}$ from the triangle)
i.e ${\omega }^2=\frac{2g}{\sqrt{3}a}$