Question

A bead $A$ can slide freely along a smooth rod bent in the form of a half circle of radius $R$. The system is set in rotation with a constant angular velocity $\omega$ about a vertical axis OO'. Find the angle $\theta$ corresponding to steady position of the bead.

• A. ${\cos }^{-1}\left(\frac{R{\omega }^2}{g}\right)$
• B. ${\cos }^{-1}\left(\frac{g}{R{\omega }^2}\right)$
• C. ${\sin }^{-1}\left(\frac{g}{R{\omega }^2}\right)$
• D. ${\sin }^{-1}\left(\frac{R{\omega }^2}{g}\right)$

Answer 1

The correct option is B ${\cos }^{-1}\left(\frac{g}{R{\omega }^2}\right)$


From right angled triangle shown $r=R\sin \theta .$
For equilibrium,
Forces acting along the tangent
$mg\sin \theta -mr{\omega }^2\cos \theta =0$
$mg\sin \theta [1-\frac{R{\omega }^2}{g}\cos \theta ]=0\theta ={\cos }^{-1}\left(\frac{g}{R{\omega }^2}\right)$.