Question

A block of mass $m$ is at rest on a smooth inclined plane, which is rotated with a constant angular velocity ${}^{\prime }{\omega }^{\prime }$ about a vertical axis as shown in figure. Then, $\omega$ will be:

• A. $\sqrt{\frac{g\tan \theta }{x}}$
• B. $\sqrt{\frac{2g\tan \theta }{x}}$
• C. $\sqrt{\frac{g\tan \theta }{2x}}$
• D. $\sqrt{\frac{3g\tan \theta }{x}}$

Answer 1

The correct option is A $\sqrt{\frac{g\tan \theta }{x}}$


Since the block is at rest with respect to the inclined plane (rotating with constant $\omega$), we can analyse it from the inclined plane's frame of reference by including a centrifugal force in radially outward direction on the block.

$F_{\text{centrifugal}}=m{\omega }^{2}x$

Then, component of $F_{\text{centrifugal}}$ along the inclined plane $=m{\omega }^{2}x\cos \theta$
and
Component of $F_{\text{centrifugal}}$ perpendicular to the inclined plane $=m{\omega }^{2}x\sin \theta$

Applying equilibrium condition for the block along the inclined plane gives:
$mg\sin \theta =m{\omega }^{2}x\cos \theta$
$\Rightarrow \tan \theta =\frac{{\omega }^{2}x}{g}$
$\therefore \omega =\sqrt{\frac{g\tan \theta }{x}}$