Question

A hemisphere ball of radius $R$ is set rotating about its axis of symmetry which is kept vertical. A small block kept in the bowl rotates with the bowl without slipping on its surface. If the surface of the bowl is smooth and the angle made by the radius through the block with the vertical is $\theta$, find the angular speed at which the bowl is rotating.

• A. $\omega =\sqrt{\frac{g\cos \theta }{R}}$
• B. $\omega =\sqrt{\frac{g\sec \theta }{R}}$
• C. $\omega =\sqrt{\frac{2g\cos \theta }{R}}$
• D. $\omega =\sqrt{2gR\sec \theta }$

Answer 1

The correct option is B $\omega =\sqrt{\frac{g\sec \theta }{R}}$

Assume the bowl is rotating with an angular velocity $\omega$.
Radius of the circular motion $=R\sin \theta$


From verticle equilibrium:
$N\cos \theta =mg$ ... (1)
By horizontal equilibrium
$N\sin \theta =m{\omega }^{2}R\sin \theta$
$N=m{\omega }^{2}R$ ... (2)
From eqn. (1) and (2)
$\frac{mg}{\cos \theta }=m{\omega }^{2}R$
$\omega =\sqrt{\frac{g\sec \theta }{R}}$