Question

A hemispherical bowl of radius $r$ is set rotating about its axis of symmetry in 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 $\theta$, then find the angular speed at which the ball is rotating.

• A. $\omega =\sqrt{rg\sin \theta }$
• B. $\omega =\sqrt{g/r\cos \theta }$
• C. $\omega =\sqrt{\frac{gr}{\cos \theta }}$
• D. $\omega =\sqrt{\frac{gr}{\tan \theta }}$

Answer 1

Answer: (B)

$\omega =\sqrt{g/r\cos \theta }$

The centripetal force that the rotation exerts on the block $=m{\omega }^2{r}^{\prime }$

The forces must balance in vertical direction.

Hence, $Ncos\theta =mg$

and also they must balance in horizontal direction,

$Nsin\theta =m{\omega }^2{r}^{\prime }$

Hence, ${\omega }^2=\frac{gtan\theta }{r^{\prime }}$ $=\frac{gtan\theta }{rsin\theta }$ $=\frac{g}{rcos\theta }$

$⟹\omega =\sqrt{\frac{g}{rcos\theta }}$