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

Answer 1

Answer: (B)

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

Answer 2

The situation can be figured as Taking horizontal direction as $X$-axis and vertical direction as $y$-axis resolving the forces along the axes, we get $N \sin \theta =m \omega^{2} r \sin \theta$ $\Rightarrow N =m \omega^{2} r$ ... (i) and $N \cos \theta =m g$ ... (ii) Dividing E (i) by E (ii), we get $\frac{1}{\cos \theta}=\frac{\omega^{2} r}{g}$ Angular speed of the ball $\omega=\sqrt{\frac{g}{r \cos \theta}}$