Question

A hemispherical bowl of radius R is rotating about its axis of symmetry which is kept vertical. A small ball 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 ball with the vertical is $\alpha .$ Find the angular speed at which the bowl is rotating.

Answer 1

Let $\omega$ be the angular speed of rotation of the bowl. Two forces are acting on the ball.
1. normal reaction N
2. weight mg
The ball is rotating in a circle of radius $r(=R\sin \alpha )$ with centre at A at an angular speed $\omega .$ Thus,
$N\sin \alpha =mr{\omega }^2$
$=mR{\omega }^2\sin \alpha$...(i)
and $N\cos \alpha =mg$ ...(ii)
Dividing Eq. (i) by (ii), we get
$\frac{1}{\cos \alpha }=\frac{{\omega }^{2}R}{g}$
$\therefore \omega =\sqrt{\frac{g}{R\cos \alpha }}$