Question

A hemisphere bowl of radius R is about its axis of symmetry which kept vertical. A small ball kept in 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 $\theta$, find the angular speed at which the bowl is rotating.

Answer 1

Lwt $\omega$ be the angular speed of the rotation of the bowl. Two forces are (a) normal reaction, N and (b) weight, mg,
The ball is rotating in a circle of radius $\left(Rsin\theta \right)$with centre at A at an angular speed $\omega$. Thus,
$Nsin\theta =mr{\omega }^2$
$=mR{\omega }^{2}sin\theta$
$N=mR{\omega }^2$ (i)
and $Ncos\theta =mg$ (ii)
Dividing Eqs. (i) by (ii), we get
$\frac{1}{cos\theta }=\frac{{\omega }^{2}R}{g}$
$\omega =\sqrt{\frac{g}{Rcos\theta }}$