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.

Explanation

Let `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 `r( = R sin theta )` with center at `A` at an angular speed `omega` Than
`N sin theta = mr omega^(2)`
` = mR omega^(2)`
` N = mR omega^(2)`…(i)
`and N cos theta = mg `….(ii)
Dividing Eqs (i) by(ii), we get
` (1)/(cos theta) = (omega ^(2)R)/(g)`
`omega = sqrt((g)/(R cos theta))`