A hemispherical bowl of radius R si set rotating abouv its axis of symmetry whichis kept vertical. A small blcok kept in the bowl rotates with the bowl without slippingn on its surface. If the surfaces of the bowl is mooth, and the abgel made by the radius through the block with the vertical is `theta`, 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))`