A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical.A small block is kept in the bowl at a position where the radius makes an angle `theta` with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl surface is `mu`. Find the range of the angular speed for which the block will not slip.

Explanation

If the block has tendency to slip downward, friction will act in upward direction along the surface.
`N cos theta+f sin theta=mg` …..`(i)`
`N sin theta-f cos theta=m omega^(2)r` …..`(ii)`
`f=f_(max)=muN`
`(omega^(2)r)/(g)=(N sin theta-mu N cos theta)/(N cos theta+mu N sin theta)=(tan theta-mu)/(1+mu tan theta)`
`omega=omega_(min)=sqrt((g)/(R sin theta)((tan theta-mu))/((1+mu tan theta)))`
If the block has tendency to slip upwards, friction will act in downward direction along the surface.
Solving as above, we have
`omega=omega_(max)=sqrt((g)/(R sin theta)((tan theta+mu))/((1-mu tan theta)))`