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A hemispherical bowl of radius R is set ...

A hemispherical bowl of radius R is set rotating about its axis of symmetry which is kept vertical. A small block kept in the bowl rotates with the bowl without slipping on its surface. If the surfaces of the bowl is smooth, and the angle made by the radius through the block with the vertical is `theta`, find the angular speed at which the bowl is rotating.

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Suppose the angular speed of rotatioin of the bowl is `omega`. The block also moves with this angular speed. The forces on the block are ure.
a. the normal force N and
b. the weight mg

The block moves in a horizontal circle with the centre at C, so that teh radius is PC=OP `sintheta=Rsintheta.` Its acceleration is therefore `omega^2Rsintheta.` Resolving the forces along PC and applying Newton's second law,
`Nsintheta=momega^2Rsintheta`
`or, N=m omega^2R` ........i
As there is vertical acceleration,
`Ncostheta=mg`..........ii
Dividing i by ii
`1/(costheta)=(omega^2R)/g`
or, `omega=sqrt(g/(Rcostheta))`
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