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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 angel made by the radius through the block with the vertical is `theta`, find the angular speed at which the bowl is rotating.

Text Solution

Verified by Experts

The correct Answer is:
A, C
Let `omega` be the angular speed of rotation of the bowl. Two force are acting on the ball.

1. normal reaction `N`
2. weight `mg`
The ball is rotating in a circle of radius `r(=Rsinalpha)` with centre at `A` at an angular speed `omega` . Thus,
and `Ncosalpha=mg`
Dividing Eq. `(i)` by Eq. `(ii)` we get
`(1)/(cosalpha)=(omega^(2)R)/(g)`
:. `omega=sqrt((g)/(Rcosalpha))`
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