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.

Options

A. `sqrt((g)/(R.cos theta))`
B. `sqrt((R cos theta )/(g))`
C. `sqrt((g sin theta )/(R ))`
D. `sqrt((R sin theta )/(g))`

Answer

A

Explanation

The vessel is rotaing with angular velocity `omega ` about the vertical axis . While the ball is rotatin in a horizontal circle of radius `OB = r = R sin theta `
The normal reaction N, is resolved into horizontal and vertical components N sin ` theta and N cos theta ` respectively .
for equilibrium `N cos theta theta = mg `
` and N sin theta = m r omega ^(2) =m R sin theta omega ^(2)`
` therefore N m R omega^(2)`
` therefore ` From (1) and (2)
` m R omega ^(2) cos theta = mg `
` therefore omega ^(2) = (g)/( R cos theta ) " " therefore = sqrt((g) /(R cos theta ))`