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. `omega = sqrt(rg sin theta)`
B. `omega = sqrt(g//r cos theta)`
C. `omega = sqrt((gr)/(cos theta))`
D. `omega = sqrt((gr)/(tan theta)`

Answer

B

Explanation

Let at an instant, the block kept in the bowl be at location `P` when bowl is rotating with angular velocity `omega` and block does not ship no bowl, Fig. Let `R` be the normal reaction of bowl on block. If is acting along `PO`. Ig `m` is the mass of the lcok, then in equilibrium position, the horizontal component of `R` will provide centripetal force and vertical component of `R` will balance weight of the block. Thus
`R sin theta = m omega^(2) (r sin theta) or R = m omega^(2) r` ..(i)
and `R cos theta = mg or m omega^(2) r(cos theta) = mg`
[from (i)]
or `omega = sqrt((g)/(r cos theta))`