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Question:

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

ω=grcosθ

ω=rgsinθ

ω=g/rcosθ

ω=grtanθ

Solution:

Correct option is B. ω=g/rcosθ
Under equilibrium conditions,
Ncosθ=mg. (1)
and Nsinθ=mω²rsinθ. (2)
where N=Normal reaction
R=Radius
ω=Angular speed
Substitute N=mgcosθ in equation (2)
mgcosθ[sinθ]=mω²rsinθ
mgcosθ=mω²r
ω²=gcosθ/r
ω=√(g/rcosθ)