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Question

# A hemisphere ball 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 surface of the bowl is smooth is smooth and the angle made by the radius through the block with the vertical is θ, find the angular speed at which the bowl is rotating.

A
ω=gcosθR
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B
ω=gsecθR
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C
ω=2gcosθR
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D
ω=2gRsecθ
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Solution

## The correct option is B ω=√gsecθRAssume the bowl is rotating with an angular velocity ω. Radius of the circular motion =Rsinθ From verticle equilibrium: Ncosθ=mg ... (1) By horizontal equilibrium Nsinθ=mω2Rsinθ N=mω2R ... (2) From eqn. (1) and (2) mgcosθ=mω2R ω=√gsecθR

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