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Question

# A hemispherical bowl of radius R is rotating about its axis of symmetry which is kept vertical. A small ball 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 ball with the vertical is α. Find the angular speed ω at which the bowl should rotate for this to happen.

A
ω=gRcosα
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B
ω=gRsinα
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C
ω=gRsinα
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D
ω=gRcosα
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Solution

## The correct option is D ω=√gRcosαWe know for the maximum velocity, the block will try to move upwards as a result friction will be acting downwards. we have givenNCosθ=FSinθ+mg.NCosθ=μNSinθ+mg.N(Cosθ−μSinθ)=mg.⇒N=mg/(Cosθ−μSinθ)Second equation,NSinθ+μNCosθ=mω²rSinθN(Sinθ+μCosθ)=mω²rSinθSubstituting the above values of N in this, We will get, mg/(Cosθ−μSinθ)×(Sinθ+μCosθ)=mω²rSinθ⇒ω²=g(Sinθ+μCosθ)/rSinθ(Cosθ−μSinθ)∴ωmax.=√[g(Sinθ+μCosθ)/rSinθ(Cosθ−μSinθ)]Now, Similarly, ωmin. can be find. For easy calculations, Substitute the Plus sign instead of minus sign and vice versa in max. so that we can get the min. angular velocity. ∴ωmin.=√[g(Sinθ−μCosθ)/rSinθ(Cosθ+μSinθ)]Hence, the option D is the correct answer.

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