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Question

A hemi spherical bowl of radius R=0.1 m is rotating about its own axis (which is vertical) with an angular velocity ω . A particle of mass 102kg is also rotating with same ω on the friction-less inner surface of the bowl. The particle is at a height h from the bottom of the bowl.



The relation between h and ω is

A
h=ω2g
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B
h=R2
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C
h=Rgω2
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D
None
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Solution

The correct option is C h=Rgω2
F.B.D of particle

Applying the equation of dynamics along the centre of circular path, for which radius is,
r=Rsinθ
N sin θ=mω2r=mω2R sin θ
On comparing we get,
N=mω2R

Applying equilbrium condition in vertical direction( particle is performing circular motion in a horizontal plane),
N cos θ=mg
cos θ=mgmRω2=gRω2
From the triangle,
cosθ=RhR
RhR=gRω2
h=Rgω2

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