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Question

A hemi-spherical bowl of radius R=0.1 m is rotating about its own axis (vertical) with uniform angular speed ω. A particle of mass 102 kg on the smooth inner surface of the bowl is also rotating with the same ω at a height h from the bottom of the bowl. What is the minimum value of ω to have a non-zero value of h, for which the particle remains at rest with respect to the bowl?
(Take g=10 m/s2)

A
20 rad/s
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B
10 rad/s
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C
5 rad/s
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D
1 rad/s
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Solution

The correct option is B 10 rad/s

Equation of dynamics along centre of circular path:
Fnet= m×a= m×rω2.......(1)
Since Fnet=Nsinθ
Nsinθ=mrω2..........(2)
For equilibrium in vertical direction,
Ncosθ=mg..........(3)
From geometry we get, r=Rsinθ...........(4)
Combining equation (2) and (4) gives:
N=mω2R
From equation (3) we get:
cosθ=mgmRω2=gRω2
From the geometry, cosθ=RhR
RhR=gRω2
h=Rgω2
For non-zero height,
h0Rgω2ωgR
ωmin=gR=100.1=10 rad/s

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