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Question

A hollow sphere is projected horizontally along a rough surface with speed v and angular velocity ω0, find out the ratio (vω0), so that the sphere stops moving after some time.

A
R
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B
4R3
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C
2R3
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D
7R3
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Solution

The correct option is C 2R3
Let the given situation be as shown in the below figure,


Torque about lowest point of a sphere.
fk×R=Iα
or, μmg×R=23mR2α
α=3μg2R, angular accelaration in opposite direction of angular velocity.
Now we have,
ω=ω0αt (final angular velocity ω=0)
ω0=3μg2R×t

t=ω0×2R3μg

Also, acceleration a=μg
so, we have
vf=vat ( final velocity vf=0)
v=μg×t
t=vμg

Now, to stop the sphere, time at which v & ω are zero, should be same.
Thus,
vμg=2ω0R3μg vω0=2R3
Hence, (B) is correct answer.

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