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Question

A hoop of radius r and mass m rotating with an angular velocity ω0 is placed on a rough horizontal surface. The initial velocity of the center of the hoop is zero. What will be the velocity of the center of the hoop when it ceases to slip?

A
rω02
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B
rω0
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C
rω04
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D
rω03
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Solution

The correct option is A rω02
Let v be the velocity of the center of the hoop, when it ceases to slip.
By applying the principle of conservation of angular momentum, we get
(mr2)ω0=mv×r+mr2ω=mv×r+mr2(vr)=2mvr
v=(mr2)ω02mr=ω0r2

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