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Question

A uniform ring of radius R is given a backspin of angular velocity Vo/2R and thrown on a horizontal rough surface with the velocity of the centre to be Vo. Find the velocity of the centre of the ring when it starts pure rolling.

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Solution

We know about point any point P on ground Angular momentum will be conserved. As torque due to friction blows off.
V=rw
Icm=Mv2vi=v0vf=vwi=v02Rwf=wLp=constantmv0rmr2×v02v=mvr+mr2wv0v02=2vv=v04
Final velocity is v04 in forward direction.


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