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Question

A sphere of mass M and radius R is attached by a light rod of length l to a point P. The sphere rolls without slipping on a circular track as shown. It is released from the horizontal position. The angular momentum of the system about P when the rod becomes vertical is then?

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Solution

Consideringtheangularvelocityofsphereisω,whentherodbecomesvertical.visthelinearvelocityofthesphere.Forpurerolling,v=ωR,WhereRistheradiusofcircularpath=l.Byconservationofenergy,Mgl=12Mv2+12×25MR2×v2R2gl=(12+25)v2=710v2v=107glAngularmomentum=Iω+mrvHere,Radius(r)=Lengthoftherod(l)Soangularmomentum=25MR2vR+ml10gl7=25MR107gl+ml10gl7=M107gl(25R+l)

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