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Question

A ring of mass 3kg is rolling without slipping with linear velocity 1 ms1 on a smooth horizontal surface. A rod of same mass is fitted along its one diameter. Find total kinetic energy of the system (in J).

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Solution

Moment of inertia of the rod about O, Io=12m(2R)2=13mR2
Moment of inertia of the ring about O, I′′o=mR2
Total moment of inertia of the system about O, IO=Io+I′′o=43mR2
Now total kinetic energy of the system, K.E=K.Etranslational+K.Erotational
K.E=12(2m)v2+12IOw2
K.E=mv2+12×43mR2×w2
Also v=Rw (rolling without slipping)
K.E=mv2+23mv2=53mv2
Given: v=1m/sm=3kg
K.E=53×3×12K.E=5J

447855_161940_ans_81248f99662c4bc983ccce33ed495930.png

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