Consider a disc of mass 5 kg and radius 2 m, rotating with an angular velocity of 10 rad/s about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together with…

Question

Consider a disc of mass 5 kg and radius 2 m, rotating with an angular velocity of 10 rad/s about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is J.

Accepted Answer

250. Moment of inertia of one disc: I=1/2MR²=1/2(5)(4)=10 kg m². The second disc is placed gently, so no external torque acts and angular momentum is conserved while the two faces rub to a common speed. Iω=(2I)ω′, so ω′=ω/2=5 rad/s. Initial kinetic energy: 1/2(10)(10)²=500 J. Final kinetic energy: 1/2(20)(5)²=250 J. Energy dissipated =500-250=250 J. Exactly half is lost. That is always the case when a body is dropped onto an identical one spinning freely, whatever the numbers.

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