[figure] Consider a Disc of mass 5 kg, radius 2 m, rotating with 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.
Answer
250
Solution
I= MR22= 5 222=10 kg m2
No external torque: I 0=2I = 102=5 rad/s
Ki= 12I 02= 12 10 102=500 J
Kf= 12(2I) 2= 12 20 52=250 J
Energy dissipated =500-250=250 J
Answer: 250