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Question:

Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities ω1 and ω2. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is:

18(ω1−ω2)2

12I(ω1+ω2)2

14I(ω1−ω2)2

I(ω1−ω2)2

Solution:

Let the angular velocity of the combination be ω
conservation of angular momentum:
Iω1+Iω2=(I+I)ω → 12(ω1+ω2)
Initial kinetic energy
ki=12Iω12+12Iω22
Final kinectic energy
kf=12(2I)ω2=I4(ω1+ω2)2
∴Loss in energy Δk=ki−kf=12I(ω12+ω22)−14(ω1+ω2)2=I4[ω12+ω22−(ω12+ω22+2ω1ω2)]=I4(ω1−ω2)2