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Question: Answered & Verified by Expert
A disc of the moment of inertia I1 is rotating in a horizontal plane about an axis passing through a centre and perpendicular to its plane with constant angular speed ω1 Another disc of the moment of inertia I2 having zero angular speed is placed coaxially on a rotating disc. Now both the disc are rotating with the constant angular speed ω2. The energy lost by the initial rotating disc is
PhysicsRotational MotionMHT CETMHT CET 2017
Options:
  • A 12I1+I2I1I2ω12
  • B 12I1I2I1-i2ω12
  • C 12I1-I2I1I2ω12
  • D 12I1I2I1+I2ω12
Solution:
1107 Upvotes Verified Answer
The correct answer is: 12I1I2I1+I2ω12
I1ω1=I1+I2ω2
ω2ω1=I1I1+I2
E1-E2
=12I1ω12-12(I1+I2)ω22
=12ω12I1-I1+I2ω22ω12 
=12ω12I1-I1+I2I12I1+I22 
=12ω12I12+I1I2-I12I1+I2=12I1I2I1+I2ω12  

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