A circular disk of moment of inertia It is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed i. Another disk of moment of inertia Ib is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed . The energy lost by the initially rotating disc to friction is
- 12 Ib2(It + Ib) i2
- 12 It2(It + Ib) i2
- Ib - It(It + Ib) i2
- 12 Ib It(It + Ib) i2
Answer
(4) 12 Ib It(It + Ib) i2
Solution
No external torque acts, so angular momentum is conserved:
It i = (It + Ib)
= It iIt + Ib
Energy lost:
E = 12It i2 - 12(It + Ib) 2
E = 12It i2 - 12 It2 i2It + Ib
E = 12 It IbIt + Ib i2
Answer: (4)