-
A circular disk of moment of inertia It is rotating in a horizontal plane, its symmetry axis, with a constant angular speed ωi. 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 ωƒ. The energy lost by the initially rotating disk to friction is:
-
-
1 Ib² ωi² 2 (It + Ib) -
It² ωi² (It + Ib) -
Ib - It ωi² (It + Ib) -
1 IbIt ωi² 2 (It + Ib)
-
Correct Option: D
By conservation of angular momentum,
Itωi =(It + Ib) ωƒ
where ωƒ is the final angular velocity of disks
| ∴ ωƒ = |  |  | ωi | |
| It + Ib |
Loss in ΔK.E., = Initial K.E. – Final K.E.
| = | Itωi² - | (It + Ib)ωƒ² | ||
| 2 | 2 |
| ⇒ Δ K = | Itωi² - | (It + Ib) | ωi² | |||
| 2 | 2 | (It + Ib)² |
| = | ωi² | (It + Ib - It) = | ωi² | ||||
| 2 | It + Ib | 2 | It + Ib |
