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A disc of moment of inertia I(a) is rota...

A disc of moment of inertia `I_(a)` is rotating in a horizontal plane about its symmetry axis with a constant angular speed ω. Another discinitially at rest of moment of inertia `I_(b)` is dropped coaxially on to the rotating disc. Then, both the discs rotate with same constant angular speed. The loss of kinetic energy due to friction in this process is,

A

`(1)/(2)(I_(b)^(2))/((I_(a)+I_(b)))omega^(2)`

B

`(I_(b)^(2))/((I_(a)+I_(b)))omega^(2)`

C

`((I_(b)-I_(a))^(2))/((I_(a)+I_(b)))omega^(2)`

D

`(1)/(2)(I_(b)I_(b))/((I_(a)+I_(b)))omega^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
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