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Two discs of same moment of inertia rota...

Two discs of same moment of inertia rotating their regular axis passing through centre and perpendicular to the plane of disc with angular velocities `omega_(1)` and `omega_(2)`. They are brought into contact face to the face coinciding the axis of rotation. The expression for loss of enregy during this process is :

A

`(1)/(4) I(omega_1 - omega_2)^2`

B

`I(omega_1 - omega_2)^2`

C

`(1)/(9) I(omega_1 - omega_2)^2`

D

`(1)/(2) I(omega_1 + omega_2)^2`

Text Solution

Verified by Experts

The correct Answer is:
A

(a) COAM, `I omega_1 = I omega_2 = 2 I omega rArr omega = (omega_1 + omega_2)/(2)`
`(K.E.)_i = (1)/(2) I omega_1^2 + (1)/(2) I omega_2^2`
`(K.E.)_f = (1)/(2) xx 2 I omega^2 = I((omega_1 + omega_2)/(2))^2`
Loss in `K.E. =(K.E)_i - (K.E.)_f = (1)/(4) (omega_1 - omega_2)^2`.
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