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 :

Options

A. `(1)/(4) 1 (omega _(1) - omega_(2)) ^(2)`
B. `1 (omega _(1) - omega_(2)) ^(2)`
C. `(1)/(8) (omega _(1) - omega_(2)) ^(2)`
D. `(1)/(2) 1(omega _(1)- omega _(2)) ^(2)`

Explanation

External torque = zero
By sonservation of angular momentum
`I _(1) omega _(1) + I_(2) omega _(2) = (I _(1) + I _(1) ) omega `
`I _(1) = I _(2) = I `
`omega = ( omega _(1) + omega _(2))/(2)`
`KE _(1) = (1)/(2) I omega _(1) ^(2) + (1)/(2)I omega _(2) ^(2)`
`KE _(2) = (1)/(2) (21) omega ^(2)`
`KE _(2) = I [ (omega _(1) + omega _(2))/( 2) ] ^(2)`
`Delta KE = (1)/(2) 1 [ omega _(1) - omega _(2) ] ^(2)`