Two discs of same moment of inertia rotating about 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 face coinciding the axis of rotation . The expression for loss of energy during this process is

Options

A. `(1)/(4) I (omega_1 - omega_2)^(2)`
B. `I (omega_(1) - omega_(2))^(2)`
C. `(1)/(8) I (omega_(1) - omega_(2))^(2)`
D. `(1)/(2) I (omega_(1) + omega_(2))^(2)`

Answer

A

Explanation

Initial angular momentum = `I omega_(1) + I omega_(2)`
Let `omega` be angular speed of the combined system .
Final angular momentum = `2 I omega`
Final angular momentum = `2 I omega`
`therefore` According to conservation of angular momentum `I omega_(1) + I omega_(2) = 2 I omega or omega = (omega_(1) + omega_(2))/(2)`
Initial rotational kinetic energy , `E_(1) = (1)/(2) I (omega_(1)^(2) + omega_(2)^(2))`
Final rotational kinetic energy
`E_(f) = (1)/(2) (2 I) omega^(2) = (1)/(2) (2I) ((omega_(1) + omega_(2))/(2))^(2) = (1)/(4) I (omega_(1) + omega_(2))^(2)`
`therefore` Loss of energy `Delta E= E_(i) - E_(f)`
`= (1)/(2) (omega_(1)^(2) + omega_(2)^(2))- (1)/(4) (omega_(1)^(2) + omega_(2) ^(2) + 2omega_(1) omega_(2))`
`= (1)/(4) [ omega_(1)^(2) + omega_(2) ^(2) - 2 omega_(1) omega_(2)] = (1)/(4) (omega_(1) - omega_(2))^(2)`