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

Two discs of same moment of inertia rotating about their regular axis

1.	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
Answer» `(1)/(4) I (omega_1 - omega_2)^(2)` `I (omega_(1) - omega_(2))^(2)` `(1)/(8) I (omega_(1) - omega_(2))^(2)` `(1)/(2) I (omega_(1)+ omega_(2))^(2)` Solution :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 omegaor 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)`