Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular

Question

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:

Accepted Answer

Let the moment of inertia of each disc be $I$.
According to the law of conservation of angular momentum, the total initial angular momentum equals the total final angular momentum (since no external torque acts on the system):
$L_i = L_f$
$I\omega_1 + I\omega_2 = (I + I)\omega_{common}$
$\omega_{common} = \frac{\omega_1 + \omega_2}{2}$

The initial kinetic energy of the system is:
$K_i = \frac{1}{2}I\omega_1^2 + \frac{1}{2}I\omega_2^2$

The final kinetic energy of the system is:
$K_f = \frac{1}{2}(2I)\omega_{common}^2 = I \left(\frac{\omega_1 + \omega_2}{2}\right)^2 = \frac{1}{4}I(\omega_1 + \omega_2)^2$

The loss of kinetic energy ($\Delta K$) is:
$\Delta K = K_i - K_f = \left(\frac{1}{2}I\omega_1^2 + \frac{1}{2}I\omega_2^2\right) - \frac{1}{4}I(\omega_1 + \omega_2)^2$
$\Delta K = \frac{1}{4}I [2\omega_1^2 + 2\omega_2^2 - (\omega_1^2 + \omega_2^2 + 2\omega_1\omega_2)]$
$\Delta K = \frac{1}{4}I (\omega_1^2 + \omega_2^2 - 2\omega_1\omega_2)$
$\Delta K = \frac{1}{4}I(\omega_1 - \omega_2)^2$

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