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