Two discs of same moment of inertia ( I ) are rotating in same...

The initial angular momentum of two discs is ${I}_{1}{\omega}_{1} + {I}_{2}{\omega}_{2} = I\left({\omega}_{1} + {\omega}_{2}\right)$.

The final angular momentum of system is \left({I}_{1} + {I}_{2}\right)\omega = 2I\omega$</p><p>From the conservation of angular momentum,$\omega = \dfrac{{\omega}_{1} + {\omega}_{2}}{2}$.</p><p>Initial kinetic energy,$K{E}_{i} = \dfrac{1}{2}$I\left({\omega}_{1}^{2} + {\omega}_{2}^{2}\right)$

and final kinetic energy, $K{E}_{f} = \dfrac{1}{2}\left(2I\right)\dfrac{{\left({\omega}_{1} + {\omega}_{2}\right)}^{2}}{4}$

Loss in kinetic energy is $\DeltaKE = \dfrac{1}{2}\dfrac{{I}_{1}{I}_{2}}{{I}_{1} + {I}_{2}}{\left({\omega}_{1} - {\omega}_{2}\right)}^{2}$

$= \dfrac{1}{2}\dfrac{{I}^{2}}{\left(2I\right)}{\left({\omega}_{1} - {\omega}_{2}\right)}^{2}$

$= \dfrac{1}{4}I{\left({\omega}_{1} - {\omega}_{2}\right)}^{2}$