Let $I_1 \text{ and } I_{}$2 be the moments of inertia of two discs have angular speed $\omega_1 \text{ and } \omega_2$. When they are brought in contact, the M.I. of the two discs system will be $I_1 + I_2$. Let $\omega$ = angular speed of the combined system. (a) ∴ total initial angular momentum of the two discs, $L_1 = I_1 \omega_1 + I_2 \omega_2$ Total final angular momentum of the combined system, $L_2 = (I_1 + I_2) \omega$ From the law of conservation of angular momentum, $L_2 = L_1 \text{ or } (I_1 + I_2) \omega = I_1 \omega_1 + I_2 \omega_2$ or $\omega = \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2} \ldots (i)$ (b) Initial K.E. of the two disc, $K_1 = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 \ldots (ii)$ Final K.E. of the combined system, $K_2 = \frac{1}{2} (I_1 + I_2) \omega^2 \ldots (iii)$ ∴ From (i) and (iii), we get $K_2 = \frac{1}{2} (I_1 + I_2) \left( \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2} \right)^2$ $K_2 = \frac{1}{2} \frac{ (I_1 \omega_1 + I_2 \omega_2)^2 }{ (I_1 + I_2)^2 } \ldots (iv)$ Eqn. (ii) – eqn. (iv) gives, $= \frac{I_1 I_2}{2 (I_1 + I_2)} ( \omega_1^2 + \omega_2^2 - 2 \omega_1 \omega_2 )$$= \frac{I_1 I_2}{2 (I_1 + I_2)} ( \omega_1 - \omega_2 )^2$which is a positive quantity i.e. $> 0$Hence, $K_1 - K_2 > 0 \text{ or } K_1 > K_2$or $K_2 < K_1$ i.e. rotational K.E. of the combined system is less than the sum of the initial energies of the two discs.Hence there occurs a loss of K.E. on combining the two discs and is the dissipation of energy because of the frictional forces between the faces of the two discs. These forces bring about a common angular speed of the two discs on combining. This however is an internal loss and angular momentum remains conserved.