Two discs of moments of inertia $I_1$ and $I_2$ about their respective axes (normal to the disc and passing through the centre) and rotating with angular speeds $\omega_1$ and $\omega_2$ are brought into contact face to face with their axes of rotation coincident.
$(a)$ Does the law of conservation of angular momentum apply to the situation? Why?
$(b)$ Find the angular speed of the two-disc system.
$(c)$ Calculate the loss in kinetic energy of the system in the process.
$(d)$ Account for this loss.

(N/A) Let the common angular velocity of the system be $\omega$.
$(a)$ Yes,the law of conservation of angular momentum applies because there is no net external torque acting on the system of the two discs.
$(b)$ By the law of conservation of angular momentum:
$L_f = L_i$
$(I_1 + I_2) \omega = I_1 \omega_1 + I_2 \omega_2$
$\omega = \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2}$
$(c)$ The initial kinetic energy is $K_i = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2$.
The final kinetic energy is $K_f = \frac{1}{2} (I_1 + I_2) \omega^2 = \frac{1}{2} (I_1 + I_2) \left( \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2} \right)^2 = \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{2(I_1 + I_2)}$.
The loss in kinetic energy is $\Delta K = K_i - K_f = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 - \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{2(I_1 + I_2)}$.
Simplifying this expression,we get $\Delta K = \frac{I_1 I_2}{2(I_1 + I_2)} (\omega_1 - \omega_2)^2$.
$(d)$ The loss in kinetic energy is due to the work done against the friction between the two discs as they adjust to a common angular velocity.