Loss of Kinetic Energy During Rotational Coupling of Discs

Based on the preceding problem context (Question 15), a rotating disc $A$ with moment of inertia $I_A = I$ rotating with an initial angular velocity $\omega_A = \omega$ is brought into contact coaxially with a stationary disc $B$ having moment of inertia $I_B = 2I$ (initial angular velocity $\omega_B = 0$).

Due to friction between the surfaces, the two discs eventually slip to a stop relative to each other and rotate together with a common angular velocity $\omega_c$.

1. Conservation of Angular Momentum:
   Since there is no external torque acting on the two-disc system along the axis of rotation, the total angular momentum is conserved:
   $L_i = L_f$
   $I_A \omega_A + I_B \omega_B = (I_A + I_B) \omega_c$
   $I \omega + 0 = (I + 2I) \omega_c = 3I \omega_c$
   $\omega_c = \frac{\omega}{3}$

(Note: This is verified by Question 15, where the change in angular momentum of disc $B$ is $\Delta L_B = I_B \omega_c = (2I)\left(\frac{\omega}{3}\right) = \frac{2}{3}I\omega$, giving an average torque $\tau = \frac{\Delta L}{t} = \frac{2I\omega}{3t}$.)

1. Calculation of Initial and Final Kinetic Energy:
   The initial kinetic energy of the system is:
   $K_i = \frac{1}{2} I_A \omega^2 = \frac{1}{2} I \omega^2$

The final kinetic energy of the combined system is:
$K_f = \frac{1}{2} (I_A + I_B) \omega_c^2 = \frac{1}{2} (3I) \left(\frac{\omega}{3}\right)^2 = \frac{1}{2} \cdot 3I \cdot \frac{\omega^2}{9} = \frac{1}{6} I \omega^2$

1. Loss of Kinetic Energy:
   The loss of kinetic energy during the coupling process is:
   $\Delta K = K_i - K_f = \frac{1}{2} I \omega^2 - \frac{1}{6} I \omega^2 = \left(\frac{3 - 1}{6}\right) I \omega^2 = \frac{2}{6} I \omega^2 = \frac{I\omega^2}{3}$

Alternatively, using the standard formula for loss of kinetic energy in a completely inelastic rotational collision:
$\Delta K = \frac{1}{2} \frac{I_A I_B}{I_A + I_B} (\omega_A - \omega_B)^2 = \frac{1}{2} \frac{I \cdot 2I}{I + 2I} (\omega - 0)^2 = \frac{1}{2} \cdot \frac{2I^2}{3I} \cdot \omega^2 = \frac{I\omega^2}{3}$

Therefore, the loss of kinetic energy during the process is $\frac{I\omega^2}{3}$, corresponding to Option (B).