Average Frictional Torque Between Two Rotating Discs

Based on the context of the problem, consider disc A having a moment of inertia $I_A = I$ rotating with an initial angular velocity $\omega$, and disc B having a moment of inertia $I_B = 2I$ initially at rest ($\omega_B = 0$).

When disc B is brought into contact coaxially with disc A, friction acts between their surfaces until both discs reach a common angular velocity $\omega_c$.

1. Common Angular Velocity ($\omega_c$)

Since no external torque acts on the two-disc system about the axis of rotation, the total angular momentum of the system is conserved:
$L_i = L_f$

$I_A \omega + I_B (0) = (I_A + I_B) \omega_c$

Substituting $I_A = I$ and $I_B = 2I$:
$I\omega = (I + 2I)\omega_c = 3I\omega_c$

$\omega_c = \frac{\omega}{3}$

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2. Average Frictional Torque

The average torque $\tau_{\text{avg}}$ exerted by friction on either disc over the time interval $t$ is equal to the magnitude of the rate of change of its angular momentum:

$\tau_{\text{avg}} = \frac{|\Delta L|}{t}$

Considering disc B:
$\Delta L_B = I_B \omega_c - 0 = (2I)\left(\frac{\omega}{3}\right) = \frac{2I\omega}{3}$

Therefore, the average frictional torque is:
$\tau_{\text{avg}} = \frac{\Delta L_B}{t} = \frac{2I\omega}{3t}$

Similarly, considering disc A:
$|\Delta L_A| = |I_A \omega_c - I_A \omega| = \left|I\left(\frac{\omega}{3}\right) - I\omega\right| = \frac{2I\omega}{3}$

$\tau_{\text{avg}} = \frac{2I\omega}{3t}$

Thus, the correct option is A.