1. Initial angular momenta and kinetic energies

Disc $A$ has moment of inertia $I$ and initial angular speed $2\omega$. So, $L_A = I(2\omega)=2I\omega$ $K_A = \frac12 I(2\omega)^2 = 2I\omega^2$

Disc $B$ has moment of inertia $2I$ and angular speed $\omega$. So, $L_B = (2I)(\omega)=2I\omega$ $K_B = \frac12 (2I)(\omega)^2 = I\omega^2$

Since both rotate in the same (clockwise) direction, total initial angular momentum is $L_i = 2I\omega + 2I\omega = 4I\omega$

Total initial kinetic energy is $K_i = 2I\omega^2 + I\omega^2 = 3I\omega^2$

2. Final common angular velocity after interaction

Since the discs are mounted coaxially, they will eventually rotate together with a common angular velocity $\Omega$.

Angular momentum is conserved: $L_i = (I+2I)\Omega$ $4I\omega = 3I\Omega$ $\Omega = \frac{4\omega}{3}$

3. Final kinetic energy

Total moment of inertia of the combined system: $I_{\text{tot}} = I+2I=3I$

Thus, $K_f = \frac12 (3I)\left(\frac{4\omega}{3}\right)^2$ $K_f = \frac12 (3I)\cdot \frac{16\omega^2}{9}$ $K_f = \frac{8I\omega^2}{3}$

4. Loss of kinetic energy

$\Delta K = K_i - K_f$ $\Delta K = 3I\omega^2 - \frac{8I\omega^2}{3}$ $\Delta K = \frac{9I\omega^2 - 8I\omega^2}{3}$ $\Delta K = \frac{I\omega^2}{3}$

5. Check with options

The loss of kinetic energy is $\boxed{\frac{I\omega^2}{3}}$ So the correct option is B.

6. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

They match.