Let the moments of inertia of the two discs be $I_1$ and $I_2$. For a uniform disc, $I = \frac{1}{2}MR^2$. $I_1 = \frac{1}{2} M_1 R_1^2 = \frac{1}{2} \times 2 \times (0.2)^2 = 0.04\text{ kg m}^2$ $I_2 = \frac{1}{2} M_2 R_2^2 = \frac{1}{2} \times 4 \times (0.1)^2 = 0.02\text{ kg m}^2$ Initial angular momentum of the system, $L_i = I_1\omega_1 + I_2\omega_2$ $L_i = (0.04 \times 50) + (0.02 \times 200) = 2 + 4 = 6\text{ kg m}^2\text{ s}^{-1}$ When the two discs are brought in contact, they will rotate together with a common final angular velocity $\omega$ due to internal friction between them. There is no net external torque on the system along the axis of rotation. By conservation of angular momentum: $L_i = L_f$ $6 = (I_1 + I_2)\omega$ $6 = (0.04 + 0.02)\omega$ $6 = 0.06\omega$ $\omega = \frac{6}{0.06} = 100\text{ rad s}^{-1}$