 Solution

Choose the smaller disc's initial sense as positive. For either uniform disk, the axial moment of inertia is $I=\frac{1}{2}m{r}^2$, so $I_1=I_2/2$. The signed initial angular velocities are ${\Omega }_1=2{\omega }_2$ and ${\Omega }_2=-{\omega }_2$, giving total axial angular momentum zero. The frictionless axle supplies no axial torque, and the pins' negligible mass supplies no extra stored kinetic energy. ThusThese equations allow the original state or its simultaneous reversal. A nontrivial elastic pin impact selects reversal of the relative contact velocity, rather than continued interpenetration. ThereforeThe requested magnitudes are unchanged: $|{\Omega }_1^{\prime }|={\omega }_1$ and $|{\Omega }_2^{\prime }|={\omega }_2$. The intersecting pin paths make the impact possible; no additional lever arm is needed once axial angular momentum and energy conservation have been imposed.