 Solution

For an elastic eccentric impact of a thin disc, take upward velocity as positive and let $J$ be the short upward impulse. The disc's moment of inertia about a horizontal diameter is $I=M{a}^2/4$. With impact offset $d=a/2$,The contact point has upward velocity $v_C+d\omega$. Unit restitution against the fixed post requires that velocity to be $+u$, soSince $I=M{d}^2$, $J=Mu$ and the centre's instantaneous speed is zero. The disc rotates with $\omega =2u/a$; its rotational kinetic energy $I{\omega }^2/2=M{u}^2/2$ equals the entire initial energy. Gravity's impulse during the impact is neglected. Using the polar-axis moment of inertia would describe the wrong rotation axis.