 Elastic eccentric impact of a thin disc

Let a nonrotating disc initially move downwards at speed $u$ and strike a stationary point support offset $d$ from its centre. If the impulse is vertical and the normal coefficient of restitution is one, its upward impulse satisfies $J(1/M+d^2/I)=2u$. The signed upward centre velocity is therefore the title formula, where $I$ is the moment of inertia about the horizontal diameter perpendicular to the impact lever arm. At $d=a/2$ for a thin circular disc, $I=M{a}^2/4=M{d}^2$, so its centre instantaneously stops and all initial kinetic energy becomes rotation. A fixed support can absorb momentum without doing work.