= Elastic eccentric impact of a thin disc
{title2=$v_C/u=(I-Md^2)/(Md^2+I)$}
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, $I=Ma^2/4=Md^2$, so its centre instantaneously stops and all initial <kinetic energy> becomes rotation. A fixed support can absorb <momentum> without doing work.
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