= Centre of percussion
{title2=$b=I_O/(Mh)$}
For a body constrained to rotate about a fixed pivot, a perpendicular <impulse> $K$ delivered at distance $b$ from the pivot produces $\omega=Kb/I_O$. If the <centre of mass> is distance $h$ from the pivot, its <momentum> is $M\omega h$. The pivot <impulse> is $J_O=M\omega h-K$, and vanishes when $b=I_O/(Mh)$. This impact location is the <centre of percussion>. The result assumes the stated planar rigid-body geometry and a short <impulse>; it is not a claim that all <forces> at the pivot vanish afterward.
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