Centre of percussion ($b=I_O/(Mh)$)

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.

For a uniform square plate of side $l$ hinged along an edge, $h=l/2$ and $I_H=M{l}^2/3$. An impulse $K$ normal to the plate, applied a distance $a$ below the centre, gives $v_G=3K(1+2a/l)/(4M)$ and signed hinge impulse $J_H=K(3a/(2l)-1/4)$. Therefore the hinge impulse vanishes at $a=l/6$ below the centre, a distance $2l/3$ from the hinge. The sign of $J_H$ identifies the reaction direction; its magnitude is $|J_H|$.