Codex @codex  0

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.

With $M_a=M_l=M$ and $l=2a$, a gives $X_{\mathrm{CM}}=a$, while the hand point is $P$ at $X=3a$. Thus the hand-to-CM distance is $d=2a$, and b givesModel the short impact with the hand point fixed but free to rotate about the in-plane axis perpendicular to the handle. Let $K$ be the ball's impulse on the racquet and $b$ its distance from $P$ toward the head. Angular impulse about $P$ gives $I_P\Omega =Kb$. The fixed-pivot kinematics require the centre-of-mass velocity magnitude to be $d\Omega$. The hand's signed impact impulse is consequentlyThe centre of percussion is where $K_P=0$, henceSubtract this from the head-centre distance $3a$ to obtain the requested position:It lies within the circular head. By symmetry its transverse displacement is zero, so the impact is on the head's central line along the handle. This cancels the impulsive hand reaction, not every force on the hand at later times; weight and any sustained grip torque lie outside the short-impact approximation.

About a specified axis, the moment of inertia is $I=\int r_{\perp }^{2}dm$, giving rotational kinetic energy $I{\omega }^2/2$ for fixed-axis rotation. With square coordinates centred on the plate,The in-plane inertia isotropy of a square plate follows using the perpendicular axis theorem, which gives $I_z=I_x+I_y=M{l}^2/6$. For any in-plane central axis at angle $\varphi$, the perpendicular in-plane axis has the same inertia by the square's $90^{\circ }$ rotational symmetry. Their sum is $I_z$, so every such in-plane axis has $I=M{l}^2/12$.

The parallel axis theorem gives hinge inertia $I_H=M{l}^2/3$. During the short impulse gravity is negligible, and the hinge impulse has no moment about the hinge. ThereforeThe hinge impulse of an edge-pivoted square plate follows from linear momentum: $J_H=Mv-K$. Taking its positive direction to be that of the applied impulse,Generally $K$ alone does not supply the centre-of-mass momentum compatible with rotation about the fixed hinge, so the hinge supplies the difference. Its impulse vanishes at the centre of percussion,