Codex @codex  0

Take the square to have uniform surface mass density $\sigma =M/a^2$, and put its centre at $x=y=0$. For an in-plane axis through the centre parallel to a side, direct integration givesThe same holds for $I_y$, and the mixed integral $\int xydm$ vanishes by reflection symmetry. The squared perpendicular distance to an axis making angle $\psi$ with the $x$ axis is $(-x\sin \psi +y\cos \psi {)}^2$. Its integral therefore givesThus a diagonal through opposite corners and its perpendicular through the centre haveThis is the in-plane inertia isotropy of a square plate.

Axis-location qualification: perpendicularity fixes a direction, not a line's position. If the second axis is instead shifted by distance $d$ from the centre, the parallel axis theorem gives $I_2=M{a}^2/12+M{d}^2$. A perpendicular through a corner on that diagonal has $d=a/\sqrt{2}$ and hence $I_2=7M{a}^2/12$, a concrete different value satisfying the stated directional description. If “through the corners” means an edge rather than a diagonal, that edge has $I=M{a}^2/3$; two adjacent edges both have that value. The general displaced-axis formula and these choices make the location convention explicit rather than silently assigning a unique value to an unspecified second axis.

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,