Past exam of the physics 1 paper of the University of Oxford / 2014 resit / 1

The parallel axis theorem relates the moment of inertia about an axis through the centre of mass to that about a parallel axis displaced by perpendicular distance $d$:The reference axis must pass through the centre of mass. To see why, write the displacement from it as $\mathbf{r}=\mathbf{R}+\mathbf{d}$ in the plane perpendicular to the axes. Expanding $\int |\mathbf{R}+\mathbf{d}{|}^{2}dm$ gives $I_G+M{d}^2$ because $\int \mathbf{R}dm=0$.

For a thin planar mass distribution, let $x,y$ be perpendicular axes in its plane and $z$ the normal axis through their same intersection. The perpendicular axis theorem statesIndeed, $I_z=\int (x^2+y^2)dm$, while $I_x=\int y^{2}dm$ and $I_y=\int x^{2}dm$. Their common intersection need not be the centre of mass, but an arbitrary three-dimensional body does not satisfy the planar equality.

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.

Use two central perpendicular axes in the plate. The perpendicular axis theorem and the preceding moments of inertia giveAlternatively, axes along adjacent edges have moments of inertia $M{a}^2/3$ each. The normal axis through their shared corner therefore has $2M{a}^2/3$. It lies a distance $a/\sqrt{2}$ from the parallel normal axis through the centre, so the parallel axis theorem yields $I_{z,G}=2M{a}^2/3-M(a^2/2)=M{a}^2/6$. This normal central result is independent of the in-plane axis-location ambiguity.