Past exam of the physics 1 paper of the University of Oxford / 2017 / 6

The parallel axis theorem relates the moment of inertia $I$ about an axis to $I_{\mathrm{CM}}$ about a parallel axis through the centre of mass, separated by perpendicular distance $d$:In the defining mass integral, expanding the squared distance produces a cross term whose first mass moment about the centre of mass is zero.

For a thin planar rigid body, the perpendicular axis theorem applies to two perpendicular axes $x,y$ in its plane and the perpendicular axis $z$, all meeting at the same point:Indeed, $I_x=\int y^{2}dm$, $I_y=\int x^{2}dm$, and $I_z=\int (x^2+y^2)dm$. The planar condition is essential; this is not an identity for an arbitrary three-dimensional rigid body.

Use the usual dumbbell geometry: the uniform thin rod joins the disc centres and is perpendicular to both disc faces. The required axis through the rod midpoint lies in a direction parallel to a diameter of each disc. For either disc, rotational symmetry and the perpendicular axis theorem giveEach disc centre lies a distance $l/2$ from the requested axis. The parallel axis theorem therefore gives a disc contribution $M{a}^2/4+M(l/2{)}^2$. The rod contributesAdd the two disc contributions and the rod's moment of inertia:Using $M{a}^2/2$ for each disc's central contribution would instead use an axis perpendicular to its face, which is parallel to the rod and is not the requested axis.