Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics IV resit / 6

By symmetry, the centre of mass is $G=(0,0,\ell /2)$. Choose horizontal axes through $G$ and the vertical symmetry axis as the third axis, writing $Z=z-\ell /2$. With uniform mass density $\rho$, the inertia tensor has entriesReflection in each coordinate plane makes all the mixed integrals $\int xydm$, $\int xZdm$, and $\int yZdm$ vanish. Rotational symmetry about the vertical axis gives $I_{11}=I_{22}=A$. Thus these are principal axes andThe axial moment of inertia of a hollow cylinder isIts mass is $M=\pi \rho \ell (b^2-a^2)$. Dividing and factoring the difference of squares gives

Choose positive $x$ downhill and positive $\theta$ in the corresponding rolling direction, with $x=\theta =0$ initially. For rolling without slipping, the contact point has zero velocity, so $\dot{x}-b\dot{\theta }=0$. Integrating the rolling constraint givesLet $F$ be the uphill static friction. Along the inclined plane, Newton's second law for the centre of mass givesThe weight passes through the centre of mass, and the normal reaction has its line of action through it. Thus only static friction supplies torque about $G$. The angular velocity is along the cylinder's symmetry axis, so the inertia tensor gives angular momentum $C\dot{\theta }$ in that direction. The torque equation is thereforeEliminating static friction and inserting $C$ givesThe extra term represents the rotational inertia that must be accelerated along with the translational mass.

For a uniform solid cylinder, set $a=0$ in the moment of inertia formula, givingFor the hollow cylinder, $a>0$ makes the denominator larger:on a downward incline with $\sin \phi >0$. Thus the solid cylinder accelerates faster, assuming rolling without slipping for both. More of the hollow cylinder's mass lies far from the rotation axis, increasing its moment of inertia relative to $M{b}^2$.