Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 4 / 11A (Dynamics and Relativity)

For the complete uniform disc, the surface density is $\sigma =M/(\pi a^2)$. An annular strip of radius $r$ and thickness $dr$ has mass $dm=2\pi \sigma rdr$, so its contribution to the axial moment of inertia is $r^{2}dm$. HenceThis derives the moment of inertia of a uniform disc rather than assuming it.

Removing a concentric disc of radius $a/2$ at the same density removes mass $M/4$ and inertia $\frac{1}{2}(M/4)(a/2{)}^2=M{a}^2/32$. The remaining annulus therefore has mass $m_A=3M/4$ andThe moment of inertia of a uniform annulus gives the same value directly.

Let $A_c$ be the downhill centre-of-mass acceleration and let $F$ denote the magnitude of the uphill static friction. Translation, rotation about the centre, and rolling without slipping giveThus $F=I_A{A}_c/a^2=(5/8)m_A{A}_c$. Eliminating $F$ gives the rolling acceleration with rotational inertiaSubstitution gives the contact forcesThe normal force balances the perpendicular component of gravity because the centre remains at a constant distance from the plane. Although the annulus accelerates downhill, friction must act uphill to supply the torque that increases its rolling angular speed. For a plane with $0\le \alpha <\pi /2$, the assumed no-slip motion requires a coefficient of static friction satisfying ${\mu }_s\ge F/N=(5/13)\tan \alpha$. With that condition, contact friction does no work because the instantaneous contact point is stationary; it redistributes the gravitational energy between translational and rotational kinetic energy.