Past exam of the physics 1 paper of the University of Oxford / 2019 / 9 / c / Solution

With the wedge fixed, let $r$ be the signed gap between the cylinder bottom and the surface, positive outwards. The centre height is $s\sin \alpha +(R+r)\cos \alpha$. Retaining all three coordinates before imposing constraints givesUse the holonomic constraints $f_1=r=0$ and $f_2=s+R\theta -s_0=0$. With the convention $d(\partial L/\partial \dot{q})/dt-\partial L/\partial q={\sum }_j{\lambda }_j\partial f_j/\partial q$, the Lagrange multiplier equations areThe constraint gradients identify ${\lambda }_1=N$, the normal reaction, and ${\lambda }_2=f$, the uphill static friction force. The latter supplies the rolling torque. Setting $\ddot{r}=0$ and $\ddot{\theta }=-\ddot{s}/R$ yieldsThe static friction condition $|f|\le {\mu }_{s}N$ therefore becomesfor $0\le \alpha <\pi /2$. Static friction is required to create rotation, but does not dissipate energy here because the contact point is instantaneously at rest on the fixed plane.