Past exam of the physics 1 paper of the University of Oxford / 2019 resit / 4 / Solution

Use Cartesian coordinates along the edges from the pivot. With surface mass density $M/L^2$, the moment of inertia isEquivalently, the moment of inertia of a uniform square plate about its central perpendicular axis is $M{L}^2/6$, and the parallel axis theorem adds $M(L/\sqrt{2}{)}^2$.

The source figure measures $\theta$ from the upward vertical to $OC$, with $OC=L/\sqrt{2}$. The centre's height above the pivot is $(L/\sqrt{2})\cos \theta$. The smooth fixed pivot does no work. Conservation of energy, with the initially upright configuration as reference, givesFor rotation in the positive $\theta$ direction on the first descent, take the positive root. Its equivalent form is $\dot{\theta }=\sqrt{3\sqrt{2}g/L}\sin (\theta /2)$ for $0<\theta <\pi$.

Strictly, release from rest at a small angle ${\theta }_0$ replaces $1-\cos \theta$ by $\cos {\theta }_0-\cos \theta$. The displayed result is the ${\theta }_0\to 0$ limiting energy relation. Exactly upright and exactly at rest is an unstable equilibrium; without a perturbation it stays there. This distinction prevents the infinitesimal displacement convention from being mistaken for spontaneous motion.