 Solution

Measure a signed angle $\theta$ from the upward radius $A\to O$ in the direction of falling. With $I_A=3m{a}^2/2$, gravitational torque givesThere is an idealization in the release condition: exactly upright and exactly at rest is an unstable equilibrium, so the exact ideal solution stays there. Uniqueness of this smooth equation with $\theta (0)=\dot{\theta }(0)=0$ prevents spontaneous departure. The usual falling calculation means an infinitesimal disturbance, with its energy tending to that of the upright rest state. For that limiting separatrix motion, conservation of mechanical energy givesOn the branch with increasing angle, $\dot{\theta }=\sqrt{8g/(3a)}\sin (\theta /2)$ for $0<\theta <2\pi$; the opposite branch has the opposite angular velocity. The nontrivial limiting motion approaches the upright position only as $t\to -\infty$, because $dt/d\theta$ has a logarithmically divergent integral near zero.

Let $e_r$ point from $A$ toward $O$, and $e_{\theta }$ in the direction of increasing angle, so $e_r=\sin \theta e_x+\cos \theta e_z$ and $e_{\theta }=\cos \theta e_x-\sin \theta e_z$. The centre of mass acceleration on the limiting falling branch isGravity has components $-mg\cos \theta e_r+mg\sin \theta e_{\theta }$. Applying Newton's second law to the whole disc givesThe signed component parallel to the radius, positive from $A$ toward $O$, is the required $mg(7\cos \theta -4)/3$. This same radial formula is consistent at the exact stationary upright state, where $\theta =0$, $a_O=0$, and $R=mg{e}_z$; nonzero falling angles require the limiting-release interpretation just stated.