= Solution
Measure a signed angle $\theta$ from the upward radius $A\to O$ in the direction of falling. With $I_A=3ma^2/2$, gravitational <torque> gives
$$
\boxed{\ddot\theta=\frac{2g}{3a}\sin\theta.}
$$
There is an idealization in the release condition: \b[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> gives
$$
\frac12I_A\dot\theta^2=mga(1-\cos\theta),\qquad
\boxed{\dot\theta^2=\frac{4g}{3a}(1-\cos\theta).}
$$
On 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 is
$$
\boxed{a_O=-\frac{4g}{3}(1-\cos\theta)e_r+\frac{2g}{3}\sin\theta\,e_\theta.}
$$
Gravity has components $-mg\cos\theta\,e_r+mg\sin\theta\,e_\theta$. Applying <Newton's second law> to the whole disc gives
$$
\boxed{R=\frac{mg}{3}(7\cos\theta-4)e_r-\frac{mg}{3}\sin\theta\,e_\theta.}
$$
The 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.
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