Half Ring Reaction Force

This was fun!

Consider a point on the ring parameterised in polar coordinates with the angular coordinate as $t$ , and the particle's location to be defined as:

$\vec{r}_p = (R\cos{t},R\sin{t},0)$ $\vec{r}_c = (2,0,0)$

The gravitational force using Newton's law acting on an elementary arc length of the ring is:

$d\vec{F} = G\left(\frac{M \ dt}{\pi}\right) m \left(\frac{\vec{r}_c - \vec{r}_p}{\lvert \vec{r}_c - \vec{r}_p \rvert ^3}\right)$

The torque due to this force about the origin is:

$d\vec{T} = \vec{r}_p \times d\vec{F}$

We are only interested in the Y component of the force and the torque is essentially only about the Z axis. The values of the force along Y and the torque about the Z axis is found after plugging in expressions, simplifying and integrating from 0 to $\pi$ :

$F_y = -\int_{0}^{\pi} \dfrac{\sin\left(t\right)}{{\pi}\left(5-4\cos\left(t\right)\right)^\frac{3}{2}} \ dt =-\frac{1}{3 \pi}$ $T_z = -2\int_{0}^{\pi} \dfrac{\sin\left(t\right)}{{\pi}\left(5-4\cos\left(t\right)\right)^\frac{3}{2}} \ dt=-\frac{2}{3 \pi}$

The above integrals can be easily solved by substitution as the numerator is the derivative of a cosine. These steps are left out. Now, applying a vertical force and torque balance about the origin, one gets the following. The vertical reactions are assumed to be along the positive Y direction.

$F_L + F_R + F_y = 0 \implies F_L + F_R = \frac{1}{3 \pi}$ $F_LR = F_RR + T_z \implies F_R - F_L = \frac{2}{3 \pi}$

Adding the above equations followed by halving both sides results in $\boxed{F_R = \frac{1}{2\pi}}$