The easiest way to solve 2.19 (Plank Weight Trough)

A plank of weight W and length $R$ lies in a smooth circular trough of radius $R$. At one end of the plank is a weight $W/2$. Calculate the angle $\theta$ at which the plank lies when it is in equilibrium.

There are already a few solutions at the Feynman’s Lectures website, but my solution is simpler.

Since the plank is in equilibrium, it must be at its lowest possible position. This means the center of mass of the plank lies directly below the center of the trough.

We can calculate the center of mass $c$ given the length of the plank $L = R$:

${\displaystyle \begin{array}{rl}{\displaystyle c}& {\displaystyle =\frac{1}{1.5W}(\frac{WL}{2}+\frac{WL}{2})}\\ {\displaystyle }& {\displaystyle =\frac{L}{1.5}=\frac{\sqrt{3}R}{1.5}=\frac{2R}{\sqrt{3}}}\end{array}}$

Since $c$ forms the hypotenuse from the left point of contact to the vertical line beneath the trough center, we have:

${\displaystyle \cos \theta =\frac{R}{c}=\frac{R\sqrt{3}}{2R}=\frac{\sqrt{3}}{2}}$

This gives $= 30^{}$