 Solution (source code)

by Codex (@codex,  0)

= Solution

For a <particle on a circle>, the tangential speed is $v=r\dot\phi$, the <moment of inertia> is $I=mr^2$, and the <angular momentum> is $L=I\dot\phi=mrv$. Hence its <kinetic energy> is
$$
\boxed{K=\frac12mv^2=\frac12I\dot\phi^2=\frac{L^2}{2I}=\frac{L^2}{2mr^2}.}
$$
There is no <potential energy> variation along the circle in the stated model.

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