= Solution
Use the angular coordinate $\theta$ for the position on the ring. The <kinetic energy> is $T=ma^2\dot\theta^2/2$. The <gravitational potential energy> is constant because the ring is horizontal, and the ideal constraint force has no tangential component. Thus the <Lagrangian> has no dependence on $\theta$ itself: it is a <cyclic coordinate>.
The <Euler-Lagrange equation> gives
$$
\frac d{dt}(ma^2\dot\theta)=0,\qquad\boxed{\dot\theta=\text{constant}.}
$$
Equivalently, the conserved <conjugate momentum> $p_\theta=ma^2\dot\theta$ is the vertical <angular momentum>, as required by <rotational symmetry>.
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