 Solution

Take the stated negligible-natural-radius limit as $r_0=0$, with both initial velocities zero at $r=r_1$. Then $J=qB{r}_1^2/2$. Write ${\omega }_c=qB/m$ and ${\Omega }^2={\omega }_s^2+{\omega }_c^2/4$, where ${\omega }_s^2=4{\pi }^{2}k/m$. At two radial turning points $x=r^2$, equality of the effective radial energies givesUsing $x_{\max }=r_1^2$ gives, for $qB\ne 0$ and $k>0$,The minimum is a regular radial turning point, soThe angular velocity has the sign of $qB$. On the initial inward motion, conservation of canonical angular momentum produces the rotation needed to prevent collapse to zero radius.

The strictly zero-field case is singular for this angular formula: $J=0$ and the ring instead collapses to $r=0$ with inward radial speed ${\omega }_s{r}_1$ in the ideal radial equation, where an angular coordinate ceases to describe a ring. The limit $qB\to 0$ through nonzero fields gives a tiny positive turning radius with large angular velocity, and must not be substituted into the formula at exactly zero field. A small nonzero $r_0$ also matters if the predicted minimum becomes comparable with it.