= Solution
Take the stated negligible-natural-radius limit as $r_0=0$, with both initial velocities zero at $r=r_1$. Then $J=qBr_1^2/2$. Write $\omega_c=qB/m$ and $\Omega^2=\omega_s^2+\omega_c^2/4$, where $\omega_s^2=4\pi^2k/m$. At two radial turning points $x=r^2$, equality of the effective radial energies gives
$$
x_{\max}x_{\min}=\frac{J^2}{m^2\Omega^2}.
$$
Using $x_{\max}=r_1^2$ gives, for $qB\ne0$ and $k>0$,
$$
\boxed{r_{\min}=r_1\frac{|\omega_c|}{2\Omega}
=\frac{|qB|\,r_1}{\sqrt{q^2B^2+16\pi^2km}}.}
$$
The minimum is a regular radial turning point, so
$$
\boxed{\dot r_{\min}=0,\qquad
\dot\phi_{\min}=\frac J{mr_{\min}^2}-\frac{\omega_c}{2}
=\frac{2\omega_s^2}{\omega_c}=\frac{8\pi^2k}{qB}.}
$$
The <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 $qB\to0$ 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.
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