= Solution
The radial and angular <Euler-Lagrange equations> are
$$
(M+m)\ddot r=mr\dot\theta^2-Mg,\qquad
\frac d{dt}(mr^2\dot\theta)=0.
$$
Writing the conserved <angular momentum> as
$$
\boxed{J=mr^2\dot\theta,}
$$
we obtain
$$
\boxed{(M+m)\ddot r=\frac{J^2}{mr^3}-Mg.}
$$
A positive constant radius requires zero radial <acceleration>. Hence the <circular motion> has
$$
\boxed{r_0=\left(\frac{J^2}{mMg}\right)^{1/3},}
$$
provided $J\ne0$ and $0<r_0<l$. Its angular <velocity> is $\dot\theta_0=J/(mr_0^2)$. No positive circular radius exists for $J=0$ with $M,g>0$.
For later use, the conserved radial <energy> is
$$
\mathcal H=\frac12(M+m)\dot r^2+U_J(r),\qquad
\boxed{U_J(r)=\frac{J^2}{2mr^2}+Mgr-Mgl.}
$$
The <effective potential> has $U_J''(r)=3J^2/(mr^4)>0$, so its stationary radius is a stable minimum. The <tension> remains positive in the ideal allowed region: the hanging <Newton's second law> gives $T_s=M(g+\ddot r)=Mm(g+r\dot\theta^2)/(M+m)>0$.
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