= Solution
The <central force> has no transverse component. Its angular equation is therefore
$$
r\ddot\theta+2\dot r\dot\theta=0
\quad\Longrightarrow\quad
\frac d{dt}(r^2\dot\theta)=0.
$$
Thus the specific <angular momentum> is the constant
$$
\boxed{l=r^2\dot\theta.}
$$
The radial equation is
$$
\ddot r-r\dot\theta^2=-V'(r).
$$
Substitute $\dot\theta=l/r^2$:
$$
\ddot r=\frac{l^2}{r^3}-V'(r)
=-\frac d{dr}\left[V(r)+\frac{l^2}{2r^2}\right].
$$
Consequently the radial motion for $r>0$ is exactly a unit-mass one-dimensional motion in the <effective potential>
$$
\boxed{V_{\rm eff}(r)=V(r)+\frac{l^2}{2r^2}.}
$$
Its conserved <energy> is $\tfrac12\dot r^2+V_{\rm eff}(r)$; the second term in the <effective potential> accounts for the angular <kinetic energy>.
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