= Solution
{c}
Define $f(r)>0$ as the magnitude of the inward <central force>, so $\mathbf F=-f(r)\widehat{\mathbf r}$ and $V'(r)=f(r)$. The conserved <angular momentum> is $J=mr^2\dot\theta$. Eliminating $\dot\theta$ from the <mechanical energy> yields
$$
\boxed{V_{\mathrm{eff}}(r)=V(r)+\frac{J^2}{2mr^2},\qquad
E=\frac12m\dot r^2+V_{\mathrm{eff}}(r).}
$$
The added term in the <effective potential> is the <centrifugal barrier>. If instead $f$ denotes the signed radial <force> component, the convention becomes $V'=-f$.
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