= Solution
Choose ordered particle angles $q_1,q_2,q_3$ with
$$
\alpha=q_2-q_1,\qquad
\beta=q_3-q_2,\qquad
\gamma=2\pi-\alpha-\beta,
$$
and put $\Phi=(q_1+q_2+q_3)/3$. Solving for the individual angles gives
$$
q_1=\Phi-\frac{2\alpha+\beta}{3},\qquad
q_2=\Phi+\frac{\alpha-\beta}{3},\qquad
q_3=\Phi+\frac{\alpha+2\beta}{3}.
$$
The <kinetic energy> is therefore
$$
T=\frac{mr^2}{2}\sum_{j=1}^3\dot q_j^2
=\frac{3mr^2}{2}\dot\Phi^2
+\frac{mr^2}{3}
\left(\dot\alpha^2+\dot\alpha\dot\beta+\dot\beta^2\right).
$$
The collective angle is an <ignorable coordinate> and decouples from the gaps, so its term may be omitted from the relative <Lagrangian mechanics>[Lagrangian]. Substituting $\gamma=2\pi-\alpha-\beta$ into the <potential energy> gives
$$
\boxed{
L_{\rm rel}
=\frac{mr^2}{3}
\left(\dot\alpha^2+\dot\alpha\dot\beta+\dot\beta^2\right)
-V_0\left(
e^{-2\alpha}+e^{-2\beta}+e^{-4\pi+2\alpha+2\beta}
\right).
}
$$
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