Equilibrium points in lagrangian mechanics Suppose we have a one particle system with generalized coordinates $q_i$. In classical mechanics, the corresponding Lagrangian is $L = T - V$. Assume $V(q)$ is time-independent. What additional conditions on the system determine whether
$$\nabla V (q) = 0 \iff q \text{ is an equilibrium point} \, .$$
For example, sometimes this condition holds only if $V$ is the effective potential.
 A: The condition for equilibrium is actually best understood using the Hamiltonian.  If the Lagrangian is time-independent, then the Hamiltonian is conserved (albeit not necessarily the energy) and the evolution must take place on a curve of $H=$ constant.
Given a Hamiltonian $H(p,q)$ (in one-dimension to keep it simple), then the condition for an equilibrium position is 
$$
\left(\frac{\partial H}{\partial p},\frac{\partial H}{\partial q}\right)=0=(\dot{q},-\dot{p})\, .
$$
Geometrically, the points that satisfy this are extrema in the $H$ landscape, i.e. when thinking of $H(p,q)$ as a surface in 3D.  Mathematically, by Hamilton's equations, the momentum and position are exactly extremal at those points.  
Mathematically, this formulation, which involves first derivatives, ties in with the rich topic of qualitative behaviour of coupled first order differential equations, which are applicable to a wide variety of systems: the study of predator-prey systems, Lanchester's model of warfare, etc (the list is very long).
An example for which this applies is the flyball governor. The Lagrangian for the system is 
$$
L=\ell^2(m_1+2m_2\sin^2\alpha)\dot{\alpha}^2+
m_1\ell^2\Omega^2\sin^2\alpha
+2(m_1+m_2)g\ell\cos\alpha\, .
$$
and it is difficult to identify a "potential" $V(\alpha)$ since the coefficient of term in $\dot{\alpha}^2$ is actually a function of $\alpha$.

The momentum $p_\alpha=\partial L/\partial \dot{\alpha}=2\ell^2(m_1+2m_2\sin^2\alpha)\dot{\alpha}$,
so the Hamiltonian is found, after straightforward manipulations, to be
\begin{align}
H&= \frac{p^2_\alpha}{4\ell^2(m_1+2m_2\sin^2\alpha)}-m_1\ell^2\sin^2\alpha\,\Omega^2
-2(m_1+m_2)g\ell\cos\alpha\, .
\end{align}
The fixed points are easily obtained. 
Clearly $\partial H/\partial p_\alpha=0$ implies $p_\alpha=0$.  On the other hand:
$$
\frac{\partial H}{\partial \alpha}\vert_{p=0}=
2\ell\sin(\alpha)\left(m_1\ell \cos(\alpha)\Omega^2-(m_1+m_2)g\right)=0\, ,
$$
which gives $\alpha=0$ but also a non-trivial equilibrium point if one can satisfy $m_1\ell \cos(\alpha_0)\Omega^2-(m_1+m_2)g=0$ for some angle $\alpha_0$.  See here for another example.
If the system is natural, so that 
$H=T+V_{\hbox{eff}}$ with $T=p^2/(2m)$, this automatically reduces to a condition on the derivative of the effective potential.
