= Solution
The bob has $\dot x=\ell\cos\theta\,\dot\theta$ and $\dot y=at-\ell\sin\theta\,\dot\theta$. Substituting these into the <Lagrangian> gives
$$
L=\tfrac12m\ell^2\dot\theta^2-ma\ell t\sin\theta\,\dot\theta+mg\ell\cos\theta+\tfrac12m(a^2+ga)t^2.
$$
The mixed term is $\frac{d}{dt}(ma\ell t\cos\theta)-ma\ell\cos\theta$. Removing this total derivative and the time-only term leaves the equivalent <Lagrangian> $L_{\rm eff}=\frac12m\ell^2\dot\theta^2+m(g-a)\ell\cos\theta$. The <Euler-Lagrange equation> becomes
$$
\boxed{\ddot\theta+\frac{g-a}{\ell}\sin\theta=0.}
$$
In free fall, $a=g$, so $\theta=\theta_0+\omega_0t$: a bob initially at rest relative to the lift remains at its initial angle, and otherwise moves around the pivot with constant angular speed. Every constant angle is then an <equilibrium>, with no restoring force. For $a\ne g$, the <equilibria> are $\theta=0$ and $\theta=\pi$ modulo $2\pi$.
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