= Solution
The particle moves on a horizontal circle of radius
$$
r=L+a\sin\phi_1.
$$
If $T$ is the tension in the light rod, vertical and radial balance give
$$
T\cos\phi_1=mg,
\qquad
T\sin\phi_1=m\omega^2(L+a\sin\phi_1).
$$
Dividing eliminates both $T$ and $m$, yielding the <rotating conical pendulum with an offset pivot> condition
$$
\boxed{
\omega=\sqrt{\frac{g\tan\phi_1}{L+a\sin\phi_1}}}.
$$
The ratio under the square root has units $T^{-2}$. The answer is independent of $m$; increasing $L$ increases the orbit radius and lowers the angular <velocity> required at fixed $\phi_1$.
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