= Solution
The polar <Euler-Lagrange equation> is
$$
\ddot\theta=\sin\theta\cos\theta\,\dot\phi^2-\frac{3g}{2a}\sin\theta.
$$
For a constant angle with $0<\theta_0<\pi$, conservation of $h$ makes $\dot\phi$ constant, say $\Omega$. The remaining equation is
$$
\cos\theta_0\,\Omega^2=\frac{3g}{2a}.
$$
If $g>0$, every nonvertical constant-angle motion is a <conical pendulum> motion:
$$
\boxed{0<\theta_0<\frac\pi2,\qquad\phi(t)=\phi_0\pm\sqrt{\frac{3g}{2a\cos\theta_0}}\,t.}
$$
The rod sweeps out a cone about the downward vertical, in either azimuthal direction. Equivalently, $h^2=(3g/(2a))\sin^4\theta_0/\cos\theta_0$. The exactly downward and exactly upward vertical rods are also static solutions; azimuth has no physical meaning at either pole. No static tilted rod is possible under nonzero gravity.
If $g=0$, the condition is $\sin\theta_0\cos\theta_0\Omega^2=0$. Thus \b[a rod may remain at rest in any orientation, or rotate uniformly in the horizontal plane] with $\theta_0=\pi/2$ and arbitrary constant $\Omega$. The vertical cases are included among static orientations; changing only their azimuth does not move the rod.
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