= Solution
{c}
Choose the vertical $z$ axis downward, and measure the <polar angle> $\theta$ from it. With <azimuth> $\phi$ and fixed rod <length> $r$, the <Cartesian coordinates> are
$$
\boxed{x=r\sin\theta\cos\phi,\qquad y=r\sin\theta\sin\phi,\qquad z=r\cos\theta.}
$$
The massless rod has no <kinetic energy>. Differentiating these <coordinates> and adding the squared components of <velocity> yields
$$
v^2=\dot x^2+\dot y^2+\dot z^2
=r^2\left(\dot\theta^2+\sin^2\theta\,\dot\phi^2\right).
$$
The <gravitational potential energy> is $V=-mgz=-mgr\cos\theta$, up to an arbitrary constant. The <Lagrangian> of this <spherical pendulum> is therefore
$$
\boxed{L=\frac12mr^2\left(\dot\theta^2+\sin^2\theta\,\dot\phi^2\right)+mgr\cos\theta.}
$$
Applying the <Euler-Lagrange equation> to the two <generalized coordinates> gives
$$
mr^2\ddot\theta-mr^2\sin\theta\cos\theta\,\dot\phi^2+mgr\sin\theta=0,
\qquad
\frac{d}{dt}\left(mr^2\sin^2\theta\,\dot\phi\right)=0.
$$
Equivalently, away from the <coordinate singularities> at the poles,
$$
\boxed{\ddot\theta=\sin\theta\cos\theta\,\dot\phi^2-\frac gr\sin\theta,\qquad
\ddot\phi+2\cot\theta\,\dot\theta\dot\phi=0.}
$$
Since $L$ is independent of $\phi$, the <azimuth> is a <cyclic coordinate>. Its <conjugate momentum>
$$
p_\phi=mr^2\sin^2\theta\,\dot\phi
$$
is the conserved component of <angular momentum> about the downward vertical axis. The <polar angle> is not a <cyclic coordinate>. Using an upward $z$ axis instead reverses the sign of the cosine term in $L$ and of the corresponding gravity term; it describes the same motion with the opposite <polar angle> convention.
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