= Solution
{c}
Let $\mathbf N$ be the <normal reaction> of the stationary smooth <surface of revolution>. Its normal lies in the span of $\mathbf e_r,\mathbf e_z$, so it has no $\mathbf e_\theta$ component. The transverse component of <Newton's second law>, $m\ddot{\mathbf r}=\mathbf N-mg\mathbf e_z$, is therefore
$$
2\dot r\dot\theta+r\ddot\theta=0,
$$
which gives $\boxed{h=r^2\dot\theta=\text{constant}}$.
The <velocity> is tangent to the stationary <surface>, so $\mathbf N\cdot\dot{\mathbf r}=0$. Take the <dot product> of <Newton's second law> with $\dot{\mathbf r}$:
$$
\frac d{dt}\left(\frac12m|\dot{\mathbf r}|^2\right)=-mg\dot z.
$$
Using $z=f(r)$ proves <conservation of mechanical energy> per unit <mass>:
$$
\boxed{E=\frac12|\dot{\mathbf r}|^2+gf(r)=\text{constant}.}
$$
The ideal <constraint> does no <work>. These equations describe motion while the particle remains constrained to the <surface>.
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