 Solution

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 thereforewhich 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 $\dot{\mathbf{r}}$:Using $z=f(r)$ proves conservation of mechanical energy per unit mass:The ideal constraint does no work. These equations describe motion while the particle remains constrained to the surface.