= Solution
{c}
For a smooth surface the <friction> force vanishes. The $\mathbf e_\theta$ component of <Newton's second law> then gives $r^2\dot\theta=h$, constant. Along the <constraint> $z=H(r)$, the <chain rule> gives
$$
\dot z=H'\dot r,\qquad\ddot z=H'\ddot r+H''\dot r^2.
$$
The <acceleration in cylindrical coordinates> is
$$
\ddot{\mathbf r}=(\ddot r-r\dot\theta^2)\mathbf e_r+(r\ddot\theta+2\dot r\dot\theta)\mathbf e_\theta+\ddot z\mathbf e_z.
$$
Take the <dot product> of <Newton's second law> with the <tangent vector> $\mathbf e_r+H'\mathbf e_z$. The <normal force> disappears, while the <gravity> term is $-mgH'$. Dividing by $m$ gives
$$
\ddot r-r\dot\theta^2+H'(H'\ddot r+H''\dot r^2)=-gH'.
$$
Using $\dot\theta=h/r^2$ yields
$$
\boxed{[1+(H')^2]\ddot r+H'H''\dot r^2-\frac{h^2}{r^3}=-gH'.}
$$
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