= Solution
{c}
On the <surface of revolution>, a <normal vector> is proportional to $-H'(r)\mathbf e_r+\mathbf e_z$. Thus the <normal force> and <gravity> have no $\mathbf e_\theta$ component. The $\mathbf e_\theta$ component of <Newton's second law>, using the given <linear drag> force, is
$$
m(r\ddot\theta+2\dot r\dot\theta)=-\alpha r\dot\theta.
$$
Multiplication by $r$ therefore gives
$$
\boxed{\dot h=-\frac\alpha m h,\qquad h(t)=h_0e^{-\alpha t/m},\qquad\dot\theta(t)=\frac{h_0e^{-\alpha t/m}}{r(t)^2}.}
$$
\b[The printed target <equation> omits the factor $1/m$.] The matching original PDF, page 5, gives both $\mathbf F_D=-\alpha\dot{\mathbf r}$ and $\dot h=-\alpha h$, so this is a source inconsistency. If its $\alpha$ in the target <equation> is intended to mean the drag <coefficient> per unit <mass>, put $\lambda=\alpha/m$ and write $\dot h=-\lambda h$. Equivalently, if the intended force was $\mathbf F_D=-m\alpha\dot{\mathbf r}$, the requested convention gives
$$
\boxed{h(t)=h_0e^{-\alpha t},\qquad\dot\theta(t)=h_0e^{-\alpha t}/r(t)^2.}
$$
These formulas determine $\dot\theta$ once $r(t)$ is known; they do not assume a constant <radius>.
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