Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics IV / 5

On the surface of revolution, a normal vector is proportional to $-H^{\prime }(r){\mathbf{e}}_r+{\mathbf{e}}_z$. 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, isMultiplication by $r$ therefore givesThe 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 givesThese formulas determine $\dot{\theta }$ once $r(t)$ is known; they do not assume a constant radius.

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 givesThe acceleration in cylindrical coordinates isTake the dot product of Newton's second law with the tangent vector ${\mathbf{e}}_r+H^{\prime }{\mathbf{e}}_z$. The normal force disappears, while the gravity term is $-mg{H}^{\prime }$. Dividing by $m$ givesUsing $\dot{\theta }=h/r^2$ yields

Initially $\dot{r}=0$ also gives $\dot{z}=H^{\prime }(b)\dot{r}=0$. Thus the initial velocity is azimuthal, with $|h|=bv$. By conservation of energy, the supplied mechanical energy divided by $m/2$ givesThe first term is nonnegative. ConsequentlyFor a positive launch speed $v>0$, introduce twice the radial effective potential on a surface of revolution per unit mass,It satisfiesso it is a strictly convex function. Also $S(r)\to \infty$ as $r\to \infty$. To justify the other endpoint carefully, convexity gives $H(r)\ge H(b)+H^{\prime }(b)(r-b)$ for $0<r<b$. This supplies a finite lower bound near $0$, and the positive $r^{-2}$ term consequently forces $S(r)\to \infty$ there.

Thus $S$ has a unique minimum at some $r_m>0$. Its derivative is strictly increasing and zero there, so $S$ decreases to $r_m$ and increases afterwards. Its sublevel set $S(r)\le B$ is a single interval $[b_1,b_2]$, with $0<b_1\le b\le b_2<\infty$. If $B>S(r_m)$, the two boundary equations $S(b_i)=B$ have distinct solutions; if $B=S(r_m)$, both boundaries are $r_m$. The conservation of energy identity therefore provesThe two radii coincide exactly when $b=r_m$, equivalently $S^{\prime }(b)=0$. This givesThe equation in b then has the constant solution $r=b$, with $\dot{\theta }=\pm v/b$: this is a circular orbit.

A positive inner bound requires $v>0$. If zero launch speed is allowed, $h=0$ and the displayed inequality only gives $r\le b$ because $H$ is strictly increasing; one may take the limiting inner bound $b_1=0$. For example, $H(r)=r^2$ satisfies all the stated hypotheses but permits inward motion from rest all the way towards the excluded axis. The two-positive-turning-radii argument therefore uses the natural interpretation of “launched with speed $v$” as $v>0$.