Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics IV / 5

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.

Take $a>0$ and the intended horizontal launch around a parallel of the surface of revolution: $\dot{r}(0)=\dot{z}(0)=0$, $a\dot{\theta }(0)=v$. Then $h=av$ and $E=v^2/2+gf(a)$. Since $\dot{z}=f^{\prime }(r)\dot{r}$, the conserved energy becomesHere $U$ is the radial effective potential on a surface of revolution. Rearrangement givesFor a nonconstant bounded radial motion with $v>0$, there must be another finite radial bound $b\ne a$ satisfying $D(b)=0$, equivalentlyThe interval between the radial bounds must be dynamically accessible: $D\ge 0$ inside, and a simple endpoint must have a forbidden region $D<0$ beyond it. Equivalently the motion must lie in a bounded part of $\{r>0:U(r)\le E\}$ separated from escape by a potential barrier. When $f$ stays finite near zero and $v>0$, the centrifugal barrier already excludes $r=0$; an outer barrier is the additional requirement. An arbitrary second algebraic root on a disconnected allowed branch would not establish boundedness.

The initial radial acceleration, obtained by projecting Newton's second law along ${\mathbf{e}}_r+f^{\prime }(a){\mathbf{e}}_z$, isThus $v^2/g>a{f}^{\prime }(a)$ starts the particle outward, and $v^2/g<a{f}^{\prime }(a)$ starts it inward. Equality gives the exact circular orbit $r=a$ and needs no second distinct radial bound. A double root at a different radius can instead be approached asymptotically: this separatrix is bounded, although the radial motion is not periodic. If $v=0$, the centrifugal barrier disappears and the particle may pass through the axis, still with bounded $r$.

The source leaves a direction ambiguity at points where $f^{\prime }(a)=0$: every horizontal tangent direction is then possible, so “projected horizontally” alone does not imply $h=av$. The displayed inequality uses the azimuthal launch specified above; at $f^{\prime }(a)\ne 0$ that direction is forced by horizontality and tangency.

Use the dimensionless variablesThe stated $R>a/2$ gives $0<A<1$. Since $f(r)=4Rx(1-x)$, the inequality from b factorizes asThe two additional possible turning points areThey are real exactly when $\beta +A\le 1$. This exposes the intended critical valuebut the ordering of the initial radius $A$ relative to these roots must also be checked.

First suppose $0<A<1/3$. Then $A<(1-A)/2$. If $\beta <1-A$, the allowed branch starting at $A$ is trapped either between $x_{-}$ and $A$ or between $A$ and $x_{-}$, according to its initial direction. At the coincident-root value ${\beta }^2=4A(1-2A)$ it is exactly a circular orbit. For $\beta =0$, the inner radial bound is the axis rather than a nonzero turning point.

At $\beta =1-A$, the additional roots coalesce at $x_{\ast }=(1-A)/2>A$. The particle starts outward and approaches $x_{\ast }$, orwithout reaching it in finite time. Indeed $D$ is proportional to $(x-A)(x-x_{\ast }{)}^2$ and the radial energy equation gives $dt/dx$ proportional to $1/|x-x_{\ast }|$ near $x_{\ast }$. Its integral diverges. This is a bounded separatrix approaching an unstable circular orbit. For $\beta >1-A$, there is no outer root and the particle escapes. Thus for $0<A<1/3$, boundedness holds exactly when $\beta +A\le 1$, with the equality case included. At $A=1/3$, the same inequality characterizes boundedness, but equality is the exact circular orbit at the starting radius.

The source's weaker assumption $A<1$ is insufficient to make this single threshold a general criterion. For $1/3<A<1/2$, the initial radius lies beyond the coalescing-root position. The transition is instead the unstable circular orbit atBelow this speed the initial radial acceleration is inward and the particle remains in the inner allowed interval. At equality it stays exactly circular; above it the initial acceleration is outward and the accessible branch extends to infinity, even if $\beta +A<1$. For $A>1/2$, the surface slopes downward outward and every such launch escapes. At $A=1/2$, only $\beta =0$ remains at rest on the summit circle.

Consequently the complete classification for the intended azimuthal launch, including zero launch speed and exact equilibria, isFor a concrete counterexample to the unqualified proposed criterion, take $A=3/4$ and $\beta =1/10$. Then $\beta +A=17/20<1$, but $f^{\prime }(a)<0$ and the outward motion has no accessible outer barrier. This satisfies the printed radius assumption and escapes. A sufficient additional radius restriction for the intended threshold is $a\le 2R/\sqrt{3}$, with the equality and circular orbit qualifications above.