Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics IV resit / 5

Use $g$ as the gravitational acceleration, so the weights are $mg$ and $\sqrt{2}mg$, consistently with the requested equations. The printed introductory phrase calling $g$ a force magnitude omits the factor $m$.

For the particle on a cone coupled to a hanging mass, write $z=-r$ and use the cylindrical coordinates unit vectors ${\mathbf{e}}_r,{\mathbf{e}}_{\theta },{\mathbf{e}}_z$. The ideal string segment from the cone particle to the hole has length $\sqrt{2}r$ and direction $(-{\mathbf{e}}_r+{\mathbf{e}}_z)/\sqrt{2}$. If $h$ is the downward length of the hanging segment and $s_0$ the fixed total length, thenThe light ideal string and smooth hole give the same tension $T$ on both sides. Applying the vertical component of Newton's second law to the hanging particle givesWrite the normal reaction on the cone particle as $N({\mathbf{e}}_r+{\mathbf{e}}_z)/\sqrt{2}$, allowing $N$ to be signed according to the side of the ideal constraining surface. The acceleration in cylindrical coordinates and Newton's second law giveThere is no azimuthal force. Multiplying the middle equation by $r/m$ proveswhere the positive angular direction is chosen along the initial velocity. This is conservation of angular momentum about the vertical axis.

Subtract the radial equation from the vertical equation to eliminate the normal reaction. Since $\ddot{r}=-\ddot{z}$, this givesSubstitute the tension and the conserved angular momentum, using $r=-z$:Multiplication by $\dot{z}$ now gives an exact time derivative:Initially $z=-a$ and $\dot{z}=0$, because the initial velocity is horizontal. ThereforeThis is also total mechanical energy divided by $m$, with an irrelevant constant removed: the two particles contribute $(1+\sqrt{2})m{\dot{z}}^2$ to the radial and vertical kinetic energy, while their combined gravitational potential energy is $-mgz$ plus a constant.

In terms of the positive cylindrical radius $r=-z$, the conserved mechanical energy becomesThe effective potential $U$ tends to infinity at both ends of $(0,\infty )$ for $v>0$, and $U^{\prime \prime }(r)=3a^2{v}^2/r^4>0$. Thus its two turning points enclose the allowed radial motion. At a turning point, $\dot{r}=0$, soBesides $r=a$, the positive root of a polynomial isthe other root of a polynomial is negative. The quadratic polynomial $Q(r)=2g{r}^2-v^{2}r-a{v}^2$ is negative at zero and has just one positive root of a polynomial. Since $Q(a)=2a(ga-v^2)$, we have $r_{\ast }<a$ when $v^2<ga$, and $r_{\ast }>a$ when $v^2>ga$.

Height is $z=-r$, so its maximum occurs at the smaller positive turning point. HenceIndeed, the initial vertical acceleration isBelow the critical speed the particle first rises to $z=-r_{\ast }$; above it the particle initially descends, and its initial height is already its greatest height. At $v=\sqrt{ga}$ the turning points merge and the particle follows a horizontal circular orbit at $z=-a$. For zero initial speed the radial conservation of energy gives motion to the apex level $z=0$; if the hole excludes the apex itself, zero is the supremum of the heights before arrival at the hole.

These are the heights under the maintained mechanical constraint used in part (a). A unilateral contact with only one side of a cone would additionally require checking the sign of the normal reaction; the ideal constrained equations alone do not impose that condition.