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

The free-body diagram shows a meridional section of the right circular cone. On the first particle, gravity acts vertically downwards, tension points along the straight line to the apex, and the normal reaction is perpendicular to the surface. The smooth mechanical constraint supplies no tangential friction. The second particle has only its downward gravitational weight and the upward tension.

Adding the forces on each particle and applying Newton's second law givesAt $z_1=-r_1$, a normal vector is ${\mathbf{e}}_r+{\mathbf{e}}_z$. Its sign depends on the supporting face of the cone; either sign has zero dot product with every permitted velocity. The diagram chooses the reaction towards the upper side of the surface.

The massless inextensible string has a common tension magnitude $T$. Since ${\mathbf{r}}_1=r_1({\mathbf{e}}_r-{\mathbf{e}}_z)$, the unit vector from the first particle towards the apex is $(-{\mathbf{e}}_r+{\mathbf{e}}_z)/\sqrt{2}$. The second particle lies below the apex, soThe string length constraint isThe velocity of the first particle in cylindrical coordinates is ${\dot{r}}_1{\mathbf{e}}_r+r_1\dot{\theta }{\mathbf{e}}_{\theta }+{\dot{z}}_1{\mathbf{e}}_z$. Therefore the powers delivered by the two tension forces areThusThe tension can transfer energy between the particles, but its total power on the pair is zero.

All forces on the first particle lie in the plane spanned by ${\mathbf{e}}_r$ and ${\mathbf{e}}_z$: this holds for gravity, tension, and the normal reaction. Hence the azimuthal component of Newton's second law vanishes:Using the product rule givesStrictly, $h$ is the axial specific angular momentum; the actual axial angular momentum is $m_{1}h$, which is conserved as well.

Differentiate the stated total energy using the chain rule and then substitute the two Newton's second law equations:The gravitational potential energy terms have canceled the power of gravity. Also $\mathbf{N}$ is proportional to ${\mathbf{e}}_r+{\mathbf{e}}_z$, whose dot product with ${\dot{\mathbf{r}}}_1$ is ${\dot{r}}_1+{\dot{z}}_1=0$. Thus the stationary smooth mechanical constraint does no work. The remaining tension powers cancel by part (b), proving $\dot{E}=0$ and conservation of total energy while the massless inextensible string remains taut and contact is maintained.

Initially $r_1=a$ and $z_2=\sqrt{2}a-l$, so the total gravitational potential energy isIf the given velocity is understood as a full tangential vector $\mathbf{v}$, put $v_z=\mathbf{v}\cdot {\mathbf{e}}_z$. The string length constraint gives ${\dot{z}}_2(0)=-\sqrt{2}{v}_z$, and henceThis formula accounts for both particles' kinetic energies, including that of the hanging particle.

If $v$ denotes only the initial speed, the printed wording does not specify its direction within the tangent plane. To make that dependence explicit, let $\beta$ be the angle from the downward generator of the cone towards the azimuthal unit vector, soThen $v_z=-v\cos \beta /\sqrt{2}$, andFor a launch along a generator, $\beta =0$ or $\pi$, this reduces to $E=\frac{1}{2}(m_1+m_2)v^2-m_{1}ga+m_{2}g(\sqrt{2}a-l)$. For a horizontal azimuthal launch, $\beta =\pi /2$, it instead gives $E=\frac{1}{2}{m}_1{v}^2-m_{1}ga+m_{2}g(\sqrt{2}a-l)$. A scalar speed alone does not determine a unique energy here. Both directions are tangent to the surface; “along the surface” does not distinguish them. The assumed configuration also requires $l>\sqrt{2}a$ so the second particle hangs below the apex.