Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 3 / 15C (Classical Dynamics)

The azimuthal coordinate is cyclic, so its canonical momentum is conserved. Time independence of the Lagrangian also conserves the energy. The two constants areFor a horizontal launch from depth $D$, take $0<D<\ell$, so $\cos {\theta }_0=D/\ell$, ${\dot{\theta }}_0=0$, and $v=\ell \sin {\theta }_0{\dot{\varphi }}_0$. Thus $h=mv\sqrt{{\ell }^2-D^2}$ and $E=\frac{1}{2}m{v}^2-mgD$. The meridional motion has effective potential $U(\theta )=h^2/(2m{\ell }^2{\sin }^2\theta )-mg\ell \cos \theta$.

Grazing the equator makes it a turning point, so $\dot{\theta }=0$ at $\theta =\pi /2$. Conservation of energy gives $\frac{1}{2}m{v}^2-mgD=h^2/(2m{\ell }^2)=\frac{1}{2}m{v}^2(1-D^2/{\ell }^2)$. ThereforeTo check that no earlier turning point prevents reaching the equator, put $c=\cos \theta$. On $0\le c\le D/\ell$, $U(c)=h^2/[2m{\ell }^2(1-c^2)]-mg\ell c$ is strictly convex. Its two endpoint values are equal for this speed, so its interior values are smaller. The trajectory can therefore move between the launch point and the equator. The equator is genuinely a turning point since $dU/d\theta =mg\ell$ there.

For fixed $\theta$, the Euler-Lagrange equation for the meridional coordinate requires ${\dot{\varphi }}^2\cos \theta =g/\ell$. Combining this with $v=\ell \sin \theta \dot{\varphi }$ and $\cos \theta =D/\ell$ givesThe resulting motion is a conical pendulum. The positive-depth condition is necessary for a nondegenerate constant-latitude orbit in gravity; the bottom point is the limiting zero-speed equilibrium.