Past exam of the mathematics course of the University of Cambridge / 2005 / ia / Paper 4 / 11C / Solution

For a particle constrained to $z=h(x,y)$, the kinetic energy and potential energy areIn the stipulated gentle-surface small-slope approximation, the second kinetic term is higher order. The leading Euler-Lagrange equations for $T-V$ therefore give the small-slope particle motion on a height graph:This is also the small-slope component of gravity along the surface, with $\sin \varphi \simeq \tan \varphi$. Geometric acceleration terms must remain negligible at the speeds considered; the approximation is not asserted for arbitrarily fast motion on a sharply curved surface.

For $h=\alpha x^2$, the equations are $\ddot{x}+2g\alpha x=0$, $\ddot{y}=0$. If $\alpha >0$, put $\omega =\sqrt{2g\alpha }$. For arbitrary initial data the motion isIt is simple harmonic motion across the trough and uniform motion along it. If $\alpha =0$, both coordinates move uniformly; if $\alpha <0$, the same formula uses $\cosh (\sqrt{-2g\alpha }t)$ and $\sinh (\sqrt{-2g\alpha }t)/\sqrt{-2g\alpha }$ instead.

For the radial height profile, $h(r)=(1-r)/r^2$ is positive for $r<1$, zero at $1$, and negative for $r>1$. Its derivative is $(r-2)/r^3$, so the minimum is $h(2)=-1/4$. It diverges to positive infinity as $r\downarrow 0$ and tends to $0$ from below as $r\to \infty$. The requested profile sketch is shown below.

The equations in polar coordinates in the planar approximation areMultiplying the second equation by $r$ gives $d(r^2\dot{\theta })/dt=0$. Thus the specific angular momentum $L=r^2\dot{\theta }$ is conserved, andAt release with negligible radial velocity, the initial specific kinetic energy is indeedIt is not the total energy. Put $A=L^2+2g$. The conserved specific total energy and effective potential areThe radial turning points for an inverse-square height well follow fromAs a function of $r$, this expression tends to $-\infty$ at zero, has its maximum at $r=A/g$, and tends to $2\mathcal{E}$ at infinity. Only regions where it is nonnegative are dynamically accessible. If $\mathcal{E}<0$, there are two positive turning radii,and the radial coordinate is confined between them. They merge at $r_0=A/g$, the circular orbit. If $\mathcal{E}=0$, the particle escapes to arbitrarily large radius with radial speed tending to zero; if $\mathcal{E}>0$, it escapes with positive limiting radial speed. Consequently the bounded-orbit condition, including the circular case, isThe circular release is at $r_0=2+L^2/g$. For $A/(2g)<r_0<A/g$, the release point is the smaller turning radius; for $r_0>A/g$, it is the larger one. These statements concern the prescribed planar small-slope model. The literal surface is steep near its central singularity, so that approximation must be valid along the particular orbit under consideration.