Past exam of the physics course of the year 4 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / Section B / B1 / Solution

In a central potential, $J=m{r}^2\dot{\theta }$ is conserved. Substituting it into $U=m({\dot{r}}^2+r^2{\dot{\theta }}^2)/2+V(r)$ givesWhile the string stays taut, $V=k(r-L{)}^2/2$. Circular force balance gives $m{r}_0{\omega }_0^2=k(r_0-L)$, hence ${\omega }_0^2=k(r_0-L)/(m{r}_0)$. A radial impulse changes no angular momentum. The radial oscillation of an elastic-string orbit follows by expanding the same fixed-$J$ effective potential about its minimum:soFor an outward impulse, one convenient linear approximation is $r=r_0+ϵ\sin \Omega t$ and $\theta ={\omega }_{0}t+[2{\omega }_0ϵ/(r_0\Omega )](\cos \Omega t-1)$. The angular modulation follows from $\dot{\theta }=J/(m{r}^2)$ and is needed for a consistent trajectory sketch.