Past exam of the physics 1 paper of the University of Oxford / 2019 resit / 10

The massless inextensible string has hanging length $y=l-r$, measured downward from the table. Thus $\dot{y}=-\dot{r}$. In polar coordinates, the tabletop mass has squared speed ${\dot{r}}^2+r^2{\dot{\theta }}^2$. Adding the two kinetic energies and setting the hanging gravitational potential energy to $-Mgy$ givesand the Lagrangian isThe canonical momenta are $p_r=(M+m)\dot{r}$ and $p_{\theta }=m{r}^2\dot{\theta }$. Performing the Legendre transform gives the HamiltonianIt equals the total mechanical energy and is conserved because $L$ has no explicit time dependence. Also $\theta$ is a cyclic coordinate, so $J=p_{\theta }=m{r}^2\dot{\theta }$ is conserved angular momentum about the vertical through the hole. The table and the string guide supply external forces; conservation of the system's total linear momentum does not follow. These are the two independent generic integrals needed for the reduced tabletop mass coupled to a hanging mass.

The radial and angular Euler-Lagrange equations areWriting the conserved angular momentum aswe obtainA positive constant radius requires zero radial acceleration. Hence the circular motion hasprovided $J\ne 0$ and $0<r_0<l$. Its angular velocity is ${\dot{\theta }}_0=J/(m{r}_0^2)$. No positive circular radius exists for $J=0$ with $M,g>0$.

For later use, the conserved radial energy isThe effective potential has $U_J^{\prime \prime }(r)=3J^2/(m{r}^4)>0$, so its stationary radius is a stable minimum. The tension remains positive in the ideal allowed region: the hanging Newton's second law gives $T_s=M(g+\ddot{r})=Mm(g+r{\dot{\theta }}^2)/(M+m)>0$.

Initially $r_i=l/2$, ${\dot{r}}_i=0$ and ${\dot{\theta }}_i=v_0/r_i$, so the conserved angular momentum is $J=ml{v}_0/2$. For $v_0\ne 0$, the effective potential contains the centrifugal barrierThe finite conserved energy therefore forbids the hole from being reached:The exceptional case $v_0=0$ has no centrifugal barrier. Then $J=0$ and $\ddot{r}=-Mg/(M+m)$; starting at rest at $l/2$, the tabletop mass reaches the hole after $t=\sqrt{(M+m)l/(Mg)}$. The model ends at that event. Thus the usual nonzero initial tangential speed assumption matters.

The initial angular momentum is $J=ml{v}_0/2$, and the initial total energy isWhen the hanging mass reaches the table, $r=l$. Conservation of angular momentum fixes the tabletop tangential speed there to $J/(ml)=v_0/2$. At the limiting case of just reaching, radial velocity is zero. Equating the two energies,yieldsMore generally, the boundary radial kinetic energy isso reaching requires $v_0^2\ge 4Mgl/(3m)$, with equality for just reaching and strict inequality for nonzero outward radial speed at contact. This condition is also sufficient: $U_J$ is strictly convex, and the allowed interval connecting the initial radius to $l$ has no intervening higher effective potential barrier. At the threshold the initial radial acceleration is outward and $r=l$ is the outer turning point.

For the initial radius $r_0=l/2$ to be circular, the radial equation requiresConsider a small radial perturbation at fixed conserved angular momentum $J$. Using $J^2/(m{r}_0^3)=Mg$ and $r=r_0+ϵ$ in the radial equation givesConsequently the linearized radial motion is simple harmonic motion:The positive curvature of the effective potential establishes radial stability. This is the small-amplitude limit, not an exact finite-amplitude sinusoid. A perturbation that changes $J$ shifts the circular radius; its radial oscillations are about the new radius to the same order. A radial displacement or radial impulse preserving $J$ gives the stated oscillations about $l/2$.