Past exam of the physics B course of the University of Cambridge / 2010 / Paper 2B / Section C

Bernoulli's theorem and conservation of mass give $t=r_1^2/(2r)$. Ignoring weight, the stated stagnant-rear-pressure model gives zero resultant at $r_2=\sqrt{2}{r}_1$. Including the leading film-weight estimate gives a cancellation radius near $2g{r}_1^2/v^2$. The printed inequality does not establish negligible force at every larger radius; the shared derivation explains this source qualification.

Full derivation: solution.

Rolling without slipping gives ${\omega }^2=g/[(L-r)(1+2R^2/(5r^2))]$. With the supplied dimensions, $T\simeq 80.5\mathrm{s}$ and the maximum centre-of-mass speed for a $10\mathrm{mm}$ rise is $0.0349\mathrm{m}{\mathrm{s}}^{-1}$.

Full derivation: solution.

The torsional wave equation in a circular shaft has speed $\sqrt{G/\rho }$. The given shaft has torque $100\mathrm{N}\mathrm{m}$, twist $9.95\times 10^{-3}\mathrm{rad}$, strain energy $0.497\mathrm{J}$ and kinetic energy $181\mathrm{J}$. The respective transit-time estimates are $786\mathrm{W}$ and $286\mathrm{kW}$; these must not be confused with independent bounds on the actual $30\mathrm{kW}$ stress-work flux.

Full derivation: solution.

Stable circular power-law orbits require $nB>0$ and $n>-2$, excluding $n=0$. Bertrand's theorem selects $n=-1,2$ when every bound orbit must close. For $n=2+ϵ$, the near-circular ellipse orientation precesses at $-ϵ\Omega /8$ to first order.

Full derivation: solution.