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

For steady inviscid incompressible flow subject to a conservative body force, Bernoulli's theorem states that $p/\rho +u^2/2+gz$ is constant along a streamline. The same constant applies across streamlines only with an additional condition, such as an irrotational connected flow with the same upstream head.

Neglect gravity and let the film's free-surface pressure be atmospheric. Its radial speed is then $v$ in the ideal model. Conservation of mass equates the incoming flux $\pi r_1^{2}v$ to $2\pi rt(r)v$, soThe latter value already warns that the thin-film approximation is poor near impact; the radial expression is most useful beyond that region. The incoming vertical momentum flux gives downward force $F_m=\rho \pi r_1^2{v}^2$ when the outgoing flow is horizontal.

In the stipulated stagnant-rear-layer idealization, the excess rear pressure is the stagnation value $\rho v^2/2$. Its upward force is $\rho v^2\pi r_2^2/2$. Thus, with downward positive and ignoring the water's weight,For a large plate, the radial-film volume is approximatelyIts weight therefore changes the model toCancellation occurs nearThe printed claim of negligible force for every radius above this scale does not follow from this model. Near the cancellation radius the force is small; for substantially larger radii the quadratic upward-pressure term dominates. The original PDF, not just its transcription, uses that inequality and wording. Moreover, the rear layer's mass and detailed gravity-dependent pressure field are unspecified. The calculation establishes a cancellation scale under the elementary stated pressure approximation, rather than a general large-radius zero-force theorem. A full free-surface flow model would be needed to extend it consistently.

For a Lagrangian $\mathcal{L}=T-U$, the Euler-Lagrange equation is $d(\partial \mathcal{L}/\partial \dot{q})/dt=\partial \mathcal{L}/\partial q$ for each unconstrained generalized coordinate. For example, $\mathcal{L}=m{\dot{x}}^2/2-k{x}^2/2$ gives $m\ddot{x}+kx=0$ for a spring oscillator.

For a uniform solid sphere, slicing into disks normal to the rotation axis givesUse total dumb-bell mass $M$ and $\ell =L-r$. Since both centres lie on its spin axis, its moment of inertia about that axis is $2M{R}^2/5$. If $\theta$ measures displacement from the bottom of the track, rolling without slipping imposes spin speed $|\dot{\varphi }|=\ell |\dot{\theta }|/r$. The rolling inside a fixed circular cylinder therefore hasThe equation of motion and its small-amplitude limit areWith $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and the given lengths, $\boxed{T=2\pi /\omega \simeq 80.5\mathrm{s}}$. If the centre rises by height $h$, conservation of energy at the bottom givesThe speed expression uses exact height-energy conversion within the ideal rolling constraint; only the period formula uses the small-angle approximation.

Let $J_p=\int r^{2}dA=\pi a^4/2$ be the polar second moment of area. A twist gradient ${\partial }_x\theta$ gives shear strain $r{\partial }_x\theta$ and shear stress $Gr{\partial }_x\theta$. Integrating its moment over a cross section yieldsThe difference in torque on an element $dx$ is $G{J}_p{\partial }_x^2\theta dx$, while its angular inertia is $\rho J_{p}dx$. Hence the torsional wave equation in a circular shaft isFor uniform rotation at $\Omega$, transmitted power is $P=C\Omega$, so the given values give $C=100\mathrm{N}\mathrm{m}$. The elastic strain energy and rotational kinetic energy areThe torsional transit time is $l/c_T=6.32\times 10^{-4}\mathrm{s}$. If the requested maximum rates mean stored-energy-over-transit-time estimates, they areThese estimates require a physical qualification. They are not separate upper bounds on transmitted mechanical power. The exact local energy flux of a torsional wave is the stress-work rateFor the steadily rotating twisted shaft its magnitude is $C\Omega =30\mathrm{kW}$, even though this exceeds $c_{T}U/l$. Writing local energy densities $u_s=G{J}_p{\theta }_x^2/2$ and $u_k=\rho J_p{\theta }_t^2/2$ gives $|\mathcal{P}|=2c_T\sqrt{u_s{u}_k}\le c_T(u_s+u_k)$. A single travelling torsional wave has equal strain and kinetic energies; the steady shaft can instead be represented by oppositely travelling components whose powers largely cancel. Its stored strain pattern is not a fuel consumed once per transit. This distinguishes an energy-content estimate from the actual stress-mediated transport.

In plane polar coordinates the radial component of Newton's second law is $m(\ddot{r}-r{\dot{\theta }}^2)=-U^{\prime }(r)$. A central force has zero torque, so angular momentum $J=m{r}^2\dot{\theta }$ is constant. ThereforeFor a nonzero-angular-momentum circular orbit at $r_0$, $nB{r}_0^{n+2}=J^2/m$. The curvature there isThus a stable circular orbit, and nearby bound orbits, require $\boxed{nB>0,n>-2}$, with $n=0$ excluded. For $n>0$, $B>0$ confines the particle at large radius; for $-2<n<0$, $B<0$ permits bound energies between the effective-potential minimum and zero. At $n=-2$ there is no isolated stable minimum.

The angular frequency is ${\Omega }^2=nB{r}_0^{n-2}/m$, and radial perturbations have frequency $\kappa =\sqrt{n+2}\Omega$. A rational ratio $\kappa /\Omega$ closes the linearized near-circular trajectory. It does not prove finite-amplitude closure. Bertrand's theorem supplies the stronger conclusion: every bound orbit is closed only for $n=-1$ and $n=2$ among these smooth attractive power-law potentials. Particular circular or resonant orbits must not be confused with that all-orbit statement.

For $n=2+ϵ$, the slowly rotating elliptical orientation has angular speedThis is retrograde for $ϵ>0$ in the near-circular approximation. Consecutive radial maxima are separated by approximately $\pi$ in azimuth; the ellipse's axis is defined modulo $\pi$, which accounts for the factor of two relative to a Kepler-style apsidal convention.