Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 71 / 5 / d

The viscosity converts to $\eta =0.01\mathrm{Pa}\mathrm{s}$, and the translation speed is $U=10^{-8}\mathrm{m}{\mathrm{s}}^{-1}$. Mechanical power is $P_{\mathrm{motor}}={\zeta }_{\mathrm{chain}}{U}^2$. The hydrodynamic drag models for a polymer show why a conformation assumption is necessary.

For a largely aligned slender filament translating along its axis in unbounded fluid, useWith $L=100\mathrm{m}$, $\log (2L/a)=\log (4\times 10^{11})=26.715$, so ${\zeta }_{\parallel }\simeq 0.240\mathrm{kg}{\mathrm{s}}^{-1}$. The required force is about $2.4\times 10^{-9}\mathrm{N}$, andTransverse translation has roughly twice the leading logarithmic drag. This estimate treats the entire contour as an aligned slender body, assumes creeping flow, and omits bead drag, wall corrections, and additional motor losses. It is not a drag formula for an unstretched coil.

For an alternative nondraining coil, approximate the hydrodynamic radius by its radius of gyration, $R_h\sim R_g\simeq \sqrt{L{l}_p/3}=8.16\times 10^{-4}\mathrm{m}$. The Stokes drag law gives $\zeta \sim 6\pi \eta R_h=1.54\times 10^{-4}\mathrm{kg}{\mathrm{s}}^{-1}$ and $P\sim 1.54\times 10^{-20}\mathrm{W}$. A freely draining toy model that sums $N$ spherical-monomer drags of radius $a$ instead gives $P=N6\pi \eta a{U}^2\simeq 9.42\times 10^{-17}\mathrm{W}$. Hydrodynamic interactions are suppressed in this last model. These distinct answers explain the need to specify conformation and hydrodynamic coupling; the stated data do not select a unique one.