Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 81 / 2 / Solution

Choose the helix handedness so its unit tangent is $\mathbf{t}=\cos \alpha \mathbf{i}+\sin \alpha {\mathbf{e}}_{\varphi }$. Let $q={\Omega }_0-\Omega$; the body rotates at $+\Omega \mathbf{i}$, while the flagellum rotates at $-q\mathbf{i}$ in the laboratory, since its relative motor rotation is $-{\Omega }_0\mathbf{i}$. Both translate at $-U\mathbf{i}$. Thus the relative fluid velocity on a flagellar element is $\mathbf{w}=U\mathbf{i}+bq{\mathbf{e}}_{\varphi }$.

Low-Reynolds-number resistive-force theory gives force per unit length on the flagellumIts axial and azimuthal components areDefine the integrated axial resistance matrix of a slender helix coefficientsThe sphere has translational resistance $A_b=6\pi \mu a$ by the Stokes drag law and rotational resistance $H_b=8\pi \mu a^3$. Its hydrodynamic force is $+A_{b}U\mathbf{i}$ and its torque $-H_b\Omega \mathbf{i}$. Integration of $f_x$ and $b{f}_{\varphi }$ gives the zero total force and torque conditionsDividing the first by $L$ and expanding the coefficients gives exactly the two requested balances. The internal motor torque cancels in the total torque balance; it is not an external applied torque.

Eliminate $U$ and use $q+\Omega ={\Omega }_0$. The effective flagellar rotational resistance is $H_e=H-C^2/(A+A_b)$, soThe determinant of helical resistance in resistive-force theory is $AH-C^2=b^2{L}^2{K}_N{K}_T>0$, hence the denominator is positive. Opposite handedness changes the sign of the translation-rotation coupling and reverses the swimming direction.

For $K_T=K$, $K_N=2K$, $\alpha =\pi /4$ and $a=b$, the coefficients are $A=3KL/2$, $C=KaL/2$, $H=3K{a}^{2}L/2$. ThereforeEquivalently, with $h=KL/(\pi \mu a)$, $U=a{\Omega }_{0}4h/(2h^2+21h+48)$. These are positive speed magnitudes for the selected handedness; the translational velocity is $-U\mathbf{i}$.

This helical microswimmer with a spherical head approximation neglects head-tail hydrodynamic interactions as permitted, treats the flagellum as slender and rigid, and averages away transverse forces associated with finite helical ends. The flow is quasistatic Stokes flow, so neither body inertia nor fluid inertia enters the balance.