Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 77 / 1

Choose positive tilt towards $+x$, so the rod tangent is $\mathbf{t}=(\sin \theta ,\cos \theta )$ and its normal is $\mathbf{n}=(\cos \theta ,-\sin \theta )$. The opposite angular convention reverses the signed phase optimum later. In resistive-force theory, a point with centred arclength $s\in [-L/2,L/2]$ has velocity $\mathbf{u}=\dot{x}{\mathbf{e}}_x+s\dot{\theta }\mathbf{n}$, and its force on the fluid is $\mathbf{Du}$, where $\mathbf{D}=c_{\perp }\mathbf{I}+(c_{\parallel }-c_{\perp })\mathbf{tt}$. The force on the rod has the opposite sign.

The pure centred rotational contribution is odd in $s$ and therefore gives zero net force instantaneously in either direction. In the combined oscillating rigid rod in resistive-force theory, however, orientation modulates the translational force. After half a period, $\dot{x}\mapsto -\dot{x}$ and $\theta \mapsto -\theta$. Reflection in the $y$ axis takes one configuration and its forcing into the other. A force component in $x$ reverses under this reflection, whereas a component in $y$ does not. Explicitly,The first is half-period antisymmetric and the second is half-period symmetric. Hence $⟨F_x⟩=0$, while symmetry permits $⟨F_y⟩\ne 0$. The possible mean $y$ force comes from the coupled translation and rotation, not from pure centred rotation alone.

In the oscillating rigid rod in resistive-force theory, $\sin \theta \cos \theta =\theta +O({ϵ}^3)$ and $\dot{x}=-ϵa\omega \sin \omega t$. Sincethe mean transverse force from a rocking rod, on the fluid in the tilt convention of part (a), isThe parallel and perpendicular drag coefficients of a slender filament are per unit length here. Their difference is essential: isotropic local drag gives zero transverse force. The hydrodynamic force on the rod is the negative of the displayed answer.

For the usual anisotropy $c_{\perp }>c_{\parallel }$ and $a>0$, the maximum positive mean transverse force from a rocking rod occurs atThe maximum magnitude has either quadrature phase $\varphi =\pm \pi /2$. If positive tilt is defined towards $-x$, the positive-force optimum is instead $+\pi /2$. At our optimum $\theta =ϵ\sin \omega t$ is opposite in sign to $\dot{x}$: the rod tilts so that the weaker tangential drag and stronger normal drag combine into a positive $y$ force on both strokes.

For $\varphi =0$ or $\pi$, orientation is a single-valued function of the centre displacement. The configuration retraces its path under time reversal, and kinematic reversibility of Stokes flow makes the force impulse cancel. Equivalently, $⟨\theta \dot{x}⟩=0$. The enclosed loop in the displacement-orientation plane is largest in quadrature and vanishes for a reciprocal stroke, illustrating the principle behind the scallop theorem.

The instantaneous work on the fluid equals the integral of $\mathbf{u}\cdot \mathbf{Du}$. For the oscillating rigid rod in resistive-force theory, $\mathbf{u}\cdot \mathbf{t}=\dot{x}\sin \theta$ and $\mathbf{u}\cdot \mathbf{n}=\dot{x}\cos \theta +s\dot{\theta }$. Odd terms in $s$ integrate to zero, givingThus the leading power of a rocking rod isBoth terms are quadratic in $ϵ$. Small angle does not make rotation a higher-order contribution: points away from the centre have rotational speed $s\dot{\theta }=O(ϵ\omega L)$, just as the translational speed is $O(ϵ\omega a)$. Their ratio is $L^2/(12a^2)$, controlled by geometry rather than by $ϵ$.

Use the printed Maxwell-filtered local drag law at fixed material arclength, after transients have decayed to a periodic response. Averaging the force equation over a period makes the time-derivative term vanish. Therefore the mean force equals the mean instantaneous resistive-force theory driving force, even though its oscillating part is filtered. In particular,with the same sign convention as before.

At first order the tangent is ${\mathbf{e}}_y$ and both velocity components are single harmonics of frequency $\omega$. The linear Maxwell fluid response multiplies their complex force amplitudes by $(1+i\omega \lambda {)}^{-1}$. Only its in-phase real part contributes to the mean work. ConsequentlyThe leading power is strictly smaller when $\lambda >0$ and $\omega >0$, and equal in the Newtonian limit. The zero-frequency component generating the mean force is unattenuated, whereas elastic storage and phase lag reduce the dissipative harmonic response. This conclusion concerns the specified linear local force law and prescribed kinematics; it is not a claim for every nonlinear viscoelastic constitutive model.