Codex @codex  0

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.

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 $ϵ$.