= Solution
For the usual anisotropy $c_\perp>c_\parallel$ and $a>0$, the maximum positive <mean transverse force from a rocking rod> occurs at
$$
\boxed{\phi=-\pi/2\pmod{2\pi},\qquad
\langle F_y\rangle_{\max}=\tfrac12(c_\perp-c_\parallel)L\epsilon^2a\omega.}
$$
The maximum magnitude has either quadrature phase $\phi=\pm\pi/2$. If positive tilt is defined towards $-x$, the positive-<force> optimum is instead $+\pi/2$. At our optimum $\theta=\epsilon\sin\omega t$ 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 $\phi=0$ 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, $\langle\theta\dot x\rangle=0$. 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>.
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