 Solution

For each incompressible Stokes flow without body force, ${\partial }_j{\sigma }_{ij}=0$ andThe divergence theorem givesIncompressibility removes the pressure term, antisymmetric velocity-gradient parts do not contract with the symmetric stress, and the remaining integrand is $2\mu e_{ij}^{(1)}{e}_{ij}^{(2)}$, symmetric under $1\leftrightarrow 2$. This proves the Lorentz reciprocal theorem for Stokes flow:

By Linearity of Stokes flow, the drag on a translating body is linear in its velocity, so $F_i=A_{ij}{U}_j$. Apply the reciprocal theorem to two translation solutions with velocities ${\mathbf{U}}^{(1)}$ and ${\mathbf{U}}^{(2)}$. Their surface velocities are constant and the far-field contribution vanishes, yieldingSince the two vectors are arbitrary,Thus the hydrodynamic resistance matrix is symmetric and, for fixed fluid viscosity, depends only on body geometry.

A straight rod has fore-aft symmetry about its centre. Gravity acts through that centre, so the translational-rotational coupling and applied torque vanish. Kinematic reversibility of Stokes flow and Uniqueness of Stokes flow then rule out spontaneous rotation: the inclination $\theta$ remains constant.

Let the speed for perpendicular translation be $U_0$. The given factor of two makes the parallel speed $2U_0$ for the same force component. Resolving gravity parallel and perpendicular to the rod givesResolving these velocities horizontally and vertically givesTherefore