= Solution
For each incompressible <Stokes flow> without body force, $\partial_j\sigma_{ij}=0$ and
$$
\sigma_{ij}=-p\delta_{ij}+2\mu e_{ij},
\qquad
e_{ij}=\frac12(\partial_i u_j+\partial_j u_i).
$$
The <divergence theorem> gives
$$
\int_Su_i^{(1)}\sigma_{ij}^{(2)}n_j\,dS
=\int_V\partial_j u_i^{(1)}\sigma_{ij}^{(2)}\,dV.
$$
Incompressibility 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 $1\leftrightarrow2$. This proves the <Lorentz reciprocal theorem for Stokes flow>:
$$
\boxed{\int_Su_i^{(1)}\sigma_{ij}^{(2)}n_j\,dS
=\int_Su_i^{(2)}\sigma_{ij}^{(1)}n_j\,dS.}
$$
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, yielding
$$
U_i^{(1)}A_{ij}U_j^{(2)}=U_i^{(2)}A_{ij}U_j^{(1)}.
$$
Since the two vectors are arbitrary,
$$
\boxed{A_{ij}=A_{ji}.}
$$
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 gives
$$
U_\parallel=2U_0\cos\theta,
\qquad
U_\perp=U_0\sin\theta.
$$
Resolving these velocities horizontally and vertically gives
$$
U_H=U_0\sin\theta\cos\theta,
\qquad
U_V=U_0(1+\cos^2\theta).
$$
Therefore
$$
\boxed{\tan\phi=\frac{U_H}{U_V}
=\frac{\sin\theta\cos\theta}{1+\cos^2\theta}.}
$$
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