= Solution
For two <Stokes flows> $(u^{(1)},\sigma^{(1)})$ and $(u^{(2)},\sigma^{(2)})$ of the same <dynamic viscosity> in the same region, with no volume <force>, the <Lorentz reciprocal theorem> states
$$
\boxed{\int_{\partial V}u^{(1)}\cdot\sigma^{(2)}n\,dS
=\int_{\partial V}u^{(2)}\cdot\sigma^{(1)}n\,dS.}
$$
Indeed, the <divergence> of their cross-work difference is
$$
\partial_j\left(u_i^{(1)}\sigma_{ij}^{(2)}-u_i^{(2)}\sigma_{ij}^{(1)}\right)
=2\mu\left(e^{(1)}:e^{(2)}-e^{(2)}:e^{(1)}\right)=0.
$$
The <pressure> terms vanish by <incompressible flow>, and the derivatives of the <Cauchy stress tensors> vanish by the <Stokes equation>. The <divergence theorem> proves the result.
Use the convention that $F,G$ are the <force> and <torque> exerted by the body on the fluid. By <Linearity of Stokes flow>, $(F,G)=\mathsf R(U,\Omega)$. Applying the <Lorentz reciprocal theorem> to two independent rigid motions gives $q_1^T\mathsf Rq_2=q_2^T\mathsf Rq_1$, so the <hydrodynamic resistance matrix> is a <symmetric matrix>. The boundary-power identity gives
$$
q^T\mathsf Rq=F\cdot U+G\cdot\Omega=2\mu\int e:e\,dV.
$$
This is strictly positive for nonzero rigid motion: equality would imply $e=0$, hence a rigid motion throughout the connected fluid, which must vanish since the fluid is at rest at infinity. The <no-slip boundary condition> would then force $q=0$. Therefore $\mathsf R$ is a <positive-definite matrix>. Reversing to the fluid-on-body <force> changes the sign of the force law, not the positive resistance coefficients.
For the two rods, take the <torque> about $O$ and use body axes. On the $x$-rod, $X=(s,0,0)$, $0\leq s\leq2L$, and its <slender-body force density> is
$$
f_x=C\left(\frac{U_x}{2},\ U_y+s\Omega_z,\ U_z-s\Omega_y\right).
$$
On the $y$-rod it is
$$
f_y=C\left(U_x-s\Omega_z,\ \frac{U_y}{2},\ U_z+s\Omega_x\right).
$$
Integrate $f$ and $X\times f$ along both rods, using $\int ds=2L$, $\int s\,ds=2L^2$ and $\int s^2\,ds=8L^3/3$. A convenient dimensionally uniform statement of the complete <right-angle two-rod resistance matrix> is
$$
\boxed{\begin{pmatrix}F_x\\F_y\\F_z\\G_x/L\\G_y/L\\G_z/L\end{pmatrix}
=CL\begin{pmatrix}
3&0&0&0&0&-2\\
0&3&0&0&0&2\\
0&0&4&2&-2&0\\
0&0&2&8/3&0&0\\
0&0&-2&0&8/3&0\\
-2&2&0&0&0&16/3
\end{pmatrix}
\begin{pmatrix}U_x\\U_y\\U_z\\L\Omega_x\\L\Omega_y\\L\Omega_z\end{pmatrix}.}
$$
In physical coordinates this means translation–translation entries scale as $CL$, the two translation–rotation blocks as $CL^2$, and rotation–rotation entries as $CL^3$. The planar block has inverse
$$
\begin{pmatrix}3&0&-2\\0&3&2\\-2&2&16/3\end{pmatrix}^{-1}
=\begin{pmatrix}1/2&-1/6&1/4\\-1/6&1/2&-1/4\\1/4&-1/4&3/8\end{pmatrix}.
$$
This verifies the printed <hydrodynamic mobility matrix> with its third <velocity> component $L\Omega_z$; the TeX aid's $\Omega_x$ is an OCR error.
Take laboratory vertical <velocity> positive upwards. The body axes are $e_1=(\cos\theta,\sin\theta)$ and $e_2=(-\sin\theta,\cos\theta)$. In quasistatic sedimentation the <force> and <torque> exerted on the fluid equal the gravitational resultants on the body:
$$
F_x=-(1+2\lambda)mg\sin\theta,\quad F_y=-(1+2\lambda)mg\cos\theta,
\quad G_z=2\lambda mgL(\sin\theta-\cos\theta).
$$
The out-of-plane block is unforced, so its positive <hydrodynamic resistance matrix> gives $U_z=\Omega_x=\Omega_y=0$. Substitution in the planar <hydrodynamic mobility matrix> gives
$$
\frac{CL}{mg}U_x=\frac{1-\lambda}{6}\cos\theta-\frac{1+\lambda}{2}\sin\theta,\qquad
\frac{CL}{mg}U_y=\frac{1-\lambda}{6}\sin\theta-\frac{1+\lambda}{2}\cos\theta,
$$
and
$$
\boxed{CL^2\dot\theta=\frac{(1-\lambda)mg}{4}(\cos\theta-\sin\theta).}
$$
For $0\leq\lambda<1$, this <ordinary differential equation> has positive right side on $0\leq\theta<\pi/4$, with a stable zero at $\pi/4$. Uniqueness prevents crossing that equilibrium in finite time. For $\lambda>1$ the initial angular <velocity> is negative, and the stable equilibrium reached from zero is $-3\pi/4$. \b[The heavier-end body turns clockwise through $3\pi/4$; it does not settle at $\pi/4$.] More explicitly, with $K=(1-\lambda)mg/(4CL^2)$,
$$
\tan\frac{\theta-\pi/4}{2}=-\tan(\pi/8)e^{-\sqrt2 Kt},
$$
where the continuous branch has $\theta-\pi/4\in(-\pi,0)$. At $\lambda=1$, $\theta$ remains zero.
Transforming the translational <velocity> back to laboratory axes gives
$$
\boxed{U_h=\frac{(1-\lambda)mg}{6CL}\cos2\theta},\qquad
U_v=\frac{mg}{CL}\left[-\frac{1+\lambda}{2}+\frac{1-\lambda}{6}\sin2\theta\right]<0.
$$
For $\lambda\ne1$, eliminate time between $U_h$ and the angular <velocity>:
$$
\frac{dx_O}{d\theta}=\frac{2L}{3}(\cos\theta+\sin\theta),\qquad
x_O(\theta)-x_O(0)=\frac{2L}{3}(1+\sin\theta-\cos\theta).
$$
At either limiting orientation $\sin\theta=\cos\theta$. Thus \b[both cases have the same net horizontal displacement],
$$
\boxed{x_O(\infty)-x_O(0)=2L/3.}
$$
For $\lambda<1$ the drift is monotonically rightwards. For $\lambda>1$ it first moves left, reaching $x_O-x_O(0)=(2L/3)(1-\sqrt2)$ at $\theta=-\pi/4$, then reverses. The total horizontal path length in this second case is $(2L/3)(2\sqrt2-1)$, whereas its net displacement is $2L/3$ rightwards. At $\lambda=1$ it falls vertically without rotation or drift; taking the infinite-time limit before $\lambda\to1$ is consequently singular.
For the requested <sedimentation drift of a weighted two-rod body> sketch, a full parametric trajectory follows by also integrating $U_v/\dot\theta$. Put $\psi=\theta-\pi/4$ and set the initial height to zero:
$$
z_O(\theta)=\frac{2\sqrt2 L(1+2\lambda)}{3(1-\lambda)}
\ln\frac{|\tan(\psi/2)|}{\tan(\pi/8)}
-\frac{2\sqrt2 L}{3}\left(\cos\psi-\frac1{\sqrt2}\right).
$$
It tends to $-\infty$ in either case while $x_O$ tends to $2L/3$. At $\lambda<1$ both rods end pointing upwards from $O$ symmetrically; at $\lambda>1$ they end pointing downwards symmetrically. The asymptotic downward speed is $(1+2\lambda)mg/(3CL)$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-68-sedimentation.png]
{title=Falling two-rod bodies: trajectory of O and successive orientations for lighter and heavier end masses}
{height=640}
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