= Solution
For rotation about the <hydrodynamic centre of a two-sphere dumbbell>, the sphere centres have <velocities> $\mathbf v_i=\boldsymbol\Omega\times(\mathbf r_i-\mathbf r_c)$. Their resultant drag <force> in the <free-draining approximation> is
$$
\mathbf F=-\sum_i\zeta_i\boldsymbol\Omega\times(\mathbf r_i-\mathbf r_c)
=-\boldsymbol\Omega\times\left(\sum_i\zeta_i\mathbf r_i-(\zeta_1+\zeta_2)\mathbf r_c\right)
=\boxed{\mathbf0.}
$$
Each sphere also experiences a spin drag <torque>, but its isolated rotation causes no net <force>. Those spin contributions therefore do not alter the cancellation.
This illustrates <translation-rotation reciprocal coupling>. About any fixed origin, write the fluid's rigid-body <force> and <torque> as
$$
\begin{pmatrix}\mathbf F\\\mathbf T\end{pmatrix}=-\begin{pmatrix}A&B\\B^T&C\end{pmatrix}\begin{pmatrix}\mathbf U\\\boldsymbol\Omega\end{pmatrix}.
$$
The transposed cross-blocks follow from the <Lorentz reciprocal theorem>. If pure translation causes zero <torque> about an origin for every $\mathbf U$, then $B^T=0$, hence pure rotation about that origin causes zero <force>. This implication is general for linear <Stokes flow>; the existence and leading location of such an origin here use the two-sphere geometry and the approximation in part <2/c>.
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