Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 329 / 2

For two body-force-free Stokes flows $({\mathbf{u}}^{(1)},{\mathbf{\sigma }}^{(1)})$ and $({\mathbf{u}}^{(2)},{\mathbf{\sigma }}^{(2)})$ in the same domain, the Lorentz reciprocal theorem for Stokes flow isIndeed, the difference of the two volume integrands isbecause both flows are incompressible and the Newtonian fluid stress tensor is symmetric. The divergence theorem proves the boundary identity.

On a rigid body, ${\mathbf{u}}^{(r)}={\mathbf{U}}^{(r)}+{\mathbf{\Omega }}^{(r)}\times \mathbf{x}$. The reciprocal theorem becomesWriting $(\mathbf{F},\mathbf{G}{)}^T=\mathsf{R}(\mathbf{U},\mathbf{\Omega }{)}^T$ and choosing arbitrary pairs of rigid velocities shows thatThus the hydrodynamic resistance matrix is symmetric.

The body is invariant under the reflections and half-turns that preserve its $z$-axis. A polar vector force $F{\mathbf{e}}_z$ can therefore produce only the polar velocity $U_z{\mathbf{e}}_z$; every rotation is excluded because angular velocity is a pseudovector. Hence only $U_z$ is nonzero when $\mathbf{F}=(0,0,F)$ and $\mathbf{G}=0$.

For $\mathbf{F}=(F,0,0)$, the surviving symmetry-allowed components arewhile $U_y,U_z,{\Omega }_x,{\Omega }_z$ vanish. The coupling between translation in $x$ and rotation about $y$ is permitted because the two horizontal rods lie at opposite vertical offsets.

For pure translation, integrate the slender-body force density over each rod and take its moment about $O$. The three coordinate directions giveThuswhen $\mathbf{\Omega }=0$.

LetThe symmetry proved in part (a) determines the force generated by rotation from the translation-generated couple. Combining this with the given rotational resistance gives the complete matrix

The zero-couple equations in the two coupled horizontal blocks areand ${\Omega }_z=0$. The force equations then becomeThereforeand

A vector fixed in space has body-frame derivative of a space-fixed vectorSubstitution of part (d) gives

If $\mathbf{\Omega }$ is constant and nonzero, then $k_x$ and $k_y$ are constant and not both zero. Their evolution equations force $k_z=0$, and the last equation then gives $k_x^2=k_y^2$. Hence

For $\mathbf{k}=(1,1,0)$, both the translational velocity and the angular velocity are parallel to the fixed vertical force, with the latter oppositely directed. The body therefore falls on a straight vertical line while spinning steadily about that line. Its body $z$-axis remains horizontal, and its $x$- and $y$-axes remain at $45^{\circ }$ to the vertical.

For the initial condition $(1,0,0)$, symmetry preserves $k_y=0$. With $c=\gamma /L$ the exact body-frame solution isThe body rotates about its $y$-axis, its fall path bends slightly because the horizontal and axial mobilities differ, and $\mathbf{k}\to (0,0,-1)$. Thus the body $z$-axis becomes vertical and the angular velocity tends to zero; asymptotically it falls without rotating.