Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 38C (Fluid Dynamics II)

Choose the vertical plane through the rod's axis. Reflection in that plane leaves the horizontal rod, the vertical gravitational force, and the Stokes equations unchanged, but reverses any velocity normal to the plane. A second reflection in the vertical plane perpendicular to the rod reverses velocity along the rod. By uniqueness of Stokes flow, both horizontal velocity components vanish. The rod therefore translates vertically.

Equivalently, the hydrodynamic resistance matrix of an axisymmetric rod has one coefficient parallel to its axis and another perpendicular to it. A vertical force on a horizontal rod lies entirely in the perpendicular eigenspace, so its velocity is vertical.

The uniform rod is invariant under inversion through its centre. This centrosymmetry makes the translation-rotation blocks of its generalized hydrodynamic resistance matrix vanish: a polar translation velocity changes sign under inversion, whereas an axial angular velocity does not, so an invariant linear coupling between them is impossible.

Gravity acts through the centre of mass and supplies no torque. The remaining rotational resistance is positive definite, so zero applied torque implies zero angular velocity. Thus an initially inclined rod keeps its orientation while settling.

At negligible Reynolds number, the Stokes equations and their no-slip boundary conditions are linear in velocity. Scaling or superposing a prescribed rigid translation therefore scales or superposes the resulting stress and total hydrodynamic force. Hence, for a fixed rod orientation,The minus sign records that viscous drag opposes motion, and $\mathbf{R}$ is the translational hydrodynamic resistance matrix.

The Lorentz reciprocal theorem for Stokes flow states that two Stokes solutions $({\mathbf{u}}^{(1)},{\mathbf{\sigma }}^{(1)})$ and $({\mathbf{u}}^{(2)},{\mathbf{\sigma }}^{(2)})$ in the same fluid domain satisfyApply it to the same rod translating with arbitrary velocities ${\mathbf{U}}_1$ and ${\mathbf{U}}_2$. The integral at infinity vanishes, and the rod-surface integrals giveSince ${\mathbf{F}}_i=-\mathbf{R}{\mathbf{U}}_i$,for every pair of vectors. Therefore

The power supplied to overcome drag equals the viscous dissipation:For a nonzero rigid translation the strain cannot vanish throughout a fluid at rest at infinity, so the integral is strictly positive. Since $\mathbf{F}=-\mathbf{RU}$,Thus the symmetric resistance matrix is positive definite.

Let $\mathbf{e}$ be the downward vertical unit vector and let the rod's weight after buoyancy be $W\mathbf{e}$. Force balance givesBecauseits matrix inverse isUsing $\mathbf{e}\cdot \mathbf{t}=\cos \theta$ givesThereforeandThe required angle is consequentlyIf $c_1=c_2$, then $\cos \alpha =1$, so the rod falls vertically. For an anisotropic rod with $c_2>c_1$, the component parallel to its long axis encounters less drag, and the settling direction is generally deflected toward that axis.