Settling of an asymmetric dumbbell in a quiescent fluid

Abstract

We compute the hydrodynamic torque on a dumbbell (two spheres linked by a massless rigid rod) settling in a quiescent fluid at small but finite Reynolds number. The spheres have the same mass densities but different sizes. When the sizes are quite different the dumbbell settles vertically, aligned with the direction of gravity, the largest sphere first. But when the size difference is sufficiently small then its steady-state angle is determined by a competition between the size difference and the Reynolds number. When the sizes of the spheres are exactly equal then fluid inertia causes the dumbbell to settle in a horizontal orientation.

Figures (1)

• Figure 1: Illustration of the dumbbell and the notation used in this article. The centres of two spheres ($\rm A$ and $\rm B$) of diameters $b+\delta b$ and $b$ are linked by a massless rigid rod of length $a$. The centre-of-mass of the dumbbell is denoted by $C$, and its velocity by $\hbox{\boldmath$v$}$. The angle of the centre-of-mass velocity with the direction of gravity (negative ${\bf e}_2$-direction) is denoted by $\beta$. As drawn the angle is negative, $\beta< 0$. Since we consider a steady state where the angular velocity of the dumbbell vanishes (see text), the centre-of-mass velocities of both spheres are also $\hbox{\boldmath$v$}$. The angle of inclination of the dumbbell (defined by the unit vector $\hat{\hbox{\boldmath$n$}}$) with respect to the plane orthogonal to gravity (the ${\bf e}_1$-${\bf e}_3$-plane) is denoted by $\alpha$.