Question

A light rod is fitted with a massive sphere as shown. In which case
the rod will fall more rapidly,
if it is placed vertically on end
B or on end A (assuming that
the end of the rod on the
ground does not slip).

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Answer 1

Answer:

As an inclined rod sediments in an unbounded viscous fluid it will drift horizontally

but will not rotate. Whkn it approaches a vertical wall, the rod rotates and so turns

away from the wall. Illustrative experiments and a slender-body theory of this phenomenon are presented. In an incidental study the friction coefficients for an isolated rod

are found by numerical solution of the slender-body integral equation. These friction

coefficients are compared with the asymptotic results of Batchelor (1970) and the

numerical results of Youngren & Acrivos (1975), who did not make a slender-body

approximation.

1. Introduction

Sedimentation of a sphere through a Newtonian liquid in the absence of inertia is

straightforward. In an unbounded fluid the sphere does not rotate and falls in the

direction of gravity. The proximity of a vertical wall induces rotation about a horizontal axis parallel to the surface but causes no drift. This well-known behaviour follows

from the linearity of the Stokes equations and the symmetry of the geometry.

With rodlike particles a more interesting behaviour can be observed. Taylor( 1969)

and others have demonstrated that the sedimentation rate of a rod depends on its

orientation: a slender cylinder falls approximately twice as fast when it is vertical as it

does when it is horizontal. Consequently a rod will drift laterally at intermediate

orientations. The absence of rotation, however, again follows from reversibility.

In a tall container the horizontal drift must eventually bring the particle close to a

side boundary

Explanation: