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Stokes drag

For a small sphere moving slowly through a viscous fluid,

$F_d=6\pi\eta r v$

where $\eta$ is viscosity, $r$ is sphere radius and $v$ is speed relative to the fluid.

Drag opposes motion

The drag force points opposite the sphere’s velocity. As speed increases, drag increases linearly in this model, reducing the resultant force when the driving force is fixed.

Approach to terminal speed

For a falling sphere, weight drives the motion and viscous drag grows with speed. When drag balances the effective weight, acceleration becomes zero and terminal speed is reached.

Common trap

Do not treat viscosity $\eta$ as the same quantity as drag force, and do not forget that the formula applies to the stated small-sphere, viscous-flow model.

[ 2 ]

Describe why the acceleration of the oil droplet changes.

Build the force model

Choose the system, draw a labelled free-body diagram, classify the interactions and resolve components. Apply Newton’s laws with the correct boundary: contact forces, field forces, friction, tension, buoyancy and restoring forces each have their own direction and conditions.

Track momentum

Use $ec p=m ec v$ , $ec J=\Delta ec p$ and momentum conservation only after checking external impulse. Distinguish elastic and inelastic collisions, explosions and energy transfer.

Track circular motion

The inward resultant provides $a_c=v^2/r=\omega^2r$ . It may come from tension, gravity, normal, friction or a field force. Angular and linear descriptions are linked by $v=\omega r=2\pi r/T$ .

Final checks

Ask: Which body is the system? Which forces are external? Is mass constant? Is acceleration uniform or radial? Is kinetic energy conserved, transferred or increased?

A.2.10—Viscous drag

Model Viscous Drag on a Small Sphere

Retrieve the A.2 Forces and Momentum Model

A.2.10 Exam Analysis