A.2.10—Viscous drag
- Syllabus
- First assessment 2025
- Objective
- —
- Level
- SL
Stokes drag
For a small sphere moving slowly through a viscous fluid,
Fd=6πηrv
where η 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 η as the same quantity as drag force, and do not forget that the formula applies to the stated small-sphere, viscous-flow model.
The evidence asks why a droplet’s acceleration changes and asks for the shape of acceleration against velocity during a fall.
Describe / Identify
As a droplet speeds up, use the given drag model to explain that drag increases, so the net force and acceleration change. At terminal speed, drag balances the driving force and acceleration is zero.
Claiming that acceleration remains constant at g even after viscous drag becomes significant.
Representative question
Describe why the acceleration of the oil droplet changes.
As the speed of the droplet increases, the drag force increases
net force changes/decreases
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