The big idea: Drag is the resistive force a fluid (air or liquid) pushes back with when an object moves through it.
Drag gets bigger the faster the object goes — and it always points against the motion.
When the drag grows enough to balance the weight, the object stops speeding up and falls at a steady speed called the terminal velocity.
Spot it: Right after release: no drag (no speed yet) → acceleration ≈ g (the largest it gets).
As speed rises: drag grows → acceleration shrinks.
Drag = weight: acceleration = 0 → terminal velocity reached.
For a small sphere moving slowly through a fluid, the drag is given by Stokes' law. 'Viscosity' (η) is just how thick or sticky the fluid is — honey has a high viscosity, water a low one.
- drag (resistive) force (N)
- viscosity of the fluid (Pa s)
- radius of the sphere (m)
- speed of the object through the fluid (m s⁻¹)
At terminal velocity the object moves at a steady speed, so the forces are balanced: the weight pulling down equals the drag pushing back up.
[Diagram: phys-free-body] - Available in full study mode
- weight (N)
- mass (kg)
- gravitational field strength (9.8 N kg⁻¹)
The terminal-velocity rule: At terminal velocity the net force is zero (steady speed = no acceleration), so set:
weight = drag → mg = 6πηrv.
Then rearrange for whatever the question wants — usually the terminal velocity v.
| Stage of the fall | Weight | Drag | Net force → acceleration |
|---|---|---|---|
| Just released (v = 0) | mg | 0 | mg → a ≈ g (largest) |
| Speeding up | mg | growing | mg − Fd → a shrinking |
| Terminal velocity | mg | = mg | 0 → a = 0 (steady speed) |
Worked example — terminal velocity of a sphere
A small sphere of weight 1.2 × 10⁻⁴ N falls at terminal velocity through oil. The drag on it is given by Stokes' law with 6πηr = 4.0 × 10⁻⁵ N s m⁻¹. Find its terminal velocity.
Solution
- At terminal velocity the forces balance — weight = drag:
- Put in the numbers (mg = 1.2 × 10⁻⁴, 6πηr = 4.0 × 10⁻⁵):
- Rearrange and solve — keep the unit:
Final answer
terminal velocity = 3.0 m s⁻¹.
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How this is tested: Drag and terminal velocity are tested as a force balance.
- Paper 1A: quick MCQs — pick the graph of acceleration (or speed) vs time as drag builds, or work out how the terminal velocity scales when the radius changes. - Paper 2: 'show that' / 'outline' — set weight = drag for an object falling at constant speed (e.g. an oil drop) and solve for a speed, a radius, or a viscosity.
Classic trap: thinking the acceleration is constant while it falls. It is not — acceleration starts near g and decreases to zero as drag grows.
Why bigger spheres fall faster (scaling): Weight grows with volume (∝ r³); Stokes drag grows with radius (∝ r). Setting weight = drag, the terminal velocity scales as v ∝ r².
So doubling the radius makes the terminal velocity 4× bigger (2² = 4).
IB-style question — terminal velocity vs radius
Two solid spheres are made of the same material and fall through the same fluid. Sphere Y has twice the radius of sphere X. Each reaches terminal velocity. Show that sphere Y's terminal velocity is 4 times sphere X's.
Solution
- At terminal velocity, weight = Stokes drag for each sphere:
- Weight uses mass = density × volume, and volume of a sphere ∝ r³, so mg ∝ r³. Put that in and cancel the constants:
- Double the radius ⇒ multiply v by 2²:
Final answer
vY = 4 vX — terminal velocity scales as the radius squared, so doubling r gives 4× the terminal velocity.