The big idea: Drop a coffee filter (or a feather) and it flutters down at a slow, steady speed — it never keeps speeding up like a stone.
That's drag: a fluid pushes back harder the faster you move. Once the drag grows to balance the weight, the object stops speeding up — that steady speed is its terminal velocity.
The weight stays the same the whole way down; the drag grows until it equals the weight — then the forces balance (net = 0) and the speed stops changing.
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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.
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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.
Free-body diagram at terminal velocity: the downward weight (mg) is exactly balanced by the upward drag, so the net force is zero.
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- 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) |
A small sphere of weight 1.2 × 10⁻⁴ N falls at terminal velocity through oil. The drag is given by Stokes' law with 6πηr = 4.0 × 10⁻⁵ N s m⁻¹.
Find its terminal velocity.
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Two balls of the same material fall through the same liquid. One is wider. How much faster does it fall?
Twice as wide: 8× heavier, but only 2× the drag: Take a ball and make it twice as wide.
Weight: 8 times bigger. Widening it makes it bigger in every direction at once — taller, wider and deeper — so the new ball holds 2 × 2 × 2 = 8 times as much material.
Drag: only 2 times bigger (at the same speed). Drag cares about how wide the ball is, not about how much material is packed inside it.
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The 30-second method: 1. Ask how many times wider the new one is. Call that number n.
2. Weight: n × n × n times bigger.
3. Drag: only n times bigger — so the speed has to make up the rest: n × n times faster.
Two solid spheres of the same material fall through the same fluid. Sphere Y has twice the radius of sphere X. Each reaches terminal velocity.
Show that Y's terminal velocity is 4× X's.
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The classic trap: Answering with the volume factor. Triple the width and the sphere is 27 times heavier — but its terminal speed is only 9 times bigger, because the drag already grew 3 times on its own.
How this is tested — drag and terminal velocity are tested as a force balance. Two flavours:
Paper 1A
- Pick the graph of acceleration (or speed) vs time as drag builds.
- How the terminal velocity scales when the radius changes.
Paper 2
- 'Show that' / 'outline' — set weight = drag for a constant-speed fall.
- Solve for a speed, a radius, or a viscosity.
The classic trap: Thinking the acceleration is constant while it falls. It is not — it starts near g and decreases to zero as drag grows.
A steel ball bearing of mass 4.5 × 10⁻⁴ kg and radius 1.1 × 10⁻³ m sinks through glycerol at a constant speed of 0.12 m s⁻¹. Take g = 9.8 N kg⁻¹.
Determine the viscosity of the glycerol.
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A droplet of paraffin is released at the bottom of a tank of water and rises at a steady speed. Its weight is 3.6 × 10⁻⁸ N and the viscous drag on it is 0.9 × 10⁻⁸ N. The density of water is 1.0 × 10³ kg m⁻³. (a) Determine the upthrust on the droplet. (b) Determine the density of the paraffin.
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