Viscosity and Stokes’ law
Viscosity: friction inside a fluid
Tip a spoon of water and a spoon of honey. The water runs off at once; the honey oozes out slowly. Honey resists flowing much more. We say it is more viscous.
Viscosity is like friction inside a fluid. When one layer of fluid slides over another, the layers drag on each other and oppose the motion.
Picture a thin layer of oil between a fixed bottom plate and a top plate moving at speed v. The oil touching the bottom plate stays at rest. The oil touching the top plate moves with it.
In between, the layers slide over each other, each a bit faster than the one below. The speed grows steadily from zero to v across the thickness l. This smooth, layered flow is called laminar flow.
To keep the top plate moving, you must keep pulling with a force F. Experiments show that the force per unit area, F over A, is proportional to v over l, the rate at which the speed changes across the layers.
The constant is the coefficient of viscosity, η: F over A equals η v over l. Its SI unit is the pascal second, or newton second per square metre. The old unit, the poise, is 0.1 pascal second.
Water at 20 degrees has η of about one thousandth of a pascal second. Blood is about three times as viscous, and glycerine is nearly a thousand times more viscous than water.
Heat a liquid and its viscosity drops quickly: warm honey and warm engine oil flow easily. Gases are the opposite: their viscosity rises with temperature.
Worked example: a block of area 0.10 square metres slides at a steady 0.085 metres per second on an oil film 0.30 millimetres thick, pulled by a 0.010 kilogram hanging mass. The pull is 0.098 newtons.
So η equals F l over A v: 0.098 times 0.30 times 10⁻³, divided by 0.10 times 0.085. That is about 3.5 times 10⁻³ pascal seconds.
Stokes’ law and terminal velocity
Drop a steel ball into a tall jar of glycerine. It speeds up for a moment, then sinks at a steady speed. Something must be balancing gravity.
As the ball moves, it drags the layers of liquid around it, and the liquid drags back on the ball. This is the viscous drag force.
George Stokes showed that for a small sphere of radius a moving slowly at speed v, the drag is F equals 6 π η a v. This is Stokes’ law. The drag grows in proportion to the speed.
Watch the forces on the falling ball. The weight pulls down. The buoyant force of the displaced liquid pushes up. And the drag pushes up, growing as the ball speeds up.
Soon the drag plus the buoyancy equals the weight. The net force is zero, so the ball stops speeding up. It now falls at a constant speed called the terminal velocity, v t.
Let the ball’s density be ρ and the liquid’s density σ. Balance the forces: 6 π η a v t equals four thirds π a cubed, times ρ minus σ, times g.
So v t equals 2 a squared times ρ minus σ times g, divided by 9 η. A bigger ball falls faster: double the radius and the terminal velocity becomes four times larger.
This is why raindrops do not hit us like bullets. Air drag gives them a modest terminal velocity. Tiny cloud droplets have such a small terminal velocity that they seem to float.
Worked example: a copper ball of radius 2.0 millimetres falls through oil at a terminal velocity of 6.5 centimetres per second. Copper’s density is 8900, the oil’s 1500 kilograms per cubic metre.
Rearranging, η equals 2 a squared times ρ minus σ times g, over 9 v t. That is 2 times 4 times 10⁻⁶ times 7400 times 9.8, over 9 times 0.065: about 0.99 pascal seconds.
Try it: a ball in a viscous liquid
- Press Drop (or Space). The ball reaches its terminal velocity within milliseconds, then falls steadily.
- At terminal velocity the drag (amber) plus buoyancy (blue) exactly balance the weight (red).
- Double the radius: vₜ becomes four times bigger, because vₜ ∝ r².
- Try a glass ball: its density is closer to the liquid’s, so it falls much more slowly.
At terminal velocity
A steel ball falls through glycerine at its terminal velocity. Which statement is correct?
- The viscous drag equals the weight of the ball
- The net force on the ball is zero, so its velocity is constant
- The ball’s acceleration is still g, but the drag slows it
- The drag is zero because the speed is no longer changing
Answer
B. At terminal velocity the forces balance: drag + buoyancy = weight. The net force is zero, so there is no acceleration and the speed is constant. The drag alone is less than the weight, because buoyancy helps hold the ball up. The drag is not zero: it is at its largest, 6πηavₜ.
A bigger ball
A small steel ball falls through oil with a terminal velocity of 1.5 cm/s. A steel ball with twice the radius is dropped into the same oil. What is its terminal velocity?
- 0.75 cm/s
- 3.0 cm/s
- 6.0 cm/s
- 12 cm/s
Answer
C. vₜ = 2a²(ρ − σ)g / 9η, so vₜ ∝ a². Doubling the radius makes vₜ 2² = 4 times larger: 4 × 1.5 = 6.0 cm/s. The weight grows as a³, but the drag only as a, so the bigger ball must go faster before they balance.
Summary
- Coefficient of viscosity:
F / A = η v / l (unit: Pa·s; 1 poise = 0.1 Pa·s) - Stokes’ law:
F = 6πηav - Terminal velocity:
vₜ = 2a²(ρ − σ)g / 9η
- Viscosity is internal friction between layers of a fluid moving at different speeds.
- The viscosity of a liquid falls as it is heated; the viscosity of a gas rises.
- A falling sphere reaches terminal velocity when drag + buoyancy = weight; then it moves at constant speed.
- vₜ ∝ a²: bigger drops and balls fall faster, which is why tiny cloud droplets seem to float.