Viscous drag and Stokes' lawEdexcel A-Level Physics: Revision notes

A fluid resists the movement of an object through it. This resistance, the viscous drag, acts in the opposite direction to the motion and increases as the speed increases. How strongly a fluid resists flow is measured by its viscosity, η, with unit Pa s (equivalently N s m⁻²). Glycerol is very viscous (about 1.4 Pa s at 20 °C), while water is far less viscous.

For a small sphere of radius r moving at speed v through a fluid of viscosity η, the drag is given by Stokes' law:

$F = 6\pi\eta r v$

Stokes' law applies only when:

• the object is a small sphere
• it moves at low speed
• the flow past it is laminar (smooth layers, no eddies)

At higher speeds the flow becomes turbulent and the drag is larger than 6πηrv. Rearranging gives η = F/(6πrv), so the unit of η is N s m⁻².

Worked example: r = 1.5 mm, v = 0.020 m s⁻¹, η = 1.4 Pa s gives F = 6π × 1.4 × 1.5 × 10⁻³ × 0.020 = 7.9 × 10⁻⁴ N.

A ball released from rest in a fluid accelerates, and the drag grows with speed. It reaches a constant terminal velocity when the forces balance:

$W = U + F$

where W is the weight, U the upthrust and F the viscous drag. Substituting W − U = (4/3)πr³(ρ_s − ρ_f)g and F = 6πηrv gives

$\eta = \frac{2r^2(\rho_s-\rho_f)g}{9v}$

Worked example: a steel ball (r = 2.0 mm, ρ = 7800 kg m⁻³) falls at 0.085 m s⁻¹ in oil of density 920 kg m⁻³. Weight − upthrust = 2.26 × 10⁻³ N, so η = 2.26 × 10⁻³ / (6π × 2.0 × 10⁻³ × 0.085) = 0.71 Pa s.

The viscosity of a liquid decreases as its temperature rises. At higher temperature the molecules have more energy and the forces between layers are overcome more easily, so the liquid flows more readily. For the same ball, a lower viscosity means a smaller drag at a given speed, so the ball must travel faster before the forces balance and its terminal velocity is higher. Since η ∝ 1/v for the same ball and liquid, a ball falling 6.5 times faster means the viscosity is 6.5 times smaller.

To find the viscosity of a liquid such as glycerol:

1. Measure the ball's diameter with a micrometer at several places and take a mean, giving r.
2. Use known densities (or find them with a balance and measuring cylinder) to calculate the weight minus upthrust.
3. Mark two lines on a tall cylinder well below the surface, so the ball has reached terminal velocity, and measure the distance between them.
4. Release the ball at the centre of the surface and time it between the marks. Repeat and average.
5. Calculate v and then η = F/(6πrv).

Improve accuracy with a wide cylinder (to reduce wall effects), a small ball (to keep flow laminar) and a recorded temperature.