Stokes’ viscosity theory
Understand stokes’ viscosity: the equations, assumptions, supporting concepts and measurement method.
01Dynamic viscosity: resistance to shear
A Newtonian liquid supports a shear stress proportional to the rate at which adjacent layers slide past one another. Dynamic viscosity η is the proportionality factor, measured in pascal seconds. It is distinct from density: liquids of similar density can have very different flow resistance. In this model the liquid is uniform, Newtonian and maintained at fixed temperature, so one constant viscosity describes every run.
Is viscosity the same physical quantity as density?
No. Density measures mass per volume; dynamic viscosity measures resistance to shearing motion.
02Three forces on a falling sphere
Gravity acts downward with the weight of the sphere. The surrounding liquid exerts an upward buoyancy equal to the weight of displaced liquid. Once the sphere moves downward, drag also acts upward. Subtracting buoyancy from weight gives the driving force associated with the difference in densities. A sphere denser than the liquid sinks; this lesson uses a positive density difference of 1240 kg/m³.
Why is sphere density alone insufficient in the calculation?
The liquid supports part of the sphere's weight through buoyancy, so the relevant driving force depends on the density difference.
03Stokes drag and creeping flow
For a smooth sphere moving slowly through a Newtonian liquid far from boundaries, Stokes drag is proportional to viscosity, sphere radius and speed. This relation assumes creeping flow, where fluid inertia is negligible compared with viscous effects. A smooth-looking fall alone is not a sufficient test: the Reynolds number must also be small. The virtual parameters keep it below 0.01 over the complete radius range.
What happens to Stokes drag if speed doubles at fixed radius and viscosity?
It doubles because Stokes drag is directly proportional to speed.
04Terminal velocity is a balance of forces
At release the sphere can accelerate. As its speed rises, the drag increases until the downward effective weight and upward drag balance. At terminal velocity its acceleration is zero while it continues moving. Equating the two forces produces a speed proportional to r² and inversely proportional to η. Timing begins only after the settling region, so the calculation uses a steady transit speed rather than an average over initial acceleration.
Does zero acceleration at terminal velocity mean that the sphere has stopped?
No. Its speed is constant and nonzero because the forces balance.
05Timing a known distance
Use the centre of the sphere to identify both mark crossings consistently. Divide the fixed mark separation by elapsed physical time. Repeat timing for each size when using real equipment to assess scatter. The animation runs faster than real transit time; its displayed timer represents the modeled physical time, not wall-clock viewing time. For the 0.100 m gate, the model takes about 493 s at radius 0.300 mm and 54.8 s at 0.900 mm.
If you accidentally time part of the acceleration region, how can viscosity be biased?
The average measured speed can be too low, leading to an overestimated viscosity when the terminal-speed formula is used.
06The straight-line graph and its gradient
Place r² in square metres on the horizontal axis and terminal speed in metres per second on the vertical axis. The gradient s has units inverse metre per second and is 2Δρg/(9η). A larger gradient means larger speeds for the same radius and hence a smaller inferred viscosity. Convert radius from millimetres to metres before squaring; squaring a millimetre value first requires a factor of one million when converting to square metres.
If radius doubles while other parameters stay fixed, how does terminal speed change?
It becomes four times as large because speed is proportional to radius squared.
07Checking the model and reporting uncertainty
Calculate a Reynolds number for each recorded speed using sphere diameter as the length scale. At the largest permitted radius the predicted value is about 0.00345, comfortably below 0.01. In a real narrow tube, nearby walls increase drag and can make an uncorrected viscosity appear too high. Temperature drift, bead size, density and timing also matter. Regression uncertainty describes scatter about the chosen line; it does not include all these systematic contributions.
Does a small Reynolds number eliminate the need to consider tube walls?
No. Creeping flow can still be strongly influenced by nearby walls; the unbounded-fluid assumption must be checked separately.
The Newtonian liquid has a model viscosity of 1.20 Pa s at a fixed temperature. An effectively unbounded liquid is assumed. The simulated Reynolds number stays below 0.004 across the allowed radii. There are no wall/end corrections or temperature changes. Pre-terminal motion is not timed; the displayed fall is accelerated and not to scale.