Wait, What? The useful part of a falling-ball viscosity experiment begins only after the ball has stopped accelerating.

A sphere released into a viscous liquid first accelerates. Its weight exceeds the upward forces. As speed rises, viscous drag rises too. Eventually drag plus upthrust balance weight and acceleration becomes approximately zero. Only then does a constant-speed measurement represent terminal velocity—the quantity needed for the familiar Stokes-law inference.

The causal model

For a small sphere moving slowly through a Newtonian fluid in the Stokes regime, viscous drag is F_d = 6πηrv. At terminal speed v_t, weight is balanced by buoyancy plus drag. Combining the force balance gives η = 2r²g(ρ_s−ρ_f)/(9v_t). This equation is powerful precisely because it is conditional.

How do we know terminal speed has been reached?

Place several equally spaced markers down the tube and measure transit times between successive markers. If equal distances take approximately equal times, speed has become approximately constant. The Institute of Physics recommends this pattern as a direct way to reveal terminal motion before advanced students infer viscosity. IOPSpark: falling through a high-viscosity liquid.

Measure diameter carefully: radius is squared

The inferred viscosity depends on r². A 2% error in radius contributes roughly 4% to the viscosity result before other uncertainties are included. Use a micrometer for small spheres, measure more than one orientation if needed, and do not treat a nominal manufacturer diameter as perfect evidence.

Density and buoyancy are not optional corrections

The sphere does not fall under its full weight alone. The liquid exerts an upthrust equal to the weight of displaced fluid. That is why the density difference (ρ_s−ρ_f) appears in the equation. Ignoring fluid density systematically overestimates the driving force.

Quantitative window

Suppose r = 1.00 mm, ρ_s = 7800 kg m⁻³, ρ_f = 1260 kg m⁻³ and terminal speed is 0.050 m s⁻¹. Then η ≈ 2(0.001)²(9.81)(6540)/(9×0.050) ≈ 0.285 Pa s. The numerical answer is meaningful only if the sphere was actually in the Stokes regime and wall effects were small.

The hidden assumption: Stokes flow

Stokes’ drag law applies when inertial effects are small relative to viscous effects, commonly associated with low Reynolds number. A large sphere falling rapidly through a low-viscosity liquid may violate this assumption. The data can still show terminal velocity, but the simple 6πηrv drag model may no longer be valid.

Walls can make the fluid look more viscous

A narrow tube constrains the flow around the sphere and can increase drag. Keep sphere diameter small relative to tube diameter or use an appropriate correction in higher-resolution work. Also release the sphere near the centre so it does not rub the wall.

Temperature is part of the measurement

Liquid viscosity can be strongly temperature-dependent. Record temperature and keep it stable. Comparing two trials at different temperatures without acknowledging this is not a clean repeat measurement.

Observation versus inference

Observation: a sphere took 0.82 s, 0.81 s and 0.83 s to cross three equal 4.0 cm intervals. Inference: its speed was approximately constant in that region. Further inference: the force resultant was approximately zero. Only after adding the Stokes-model assumptions may we infer viscosity from that terminal speed.

Misconceptions and failure modes

• Timing immediately after release, while the sphere is still accelerating.
• Assuming terminal velocity means no forces act; it means the forces balance.
• Ignoring buoyancy.
• Using a tube so narrow that wall effects dominate.
• Allowing air bubbles to cling to the sphere and change effective drag/buoyancy.
• Changing temperature between trials.
• Applying Stokes’ law outside its low-Reynolds-number regime.

Unfamiliar transfer: raindrops and sediment

Raindrops, pollen, sediment and aerosol particles can also approach terminal speeds, but their drag laws need not be Stokes-like. The transferable reasoning is: identify forces, establish whether speed is steady, choose a drag model appropriate to the regime, and state the model boundary.

Secondary → JC → deeper Science

Secondary: recognise balanced forces and terminal speed. JC: derive the viscosity relation from weight, upthrust and Stokes drag; propagate measurement uncertainty and test assumptions. Deeper Science: connect to Reynolds number, non-Newtonian rheology, wall corrections, particle settling and microfluidics.

Checkpoint

A ball takes progressively less time across the first three equal intervals but equal times across the next three. Which data should be used for a Stokes-law viscosity estimate?

Answer key and WHY reasoning

Use the later equal-time intervals. The decreasing early times show acceleration; the later constant transit times support approximately constant terminal speed, which is required by the force-balance derivation.

Evidence boundaries

This experiment estimates viscosity under the tested temperature, sphere size and flow regime. Agreement with Stokes’ law does not establish that the fluid is Newtonian under every shear rate or temperature, nor that the same drag law applies to large or irregular objects.

Authoritative next steps

• Institute of Physics: Falling through a high-viscosity liquid
• Institute of Physics: A-level practical requirements

Teaching Guide

Do not begin with the viscosity formula. Begin with interval times. Ask students to locate where the motion becomes constant, draw the force diagram there, and only then derive the measurement equation. The experiment becomes a chain of evidence rather than a substitution exercise.