Viscosity, drag and a falling sphere

Viscosity describes resistance to flow

Viscosity describes resistance to relative motion between adjacent layers of a fluid. A fluid moving past a solid exerts a drag force opposing their relative motion. Dynamic viscosity η (eta) is measured in Pa s, equivalent to N s m⁻². Density and viscosity describe different properties: one compares mass and volume, the other describes resistance to flow. Neither can be inferred reliably from the other.

For a small sphere moving slowly through a Newtonian fluid, with smooth laminar flow and negligible nearby wall effects, Stokes’ law gives the drag magnitude D = 6πηrv. Here r is the sphere radius and v is its speed relative to the fluid. At fixed r and η, drag is proportional to speed; at fixed v and η, it is proportional to radius. This is not a universal drag law. Large or fast objects that create turbulent wakes require a different model, and a narrow tube can increase drag even without turbulence.

Temperature must be stated because viscosity depends on it. For common liquids, warming usually reduces viscosity, so the same sphere moving at the same speed experiences less viscous drag. Gas viscosity often increases with temperature, so do not turn the liquid trend into a rule for every fluid. A fair comparison of two liquids controls temperature, sphere radius and relative speed; watching a bead fall faster without controlling these quantities does not isolate viscosity.

D = 6πηrv η in Pa s v is speed relative to the fluid

Symbols and units

• D is viscous drag magnitude in N; η is dynamic viscosity in Pa s; r is sphere radius in m.
• v is speed relative to fluid in m s⁻¹. Stokes’ law assumes slow laminar flow around an isolated sphere.

How the equation explains the idea

1. Stokes’ law D = 6πηrv is directly proportional to v if η and r stay fixed. Doubling either speed or radius doubles the drag in this regime.
2. Check units: (Pa s) × m × (m s⁻¹) = Pa m² = N. Use fluid-relative speed, not necessarily speed relative to the laboratory.
3. Nearby walls or turbulent flow can invalidate the relation even when the arithmetic is correct.

Worked example

Calculate a small drag

1. A sphere has radius 0.50 mm = 5.0 × 10⁻⁴ m and speed 0.020 m s⁻¹.
2. The fluid viscosity is 0.80 Pa s. Assume Stokes’ conditions are satisfied.
3. D = 6π × 0.80 × 5.0 × 10⁻⁴ × 0.020 = 1.508 × 10⁻⁴ N.

Drag is about 1.51 × 10⁻⁴ N, opposite the relative motion.

Keep this idea. State the flow assumptions before applying the linear Stokes relationship, and control temperature.

Keep upthrust in the terminal balance

Choose a sphere falling downward through stationary liquid. Its downward weight is ρsphere Vg. Upthrust ρliquid Vg and viscous drag 6πηrv act upward. Downward resultant force is therefore (ρsphere − ρliquid)Vg − 6πηrv. Immediately after release from rest the drag is zero in this simple model, but upthrust is already present. The initial acceleration is less than g when the liquid density is significant.

At terminal speed vt the acceleration is zero. For a sphere of volume 4πr³/3, balancing forces and cancelling common factors gives vt = 2r²g(ρsphere − ρliquid)/(9η). This result describes a denser sphere falling under Stokes’ conditions. At fixed fluid properties, doubling radius predicts four times the terminal speed, but the larger sphere may no longer satisfy the slow-flow assumption. A mathematical scaling prediction must be checked against the physical model.

Rearrange to η = 2r²g(ρsphere − ρliquid)/(9vt) for a viscosity measurement. Neglecting upthrust replaces the density difference by the sphere density and overestimates viscosity. A speed measured before terminal motion is too small and also gives an overestimate. Neither error is repaired merely by repeating the same measurement. If sphere and fluid densities were equal, the ideal steady settling speed would be zero; the formula would not provide a useful falling-time experiment.

At terminal speed: ρsphere Vg = ρliquid Vg + 6πηrvt vt = 2r²g(ρsphere − ρliquid) / (9η)

Symbols and units

• vt is terminal speed in m s⁻¹; r is sphere radius in m; η is viscosity in Pa s.
• ρsphere and ρliquid are densities in kg m⁻³; V = 4πr³/3 is sphere volume in m³; g is field magnitude in m s⁻².

How the equation explains the idea

1. At terminal descent, weight equals upthrust plus drag: (ρsphere − ρliquid)Vg = 6πηrvt.
2. Substitute V = 4πr³/3 and divide by 6πηr to get vt = 2r²g(ρsphere − ρliquid)/(9η).
3. Thus vt is proportional to r² for unchanged fluid and sphere material, but increasing radius can eventually break the slow-flow assumption.

Worked example

A bead settling in a viscous liquid

1. Radius = 0.50 mm, sphere density = 7800 kg m⁻³, liquid density = 1200 kg m⁻³.
2. Use g = 9.81 N kg⁻¹ and η = 0.90 Pa s; the density difference is 6600 kg m⁻³.
3. vt = 2 × (5.0 × 10⁻⁴)² × 9.81 × 6600 / (9 × 0.90) = 0.003997 m s⁻¹.

The predicted terminal speed is about 4.00 mm s⁻¹, provided Stokes’ conditions hold.

Keep this idea. Write the three-force balance before substituting radius, density difference and terminal speed.

Core practical 2: measure and challenge the model

Use a stable, transparent, wide cylinder containing a suitable viscous liquid and small spherical beads of known density. Measure each bead’s diameter in several directions with a zero-checked micrometer and calculate its radius. Determine the liquid density from a measured mass and volume at the experimental temperature, or use a justified supplied value. A thermometer records that temperature. Measure vertical spacing between timing marks with a ruler; use a longer timed distance when possible to reduce the fractional effect of timing uncertainty.

Release one clean bead near the centre without a downward push or attached air bubbles. Put the first timing mark far enough below the surface to allow acceleration to end, and the last mark well above the bottom. Time the bead between two equal-length intervals lower in the liquid, using video or light sensors if available. Equal interval speeds within uncertainty support terminal motion; visibly decreasing interval times show the bead is still accelerating. Wait for the liquid to settle between runs and avoid falling close to previous beads or the wall.

Calculate vt = s/t and then η from the force balance. Repeat using the same bead size to estimate scatter, and compare several suitably small radii. A plot of vt against r² should be approximately straight through the origin under the stated constant-density, constant-temperature model. Systematic curvature or different inferred viscosities with size indicates a limitation, such as wall drag or failure of slow flow. Agreement is supporting evidence, not proof that every assumption is exact.

As an approximate worst-case uncertainty estimate, add fractional uncertainties for products and quotients: η is proportional to r², the density difference, and t/s, so the radius contribution is doubled. Radius inferred from a measured diameter has the same fractional uncertainty as that diameter. For the density difference, combine the absolute uncertainties of the two densities before dividing by the difference. Secure the cylinder, catch spills promptly, keep small beads away from mouths, use a retrieval tool, and have a teacher assess the liquid, glassware and any temperature control. These planning notes do not replace a supervised practical or its local risk assessment.

vt = s/t η = 2r²g(ρsphere − ρliquid)t / (9s)

Symbols and units

• s is timed travel distance in m; t is travel time in s; vt = s/t is terminal speed in m s⁻¹.
• η is viscosity in Pa s; r is sphere radius in m; the density difference is in kg m⁻³ and g in m s⁻².

How the equation explains the idea

1. Time a region after the sphere has reached steady speed, so vt = s/t is a valid estimate.
2. Rearrange the terminal-speed law for η, then substitute s/t: η = 2r²g(ρsphere − ρliquid)t/(9s).
3. At fixed s, radius and densities, greater viscosity gives a longer time. Control temperature because viscosity can depend strongly on it.

Worked example

Evaluate a timed run

1. A 0.50 mm radius bead with density difference 6600 kg m⁻³ travels 0.080 m in 20.0 s after terminal motion is checked.
2. vt = 0.080/20.0 = 0.00400 m s⁻¹.
3. η = 2 × (5.0 × 10⁻⁴)² × 9.81 × 6600 / (9 × 0.00400) = 0.89925 Pa s.
4. With diameter uncertainty 1%, density-difference uncertainty 1%, time uncertainty 1% and distance uncertainty 0.5%, the approximate worst-case total is 2 + 1 + 1 + 0.5 = 4.5%.

Report about 0.90 Pa s at the recorded temperature, with about 5% estimated measurement uncertainty; model bias is additional.

Keep this idea. Measure terminal speed, test whether the model remains appropriate, and distinguish measurement uncertainty from model bias.