Falling-ball viscometry: is the discrepancy just 'experimental error'?
A student drops a steel ball of diameter 20.0 mm and density 7.8 × 10³ kg m⁻³ into a 1.00 m tall measuring cylinder filled with glycerol of density 1.26 × 10³ kg m⁻³. Timing marks are 0.10 m and 0.90 m below the surface, and the ball passes them 1.32 s apart. The student notes that the gap between successive 0.10 m marks is the same over the lower half of the fall. Using Stokes' law, F = 6πηrv, the student calculates η = 2.3 Pa s to two significant figures. The accepted value at the laboratory temperature of 21 °C is 1.4 Pa s. The cylinder's internal diameter is 0.100 m. The student's partner argues that the discrepancy is simply 'experimental error' and that a second run with a larger steel ball would remove it. The student also has a stopwatch reading to 0.01 s, a metre rule reading to 1 mm, and no thermostat.
1. Show that the ball is at terminal velocity at the first timing mark, and use the data to obtain a value for η, stating the assumptions you make.
Read the worked answer
Terminal velocity. The separation of the ball from successive 0.10 m marks is constant in the lower half of the fall, so the speed is constant. Constant speed with a constant weight and upthrust means the resultant force is zero, i.e. the ball has reached terminal velocity before the first mark.
v = 0.80 m / 1.32 s = 0.606 m s⁻¹.
Force balance at terminal velocity: weight = upthrust + viscous drag (4/3)πr³ρ_s g = (4/3)πr³ρ_f g + 6πηrv η = 2r²(ρ_s − ρ_f)g / (9v)
With r = 0.0100 m and ρ_s − ρ_f = (7.80 − 1.26) × 10³ = 6.54 × 10³ kg m⁻³: η = 2(1.00 × 10⁻²)²(6.54 × 10³)(9.81) / (9 × 0.606) η = 12.83 / 5.454 = 2.35 Pa s ≈ 2.3 Pa s.
This reproduces the student's value.
Assumptions: the ball falls along the axis of the cylinder; the flow round it is laminar and steady; there is no wall effect; Stokes' law applies to a smooth sphere; the densities quoted apply at 21 °C. The force balance also assumes zero acceleration, which the constant spacing supports.
2. Assess the partner's claim. Use a Reynolds-number estimate and the finite cylinder diameter to decide whether either effect can account for the factor of about 1.6, and state which change to the experiment would genuinely reduce the discrepancy.
Read the worked answer
Reynolds number: Re = ρ_f v d / η = (1.26 × 10³)(0.606)(0.0200) / 2.35 ≈ 6.5.
Stokes' law is only reliable for Re well below about 1. At Re ≈ 6.5 the wake contributes extra inertial (form) drag, so the real drag exceeds 6πηrv. A larger drag at the observed terminal speed means the true η is smaller than the value obtained by assuming Stokes' law: this error inflates η.
Wall effect: r/R = 0.0100/0.0500 = 0.20, so the cylinder walls retard the ball. The Ladenburg correction v_∞ ≈ v(1 + 2.4r/R) gives a factor 1.48, i.e. η ≈ 2.35/1.48 ≈ 1.6 Pa s. Because r/R = 0.20 is far from the r/R ≪ 1 limit, the correction is itself only approximate, but its direction is reliable - it also inflates η. Combining both effects with the strong temperature dependence of glycerol's viscosity moves the result towards 1.4 Pa s.
Verdict: the partner is wrong. Both corrections push η downwards, so this is systematic, not random, error. A larger ball raises both Re and r/R and makes matters worse. A smaller, less dense ball in a wider cylinder (with both corrections applied and the temperature controlled) is the correct change.
3. The student wants to quote a percentage uncertainty for η. Given diameter ±0.05 mm, fall distance ±2 mm, time ±0.05 s and each density to ±1%, decide which quantity should be improved first, and identify the uncertainty that the budget omits.
Read the worked answer
Since η ∝ r²(ρ_s − ρ_f)/v and v = s/t:
- radius: 2 × (0.05/20.0) = 0.5%
- density difference: Δρ = 6540 kg m⁻³ with δ(Δρ) ≈ 78 + 13 = 91 kg m⁻³, i.e. 1.4%
- distance: 2/800 = 0.25%
- time: 0.05/1.32 = 3.8%
- g: negligible
Adding in quadrature: √(0.5² + 1.4² + 0.25² + 3.8²) ≈ 4.1%.
Timing dominates, so the first improvement is to replace the stopwatch by light gates (or frame-by-frame video), which also removes reaction-time bias; a longer fall between the marks would help further. The next largest term is the density difference, which is limited by the glycerol temperature - a thermostat would fix both ρ_f and the viscosity used for comparison.
The budget omits the dominant term. The wall and Reynolds-number corrections are systematic and together are worth tens of per cent - larger than every random term combined. Quoting a percentage uncertainty from random terms alone would badly understate the total uncertainty in η.