Determining viscosity by a falling-ball method

Practical method

Aim

Determine the viscosity of a liquid by measuring the terminal velocity of a ball falling through it and applying Stokes' law.

Variables

Independent variable: ball radius (optional)
Dependent variable: terminal velocity
Control variables: liquid temperature; the same liquid and tube; balls released centrally.

Apparatus

• Tall measuring cylinder of glycerol
• Steel ball bearings
• Micrometer
• Rubber bands as markers
• Stopwatch or video
• Thermometer
• Magnet to retrieve balls

Method

1. Measure the ball's diameter with a micrometer and look up the densities of steel and glycerol.
2. Place rubber bands around the cylinder: one well below the surface (so the ball has reached terminal velocity) and others 10 cm apart below it.
3. Release the ball at the centre of the surface and time it between the bands. Check that the times between successive pairs of bands are equal.
4. Calculate terminal velocity v = distance ÷ time. Repeat with several balls.
5. Use Stokes' law at terminal velocity: weight = upthrust + drag, so η = \(\dfrac{2r^{2}(\rho _{s} - \rho _{l})g}{9v}\).

Sample results

Steel ball, radius 1.00 mm; glycerol at 20 °C; \(\rho_{\text{steel}}\) = 7800 kg m⁻³, \(\rho_{\text{glycerol}}\) = 1260 kg m⁻³.

Analysis

The equal times show the ball is at terminal velocity, v = 0.0101 m s⁻¹. η = \(\dfrac{2 \times (1.00 \times 10^{-3})^{2} \times (7800 - 1260) \times 9.81}{9 \times 0.0101}\) = 1.41 Pa s, close to the accepted value for glycerol at 20 °C.

Uncertainty

The radius (1.00 ± 0.01 mm, 1%) appears squared (2%). Timing about 10 s with a reaction-time uncertainty of 0.2 s gives 2%. Viscosity falls sharply with temperature, so a 1 °C change alters η by several percent — record the temperature.

Errors and improvements

Risk assessment

Common mistakes

• Forgetting the upthrust term (\(\rho_{s}\) − \(\rho_{l}\)).
• Using the diameter instead of the radius.
• Starting timing at the surface.

Worked analysis

v = \(\dfrac{0.100}{9.9}\) = 0.0101 m s⁻¹. η = \(\dfrac{2r^{2}(\rho _{s} - \rho _{l})g}{9v}\) = \(\dfrac{2 \times 1.00 \times 10^{-6} \times 6540 \times 9.81}{9 \times 0.0101}\) = 1.41 Pa s.

Common misconception

At terminal velocity the ball still moves; the forces on it (weight, upthrust and drag) are balanced, so its acceleration is zero.