A brief note on Viscosity and Surface Tension . Read it before a test. Try each question at the end before you open its answer.
The big ideas Viscosity : F = η A v d F = \eta A\dfrac{v}{d} F = η A d v , with η \eta η in Pa s \text{Pa s} Pa s . Warm liquids are thinner; warm gases are thicker. Stokes' law : 6 π η r v 6\pi\eta rv 6 π η r v ; terminal speed v t = 2 r 2 ( ρ − σ ) g 9 η v_t = \dfrac{2r^2(\rho - \sigma)g}{9\eta} v t = 9 η 2 r 2 ( ρ − σ ) g . Reynolds number : R e = ρ v d η Re = \dfrac{\rho vd}{\eta} R e = η ρ v d ; small means smooth flow. Surface tension : force per unit length, or energy per unit area; drops 2 S r \dfrac{2S}{r} r 2 S , soap bubbles 4 S r \dfrac{4S}{r} r 4 S . Capillary rise : h = 2 S cos θ ρ g r h = \dfrac{2S\cos\theta}{\rho gr} h = ρ g r 2 S cos θ ; wetting liquids rise, others fall. The facts at a glance Quantity Formula Viscous force η A v d \eta A\dfrac{v}{d} η A d v Stokes' drag 6 π η r v 6\pi\eta rv 6 π η r v Terminal speed 2 r 2 ( ρ − σ ) g 9 η \dfrac{2r^2(\rho - \sigma)g}{9\eta} 9 η 2 r 2 ( ρ − σ ) g Work on a film 2 S Δ A 2S\,\Delta A 2 S Δ A Drop; soap bubble 2 S r \dfrac{2S}{r} r 2 S ; 4 S r \dfrac{4S}{r} r 4 S Capillary rise 2 S cos θ ρ g r \dfrac{2S\cos\theta}{\rho gr} ρ g r 2 S cos θ
A worked example A grain of sand of radius 0.050 mm 0.050\ \text{mm} 0.050 mm (density 2650 kg/m 3 2650\ \text{kg/m}^3 2650 kg/m 3 ) settles through still water ( η = 1.0 × 10 − 3 Pa s \eta = 1.0 \times 10^{-3}\ \text{Pa s} η = 1.0 × 1 0 − 3 Pa s ).
Terminal speed. 2 r 2 ( ρ − σ ) g 9 η \dfrac{2r^2(\rho - \sigma)g}{9\eta} 9 η 2 r 2 ( ρ − σ ) g , with 2 r 2 ( ρ − σ ) g = 8.09 × 10 − 5 kg/s 2 2r^2(\rho - \sigma)g = 8.09 \times 10^{-5}\ \text{kg/s}^2 2 r 2 ( ρ − σ ) g = 8.09 × 1 0 − 5 kg/s 2 and 9 η = 9.0 × 10 − 3 Pa s 9\eta = 9.0 \times 10^{-3}\ \text{Pa s} 9 η = 9.0 × 1 0 − 3 Pa s , gives 8.98 × 10 − 3 m/s 8.98 \times 10^{-3}\ \text{m/s} 8.98 × 1 0 − 3 m/s . Is Stokes' law fair? R e = 1000 kg/m 3 × 0.009 m/s × 1.0 × 10 − 4 m 1.0 × 10 − 3 Pa s = 0.9 Re = 1000\ \text{kg/m}^3 \times 0.009\ \text{m/s} \times \dfrac{1.0 \times 10^{-4}\ \text{m}}{1.0 \times 10^{-3}\ \text{Pa s}} = 0.9 R e = 1000 kg/m 3 × 0.009 m/s × 1.0 × 1 0 − 3 Pa s 1.0 × 1 0 − 4 m = 0.9 , small, so yes. Settling 2.0 m 2.0\ \text{m} 2.0 m . 2.0 m 0.009 m/s = 223 s \dfrac{2.0\ \text{m}}{0.009\ \text{m/s}} = 223\ \text{s} 0.009 m/s 2.0 m = 223 s , nearly four minutes. Practice with new numbers
A program to try The program works out the grain of sand of the worked example.
Read it, then run it.
Try a grain of radius 0.020 mm 0.020\ \text{mm} 0.020 mm . How long does it take to settle now?
Mistakes to avoid Thickness in millimetres in η A v d \eta A\dfrac{v}{d} η A d v . Change to metres. r r r for r 2 r^2 r 2 in the terminal speed. It grows as the square of the radius. The ball's density alone. Use ρ − σ \rho - \sigma ρ − σ . One surface for a soap film or bubble. It has two. A positive rise for mercury. Its cos θ \cos\theta cos θ is negative: it falls. Check yourself Two drops of oil are falling through air at their terminal speeds; one has 1.5 1.5 1.5 times the radius. How do their speeds compare? Stuck? Show a hint Show the answerWhat is the excess pressure inside a soap bubble of radius 2.0 cm 2.0\ \text{cm} 2.0 cm , with S = 0.025 N/m S = 0.025\ \text{N/m} S = 0.025 N/m ? Stuck? Show a hint Show the answerWater rises 3.0 cm 3.0\ \text{cm} 3.0 cm in a tube. How high in a tube of a third of the radius? Stuck? Show a hint Show the answerWhy does a detergent help clean a greasy plate? Stuck? Show a hint Show the answerWhy does honey pour faster on a hot day? Stuck? Show a hint Show the answer