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International A LevelPhysicsEdexcelRevision Notes1. Mechanics & MaterialsDensity, Upthrust & Viscous DragViscous Drag

Viscous Drag (Edexcel International A Level Physics)

Revision Note

Lindsay Gilmour

Last updated

Stoke's Law

Viscous Drag

  • Viscous drag is defined as:

    The frictional force between an object and a fluid which opposes the motion between the object and the fluid

  • Viscous drag is calculated using Stoke’s Law;

F = 6πηrv

  • Where

    • F = viscous drag (N)

    • η = coefficient of viscosity of the fluid (N s m−2 or Pa s)

    • r = radius of the object (m)

    • v = velocity of the object (ms−1)

4-3-viscous-drag-stokes-law-equation_edexcel-al-physics-rn

  • The viscosity of a fluid can be thought of as its thickness, or how much it resists flowing

    • Fluids with low viscosity are easy to pour, while those with high viscosity are difficult to pour

4-3-viscous-drag-water-ketchup_edexcel-al-physics-rn

  • The coefficient of viscosity is a property of the fluid (at a given temperature) that indicates how much it will resist flow

    • The rate of flow of a fluid is inversely proportional to the coefficient of viscosity

Drag Force at Terminal Velocity

  • Terminal velocity is useful when working with Stoke’s Law since at terminal velocity the forces in each direction are balanced

Ws = Fd + U (equation 1)

  • Where;

    • Ws = weight of the sphere

    • Fd = the drag force (N)

    • U = upthrust (N)

4-3-viscous-drag-free-body-stokes-law_edexcel-al-physics-rn

At terminal velocity forces are balanced: W (downwards) = Fd + U (upwards)

  • The weight of the sphere is found using volume, density and gravitational force

Ws =vsρsg

W subscript s space equals space 4 over 3 pi r cubed rho subscript s g (equation 2)

  • Where

    • vs = volume of the sphere (m3)

    • ρs = density of the sphere (kg m–3)

    • g = gravitational force (N kg−1)

  • Recall Stoke’s Law

Fd = 6πηrvterm (equation 3)

  • Upthrust equals weight of the displaced fluid

    • The volume of displaced fluid is the same as the volume of the sphere

    • The weight of the fluid is found from volume, density and gravitational force as above

U space equals space 4 over 3 pi r cubed rho subscript f g (equation 4)

  • Substitute equations 2, 3 and 4 into equation 1

4 over 3 pi r cubed rho subscript s g italic space italic equals italic space italic 6 pi eta r v subscript t e r m end subscript italic space italic plus italic space italic 4 over italic 3 pi r to the power of italic 3 rho subscript f g italic space

  • Rearrange to make terminal velocity the subject of the equation

v subscript t e r m end subscript space equals space fraction numerator italic 4 over italic 3 pi r to the power of italic 3 g italic left parenthesis rho subscript s italic space italic minus italic space italic space rho subscript f italic right parenthesis over denominator italic 6 pi eta r end fraction space equals space fraction numerator 4 pi bold italic r to the power of italic 3 g italic left parenthesis rho subscript s italic space italic minus italic space italic space rho subscript f italic right parenthesis over denominator italic 18 pi eta bold italic r end fraction space

  • Finally, cancel out r from the top and bottom to find an expression for terminal velocity in terms of the radius of the sphere and the coefficient of viscosity

v subscript t e r m end subscript space equals fraction numerator 2 pi r squared g italic left parenthesis rho subscript s italic space italic minus italic space italic space rho subscript f italic right parenthesis over denominator italic 9 pi eta end fraction space

 

  • This final equation shows that terminal velocity is;

    • directly proportional to the square of the radius of the sphere

    • inversely proportional to the viscosity of the fluid

Understanding Viscosity & Stoke's Law

Conditions for Stoke’s Law Equation

  • The equation can only be used when certain conditions are met;

    • The flow is laminar

    • The object is small

    • The object is spherical

    • Motion between the sphere and the fluid is at a slow speed

 

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Laminar and Turbulent Flow

  • As an object moves through a fluid, or a fluid moves around an object, layers in the fluid are created

  • In laminar flow all the layers are moving in the same direction and they do not mix

  • This tends to happen for slow moving objects, or slow flowing liquids

  • The equation above only applies for laminar flow

  • In turbulent flow the layers move in different directions and the layers do mix

4-3-viscous-drag-laminar-turbulent_edexcel-al-physics-rn

Changing Viscosity

  • Viscosity is temperature-dependent

    • Liquids are less viscous as temperature increases

    • Gases get more viscous as temperature increases

Worked Example

A ball bearing of radius 5.0 mm falls at a constant speed of 0.030 ms–1 through a oil which has viscosity 0.3 Pa s and density 900 kg m–3.

Determine the viscous drag acting on the ball bearing. 

Answer:

Step 1: List the known quantities in SI units

  • Radius of the sphere, rs = 5.0 mm = 5.0 × 10-3 m

  • Terminal velocity of the sphere, v = 0.03 m s-1

  • Viscosity of oil, η = 0.3 Pa s

  • Density of oil, ρf = 900 kg m−3

Step 2: Sketch a free-body diagram to resolve the forces at constant speed

Ws = Fd + U

4-3-viscous-drag-worked-example-answer_edexcel-al-physics-rn

Step 3: Calculate the value for viscous drag, Fd

Fd = 6πηrv = 6 × π × 0.3 × 5.0 × 10-3 × 0.03 = 0.008482

Step 4: Write the complete answer to the correct significant figures and include units

  •  The viscous drag, Fd  = 8.5 × 10-4 N

Examiner Tips and Tricks

You may need to write out some or all of the derivation given in the first part above.

It is really important to keep clear whether you are talking about density of the sphere or the fluid, and mass of the sphere or the fluid.

Practice using subscripts and do try this at home. It isn’t one to do for the first time in an exam!

4-3-viscous-drag-calculating-stokes-law_edexcel-al-physics-rn

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    Lindsay Gilmour

    Author: Lindsay Gilmour

    Expertise: Physics

    Lindsay graduated with First Class Honours from the University of Greenwich and earned her Science Communication MSc at Imperial College London. Now with many years’ experience as a Head of Physics and Examiner for A Level and IGCSE Physics (and Biology!), her love of communicating, educating and Physics has brought her to Save My Exams where she hopes to help as many students as possible on their next steps.