International A Level (IAL)PhysicsEdexcelRevision Notes1. Mechanics & MaterialsDensity, Upthrust & Viscous DragViscous Drag

Viscous Drag (Edexcel International A Level (IAL) Physics): Revision Note

Exam code: YPH11

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

  • The viscous drag on a small sphere can be calculated using Stokes’ law:

F space equals space 6 straight pi eta r v

  • Where

    • F = viscous drag force (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 (m s−1)

  • 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

Terminal velocity of a sphere in a fluid

  • When an object falls through a fluid (e.g. a skydiver falling through the air), it reaches terminal velocity

  • This occurs when the weight of the object balances with the upthrust and the viscous drag force

weight = upthrust + viscous drag

W space equals space U space plus space F subscript d

Forces acting on a sphere falling through a fluid

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

  • Consider a small, solid sphere of radius r moving slowly at terminal velocity through a fluid of viscosity eta

  • The upthrust U on the sphere is equal to the weight of the fluid displaced

W subscript s space equals space W subscript f space plus space 6 straight pi eta r v subscript term

m subscript s g space equals space m subscript f g space plus space 6 straight pi eta r v subscript term

  • Where

    • m subscript s = mass of the sphere (kg)

    • m subscript f = mass of the fluid displaced (kg)

    • g = acceleration due to gravity (m s−2)

    • v subscript term = terminal velocity of the sphere (m s−1)

  • Using the density equation, the mass m subscript s of the sphere is given by:

m subscript s space equals space rho subscript s V space equals space 4 over 3 straight pi r cubed rho subscript s

  • Where:

    • rho subscript s = density of the sphere (kg m–3)

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

  • Therefore, the mass m subscript f of displaced fluid is given by:

m subscript f space equals space rho subscript f V space equals space 4 over 3 straight pi r cubed rho subscript f

  • Where:

    • rho subscript f = density of the fluid (kg m–3)

  • Substituting the expressions for mass back into the original equation:

open parentheses 4 over 3 straight pi r cubed rho subscript s close parentheses g space equals space open parentheses 4 over 3 straight pi r cubed rho subscript f close parentheses g space plus space 6 straight pi eta r v subscript term

  • Rearrange to make terminal velocity the subject of the equation

6 straight pi eta r v subscript term space equals space 4 over 3 straight pi r cubed g open parentheses rho subscript s minus rho subscript f close parentheses

v subscript term space equals space fraction numerator 4 over 3 straight pi r cubed g open parentheses rho subscript s minus rho subscript f close parentheses over denominator 6 straight pi eta r end fraction

v subscript term space equals space fraction numerator 2 straight pi r cubed g open parentheses rho subscript s minus rho subscript f close parentheses over denominator 9 straight pi eta r end fraction

  • 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 term space equals space fraction numerator 2 r squared g open parentheses rho subscript s minus rho subscript f close parentheses over denominator 9 eta end fraction

  • 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 flow 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 to 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 the temperature increases

    • Gases get more viscous as the temperature increases

Worked Example

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

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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 the density of the sphere or the fluid, and the 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

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    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.