Revision NotesExam QuestionsPast Papers
ASPhysicsEdexcelRevision Notes4. MaterialsDensity, Upthrust & Viscous DragViscous Drag

Viscous Drag (Edexcel AS Physics): Revision Note

Exam code: 8PH0

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 Stokes’ Law:

F space equals space 6 straight pi eta r v

  • 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 Stokes’ Law since at terminal velocity the forces in each direction are balanced

W space equals space F subscript d space plus space U (equation 1)

  • Where;

    • W = 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

W space equals space v subscript s rho subscript s g

W space equals space 4 over 3 straight 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

F subscript d space equals space 6 straight pi eta r v subscript t e r m end subscript (equation 3)

  • Upthrust equals the weight of the displaced fluid

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

  • The weight of the displaced fluid is also found from volume, density, and gravitational force:

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

  • Substitute equations 2, 3, and 4 into equation 1:

4 over 3 straight pi r cubed rho subscript s g space equals space 6 straight pi eta r v subscript t e r m end subscript space plus space 4 over 3 straight pi r cubed 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 4 over 3 straight pi r cubed g open parentheses rho subscript s italic space italic minus italic space italic space rho subscript f close parentheses over denominator 6 straight pi eta r end fraction space equals space fraction numerator 4 straight pi r cubed g open parentheses rho subscript s italic space italic minus italic space italic space rho subscript f close parentheses over denominator 18 straight pi eta r end fraction space

  • Simplify 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 space fraction numerator 2 straight pi r squared g open parentheses rho subscript s italic space italic minus italic space italic space rho subscript f close parentheses over denominator 9 straight 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

  • Stokes' law equation can only be used when the following 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 

4-3-viscous-drag-small-spherical-slow_edexcel-al-physics-rn

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 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

  • The relationship between viscosity and temperature is different for liquids and gases

  • In liquids, viscosity:

    • depends on intermolecular forces

    • tends to decrease (become less viscous) as temperature increases

  • In gases, viscosity:

    • depends on intermolecular collisions

    • tends to increase (become more viscous) as 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

W space equals space F subscript d space plus space U

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

Step 3: Calculate the value for viscous drag, Fd

F subscript d space equals space 6 straight pi eta r v

F subscript d = 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

  •  Viscous drag, F subscript d  = 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

You've read 0 of your 5 free revision notes this week

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourself

    Did this page help you?

    Previous:UpthrustNext:Core Practical 4: Investigating Viscosity
    Lindsay Gilmour

    Author: Lindsay Gilmour

    Expertise: Physics Content Creator

    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.