IBPhysicsSLRevision NotesSpace, Time & MotionForces & MomentumStoke's Law

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First teaching 2023

First exams 2025

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Stoke's Law (SL IB Physics)

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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

  • This drag force is often from air resistance
  • Viscous drag is calculated using Stoke’s Law:

 

F subscript d space equals space 6 pi eta r v

  • Where
    • Fd = viscous drag force (N)
    • η = fluid viscosity (N s m−2 or Pa s)
    • r = radius of the sphere (m)
    • v = velocity of the sphere through the fluid (ms−1)

1-2-9-stokes-law

A sphere travelling through air will experience a drag force that depends on its radius, velocity and the viscosity of the liquid

  • 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

Water has a lower viscosity than ketchup as it is easier to pour and flow

  • 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
  • The size of the force depends on the:
    • Speed of the object
    • Size of the object
    • Shape of the object

Worked example

A spherical stone of volume 2.7 × 10–4 m3 falls through the air and experiences a drag force of 3 mN at a particular instant. Air has a viscosity of 1.81 × 10-5 Pa s. Calculate the speed of the stone at that instant.

Answer:

Step 1: List the known quantities

  • Volume of stone, V = 2.7 × 10–4 m3
  • Drag force, Fd = 3 mN = 3 × 10–3 N
  • Viscosity of air, η = 1.81 × 10-5 Pa s

Step 2: Calculate the radius of the sphere, r

  • The volume of a sphere is

V space equals 4 over 3 πr cubed

  • Therefore, the radius, r is:

r space equals space cube root of fraction numerator 3 V over denominator 4 straight pi end fraction end root space equals space cube root of fraction numerator 3 space cross times space open parentheses 2.7 space cross times space 10 to the power of negative 4 end exponent close parentheses space over denominator 4 straight pi end fraction end root space equals space 0.04 space straight m

Step 3: Rearrange the Stoke's law equation for the velocity, v

F subscript d space equals space 6 pi eta r v

v space equals space fraction numerator space F subscript d over denominator space 6 pi eta r end fraction

Step 4: Substitute in the known values

v space equals space fraction numerator 3 space cross times space 10 to the power of negative 3 end exponent space over denominator 6 straight pi space cross times space open parentheses 1.81 space cross times space 10 to the power of negative 5 end exponent close parentheses space cross times space 0.04 end fraction space equals space space 220 space straight m space straight s to the power of negative 1 end exponent

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Ashika

Author: Ashika

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.