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Stokes’ Law Derivation : Assumptions, Applications and Solved Examples

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Physics Topics such as mechanics, thermodynamics, and electromagnetism are fundamental to many other scientific fields.

What is the Formula for Terminal Velocity?

Terminal Velocity of a Body in a Viscous Medium and Stokes’ Law

When a body falls through a viscous medium (liquid or gas), it drags a layer of the fluid adjacent to it due to adhesion. But fluid layers at a large distance from the body are at rest. As a result, there is relative motion between different layers of the fluid at different distances from the body. But the viscosity of the fluid opposes this relative motion. The opposing force due to viscosity increases with increase in the velocity of the body due to the gravitational acceleration g.

If the body is small in size, then after an interval of time the opposing upward force (i.e., viscous force and buoyant force) becomes equal to the downward force (weight of the body). Then the effective force acting on the body becomes zero and the body begins to fall through the medium with a uniform velocity, called the terminal velocity. A graph representing the change in velocity of a falling object with time is shown in Fig.
Stokes' Law Derivation Assumptions, Applications and Solved Examples 1

Stokes’ low: Stokes proved that, if a small sphere of radius r is falling with a terminal velocity v through a medium of coefficient of viscosity 77, then the opposing force acting on the sphere due to viscosity is
F = 6πηrv ………. (1)
Equation (1) expresses Stokes’ law.

To establish Stokes’ law, the following assumptions are made.

  1. The fluid medium must be infinite and homogeneous.
  2. The sphere must be rigid with a smooth surface.
  3. The sphere must not slip when falling through the medium.
  4. The fluid motion adjacent to the falling sphere must be streamlined.
  5. The sphere must be small in size, but it must be greater than the intermolecular distance of the medium.

Equation for terminal velocity: if the density of the material of the sphere is ρ, then the weight of the sphere \(=\frac{4}{3}\)πr3ρg.
If the density of the fluid medium is σ, then the upward buoyant force acting on the sphere \(=\frac{4}{3}\)πr3σg.
∴ The resultant downward force acting on the sphere
\(=\frac{4}{3}\)πr3ρg – \(\frac{4}{3}\)πr3σg \(=\frac{4}{3}\)πr3(ρ – σ)g ……….. (2)
If the sphere attains terminal velocity, then
6πηrv = \(=\frac{4}{3}\)πr3(ρ – σ)g or, v = \(\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}\) ………. (3)
So, from equation (3), we see that the terminal velocity obeys the following rules.

  1. Terminal velocity is directly proportional to the square of the radius of the sphere.
  2. It is directly proportional to the difference of densities of the material of the sphere and that of the medium.
  3. It is inversely proportional to the coefficient of viscosity of the medium.

If the density of the body is less than the density of the medium, i.e., ρ < σ, then it is clear that the terminal velocity becomes negative. Hence, the velocity of the body will be in the upward direction. For this reason, air or other gas bubbles move upwards through water.

Applications of Stokes’law:

i) Falling of rain drops through air: Water vapour condenses on the particles suspended in air far above the ground to form tiny water droplets. The average radius of these tiny water droplets is 0.001 cm (approx.) Assuming the coefficient of viscosity of air as 1.8 × 10-4 poise (approx.) the terminal velocity of these droplets is calculated as 1.2 cm ᐧ s-1 (approx.) which is negligible. So, these water droplets float in the sky. Collectively these droplets form clouds. But as they coalesce to form larger drops, their terminal velocities increase. For example, the terminal velocity of a water droplet of radius 0.01 cm becomes 120cm ᐧ s-1 (approx.). As a result, they cannot float any longer and so they come down as rain.

ii) Coming down with the help of a parachute: When a soldier jumps from a flying aeroplane, he falls with acceleration due to gravity but due to viscous drag in air, the acceleration goes on decreasing till he acquires terminal velocity. The soldier then descends with constant velocity and opens his parachute close to the ground at a pre-calculated moment, so that he may land safely near his destination.

Numerical Examples

Example 1.
An oil drop of density 950 kg ᐧ m-3 and radius 10-6m is falling through air. The density of air is 1.3 kg ᐧ m-3 and its coefficient of viscosity is 181 × 10-7 SI unit. Determine the terminal velocity of the oil drop, [g = 9.8 m ᐧ s-2]
Solution:
Terminal velocity,
v = \(\frac{2}{9} \cdot \frac{r^2(\rho-\sigma) g}{\eta}\)
[Here, ρ = 950 kg ᐧ m-3 ; r = 10-6 m;
σ = 1.3 kg ᐧ m-3; η = 181 × 10-7 SI]
= \(\frac{2}{9} \cdot \frac{\left(10^{-6}\right)^2(950-1.3) \times 9.8}{181 \times 10^{-7}}\)
= 1.14 × 10-4 m ᐧ s-1.

Example 2.
An air bubble of radius 1 cm is rising from the bottom of a long liquid column. If its terminal velocity is 0.21 cm ᐧ s-1, calculate the coefficient of viscosity of the liquid. Given that the density of the liquid is 1.47 g ᐧ cm-3. Ignore the density of air.
Solution:
Coefficient of viscosity of the liquid,
η = \(\frac{2}{9} \cdot \frac{r^2(\rho-\sigma) g}{v}\)
[Here, r = 1 cm ; v = -0.21 cm ᐧ s-1; ρ = 0 ;
σ = 1.41 g ᐧ cm-3]
= \(\frac{2}{9} \times \frac{(1)^2(0-1.47) \times 980}{-0.21}\) = 1524.4 poise.