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# Derivation of Stoke’s Law

Have you wondered why the raindrops falling from a great height do not harm humans? Stoke’s law can explain this. In this article, you will understand Stoke’s law and its derivation in detail.

## What is Stoke’s Law?

Stoke’s Law is a mathematical equation that expresses the settling velocities of the small spherical particles in a fluid medium. The law is derived considering the forces acting on a particular particle as it sinks through the liquid column under the influence of gravity. The force that retards a sphere moving through a viscous fluid is directly proportional to the velocity and the radius of the sphere, and the fluid’s viscosity.

### Stoke’s Law Equation

Sir George G. Stokes, an English scientist, clearly expressed the viscous drag force F as:

 $$\begin{array}{l}F=6\pi \eta rv\end{array}$$

Where r is the sphere radius, η is the fluid viscosity, and v is the sphere’s velocity.

## Stoke’s Law Derivation

From Stoke’s Law viscosity equation, we know that viscous force acting on a sphere is directly proportional to the following parameters:

• the radius of the sphere (r)
• coefficient of viscosity (η)
• the velocity of the object (v)

Mathematically, this is represented as

$$\begin{array}{l}F\propto\eta^ar^bv^c\end{array}$$

Now let us evaluate the values of a, b and c.

Substituting the proportionality sign with an equality sign, we get

$$\begin{array}{l}F=k\eta^ar^bv^c…(1)\end{array}$$

Here, k is the constant of proportionality, a numerical value with no dimensions.

Writing the dimensions of parameters on either side of equation (1), we get

[MLT–2] = [ML–1T–1]a [L]b [LT-1]c

Simplifying the above equation, we get

[MLT–2] = Ma ⋅ L–a+b+c ⋅ T–a–c… (2)

According to classical mechanics, mass, length and time are independent entities.

Equating the superscripts of mass, length and time, respectively from equation (2), we get

a = 1… (3)

–a + b + c = 1… (4)

–a –c = -2 or a + c = 2… (5)

Substituting (3) in (5), we get

1 + c = 2

c = 1 (6)

Substituting the value of (3) & (6) in (4), we get

–1 + b + 1 = 1

b = 1 (7)

Substituting the value of (3), (6) and (7) in (1), we get

$$\begin{array}{l}F=k\eta rv\end{array}$$

The value of k for a spherical body was experimentally obtained as

$$\begin{array}{l}6\pi\end{array}$$

Therefore, the equation gives the viscous force on a spherical body falling through a liquid.

$$\begin{array}{l}F=6\pi \eta rv\end{array}$$

### Terminal Velocity Formula

In the case of raindrops, initially, it is due to gravity that it accelerates. As the velocity increases, the retarding force also increases. Finally, when viscous force and the buoyant force is equal to the force due to gravity, the net force becomes zero, and so it does the acceleration. The raindrop then falls with a constant velocity, known as terminal velocity. Thus, in equilibrium, the terminal velocity vt is given by the equation

$$\begin{array}{l}v_t=\frac{2r^2(\rho -\sigma )g}{9\eta }\end{array}$$

ρ and σ are sphere and fluid mass densities, respectively.

From the equation above, we can infer that the terminal velocity depends on the square of the radius of the sphere and is inversely proportional to the viscosity of the medium.

## Stoke’s Law Example

A solid metal ball is falling in a long liquid column and has attained a terminal velocity of 4 m/s. What is the viscosity of the liquid if the radius of the metal ball is r = 5 cm and its density is

$$\begin{array}{l}\rho _{s}=8050\,kg/m^{3}\end{array}$$
. (The density of liquid is 1000 𝑘𝑔/𝑚3 and g is 10 m/s2.)

Solution:
The radius of the sphere is r = 0.05 m.
The density of the sphere is 𝜌s = 8050 kg/m3
The density of the liquid is 𝜌s= 1000 kg/m3
The terminal velocity is 4 m/s

Let the viscosity of the liquid be 𝜂.

Substituting the values in the terminal velocity equation, we get

$$\begin{array}{l}4=\frac{2}{9}\frac{(10\times (0.05)^2)(8050-1000)}{9\eta }\end{array}$$

$$\begin{array}{l}18=\frac{(25\times 10^{-3}\times 7050)}{9\eta }\end{array}$$

$$\begin{array}{l}\eta =\frac{(25\times 10^{-3}\times 7050)}{162 }=1.08\,kg/ms\end{array}$$

$$\begin{array}{l}\eta =1.08\,kg/ms\end{array}$$

## Stoke’s Law Applications

Stokes’s law finds application in several areas such as:

• Settling of sediment in freshwater
• Measurement of the viscosity of fluids

## Frequently Asked Questions – FAQs

### What is Stoke’s Law?

Stoke’s Law is an equation that expresses the drag force resisting the fall of small spherical particles through a fluid medium.

### What does the Stoke’s Law state?

Stoke’s Law states that the force that retards a sphere moving through a viscous fluid is directly proportional to the velocity and the radius of the sphere, and the viscosity of the fluid.

### When do we use Stoke’s Law?

We use Stoke’s law to determine the terminal velocity, the size and the density of sphere and liquid, respectively. Stoke’s law can also be used to calculate the viscosity of the fluid.

### When is Stoke’s Law valid?

Stoke’s Law is valid for determining viscosity when velocity is constant.

### What are the applications of Stoke’s Law?

Settling of sediment in freshwater and measurement of the viscosity of fluids are some of the applications of Stoke’s Law.