Here we will derive the formula of viscosity using Stoke’s law. We have already derived one expression of viscosity using the flow of a viscous fluid between two parallel plates. Now we derive a different equation of viscosity or coefficient of viscosity in terms of terminal velocity.

When a sphere is released and allowed to fall freely in a fluid, it is subjected to three forces: its weight, *W*, the upthrust, *U*, and the viscous drag, *F*.

Initially, a resultant force, R = *W *– (*U *+ *F*) will make the sphere accelerate downward. As the velocity of the sphere increases, the viscous drag increases according to Stokes’ law until (*U *+ *F*) = *W. *The resultant force then becomes zero, and the sphere continues to fall at a constant velocity known as the terminal velocity.

## Derive formula of Viscosity using Stokes’ formula & terminal velocity

By measuring the terminal velocity of a sphere falling through a fluid it is possible to determine the **coefficient of viscosity **of the fluid. In other words, when terminal velocity is reached we can use the equilibrium equation in this way: upthrust + viscous drag = weight. This equation gives us the formula of viscosity.

For a sphere of volume V, radius *r *and density *ρ*_{s }falling through a fluid of density *ρ*_{f }and viscosity *η *with a terminal velocity * v*, we get the following equilibrium equation:

*U + F = W* …………….. (1)

where

U= weight of displaced fluid = m_{f }g = Vρ_{f} g = (4/3) πr^{3} ρ_{f} g

F = viscous drag = 6πηr**v**

W = weight of sphere = ms g = Vρ_{s }g = (4/3)πr^{3} ρ_{s} g

[ V = volume of the sphere = volume of displaced fluid ]

Equation (1) now can be written as:

(4/3) πr^{3}ρ_{f }g + 6πηr** v** = (4/3)πr

^{3}ρ

_{s}g …………………….. (2)

**which gives the formula of viscosity as:**

viscosity ** η = 2( ρ_{s} – ρ_{f}) gr^{2} / (9v)** …………… (3)

### Terminal velocity equation

We also get *one equation of terminal velocity* in terms of density of falling object and density of the fluid from equation (3):

**Terminal velocity**** v = 2( ρ_{s} – ρ_{f}) gr^{2} / (9 η )** …………….. (4)

This also shows us that the *terminal velocity of a falling sphere in a fluid depends on the* *square of its radiu**s*. Hence very small drops of rain and the minute droplets from an aerosol fall slowly through the air.

## Related study:

expression of viscosity using the flow of a viscous fluid between two parallel plates