[Physics Class Notes] on Stokes Law Derivation Pdf for Exam

Stokes law discusses the active force applied on a body when it is dropped in a liquid. Initially, because of low viscous force, this velocity of the falling body remains low. But, as the spherical body falls with its effective weight, it gains acceleration, and this velocity of the body increases gradually. 

This also makes the liquid in contact move with a velocity same as that of this body. The movement of an object in this fluid with increasing velocity causes motion in liquid layers resulting in the development of viscous force. This force increases with the increasing velocity until the point of time when it matches this effective force with which this body moves.

In such a situation, the net force on the body becomes zero, and it attains a constant velocity named terminal velocity (Vt). This idea helped state Stokes law and this derivation of frictional force or Stokes’ drag applies to the interface between the body and fluid. 

According to Sir George Stokes, “the force acting between the liquid and falling body interface is proportional to velocity and radius of the spherical object and viscosity of fluid”. 

Stoke’s law explains the main reason why raindrops falling from the sky do not harm us on the ground. Stokes law is an extremely important concept that is a part of physics

Stokes Law Formula

Stokes came up with this formula in 1851 to calculate this drag force or frictional force of spherical objects immersed in viscous fluids. Here, look at the formula mentioned below.

F = 6 * πηrv 

Where,

• F is the drag force or frictional force at the interface 

• η is the viscosity of a liquid

• r is the radius of the spherical body

• V is the velocity of flow 

Stokes Law Derivation 

Stokes’ proposition regarding this immersion of the spherical body in a viscous fluid can be mathematically represented as, 

F ∝ ηa rb vc

By solving this proportional expression, we can get the Stokes law equation. To change this proportionality sign into equality, we must add a constant to the equation. Let us consider that constant as ‘K’, so the transformed equation becomes 

F = K * ηa rb vc  ……….. (1)

Now, Let us write down the dimensions of this equation on both sides. Please note, K is a constant which has no dimension.

 [MLT-2] = [[ML-1T-1]a . Lb . [LT-1]c ] 

We need to simplify this expression by separating each variable as follows

[MLT-2] = [Ma ] [L-a+b+c]  [T-a-c]

Comparing the value of length, mass and time, following equations can be found.

a = +1 ……….(2)

-a+b+c = +1 ……….(3)

-a-c= -2 ……………..(4)

On solving equations 2,3, and 4, we get a =1, b=1, c=1.  On substituting these values in equation 1, we have

 F = K * η1 r1 v1 = K ηrv

Further, the value of K was found to be 6π for spheres through experimental observation. The above calculations helped in Stokes equation derivation along with its fundamental formula. 

Terminal Velocity Formula

As explained earlier, terminal velocity is attained at an equilibrium position when the net force acting upon the spherical body and acceleration becomes zero. Here is the formula for terminal velocity derived from Stokes law definition.

 Vt = 2a2 (ρ−σ) g / 9η

Where ρ is the mass density of a spherical object and σ is the mass density of a fluid. 

Assumptions made in the Stokes Law

Some assumptions were made in the Stokes law. The assumptions made in the Stokes law include the following:

Applications of Stoke Law 

Stokes law is applicable in many areas. Some applications of stokes law include the following:

It helps in finding the settling of sediment in freshwater

It is also used to measure the viscosity of fluids.