Question

Derive stokes law by dimensional analysis

Answer 1

Answer:

Suppose that the sphere has radius r and falls through a fluid of viscosity η. Let the terminal velocity be v (we can calculate the viscous drag F on the sphere by dimensional analysis. Solving this gives x = 1, y = 1 and z =1.

Explanation:

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Answer 2

Answer:

Explanation:

The viscous force acting on a sphere is directly proportional to the following parameters:

the radius of the sphere

coefficient of viscosity

the velocity of the object

Mathematically, this is represented as

F∝ηarbvc

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

Substituting the proportionality sign with an equality sign, we get

F=kηarbvc (1)

Here, k is the constant of proportionality which is a numerical value and has 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

F=kηrv

The value of k for a spherical body was experimentally obtained as 6π

Therefore, the viscous force on a spherical body falling through a liquid is given by the equation

F=6πηrv