Derive Stoke’s law. Hence find the expression of terminal velocity.
Answers
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
Terminal Velocity Formula
In the case of raindrops, initially, it is due to the 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 does the acceleration. The raindrop then falls with a constant velocity. Thus, in equilibrium, the terminal velocity vt is given by the equation
vt=2a2(ρ−σ)g9η
where ρ and σ are mass densities of sphere and fluid respectively.
From the equation above, we can infer that the terminal velocity depends on the square of the radius of the sphere and inversely proportional to the viscosity of the medium.