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.
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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:
[latex]F=6\pi \eta rv[/latex] |
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
- coefficient of viscosity
- the velocity of the object
Mathematically, this is represented as
[latex]F\propto\eta^ar^bv^c[/latex]Now let us evaluate the values of a, b and c.
Substituting the proportionality sign with an equality sign, we get
[latex]F=k\eta^ar^bv^c[/latex] (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^{–1}T^{–1}]^{a} [L]^{b} [LT^{-1}]^{c}Simplifying the above equation, we get
[MLT^{–2}] = M^{a }⋅ 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
[latex]F=k\eta rv[/latex]The value of k for a spherical body was experimentally obtained as [latex]6\pi[/latex]
Therefore, the viscous force on a spherical body falling through a liquid is given by the equation.
[latex]F=6\pi \eta rv[/latex]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 v_{t} is given by the equation
[latex]v_t=\frac{2a^2(\rho -\sigma )g}{9\eta }[/latex][latex]\rho[/latex] 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.
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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 [latex]\rho _{s}=8050\,kg/m^{3}[/latex]. (The density of liquid 1 is 1000 𝑘𝑔/𝑚^{3} and g is 10 m/s^{2}.)
Solution:
The radius of the sphere is r = 0.05 m.
The density of the sphere is 𝜌_{s} = 8050 kg/m^{3}
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
[latex]4=\frac{2}{9}\frac{(10\times (0.05)^2)(8050-1000)}{9\eta }[/latex][latex]18=\frac{(25\times 10^{-3}\times 7050)}{9\eta }[/latex]
[latex]18=\frac{(25\times 10^{-3}\times 7050)}{9\eta }[/latex]
[latex]\eta =\frac{(25\times 10^{-3}\times 7050)}{162 }=1.08\,kg/ms[/latex]
[latex]\eta =1.08\,kg/ms[/latex]
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
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