Question

By using Stokes law, derive an expression for terminal velocity. On what factors does it depend?

Answer 1

Terminal velocity: It is maximum constant velocity acquired by the body while falling freely in a viscous medium.

When a small spherical body falls freely through a viscous medium, three force act on it

(i) Weight of the body acting vertically downwards.

(ii) Upward thrust due to buoyancy equal to weight of liquid displaced.

Viscous drag acting in the direction opposite to the motion of body. According to Stoke’s law, F ∝ v, i.e., the opposing viscous drag goes on increasing with the increasing velocity of the body.

As the body falls through a medium, its velocity goes on increasing due to gravity. Therefore, the opposing viscous drag which acts upwards also goes on increasing. A stag reaches when the true weight of the body is just equal to the sum of the upward viscous drag. At this stage, there is no net force to accelerate the body. Hence it starts falling with a constant velocity, which is called terminal velocity.

Let ρ be the density of the material of the spherical body of radius r and ρ₀ be the density of the medium.

∴ The weight of the body,

W = Volume × Density × g

= \(\frac{4}{3}\)πr³ρg

Upward thrust due to buoyancy,

F_T = Weight of the medium displaced

∴ F_T = Volume of the medium displaced × Density × g

= \(\frac{4}{3}\)πr³ρ₀g

If v is the terminal velocity of the body, then according to Stoke’s law, upward viscous drag,

F_v = 6πηrv

When body attains terminal velocity, then

F_T + F_v = W

∴ \(\frac{4}{3}\)πr³ρ₀g + 6πηrv = \(\frac{4}{3}\)πr³ρg

or 6πηrv = \(\frac{4}{3}\)πr³(ρ − ρ₀)g

or v = \(\frac{2r^2(ρ − ρ0)g}{9η}\)

It depends upon the terminal velocity as it varies directly as the square of the radius of the body and inversely as the coefficient of viscosity of the medium. It also depends upon densities of the body and the medium.