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 forces act on it.

(i) Weight of the body acting vertically downwards.

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

Let s = Density of material  r = Radius of spherical body  

So = Density of Medium.  

True weight of the body = w = volume x density x g 

= \(\frac{4}{3} \pi r^3 \rho g\)

Upward thrust due to buoyancy,

F_T = Weight of the medium displaced

\(\therefore\) F_T = Volume of the medium displaced x Density x g

= \(\frac{4}{3} \pi r^3 \rho_0 g1\)

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

\(F_V = 6 \pi \eta rv\)

When body attains terminal velocity, then

It depends upon the terminal velocity 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.

Comments

Thank you so, much akka