Physics > Fluid Mechanics > 7.0 Stoke's law
Fluid Mechanics
1.0 Introduction
1.1 Ideal liquid
1.2 Density of a liquid $\left( \rho \right)$
1.3 Relative density of a liquid $(RD)$
1.4 Density of a mixture of two or more liquid
1.5 Density variation with temperature
1.6 Density variation with pressure
2.0 Fluid pressure
2.1 Atmospheric pressure
2.2 Pressure variation with depth
2.3 Measurement of pressure
2.4 Pressure difference in accelerating fluids
3.0 Pascal's law
4.0 Buoyant force
5.0 Flow of fluids
6.0 Viscosity
7.0 Stoke's law
8.0 Intermolecular forces
9.0 Angle of contact
7.1 Terminal velocity $\left( {{v_T}} \right)$
1.2 Density of a liquid $\left( \rho \right)$
1.3 Relative density of a liquid $(RD)$
1.4 Density of a mixture of two or more liquid
1.5 Density variation with temperature
1.6 Density variation with pressure
2.2 Pressure variation with depth
2.3 Measurement of pressure
2.4 Pressure difference in accelerating fluids
When a spherical body fall through a viscous medium its velocity increases till the sum of viscous force and upward thrust force becomes equal to the weight of the body. After that the body moves with constant velocity known as terminal velocity.
Terminal velocity of a spherical body of radius $r$, density $\rho $ while falling freely in a viscous medium of viscosity $\eta $ and density 
$\sigma $ is given by, $${v_T} = \frac{{2{r^2}\left( {\rho - \sigma } \right)g}}{{9\eta }}$$ where,
$\sigma $ is given by, $${v_T} = \frac{{2{r^2}\left( {\rho - \sigma } \right)g}}{{9\eta }}$$ where,$\eta $: Coefficient of viscosity
$\sigma $: Density of fluid
$\rho $: Density of the body
$r$: Radius of the spherical body
Variation of terminal velocity $\left( {{v_T}} \right)$ with time $(t)$ is given by,
Derivation of terminal velocity
Consider a sphere falling from rest through a large column of viscous liquid.
The forces acting on the sphere are,
1. Weight $W$ of the sphere acting vertically downwards
2. Thrust force acting vertically upwards
3. Viscous force $F_T$ acting vertically upwards
Note: Viscous force acts in direction opposite to the velocity of the sphere
So, $$\begin{equation} \begin{aligned} W = {F_T} + {F_V} \\ \frac{4}{3}\pi {r^3}\rho g = \frac{4}{3}\pi {r^3}\sigma g + 6\pi \eta r{v_T} \\ {v_T} = \frac{{2{r^2}\left( {\rho - \sigma } \right)g}}{{9\eta }} \\\end{aligned} \end{equation} $$
Note:
- The terminal velocity $\left( {{v_T}} \right)$ is directly proportional to the square of the radius of the body $${v_T} \propto {r^2}$$
- It is inversely proportional to the coefficient of viscosity of the medium
- It depends on the density of the body and the medium
- If $\rho < \sigma $ the value $v$ is negative i.e. the body will move up with constant velocity. It is due to this reason that the gas bubbles rise up through soda water bottle
