Stoke's Law

Statement of Stokes' law

Stokes' law states that the viscous force $F_{v}$ acting on a small sphere falling through a viscous medium is directly proportional to the radius of the sphere $r$, its velocity $v$ through the fluid, and the coefficient of viscosity $\eta$ of the fluid:

$$F_{v} \propto \eta\, r\, v$$

The empirically obtained constant of proportionality is $6\pi$, so

$$F_{v} = 6\pi\eta r v$$

This is the expression for the viscous force acting on a spherical object moving through a viscous medium; it can also be obtained by dimensional analysis.

2.7.1 Terminal velocity

The three forces

A spherical object falling through a viscous fluid experiences:

• the viscous force $F_{v}$, directed upwards, whose magnitude goes on increasing as the velocity increases,
• the gravitational force, its weight $F_{g}$, directed downwards,
• the buoyant force or upthrust $F_{u}$, directed upwards.

The net downward force $f = F_{g} - (F_{v} + F_{u})$ is responsible for the initial increase in velocity. Of these, $F_{g}$ and $F_{u}$ are constant while $F_{v}$ increases with velocity, so a stage is reached when the net force $f$ becomes zero, that is, when $F_{g} = F_{v} + F_{u}$. After that the downward velocity remains constant. This constant downward velocity is called the terminal velocity, and from then on the viscous force is also constant. The whole discussion applies to streamline flow only.

Diagram to draw

A sphere of radius $r$ falling through a column of viscous liquid, with three arrows at its centre: weight $mg$ downwards, viscous force $F_{v} = 6\pi\eta r v$ upwards, and buoyant force $F_{u}$ upwards.

Derivation

Let the sphere have radius $r$ and density $\rho$, and let the medium have density $\sigma$ and coefficient of viscosity $\eta$. At the terminal velocity $v$,

Total downward force = Total upward force

Weight of sphere $(mg)$ = viscous force + buoyant force due to the medium

$$\dfrac{4}{3}\pi r^{3}\rho g = 6\pi\eta r v + \dfrac{4}{3}\pi r^{3}\sigma g$$

$$6\pi\eta r v = \left(\dfrac{4}{3}\pi r^{3}\rho g\right) - \left(\dfrac{4}{3}\pi r^{3}\sigma g\right)$$

$$6\pi\eta r v = \left(\dfrac{4}{3}\right)\pi r^{3} g\,(\rho - \sigma)$$

$$v = \left(\dfrac{4}{3}\right)\pi r^{3}g(\rho - \sigma)\times\dfrac{1}{6\pi\eta r}$$

$$v = \dfrac{2}{9}\,\dfrac{r^{2}g(\rho - \sigma)}{\eta}$$

This is the expression for the terminal velocity of the sphere. Rearranged, it also gives the coefficient of viscosity of the fluid:

$$\eta = \dfrac{2}{9}\,\dfrac{r^{2}g(\rho - \sigma)}{v}$$

Reading the result

• $v \propto r^{2}$: two bodies of the same material with radii in the ratio $1:4$ have terminal velocities in the ratio $1:16$.
• If $\rho < \sigma$, as for an air bubble in a liquid, the calculated $v$ is negative. The negative sign means the body rises rather than falls.
• The velocity with which an object can move through a viscous fluid is always less than or equal to the terminal velocity in that fluid for that object.

Standard results

• $F_{v} = 6\pi\eta r v$.
• $v = \dfrac{2}{9}\dfrac{r^{2}g(\rho - \sigma)}{\eta}$, with $\rho$ the density of the body and $\sigma$ that of the medium.
• Weight $= \dfrac{4}{3}\pi r^{3}\rho g$; buoyant force $= \dfrac{4}{3}\pi r^{3}\sigma g$.