Stokes' Law, Drag Force, and Terminal Velocity
Laws
Stokes’ law
A spherical particle with velocity $v$ and radius $r$ moving in a fluid with kinematic viscosity $\mu$ experiences the following viscous drag force $F$: $$ F = - 6 \pi \mu r v $$ This is called Stokes’ law.
Terminal velocity
The terminal velocity $v_{\infty}$ of a particle falling in a fluid is: $$ v_{\infty} = {\frac{ 2 r^{2} g \left( \rho - \rho_{f} \right) }{ 9 \mu }} $$ Here $\rho$ is the particle’s density, $\rho_{f}$ is the fluid density, and $g$ is the acceleration due to gravity.
Explanation
Stokes’ law holds under the assumption that the Reynolds number is low so the flow is laminar and the surface is smooth.
At first glance the terminal velocity formula may look odd because of the many constants, but the derivation shows it is not particularly so.
Derivation
At terminal velocity the particle no longer accelerates. From Newton’s laws of motion $F = m a$, and since the particle’s mass is the volume of a sphere $\frac{4}{3} \pi r^{3}$ multiplied by the density $\rho$, the gravitational force $F_{g}$ is: $$ F_{g} = m g = \frac{4}{3} \pi r^{3} \rho g $$ The buoyant force acting in the opposite direction $F_{f}$ is: $$ F_{f} = \frac{4}{3} \pi r^{3} \rho_{f} g $$ That the particle does not accelerate means the net force of these balances the drag, and mathematically this is expressed as: $$ \begin{align*} F_{g} - F_{f} =& - F \\ \implies \frac{4}{3} \pi r^{3} \rho g - \frac{4}{3} \pi r^{3} \rho_{f} g =& 6 \pi \mu r v \end{align*} $$ Solving this for $v$ yields the terminal velocity formula: $$ v = {\frac{ 2 r^{2} g \left( \rho - \rho_{f} \right) }{ 9 \mu }} $$
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