Viscosity is the internal resistance of a fluid to relative motion.
When an object moves through a viscous fluid, the fluid exerts a drag force that opposes the motion.
For a small sphere moving slowly through a viscous fluid, the drag force is described by Stokes’ law.
1. Stokes’ law
For creeping flow around a sphere, the viscous drag force is
\[
\begin{aligned}
F_d &= 6\pi\eta rv.
\end{aligned}
\]
Here:
- \(F_d\) is the viscous drag force.
- \(\eta\) is dynamic viscosity.
- \(r\) is the sphere radius.
- \(v\) is the speed of the sphere relative to the fluid.
Stokes’ law is a linear drag model. If speed doubles, drag force doubles.
2. Conditions for Stokes’ law
Stokes’ law is accurate only for very slow, smooth, laminar flow.
The usual check is the Reynolds number:
\[
\begin{aligned}
Re &= \frac{2\rho_frv}{\eta}.
\end{aligned}
\]
Stokes’ law is best when \(Re\ll 1\). It may still be used as an approximation near \(Re\approx 1\), but it becomes poor for ordinary large raindrops, fast motion, turbulence, or high-speed flow.
3. Weight and buoyant force
A sphere of density \(\rho_s\) and radius \(r\) has volume
\[
\begin{aligned}
V &= \frac{4}{3}\pi r^3.
\end{aligned}
\]
Its weight is
\[
\begin{aligned}
W &= \rho_s Vg.
\end{aligned}
\]
The buoyant force from the displaced fluid is
\[
\begin{aligned}
F_b &= \rho_f Vg.
\end{aligned}
\]
The net downward driving force is therefore
\[
\begin{aligned}
W-F_b &= (\rho_s-\rho_f)Vg.
\end{aligned}
\]
4. Terminal velocity
Terminal velocity occurs when the net force becomes zero.
For a falling sphere:
\[
\begin{aligned}
W-F_b-F_d &= 0.
\end{aligned}
\]
Substitute the formulas:
\[
\begin{aligned}
(\rho_s-\rho_f)\left(\frac{4}{3}\pi r^3\right)g
&=
6\pi\eta rv_t.
\end{aligned}
\]
Solve for \(v_t\):
\[
\begin{aligned}
v_t
&=
\frac{2r^2g(\rho_s-\rho_f)}{9\eta}.
\end{aligned}
\]
If \(\rho_s>\rho_f\), the sphere settles downward.
If \(\rho_s<\rho_f\), the sphere rises upward.
If \(\rho_s=\rho_f\), the object is neutrally buoyant.
5. Motion toward terminal velocity
Because Stokes drag is linear in speed, the equation of motion can be written as
\[
\begin{aligned}
m\frac{dv}{dt} &= W-F_b-bv,
\end{aligned}
\]
where
\[
\begin{aligned}
b &= 6\pi\eta r.
\end{aligned}
\]
If the sphere starts from rest, its speed approaches terminal speed according to
\[
\begin{aligned}
v(t) &= v_t\left(1-e^{-t/\tau}\right).
\end{aligned}
\]
The time constant is
\[
\begin{aligned}
\tau &= \frac{m}{b}.
\end{aligned}
\]
After one time constant, the sphere has reached about \(63\%\) of terminal speed.
After several time constants, it is very close to terminal speed.
6. Solving for radius
If terminal speed is known and the radius is unknown, rearrange
\[
\begin{aligned}
v_t &= \frac{2r^2g|\rho_s-\rho_f|}{9\eta}.
\end{aligned}
\]
Then
\[
\begin{aligned}
r
&=
\sqrt{
\frac{9\eta v_t}{2g|\rho_s-\rho_f|}
}.
\end{aligned}
\]
7. Solving for viscosity
If terminal speed is measured experimentally and viscosity is unknown:
\[
\begin{aligned}
\eta
&=
\frac{2r^2g|\rho_s-\rho_f|}{9v_t}.
\end{aligned}
\]
This is one common method for estimating the viscosity of a fluid using falling spheres.
8. Viscous force between parallel plates
Viscosity also produces shear force when fluid is trapped between two plates.
If one plate moves relative to another and the velocity profile is linear, the viscous force is
\[
\begin{aligned}
F &= \eta A\frac{\Delta v}{L}.
\end{aligned}
\]
Here \(A\) is plate area, \(\Delta v\) is the relative plate speed, and \(L\) is the thickness of the fluid layer.
The shear stress is
\[
\begin{aligned}
\tau &= \frac{F}{A}.
\end{aligned}
\]
For this linear model,
\[
\begin{aligned}
\tau &= \eta\frac{\Delta v}{L}.
\end{aligned}
\]
9. Worked example: raindrop in air
Suppose a small water droplet of radius
\[
\begin{aligned}
r &= 0.5\ \mathrm{mm}=5.0\times10^{-4}\ \mathrm{m}
\end{aligned}
\]
falls through air with speed
\[
\begin{aligned}
v &= 5\ \mathrm{m/s}.
\end{aligned}
\]
Use
\[
\begin{aligned}
\eta &= 1.81\times10^{-5}\ \mathrm{Pa\,s}.
\end{aligned}
\]
The Stokes drag estimate is
\[
\begin{aligned}
F_d
&=
6\pi\eta rv\\
&=
6\pi(1.81\times10^{-5})(5.0\times10^{-4})(5)\\
&\approx 8.53\times10^{-7}\ \mathrm{N}.
\end{aligned}
\]
However, an ordinary raindrop often has a Reynolds number too large for Stokes’ law to be accurate.
This is why the calculator includes a Reynolds-number warning.
10. Formula summary
| Goal |
Formula |
Use |
| Stokes drag |
\(F_d=6\pi\eta rv\) |
Drag on a slowly moving sphere |
| Sphere volume |
\(V=\frac{4}{3}\pi r^3\) |
Find mass and buoyancy |
| Weight |
\(W=\rho_sVg\) |
Downward gravitational force |
| Buoyant force |
\(F_b=\rho_fVg\) |
Upward force from displaced fluid |
| Terminal velocity |
\(v_t=\frac{2r^2g(\rho_s-\rho_f)}{9\eta}\) |
Settling or rising speed in Stokes flow |
| Reynolds number |
\(Re=2\rho_frv/\eta\) |
Check whether Stokes’ law is valid |
| Plate viscous force |
\(F=\eta A\Delta v/L\) |
Couette flow between plates |
| Shear stress |
\(\tau=F/A\) |
Force per unit area |
11. Assumptions and limitations
- The sphere is smooth and spherical.
- The flow is steady and laminar.
- The Reynolds number is small.
- The fluid is Newtonian, with constant dynamic viscosity.
- The sphere is far from walls or boundaries.
- For larger Reynolds numbers, quadratic drag is usually more appropriate.
- For plate flow, the velocity profile is assumed linear.
Key idea: Stokes’ law is a linear-drag model, so drag grows directly with speed and terminal velocity occurs when drag balances the net weight.
