2
$\begingroup$

In high school, I was taught that Stokes' law is dependent on assumption that drag force is proportional to velocity, viscosity and radius of the sphere (and the powers/exponents are evaluated using dimensional analysis). Is Stokes' law proven or is it just an assumption?

$\endgroup$
3
  • 1
    $\begingroup$ There is a derivation for the Stokes' law. However, if you're in high school, then you would have a hard time understanding it. $\endgroup$
    – user258881
    Mar 31, 2020 at 13:18
  • 1
    $\begingroup$ What @FakeMod means is that there is a fundamental derivation of Stokes law based on a force equilibrium of the sphere in conjunction with Newton's law of viscosity and conservation of mass. $\endgroup$ Mar 31, 2020 at 20:45
  • $\begingroup$ As pointed out by the others it can be derived analytically for low Reynolds number flow. I have written down a derivation here. I tried to keep it simple but I think I was not successful in that regard... Nonetheless I think it is one of the most exhaustive explanations you can find online. $\endgroup$
    – 2b-t
    Apr 1, 2020 at 15:49

3 Answers 3

5
$\begingroup$

Stokes' law only applies when the inertia forces in the fluid (caused by its acceleration or non-uniform motion) are negligible compared with the viscous forces.

The ratio of the two types of forces is described by a non-dimensional number called Reynolds number (usually written as Re). Stokes' law applies when Re is much smaller than $1$.

This is only true for very slow "creeping" flows, or for very small objects moving in typical fluids like air or water - for example dust particles "floating" in the air or single-celled animals "swimming" in water.

For comparison, a ball being thrown in most sports will have Re of the order of $10^5$ to $10^6$ and a large ship travelling at sea may have Re of the order of $10^9$ to $10^{10}$.

Stokes' law can be used to measure the viscosity of fluids, so long as the experiment only involves small Reynolds numbers.

For larger Reynolds numbers, the drag force is approximately proportional to the velocity squared, not to the velocity.

$\endgroup$
0
$\begingroup$

As indicated, Stokes' law applies when the inertial effects of the fluid are negligible. The pressure forces compensate the viscosity forces. It can be shown that this is true when the Reynolds number is less than 1.

In the general case, with reasonable assumptions, dimensional analysis can justify that the drag force on a sphere is of the form: $F=(1/2)\mu v^2 \pi a^2C_x(Re)$ with $C_x(Re)$ a dimensionless function of the Reynolds number $Re=\mu v a/\eta$.

If we add the idea that inertia does not intervene, the density must disappear from this relation and the only solution is a function of the form $C_x(Re)=c/Re$ with $c$ a constant.

It remains then $F=c'v\pi a \eta$ which is Stokes' law. Of course, the coefficient $c'=6\pi$ cannot be obtained by dimensional analysis.

$\endgroup$
-2
$\begingroup$

The mysterious six in Stokes' formula can be derived from adding the surface area of a sphere to the surface area of its lower hemisphere: 4pi r^2 + 2pi r^2.

When each term is multiplied by v/r, the number of radial lengths per second of the ball's descent through the fluid, we obtain the rate at which oil is sliding off of the "duck's back", a meters squared per second quantity. This is also the rate of change of volume over time (the chain rule requires the v/r factor).

When multiplied by the viscosity this yields a fluid force. A drop of oil moving over the surface is pushing against the viscosity: meters squared per second multiplied by a (kilogram per meter-second) nets out to newtons of force. The greater the drop's speed and the greater the resistance it encounters, the greater the fluid force.

The 2pi rv mu expression (using 'mu' for viscosity) represents the drag on the upper hemisphere.

The 4pi rv mu represents the streamlines pushing upwards on the lower hemisphere. One way of coming at this is to view the speed of the streamlines pushing up on the lower hemisphere as v-(-v) or 2v against a speed of only v for the overall descent of the ball. The lower hemisphere is like the biker in front of his passenger catching the "wind", which is just produced by the bike's motion, and shielding his passenger riding double. The 4pi term would refer to the lower hemisphere and the 2pi term would apply to the upper hemisphere.

This also bears comparison to the change in momentum of a ball bouncing off a wall: mv-(-mv) = 2mv.

Drag force = 4pi rv mu plus 2pi rv mu = 6pi rv mu.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.