0
$\begingroup$

I am at an absolute novice level--I took physics in high school four years ago, and am taking it again in university.

My physics professor says that fluid friction is different at low and high speeds. He says that at low speeds, it's just a constant multiplied by the velocity -- kv -- but, at high speeds, we square that velocity -- DV^2. He also told me that these constants can be different for the same material.

I'm trying to understand this basic concept. Let's say I have a ball accelerating through the air in a straight X line with a constant acceleration. At what point does linearly-increasing friction--kv--become quadratically increasing friction--DV^2--and how does the constant k compare to the constant D for the same object?

$\endgroup$
2
$\begingroup$

As your professor says, fluid friction is ... complicated.

Fluid Drag

The fluid friction force depends very sensitively on the exact geometry of the object which is moving through the fluid. By exact, I mean that if you took two perfectly identical spheres and then rubbed one with sandpaper to scuff the surface, then the fluid friction force might change dramatically. You can see a demonstration of this fact here.

This is why golf balls have little dimples in them. There are many videos on youtube (like this one) which demonstrate the difference as air passes around a golf ball and a smooth sphere. The presence of the dimples changes the so-called boundary layer of air around the sphere, and reduces the shedding of large, turbulent vortices in the fluid wake, which makes the ball fly more smoothly.

With that in mind, it is functionally impossible to calculate the exact fluid drag (as a function of velocity) on any object. The best we can do is work with a series of approximations, and try to understand which ones will apply to a given scenario.

Reynolds Number

There are several dimensionless (i.e. unit-less) numbers which one can use to characterize the flow of a fluid, and therefore predict how the drag on an object will behave. One such quantity is called the Reynolds number, and is defined as follows: $$ Re = \frac{\rho v L}{\mu}$$ where $\rho$ is the density of the fluid, $v$ is the velocity of the object through the fluid, $L$ is the characteristic size of the object (e.g. the diameter, if the object is a sphere) and $\mu$ is called the dynamic viscosity of the fluid. Roughly speaking, the viscosity is a measure of the "goopy-ness" of the fluid - you can see a more illustrative definition here.

Linear Drag - $Re<<1$

For a given flow, you can check to see whether the Reynolds number is large ($Re>>1$), small ($Re<<1$), or somewhere in between ($Re\approx 1$). If the Reynolds number is very small, then the fluid flow tends to be very smooth. Such a flow is called laminar, and is comparatively simple to describe. The drag experienced by a spherical object immersed in such a flow can be described by Stokes' Law:

$F_D = 6\pi \mu R v$

where $R$ is the radius of the sphere.

For objects which are not spherical, it's sometimes possible to apply a generalized version of Stokes' law where the constant factors are modified but $F_D \propto v$ - however, if the shape is significantly different from a sphere, this may not be true.

Quadratic Drag - $Re>>1$

On the other hand, if the Reynolds number of the flow is very large, then different physics applies. In this case, the flow tends to become turbulent rather than laminar. In such a case, the drag experienced by an object is typically proportional to the square of the velocity.

Again, this may or may not be particularly accurate, depending on the shape of the object in question. For a perfect sphere, $F_D \propto v^2$ for Reynolds numbers on the order of a few thousand or more; for objects with sharp corners (like a cube) or more complicated shapes (like an airplane wing), this number may be modified significantly.

What if $Re \approx 1$?

When neither of these approximations apply, the drag on an object cannot be easily described in general. In such cases, computer simulations can be used to calculate the expected drag on a particular object, but doing so with pencil and paper is typically not possible.


So to wrap things up, fluid drag is heavily dependent on the geometry of the object in question, but one can often get a reasonable idea of how the drag will behave by calculating the Reynolds number. A very low Reynolds number usually means the drag will be linear, whereas a very high Reynolds number usually means that the drag will be quadratic. A Reynolds number which is neither particularly large or particularly small makes prediction difficult in general, and where the cutoffs happen to be depends on the fluid and on the object moving through it.

It is crucial to remember that $Re$ is a property of a flow, not of a fluid (or an object) - in particular, if we increase the object's speed, then the Reynolds number will change accordingly. We can consider, for example, a sphere the size of a baseball moving through dry air.

In this case, we have that $\rho = 1.275\ kg/m^3$, $L = 0.074 \ m$, and $\mu = 1.845\times 10^{-5} \ Pa \cdot s$. If $v=105.1 mph$ (the fastest recorded MLB pitch), then the resulting Reynolds number is

$$Re \approx 240,000$$

This is very large, and as such the drag on the sphere is approximately quadratic. It's worth noting, however, that the stitching and rotation on an actual baseball can change the dyanamics of the drag considerably - this is the origin of some of the crazy pitches that can be thrown, and why the precise specifications of baseballs needs to be regulated.

On the other hand, consider the movement of a grain of dust ($L\approx 10^{-5} \ m$) through the air. Assuming a speed of $5 \ cm/s$, the resulting Reynolds number is

$$Re = 0.035$$

This is a small Reynolds number (though not actually that small, as air is not particularly viscous), so this is an example of a laminar, viscosity-dominated flow, and the drag on such an object would be primarily linear.

| cite | improve this answer | |
$\endgroup$

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.