Purpose:

Is the air resistance force on a falling object proportional to its velocity, or to its velocity squared?

Discussion:

When an object falls in air, the air exerts an upward air
resistance force (sometimes called a *"drag force"* ) on it.
As the speed of the falling object increases, so does the air
resistance force. When the air resistance force equals the weight of
the object, it stops accelerating and falls with a constant
**terminal velocity**. Light objects - feathers, pieces
of paper, and coffee filters exhibit this behavior quite readily -
since they have a large surface area and a small weight, they reach
their terminal velocity very quickly.

Certainly, the air resistance force on an object depends on its speed - but how? Suppose that the air resistance force is proportional to the object's velocity. Then, at terminal velocity:

F_{air}= k v = W

where k is a constant that depends on the shape of the object and the density of the air, or:

v = W/k

since the distance that an object moves (at constant velocity) equals velocity times time,

This equation says, "at terminal velocity, distance fallen is proportional to weight" - in other words, an object twice as heavy will fall twice as far in the same time, or an object three times as heavy will fall three times as far in the same time, etc.

On the other hand, what if the air resistance force is proportional to the square of the speed? In this case, at terminal velocity:

F_{air}= kv^{2}= W

(where k is a constant) or:

which leads to:

So if air resistance is proportional to v^{2}, then
distance fallen is proportional to the square root of the object's
weight, so an object twice as heavy would fall about 1.4 times as far
in the same time (instead of twice as far), and object three times as
heavy would fall about 1.7 times as far in the same time (instead of
3 times as far), etc.

So, is the air resistance force on a typical falling object proportional to its velocity, or the square of the velocity?

Equipment:

several coffee filters |
meter sticks |

stopwatch (?) |

Procedure:

First of all, notice that two coffee filters placed together have twice the weight as one coffee filter, but about the same surface area. Three coffee filters have three times the weight as one coffee filter, etc. This means that the appropriate unit of weight in this lab would be "coffee filters", as in "1 coffee filter", "2 coffee filters", etc.

From here, there are a couple of reasonable ways to proceed. The "straightforward" way would be to drop coffee filters a known distance and time their fall. A graphical analysis (starting with weight vs. time of fall, perhaps) of carefully collected and recorded data might reveal whether the drag force is proportional to velocity or velocity squared.

Another approach would be to drop a single coffee filter a known
distance (a meter, say) each time, and determine the height from
which a double (or triple, whatever) coffee filter must be dropped**
to** **hit the floor in the same time**. (If
F_{air} is proportional to v, this would be 2 meters...) You
wouldn't have to mess with a stopwatch this way, for one thing. Here,
you would be analyzing weight vs. distance fallen.

Analysis:

Which model for air resistance (if any) does your experiment support? Why do you think so? How confident can you be in your conclusion? Discuss the major sources of error in your experiment.

last update November 10, 2007 by JL Stanbrough