Purpose:

If the air resistance force on a falling coffee filter is proportional v^{n}, then what is n?

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, v _{t} **. 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 some power of the object's velocity, v^{n}. Then, at terminal velocity, v_{t}, the net force on the falling object is zero, so the upward forces and downward forces must balance:

F_{air}= k v^{n}= W

where k is a constant that depends on the shape of the object and the density of the air. Taking the logarithm of both sides of this equation gives:

Applying the properties of logarithms gives:

rearranging gives:

This equation says (compare to y = mx + b) that if you plot the logarithm of W versus the logarithm of v, the graph will be a straight line with slope n (and "y"-intercept ln(k)). You could easily determine the exponent n from this graph. Note that the conventional approach would be to plot the "weight number" on the horizontal axis and the "velocity number" on the vertical axis, since the weight of the stack of coffee filters is what you change in the experiment, and the velocity changes by itself. However, it simplifies the calculations a lot to plot the "weight number" on the vertical axis in this situation.

Equipment:

several coffee filters | Pasco motion detector | Pasco computer interface |

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.

You could place a motion detector on the floor, oriented vertically, and drop coffee filters on it. By analyzing the position vs. time graph of the motion, you should be able to determine the terminal velocity for the dropped coffee filters. After collecting sufficient data, you could plot a graph of ln(W) versus ln(v). If this graph is a straight line, you can determine the exponent, n, from the graph.

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 13, 2006 by JL Stanbrough