Download PHYSICS 110 Laboratory

Physics 101 Laboratory
Air Resistance and Terminal Speed
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Introduction:
In this exercise we are going to determine a relationship between air resistance and terminal speed of
a falling object. We are going to develop this concept by using some of the short hand notations (equations) we
have been studying in the lecture class.
Equipment needed:
Special air resistance devices (coffee filters), a 2-meter stick and stop watch.
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Think about an object falling under the influence of gravity. In this case, instead of ignoring the effects of air resistance
we want to include them. So, in this case, we can think of two forces acting on the object. The first is gravity,
which always acts downward. We know that we can identify the force of gravity with the value mg, where m is the
mass of the object and g is the acceleration do to gravity. The second force acting on the object is due to air
resistance. We will write this force simply as R.
From our study of Newton's Second Law we can write the net force equation as
Fnet = (mg - R) = m a
(1)
In this expressing, Fnet is the net force acting on the falling object of mass m and R is the force due to the air
resistance.
We recognize that mg is the weight, W of the object so we can write the force expression as
W–R =ma
(2)
Now, look what happens when the air resistance (an upward force) equals the weight (a downward force). Of course,
the acceleration must be zero! That is why we call terminal speed what we do, the change in speed (acceleration)
has terminated. So, in our shorthand notation we can say
R = W ( at terminal speed )
(3)
This is all just what we decided in our lecture class except with our shorthand notation ( the equations). In other words,
when the air resistance comes to equal the weight, the acceleration ceases (the forces balance out) and this
means the velocity is constant and terminal speed has been reached. These two equations say a whole lot, don't
they?
But we also know something else about this object falling at terminal velocity. We learned in our study of kinematics
that, because the velocity VT is constant, the distance, D, traveled in time t at terminal speed, VT, is given by
D = VT t
(4)
What we want to do now is to use our simple shorthand expressions (equations?) to find out something about air
resistance. We know that the air resistance R depends on the speed in some way but we don't really know just
what that dependence may be. We consider in this experiment two possible relationships.
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101 Laboratory
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Air Resistance and Terminal Speed
First assumption (hypothesis)
One relationship is to assume the air resistance force R is proportional to the speed. We use a symbol ~ to mean
"proportional to" so that if R is proportional to terminal speed (R ~ VT) then we know that the weight, W and
terminal speed VT must also be proportional to each other (VT ~ W). From our equation (3), R = W, so that what
goes for R must also go for W. But that's not all. According to our equation (4) we know that D and VT are related
so we can also say that D is proportional to weight according to
D~Wt
If we did an experiment with an object falling a distance D at terminal speed and if we then doubled the weight we
would expect the object to fall the distance D in half the time as for a single weight (note: if we double W then we
must cut t in half in order for the product W t to stay the same value, which is D). If we use triple weight, then the
time should be 1/3 the time for single weight.
2.
Second assumption (hypothesis)
Another relationship is to assume the air resistance force R is proportional to the square of the speed. Assuming R is
proportional to square of the terminal speed ( R ~ VT2 ) we see from equation (3) again that the square of the
terminal speed would also be proportional to the weight W (VT2 ~ W). What this means is that the terminal speed
VT should be proportional to the square root of the weight, VT ~ (W)1/2. And once again from our equation (4) we
should expect that the falling distance must also be proportional to the square root of the weight according to,
D ~ (W)1/2 t.
In this case, if we double the weight we would expect the object to fall a fixed distance D in about 70% (actually 1/√2)
the time of a single weight. If we triple the weight the time should be about 60% ( 1/√3 ) the time of a single
weight and finally if we quadrupled the weight the time should be 1/2 ( 1/√4 ) the time for one weight.
The experiment we perform to determine which of these assumptions is the better is really very simple. Let's do it.
The Procedure:
1.
Measure the time for a stack of 16 coffee filters to drop 2m and record the time in the data table. Repeat each
measurement process five times to establish an average value.
2.
Repeat step 1 with 9, 4, 2, and 1 coffee filters, taking care to preserve the shape of the filters.
3.
From the data in the table calculate each speed. Because the coffee filters reach terminal speed so quickly these
calculated speeds are a very good approximation to the final, terminal speed.
Time (seconds) for coffee filters to drop 2m
# of filters->
1
2
4
Drop 1
Drop 2
Drop 3
Drop 4
Drop 5
Avg. time
V(m/s)
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Physics 101 Laboratory
Air Resistance and Terminal Speed
Now, we wish to determine if the experimentally found velocity is described best by the weight ( W ) or the square root
of the weight ( (W)1/2 ) and we do this by a graphical method. Plot Terminal velocity ( V ) vs. weight and vs.
square root of weight and comment on what you discover. Assume that all filters have the same weight so one
filter weighs 1 unit, two filters weighs 2 units etc.
Discuss what you notice from the data and the plots. Include in the discussion what information you discover from the
data and plots regarding terminal velocity?
As an interesting check on the procedure, simultaneously drop one filter from a height of one meter and two filters from
a height of 1.4-meters. Then simultaneously drop one filter from a height of one meter and two filters from a
height of 2-meters. Also try dropping four filters from two meters and one filter from one meter at the same time.
The purpose is to see which case(s) best fits the description "both hit at the same time." Note that this is looking
at our equation (4) with time fixed rather than D fixed. Discuss these results, mentioning such things as the
reasoning behind the procedure, what results you obtained and if these results are consistent with the "time of
flight" measurements above?
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