This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Laboratory 3
Motion with Nonuniform Acceleration
Introduction
In Laboratories 1 and 2, we measured the displacement of an object that was subject to a constant acceleration, which led to a quadratic dependence of the object’s position on time. In this
laboratory, we will analyze motion where the acceleration is not uniform; this will demonstrate
the power of Newton’s Second Law to predict behavior for more complicated physical situations.
The example we will study is the acceleration of an object subject to a constant gravitational force
combined with the retarding inﬂuence of air drag. We will also explore some of the features of
motion with air drag, which will prepare you for the study of ﬂuid mechanics in the future. Figure 1: Galileo Galilei (15641642) was an Italian physicist, mathematician, and astronomer
who established that gravitational acceleration is constant at Earth’s surface. He also supported
heliocentrism, which the Inquisition did not appreciate. As a result, he spent the last ten years of
his life under house arrest.
One famous experiment involving gravity and air resistance is usually attributed to Galileo Galilei
(see Figure 1), although recent scholarship has shown that it was probably anecdotal. The popular
story describes how Galileo, in an effort to get the intellectual establishment of the time to accept
that objects of different masses  in the absence of air resistance  fall through the same distance
in the same amount of time, dropped objects of various masses off the leaning tower of Pisa and
20 timed their falls. The story is almost certainly false. But it is true that, at the time, most intellectuals believed in the teachings of Aristotle, who claimed that heavy objects fall faster than lighter
objects in direct proportion to their weight. Although Aristotle was correct under conditions of air
resistance as you will see in a moment, Galileo was correct on principle: gravitational acceleration
does not depend on the mass of the falling object. Theoretical Background: Gravity and Mass
Mass is a measure of how hard it is to get something to move. Subject to the same force, a more
massive object will accelerate more slowly than a less massive object. Imagine that the force of
gravity were constant for all objects. If Fg = C, we would expect that a heavier object in freefall would accelerate more slowly downward than a lighter object. From Newton’s second law,
F = ma, we would obtain the result that
ag = C
,
m which, experiments have shown, is wrong.
Why? Because the force of gravity is not constant. In fact, the force of gravity depends linearly on
mass, so heavier objects are subject to a greater force than lighter objects. In reality, Fg = m C, so
that
Cm
ag =
= C.
m
The acceleration is constant, and the constant, C, is the familiar g (on Earth)  the constant acceleration due to gravity. This was Galileo’s result.
Many students get tripped up because most of the forces we encounter in lecture  people pushing
on boxes, horses pulling on carts, and the like  are constant. The horse can only pull so hard, so
a heavier cart will accelerate more slowly than a lighter cart. But gravity is special in this regard:
the heavier an object is, the harder the Earth’s gravitational ﬁeld pulls on it, so the acceleration is
always the same. Theoretical Background: Air Resistance
One reason why people got tripped up on this mass dependence problem for so long is that it is
very rare for an object at Earth’s surface to actually be in freefall. It is true that, in vacuum, a
feather would fall to the ground as fast as a bowling ball dropped from the same height, but this is
obviously not true in the real world. The culprit, of course, is air resistance.
We know from a variety of experiences that moving air can produce considerable force. However,
there is no way to characterize air resistance with the same simplicity and precision as we can with
21 TurbulentFlow LaminarFlow Figure 2: A diagram illustrating the differences between laminar ﬂow and turbulent ﬂow around a
Mylar cone. the gravitational force, because it depends both on an object’s geometry  how “aerodynamic” or
“streamlined” it is  and its velocity. For this reason, virtually no problem you encounter in lecture
will take air resistance into account. However, it is a very important factor that people who deal
with “real” situations (such as engineers and, in particular, aerospace engineers) need to consider
in their calculations, and we will see why in this laboratory.
Problem 3.1 Describe how the force of air resistance is unlike the gravitational force. Speciﬁcally, what quantities does it depend on that the gravitational force does not? What quantity does the gravitational force depend on
that the force of air resistance does not?
There are two major types of air ﬂow that contribute to drag on a falling object. The ﬁrst is
laminar ﬂow. Laminar ﬂow occurs when layers of air move past an object at the same speed.
The layers do not interact, and no air molecules are exchanged among them. The other type of
ﬂow is turbulent ﬂow, where the ﬂow lines are disorganized and distributed in eddies that swirl
around and exchange air molecules. In turbulent ﬂow, air molecules traveling with different speeds
bump into each other, and their speeds are constantly changing. Figure 2 contains a diagram that
illustrates the differences between laminar and turbulent ﬂow.
As might be expected, the mathematics of laminar and turbulent ﬂow are also distinct. A sphere in
laminar ﬂow experiences a force given by
F = 6πη rv, (1) where r is the radius of the sphere, η is the viscosity of the air, and v is the object’s velocity relative
to the air. Note that this equation only holds for a sphere  the calculations for other objects are too
complex for this manual. The next equation, however, is true for any object. An object in turbulent
22 ﬂow experiences a force [approximately] given by
1
F = ρ Cd Av2 ,
2 (2) where ρ is the density of the air, A is the crosssectional area of the object, and Cd is the object’s
drag coefﬁcient (which depends on its geometrical properties). Note that, in the laminar regime,
air resistance is proportional to v, whereas in the turbulent regime it is proportional to v2 . In
general, turbulent ﬂow is mathematically difﬁcult to describe, and the real expression for the force
due to this ﬂow is much more complicated. We will generally say that ﬂow is “turbulent” if the
exponent on v is greater than one; there may be v3 terms, v4 terms, etc. Equation 2 is just a good
approximation.
As an object falls and its velocity increases, the force from the air drag increasingly opposes gravity
until there comes a point when the net force, and hence the acceleration, is zero. This occurs when
Mg = F
= cvn
where c is the constant term in Equation 1 or 2 and vn is the velocity raised to some power n, which
varies depending on whether the air ﬂow is laminar (n = 1) or turbulent (n > 1). The velocity
under these conditions  the terminal velocity, which we will call vT  is therefore a constant. We
can obtain the terminal velocity in both the laminar and turbulent ﬂow regimes by solving for v in
the equation above. For laminar ﬂow,
Mg
vT =
,
(3)
6πη r
while for turbulent ﬂow
vT = Mg
c (1/n) . (4) This is an important result, because the terminal velocity is something we can easily measure.
Equation 4 is what’s called a power law. Later, you will see it’s easier to pin down n in this model
if we apply a few simple mathematical tricks to our experimental data.
Problem 3.2 Is the force of air resistance greater for a fastermoving or
slowermoving object, assuming that the two objects have the same mass
and geometric shape?
Problem 3.3 How can Equations 3 and 4 (and Figures 3 and 4) help
explain the fact that, in situations where air resistance plays a role, a heavier
object dropped from the leaning tower of Pisa would actually hit the ground
before a lighter object of the same geometry? 23 Figure 3: Position vs. time data for objects with different masses. The position of the heaviest
object follows a nearlyparabolic path, while the lighter objects approach a linear path. Figure 4: Velocity vs. time data for objects with different masses. The lightest objects clearly
reach limiting velocities soon after being dropped, while heavier objects take much longer to attain
their limiting velocities. Theoretical Background: LeastSquares Fitting
Leastsquares ﬁtting is a technique, based on the statistical concepts we discussed in Laboratory 1, that allows experimenters to determine how well a particular model ﬁts their data. In addition, the technique can be used to adjust the parameters of the model in such a way as to provide
the closest possible ﬁt to all of the data points simultaneously.
24 Figure 5 shows a picture of some noisy data with a leastsquares ﬁt line through it. LoggerPro ﬁnds
bestﬁt lines automatically via its user interface, which you have used in previous laboratories.
Leastsquares ﬁtting can be used, as mentioned, with any type of model. It depends on the concept
of a residual, which is the distance between an individual data point and the corresponding data
point predicted by the model. Although this distance can be hard to calculate for a multidimensional model, we will typically be dealing with models that can be plotted on a y vs. x graph. In
this case, the formula for a residual can be expressed mathematically as
residual for data point i = actual value of point i − ﬁt value for point i,
δi = yi − y f iti . (5) A magniﬁed picture of a region of Figure 5 is shown in Figure 6. The lengths of the green lines are
the values of the residuals for some of the data points. Points below the ﬁt line will have negative
residuals, while points above the ﬁt line will have positive residuals. Figure 5: An example of noisy data with a leastsquares ﬁt line that best describes the global
behavior. Figure 6: Diagram of how residuals are measured  not all residuals are shown. 25 The essence of the leastsquares method is that it minimizes the sum of the squared residuals, ∑
i y f iti − yi 2 . (6) We square the residuals before we add them up because we do not want negative and positive
residuals to cancel each other out.
The leastsquares method, as performed by LoggerPro, varies the parameters of the model y f it (x)
until the sum in Equation 6 is minimized. For example, in Figure 5, the ﬁt model is a line. The
individual data points have the form (x, y), and the ﬁt line has the formula
y f it (x) = m x + b.
LoggerPro varies the parameters m and b until the sum of the squared residuals is as small as
possible, and it then plots the resultant line on your graph. If you were using a quadratic model
instead, where y f it (x) = A x2 + B x + C, LoggerPro would have to vary three parameters instead of
two to get the best ﬁt.
Problem 3.4 Draw a graph with some madeup data points on it. Now
draw an approximate ﬁt line/curve through those data points. Label three of
the residuals. How does the leastsquares method adjust the ﬁt line to give
you the best possible ﬁt? Use your own words.
Problem 3.5 Would summing the residuals and then squaring the sum give
you the same result that you obtain by squaring the residuals ﬁrst? Why or
why not? Power Law Model Fitting
The expression in Equation 4 is a power law, that is to say n could in principle take on any real
number other than 0. We could try distinguishing between the laminar and turbulent ﬂow regimes
by plotting vT vs. m. If the ﬂow is laminar, then it will have a linear trend (since n = 1 for laminar
ﬂow). On the other hand, If the ﬂow is turbulent, then there is a continuum of values n can take,
and hence a continuum of model equations to which the trend could ﬁt. This means that we can’t
ﬁnd the best model without already knowing the precise value of n, but it is the value of n that we
are trying to ﬁnd by ﬁtting the data to a model! To further complicate matters, this method is not
very helpful when using a limited range of masses  reason being is that a linear trend will ﬁt most
data sets over a limited domain (think tangent line to a curve).
What we do instead is use mathematical properties of the logarithm to force our data to follow a
linear trend, where the slope of the linear trend gives us information about the value of n. Now
instead of having an inﬁnite number of possible models where the information about the ﬂow is
26 encoded in the exact model equation used to ﬁt the data, we have a single model equation for all
ﬂow types, and the information about the type of ﬂow is encoded in one of the ﬁtting parameters
of the model itself.
To get this ‘allencompassing’ model, you must make ﬁrst take the logarithm of Equation 4, then
use of the following two log properties
ln(a + b) = ln(a) + ln(b),
ln(a)b = b ln(a) (7)
(8) to further disect the right hand side of the resulting equation. Collect terms in the appropriate
fashion, and you’re left with the expression
ln vT = 1
ln M + b,
n (9) where b ≡ (1/n) ln(g/c). Equation 9 is an equation for a line, where 1/n is the slope of the line
and b is the yintercept. To see a linear trend in your experimental data, the above form suggests
we plot ln vT vs ln M , and the slope of the ﬁtted line will relate to the ﬂow parameter, n. This is
exactly what you will do in this experiment to determine if your objects are experiencing laminar
or turbulent ﬂow.
Problem 3.6 Derive Equation 9 by ﬁrst taking the natural logarithm of
Equation 4, and then use the log properties in Equations 7 and 8 to get it
into the proper linear form. Experimental Background: Dropping Different Objects
The experimental apparatus for this experiment consists of an ultrasonic ranging system and a
vacuum release system identical to the one used in Laboratory 1. Please see that laboratory for
a photograph of the apparatus. In this laboratory, you will drop four Mylar cones with different
masses.
As shown in Figure 7, there are three different interchangeable release tubes with differentsized
conical cavities that will hold the various objects to be dropped. These can be exchanged by
depressing the small metal release on the upper plastic ﬁtting that holds the release tube in place.
You will use the two smaller adapters to drop the cones. 27 Figure 7: The vacuum release system requires different release adapters for different object geometries. Figure 8: The proper positioning of a Mylar cone on the vacuum release system. Experiments
Experiment 1: Motion of a Conical Object with Air Resistance
Here we drop four cones with different masses. Each cone consists of a sheet of Mylar folded into a
cone with a ball bearing at its apex. When dropped from a height of approximately 1.5 meters, the
cones should reach their terminal velocities well before they hit the ground. We can use multiple
measurements of mass vs. terminal velocity to determine whether the cones are experiencing
laminar or turbulent air resistance.
28 1. Use the smallerdiameter vacuum release adapters to hold the cones in place above the ultrasonic sensor, exactly as you did for the rubber ball in Laboratory 1. (See Figure 8). Start
with the heaviest cone. Its mass is written on it in pen.
2. Set up LoggerPro to take 4 seconds of data at 10 samples per second. You can do this using
the template ﬁle nonuniform.cmbl from the “LoggerPro Files” folder.
3. Use the trigger to release the cone, and take position vs. time data on its way down.
4. The graph should ﬂatten out near the end of the cone’s fall. The slope of the graph in
this region is the terminal velocity. Calculate the cone’s terminal velocity by performing
a linear ﬁt. See Appendix A Section 1.2 for how to perform a linear ﬁt in LoggerPro.
Problem 3.7 In general, how can you determine a reasonable region (on
the p vs t graph) to perform the ﬁtting in order to get a somewhat accurate
value of terminal velocity?
5. Copy the graph with the linear ﬁt into your lab report. Also list the mass of the cone and its
terminal velocity on the data table in your report.
6. Repeat this procedure three more times with the same cone. Then repeat everything again
with the three other cones. In the end, you should have sixteen sets of mass vs. terminal
velocity data (4 cones × 4 trials each). Note that you do not need to paste the graphs for the
ﬁnal three trials of each cone into your lab report.
7. Calculate ln(mass) and ln(v¯ ) for each data point.
T
8. Calculate the mean, standard deviation, error of the mean, and 95% conﬁdence interval for
each of your four measurements of terminal velocity (one conﬁdence interval for each cone).
See Appendix A Section 2.2 if you get stuck. Experiment 2: Model Fitting and Analysis
Deciding whether an object is experiencing laminar or turbulent air resistance is not a trivial task.
We want to see which model of air ﬂow (laminar or turbulent) best matches our observed data.
How do we determine which model is best? That is what we will investigate in this section.
1. Below the Experiment 2 heading in your lab report, you will see a graph. This graph contains
the data on ln(mass) and ln(terminal velocity) for the cones.
2. Fit a straight line to the ln(mass) vs. ln(terminal velocity) graph using Excel’s feature Trendline, which is similar to LoggerPro’s Linear Fit. See Appendix A Section 2.4 for instructions
on how to use Trendline. Include the ﬁt equation on the graph. What does the slope of this
line correspond to? Make sure you understand this before moving on.
29 Figure 9: The position vs. time curve for an object experiencing air resistance. Observe that the
position vs. time graph begins to approach a straight line for this object rather than continuing to
follow a parabolic arc, as it did with the rubber balls we dropped in Laboratory 1. This is what one
would expect for objects that have reached their limiting velocities.
Problem 3.8 When you perform a linear ﬁt on data in LoggerPro, a text
box appears with some information about your ﬁt. One of the quantities
in the box is the RMSE, or rootmeansquared error, which is related to
Equation 6. What does this quantity tell you about the quality of your ﬁt?
Use terms and concepts from the theoretical section on leastsquares ﬁtting.
Problem 3.9 Does terminal velocity appear to increase or decrease with
mass? Does this conﬁrm what you learned in the theoretical section of this
lab? How does this help explain the observation that a bowling ball hits
the earth before a styrofoam ball of the same shape, dropped from the same
height?
3. Record your value for n for the cones using the online data collection tool (see Appendix A
Section 3).
4. When all of the class data has been collected, calculate a 95% conﬁdence interval for n for
the air resistance experienced by the cones (see Appendix A Section 2.1).
Problem 3.10 Using your graph of ln(mass) vs. ln(terminal velocity),
decide whether the cones experience laminar or turbulent air resistance.
How can you tell? Does the conﬁdence interval of the class data include
the value 1.0? Supposing it did include 1, what would be your conclusion
about whether the object was experiencing laminar or turbulent air ﬂow?
Be careful about your wording. In statistics, we can only reject hypotheses,
never prove them to be true.
30 Problem 3.11 How would the graph of position vs. time for a single cone
change under each of the following conditions?
(a) The cone’s mass was increased but its shape was kept the same.
(b) The cone’s mass was kept the same but its diameter was increased.
(c) The cone fell through vacuum instead of through air.
(d) The air density in the classroom was increased. 31 ...
View
Full
Document
This note was uploaded on 09/13/2011 for the course ECONOMICS 101 taught by Professor Gerson during the Spring '11 term at University of Michigan.
 Spring '11
 Gerson

Click to edit the document details