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Unformatted text preview: Laboratory 6
Rotational Motion I: The Inclined Plane
Introduction
With this laboratory, we begin a threepart study of rotational motion that includes today’s experiment and Laboratory 8 (Rotating Bar), and ﬁnally culminates in a study of gyroscopic motion in
Laboratory 9. This laboratory will also help you to understand twodimensional inelastic collisions
in Laboratory 7. You will ﬁnd that while all of the rotating systems we investigate initially appear
very different, they can all be analyzed using the same basic principles of energy and momentum
conservation that we have seen in previous laboratories. The equations will look slightly different,
but they are very similar in structure to the equations for translational (“linear”) motion because
they are built around these same two powerful conservation laws.
Until this point, we have analyzed all linear motion by treating the objects involved as point particles  that is, we assume that all forces acting on the object act at exactly the same spot. But in
rotational motion, it matters where on the object the forces act. Forces acting on one part of the
object may cause it to rotate, but those same forces acting on another part may cause no rotation at
all. Theoretical Background: Mass vs. Moment of Inertia
We can think of mass as a measure of how hard it is to get something to move (translationally, that
is). Moment of inertia, in contrast, is a measure of how hard it is to get something to rotate. Both
are scalars. This fact can make your life a lot easier when solving problems because, if you have a
system of multiple rotating objects, you can break that system into its component parts, calculate
their individual moments of inertia, and add them up to get the combined moment of inertia.
Have a look at the moments of inertia listed in Table 1 for some different object geometries, where
the same total mass is distributed in different ways about the axis of rotation.
One more important moment of inertia to consider is that of a point mass of mass m, located at a
distance R from the axis of rotation. Its moment of inertia is just
m R2 .
Problem 6.1 Why is the expression for moment of inertia of a point mass
the same as that for the moment of inertia of a thinwalled cylinder?
48 (1) Table 1: Moments of inertia for some common object geometries. From these expressions, it is easy to see that adding additional mass to an object will increase I .
But it is more important to realize that you can also increase I by keeping the mass the same and
just changing how it is distributed around the axis of rotation.
In our discussion, we will focus speciﬁcally on the thinwalled and solid spheres and cylinders.
Notice that all of these objects have similarlooking expressions for their moments of inertia,
I = fI m R2 , (2) where m is the object’s mass, R is its radius (the maximum distance it attains from the axis of
rotation), and fI is a variable factor, called the geometric factor, which depends on how the
object’s mass is distributed around the axis of rotation. We deﬁne the geometric factor as
fI ≡ I
.
m R2 (3) Notice also that fI gets larger as more of the object’s mass is placed further away from the axis of
rotation. It reaches its maximum, 1, in the case of the thinwalled cylinder.
Problem 6.2 What is one way you can change a cylindrical object’s moment of inertia without changing its mass or its radius?
The message here is that in rotational motion, how far something is from the axis of rotation
matters, whether that thing be a force or a bit of mass. Moment of inertia makes this adjustment
for mass, and the concept of torque does the same for force. We will discuss torque further in
Laboratory 8.
49 Theoretical Background: Rotational Kinetic Energy
Because a rotating object has mass, and that mass is moving, we intuitively sense that the object
should have some kinetic energy. However, we are used to thinking of kinetic energy in terms of
a point object moving along a straight path, so it is not immediately obvious how we can quantify
rotational kinetic energy or understand it in terms of concepts we have learned before. The trick,
it turns out, is to look at a rotating object as a large set of point particles, each of which moves
around the axis of rotation at a certain radius.
The speed of the ith particle, which we will call vi , is given by vi = ri ω , where ω is the angular
speed of the rotating object.
Problem 6.3 If we assume our rotating object is rigid, such that all of its
component particles have the same ω , do particles whose distances ri from
the axis of rotation are large have greater or smaller vi than particles closer
to the axis? Why?
Problem 6.4 Show that the units on ri ω are the same as those for vi (m/s).
If we use the standard formula for the kinetic energy of a point particle, 1/2 m v2 , we can write the
kinetic energy of the ith particle as
1
1
mi v2 = mi ri2 ω 2 .
i
2
2
If we then add up all of the different particles’ kinetic energies, we obtain
1
mi ri2 ω 2 .
2 Krot = ∑
i Pulling the constant 1/2 and ω 2 out of the sum, we ﬁnd that the rest of the sum is just a set of
masses (of the individual point particles) times their radii squared. This is actually the formal
deﬁnition of moment of inertia
I ≡ ∑ mi ri2 .
i Compare this formula to that for the moment of inertia of a point mass (Equation 1). It turns out that
the common formulas shown in Table 1 for the moments of inertia of various objects are actually
simpliﬁcations of this sum  someone did the sum in advance for different object geometries. We
can therefore rewrite the rotational kinetic energy formula as
Krot = 1
I ω 2,
2 which is the formula you have seen in class for rotational kinetic energy.
50 (4) The important thing to remember when you are doing problems involving rotational kinetic energy
is that Equation 4 already takes the distance of each part of the object from the rotational axis into
account, via its dependence on the moment of inertia. Using the moment of inertia allows one to
formulate a rotation problem in terms of ω , rather than constantly having to think about what all
of those individual point masses that make up the extended object are doing.
Problem 6.5 If mass is measured in kilograms and distance is measured
in meters, what are the units for I (moment of inertia)?
Problem 6.6 What are the units for rotational kinetic energy? Using Equation 4 and your answer to Problem 6.5, show that the units on I ω 2 are the
same as the units for translational kinetic energy. Experimental Background: The Inclined Plane and Photogates
Now we arrive at the inclined plane, down which we will roll a variety of objects. The goal of
our experiment will be to ﬁrst verify that energy conservation holds in situations where rotational
kinetic energy is included, and then to use this fact to measure some unknown moments of inertia.
To do this, we will need the theoretical tools we just discussed (rotational kinetic energy and
moment of inertia) as well as one from a previous laboratory (conservation of energy). A picture
of the experimental setup is shown in Figure 1. Figure 1: Photograph of the experimental setup.
The two photogates at the bottom of the track work a bit like the safety mechanism on an automatic garage door opener. There is a laser beam that passes from one side of the photogate to
the other, and if the beam is broken, the computer records the time at which this occurred. Here
two photogates are used in series, and LoggerPro can automatically detect the amount of time that
elapses between when an object breaks the beam of the ﬁrst photogate and when it breaks the
51 beam of the second photogate. This elapsed time, combined with a knowledge of how far apart the
photogates are, allows us to determine the object’s average velocity in the region between the photogates. We will, for the purposes of this laboratory, assume that this average velocity is actually
the object’s instantaneous velocity at the midpoint of the two photogates  admittedly, a slightly
crude approximation.
Consider a spherical/cylindrical object of mass m and radius R sitting (at rest) at the raised end of
a plane of length d which is inclined at angle φ . There are two ways in which this object can move
to the bottom of the plane, as shown in Figure 2. Figure 2: The two object positions we are concerned with for conservation of energy calculations. Way #1 The object can slide. This is the situation we have encountered so far in energy conservation problems. It is very straightforward: the object loses a certain amount of potential energy
by sliding to the bottom, and that potential energy is converted into translational kinetic energy.
Thus,
change in potential energy = ﬁnal kinetic energy,
1
m g ∆h =
m v2 .
2 (5) Way #2 The object can roll. This is slightly more complicated. The potential energy that is lost
by moving to the bottom is still converted into kinetic energy, but now there are two types of kinetic
energy: translational and rotational. Thus,
change in potential energy = ﬁnal kinetic energy
1
1
m g ∆h =
m v2 + I ω 2 .
2
2
We can simplify this equation even further if we say that the object rolls without slipping, so that
ω = v/R. Then we have
1 2
m g ∆h =
v m + I /R2 ,
(6)
2
12
=
mv (1 + fI ) ,
(7)
2
52 where fI is, again, the geometric factor.
Problem 6.7 What force causes a round object to roll down an inclined
plane instead of sliding down the plane?
Problem 6.8 How is velocity measured in this laboratory? Why is our
measurement of velocity considered to be the velocity at the midpoint of the
two photogates, rather than at the second photogate? (Hint: Consider the
deﬁnition of “average velocity”.) Experiments
Experiment 1: Rotational Kinetic Energy
In the ﬁrst experiment, we use energy conservation principles to characterize the motion of a rolling
object as it travels down an inclined plane.
Object
Radius (cm)
Steel Ball
2.540
Solid brass disk
2.540
Solid nylon disk
2.540
Disk (brass rim, nylon center)
2.515
Disk (nylon rim, brass center)
2.515
Table 2: Radii (R) for the objects used in all three experiments 1. Type the number on your inclined plane into the appropriate cell in your lab report. A value
for ∆x, the distance between the two photogates, will appear. Also record D, the length of
the plane (89 cm) on your template. Finally, calculate θ , the angle of incline of the plane, in
radians (θ = 25◦ ).
2. Weigh the steel ball and record its mass. Be sure to convert to kg.
3. Calculate ∆h, the vertical distance the ball travels from the top of the plane to the midpoint
of the two photogates, using the expression for d in Figure 3 (d = D − R − ∆x/2).
4. In your lab report, calculate the amount of potential energy the steel ball loses as it travels
from the top of the inclined plane to the midpoint of the two sensors.
5. Open LoggerPro and set it up to record the time delay between the object’s passage through
one sensor and the other. You can do this using the LoggerPro template inclined plane.cmbl.
53 6. Hold the steel ball carefully against the aluminum back plate at the top of the inclined plane,
centering it in the groove.
7. Start the data acquisition. Wait several seconds for the software to initialize the photogates.
Release the ball cleanly so that it does not wobble on its way down the track.
8. Repeat the rolling procedure three more times. LoggerPro should be recording the times it
takes the steel ball to roll between the two photogates.
9. Copy and paste these data into your lab report.
10. Calculate the mean ∆t for the four times you just measured (see Appendix A Section 2.2).
11. Using the approximation v = ∆x/∆t , where ∆t is the mean of the times you just measured,
calculate the translational velocity of the ball at the midpoint of the two photogates.
12. Calculate the ball’s translational kinetic energy at the midpoint of the two photogates.
13. Using your measurement of v and the ball’s radius (Table 2), calculate ω for the ball at the
bottom of the plane.
14. Using the ball’s moment of inertia as calculated in Table 1, calculate an experimental value
for the ball’s total translational and rotational kinetic energy at the bottom of the plane.
15. Repeat this procedure for the two solid disks (brass and nylon). Use the cell cloning procedure in Appendix A Section 2.3 to avoid having to mindlessly recopy formulas. Figure 3: Diagram of the experimental setup. In this diagram, R is the radius of the rolling object
(see Table 2). Note that you can get ∆h easily if you know d and φ by using ∆h = d sin(φ ). 54 Problem 6.9 What force does work on the ball/disks in these experiments?
Explain why the other forces these objects experience do not do work.
Problem 6.10 Compare the ﬁnal K with the ∆U for each of the rolling
objects. Are the ﬁnal K s generally greater or less than the ∆U ? What is one
factor that could explain the direction of the deviation you observe?
Problem 6.11 How would the following sources of error affect the ball’s
translational kinetic energy? Write a onesentence explanation of each.
(a) The ball slips as it rolls down the plane and losses due to friction are
negligibly small.
(b) The ball is hollow instead of solid, but total mass remains unchanged.
(c) The angle of the plane is actually less than 25◦ .
(d) The coefﬁcient of static friction of the plane is thought to be 0.2, but it
is actually 0.21.
(e) The scale you use to weigh the solid ball is miscalibrated, and measures
mass consistently too low. Experiment 2: Measuring the Geometric Factor
Here we compare the motion of objects with the same mass but different geometric factors. The
mass distributions of these objects are more complicated than those of the solid disks and ball from
Experiment 1. Here we use a procedure similar to that of Experiment 1 to measure the geometric
factors of two disks with the same mass but more complicated geometries.
1. Repeat the data acquisition procedure from Experiment 1 for the two composite disks (one
has a brass rim; the other has a nylon rim).
2. Calculate the amount of potential energy lost by the disks as they move from the top of the
plane to the point midway between the two photogates.
3. Collect four measurements of ∆t for each of the two disks and paste them into your lab
report.
4. Calculate the mean ∆t for the four times you just measured.
5. Using the approximation v = ∆x/∆t , where ∆t is the mean of the times you just measured,
calculate the translational velocities of the disks at the midpoint of the two photogates.
6. Calculate the disks’ translational kinetic energies at the midpoint of the two photogates. 55 7. Using your measurement of v and the disks’ radii (Table 2), calculate ω for each of the disks
at the bottom of the plane.
8. Using your data on the change in potential energy and the translational and rotational velocities of the disks, calculate their moments of inertia.
9. Obtain experimental values for the geometric factors for the two disks.
Experiment 3: Model Fitting and Statistics
1. Using the online data collection tool, record your values for the geometric factors of the two
composite disks (see Appendix A Section 3 for stepbystep instructions).
2. When all of the class data has been collected, paste it into your lab report and calculate a
95% conﬁdence interval for the geometric factors of both disks. To do this, you will ﬁrst
need to calculate the mean, number of trials, standard deviation, and error of the mean in the
spaces provided (see Appendix A Section 2.2).
Problem 6.12 Which composite disk has the greater geometric factor?
Why does this make sense? Which composite disk has a wider conﬁdence
interval for its geometric factor? What does this mean and what is one
feature of our experimental setup that might explain it?
Problem 6.13 In your own words, explain why a rod rotating about its
end has a different moment of inertia than the same rod rotating about its
midpoint (see Table 1 for an illustration of this.) 56 ...
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 Spring '11
 Gerson

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