Physics 161
Rotational Motion & Moment of Inertia
Introduction
In this experiment we will study motion of objects is a circular path as well as the effect of a
constant torque on a symmetrical body. In Part I of this lab, you will use World In Motion to
examine two-dimensional circular motion of an object. You will investigate the relationships
among acceleration, velocity and position of the object. In Part II you will determine the angular
acceleration of a disk when a constant torque is applied to the disk. From this we will measure its
moment of inertia, which we will compare with a theoretical value. In Part III, you will observe
the relationship between torque, moment of inertia and angular acceleration for a rotating rod
with two masses on either end. You will vary the mass connected (and therefore the torque
applied) to the rod by two pulleys. You will also change the moment of inertia of the rod system
by changing the distance of the masses from the center of mass of the rod.
Reference
Young and Freedman, University Physics, 12th Edition: Chapter 3, section 4; Chapter 9, sections
1-4
Theory
Part I
If an object is moving around a circle at any given instant the distance of the object to the origin
must be constant. Therefore
r = x2 + y2
(1)
Where x and y are the coordinates of the center of the object and r is the radius of the circular
path. Uniform circular motion can be described as the motion of an object in a circle at a
constant speed. The speed at any given instance can also be calculated as:
v = v x2 + v y2
(2)
Where vx is the velocity in x direction, vy is the velocity in y direction and v is the tangential
velocity of the object. There is another quantity that is called angular velocity and is given by:
(3)
ω = v/r
The acceleration of the object is called centripetal acceleration which points towards the center
of the circle. This acceleration can be calculated using the following equation:
v2
(4)
= rω 2
r
You will use these equations to analyze the video of an object moving around a circle.
a=
Part II-III
Moment of inertia is a measure of the distribution of mass in a body and how difficult that body
is to accelerate angularly. For both parts of the experiment, a falling mass will accelerate a
rotating object in the horizontal plane. In Part I, the object will be a disk. For Part II, you will add
a ring to the system and find the moment of inertia of the system. In Part III, you will find the
moment of inertia of a rod with two masses attached to it.
The basic equation for rotational motion is:
∑τ = Iα
(1)
where α is rotational acceleration in units of rad/s2, τ is applied torque in N m, and finally I is the
moment of inertia or rotational inertia in units of kg m2. For a uniform disk pivoted about the
center of mass, the theoretical moment of inertia is
I disk =
1
MR 2
2
(2)
where M is the mass of the disk and R is the radius of the disk. In Part I we measure α and use
this to calculate I, which we will compare with the theoretical value of I. For a ring pivoted about
the center of mass, the theoretical moment of inertia is
I ring =
1
M ( R12 + R22 )
2
(3)
where M is the mass of the disk and R1 is the inner radius and R2 is the outer radius of the ring.
In Part II, we add a ring to the disk in Part I and measure I experimentally. Then we will compare
the theoretical and experimental values of I for a ring.
In Part III the moment of inertia is the sum of the moments of inertia of the two masses and the
rod. For the masses that slip onto the rod, we will assume point masses. Thus, the moment of
inertia for one of the two masses is:
I mass = mr 2
(4)
where r is the distance of the center of mass from the axis of rotation located at the center of the
rod. Because the masses can be moved along the rod, r will be adjusted to change their moment
of inertia. The moment of the inertia of the rod with mass M and a length L is:
2
1 M
I rod = 12
rod L
(5)
The moment of inertia for a rod with length L and two masses on each end at a distance r is
simply the sum of the components as defined by Equations 4 and 5:
2
2
2
1 M
I = 2mr + 12
rod L
(6)
The second term is multiplied by two because there are two point masses. Given the moment of
inertia of the entire system, I, and the torque, τ , applied by the mass M hanging from the pulley,
angular acceleration, α can be found. Just as F = ma for translational motion,
τ = Iα
(7)
Torque, τ , is an applied force that causes rotation with an angular acceleration, α , in an object
with a moment of inertia, I. The rotating object in all three parts will be attached to a pulley
placed at the center of the rotary motion sensor which has a string wrapped around it. The string
has a mass tied to one end (that will vary) and is laid over an additional pulley which allows the
falling motion of the mass to be converted to a torque on the rotary motion sensor. For the
purposes of this lab, the applied torque will be due to a mass accelerated by gravity acting on the
pulley of the rotary motion sensor. We will use the expression:
τ = MgR − MR 2α
(8)
where M is the mass hanging on the pulley and R is the radius of the Rotary Motion pulley. The
full derivation of the above torque is given in the Appendix, The experimental moment of inertia
can be found using the following equation:
I experiment = τ / α
(9)
In Part IV, by adjusting the masses on the rod, we can observe how an increased moment of
inertia (where either the mass is distributed farther from the center of mass or the total mass is
increased) will result in a decreased angular acceleration for the same torque. It is the same type
of relationship as the one you observed for Newton's Second Law.
Procedure
Part I: Object moving around a circle.
On your lab computer, each student group should read the UCM.avi file into the “World-in-Motion”
program. This is the video of a turntable with two dimes placed on top of it. The video was taken
with a camera on a tripod pointing down at the turn table. The file is located under z drive>videos>
UCM.avi. Both x and y are real world horizontal. As the turntable rotates, the dimes will travel in
circular paths.
1.
Click on the World in Motion icon on your screen; you may need to go to the Start bar
and select Programs>World in Motion.
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2.
Once you have opened World in Motion, select Video Analysis to open the video file.
Your instructor will tell you where it is located and which file to select.
3.
Depending on the file you are using, you may need to establish a new origin. To do this,
select Video>New Origin and follow the instructions given. The points you mark should
be the tip of the shaft that comes out of the middle of the turn table.
4.
Hit Play to return to the beginning of the motion. Place your cursor over the center of
mass of the dime closer to the origin in each frame. Mark the center of the dime in each
frame by clicking the mouse, then hit Step to proceed to the next frame. Advance the
video in this fashion, stepping and clicking for a data point until the dime makes one
complete rotation. Click on Dataset 1 and notice that the video will return to its original
frame. Repeat the above steps for the second dime and click on Dataset 2 at the end.
Click on Save and choose Copy Data to Clipboard.
5.
Before you leave the World in Motion program, estimate the uncertainty in the x and y
components of the position of the dimes that you have recorded. This estimate should be
based on your ability to accurately mark the center of the dimes. Assume that the
uncertainties in the x and y components are equal.
6.
Open Excel and Paste the data into the spreadsheet.
7.
Your spreadsheet should have separate columns for t, x1, y1, x2, and y2. The time in
seconds is in the first column. The second and third columns include the xy positions for
the first dime, and the xy positions of the second dime are listed in the forth and fifth
columns.
8.
In Excel calculate the radius of the circular path for the first dime using equation 1. Find
r by taking the average of all the columns containing the r values and σr by taking the
standard deviation of the r column. Repeat this for the second dime.
9.
Now calculate the x and y components of the velocity for both dimes just as you did in
the Projectile Motion lab. Calculate v1 and v2 for any given time using equation 2. Is v1
constant? Is v2 constant? Is v1 equal to v2? Determine the uncertainty of the velocities v1
and v2 by simply determining the standard deviation of them. Produce graphs of xposition, y-position and r vs. time (all in one graph) and vx, vy, and v vs. time (all in one
graph) for the first dime.
10.
Calculate the angular velocity ω1 for the first dime and ω2 for the second dime using
equation 3. Is ω1 equal to ω2? Calculate the σω1 and σω2 using error propagation. How
many σs are these two angular velocities apart? Include the calculation in your sample
calculation.
11.
Calculate the centripetal acceleration using equation 4 for both dimes. Using error
propagation, find the uncertainty for both accelerations. Include the calculation in your
sample calculation.
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Part II: Moment of Inertia for a Disk
The experiment uses a mass hanging on a string over a pulley to exert a constant torque on the
system resulting in a constant angular acceleration on the system.
1. Set up the apparatus as shown in Figure 9.1, making sure to connect the rotary motions sensor
to the interface box. Connect a disk to the rotary motion sensor. Although in the picture, you will
not use the ring in this part of the experiment.
Figure 9.1
2. Make sure that the pulley is set up to give positive values for angular position. This means that
the rotary motion sensor will turn in a counterclockwise direction as the mass on the pulley
drops. (The motion sensor may have an indication of which direction is positive taped to it.)
3. Adjust the measurement rate to 20 Hz on the Rotary Motion Sensor.
4. Choose angular velocity, ω , in radians/s under the Measurement tab only.
5. Measure and record the radius of the rotary motion sensor pulley around which the string is
wound. Hang a mass of 50 grams from the pulley, press Start, and release it.
6. Press Stop just before the hanging mass reaches the ground.
7. Create a graph of angular velocity vs. time.
8. Using a linear trendline fit determine the angular acceleration α .
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7. Make the following table and find the experimental moment of inertia for the disk and
compare it to its theoretical value.
Rdisk(m) Rpulley(m) mhang(kg) mdisk(kg)
α
τ
I experiment = τ / α
I theory = 0.5mdisk rdisk
2
Part III: Moment of Inertia of a Rod with Two Masses Attached
The experiment uses a mass hanging on a string over a pulley to exert a constant torque on the
system resulting in a constant angular acceleration on the system.
1. Set up the apparatus as shown in Figure 9.2, making sure to connect the rotary motions sensor
to the interface box.
Figure 9.2
2. Make sure that the pulley is set up to give positive values for angular position. This means that
the rotary motion sensor will turn in a counterclockwise direction as the mass on the pulley
drops. (The motion sensor may have an indication of which direction is positive taped to it.)
3. Adjust the measurement rate to 20 Hz on the Rotary Motion Sensor.
4. Measure angular position, θ , in radians, angular velocity, ω , in radians/s and angular
acceleration, α , in radians/s². Create graphs of these quantities vs. time.
5. Measure and record the radius of the rotary motion sensor pulley around which the string is
wound. Hang a mass of 50 grams from the pulley, press Start, and release it.
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6. Press Stop just before the hanging mass reaches the table. You will repeat this for four trials
with different masses and create a table with the following format:
Mhang(kg)
I masses = 2mr 2
r = r0
0.05
0
r = r0
0.05
r = .8r0
0.05
r = 0.7 r0
0.05
Run
r(m)
rpulley(m)
I theory = 2mr 2 + I rod
α
τ
I experiment = τ / α
IMPORTANT: The difference between Mhang and m: the first is the hanging mass on the string
which accelerates the rod; the second represents the point masses on each end of the rod.
Another important distinction: r is the distance between the center of mass of the point mass and
the axis of rotation (the center of the rod), rpulley is the radius of the pulley on the rotary motion
sensor around which the string is wrapped.
For Your Lab Report:
For Part I, include a sample calculation of r, v , ω and a with uncertainties. In your report,
explain why acceleration is not zero, even though the speed was constant. Was v1=v2? ω1= ω2?
Within how many sigmas?
Include a sample calculation of the torques τ = MgR − MR 2α and moments of Inertia Itheoretical
and Iexperimental for Part II. Estimate the uncertainty of the measurements for the distance from the
axis of rotation (r), and the angular acceleration as recorded by Data Studio. Assume 3%
uncertainty for the mass and 2% uncertainty for g. Use these values to propagate your
uncertainty to find uncertainties for Itheoretical and Iexperimental for Part II. Compare Itheoretical to
Iexperimental within their uncertainties.
Appendix
The calculation of the torque applied to the rotary motion sensor is as follows:
The rotary motion sensor has a radius, R, which is the distance from the axis of rotation at which
the force, T, acts. T is the tension in the string attached to the pulley and is due in part to the
weight on the string, Mg, where M is the mass of the hanging mass and g is the acceleration due
to gravity. Figure 9.3 below shows a rough sketch of the set up.
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Figure 9.3
The application of Newton's Second Law to the hanging mass results:
ΣFy = Ma
ΣFy = Mg − T = Ma
The downward direction is taken to be positive. We can substitute for the linear acceleration, a:
a = Rα
Where α and R are the angular acceleration and the radius of the rotary motion sensor,
respectively. Continuing to solve the equation gives us:
Mg − T = MRα
T = Mg − MRα
T is the tension in the string. It is a force, not a torque. To find the torque acting on the pulley,
we must multiply by the distance from the axis of rotation at which the force acts:
τ = RT
τ = MgR − MR 2α
For the set up of this experiment, take the following example where M=0.06 kg, R ≈ 0.025 m
and α ≈ 2.58 rad/sec². Substituting these values results in:
MgR = 14.7 × 10 −3 Nm
MR 2α = 0.0969 × 10 −3 Nm
From this example, we see that MgR⟩⟩ MR 2α so we can approximate τ = MgR .
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