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Rotational Inertial
Equipment
Qty Item Parts Number
1 Rotational Motion Sensor CI- 6538
1 Rotational Inertia Set: ring and disk ME- 8953
1 Large Rod ME- 8977
1 Universal Table Clamp ME-9376B
1 Detectable Pulley ME-9448B
1 Vernier Caliper
1 Mass and Hanger Set ME- 8979
1 Padding
Purpose
The purpose of this exercise is to examine the moment of inertia of both a ring and disk, and to
experimentally confirm that the moment of inertia of an object is a function of both its mass and how
that mass is spatially distributed.
Theory
Let us assume there is a mass m, initially at rest,
that is attached to one end of a massless rod of
length r, and the other end of the rod is attached to
a frictionless pivot that is free to rotate 360
o
. If one
were to apply a net force F to the mass it would
induce rotational motion about the pivot point, but
only the component of the applied for force that is
tangential (at a right angle) to the length r would
contribute to the change in motion of the mass. By
Newtonโs Second Law we know the tangential force
would be related to the tangential acceleration by the following equation:
๐
๐
The torque acting about the point of rotation that is associated with the tangential component of the
applied force would be given by:
๐
๐
We also know that the angular acceleration that the mass is experiencing is related to its tangential
acceleration by ๐
๐
= ๐๐ผ, so we can insert that into the equation, giving us:
2
rev 05 /201 9
What would happen if you would apply a force to a rigid body (solid object) that is fixed in location, but
free to rotate about an axis? Now a rigid body is just a collection of the point masses that make the total
mass ๐ of the rigid body. Each of those point masses ๐
๐
will have its own direct line distance ๐
๐
from
the axis of rotation and itself. So for each point mass we could go through the exact same argument as
we just went through to arrive at a similar equation;
๐
๐
๐
2
The infinitesimal amount of toque ๐
๐
for each point mass is the product of the term ๐
๐
๐
2
and ๐ผ the
angular acceleration of the rigid body. The angular acceleration does not have a subscript on it because
as a rigid body all the point masses rotate together, so all the point masses have the exact same angular
acceleration. To find the total toque acting on the entire rigid body you simply sum up all the torques
acting on all the point masses.
๐
๐
2
๐
๐
The summation term is given a name. It is called the moment of inertia of the rigid body, and its SI units
are kgยทm
2
๐
๐
2
๐
๐
This allows us to rewrite our equation as
The moment of inertia is a quantification of how difficult (or easy) it is to get an object to change its
current state of rotational motion about a particular axis or rotation. The value of the summation of the
term ๐ ๐
๐
2
will depend on the total mass of the rigid body, its shape, and the axis of rotation that is
picked. However, the summation will always have the following basic algebraic form:
๐
๐
2
๐
๐
2
Here ๐ is the total mass of the object, ๐ is the โradiusโ of the object, and ๐ถ is a coefficient dependent on
the shape of the object. A chart of the moment of inertia of some basic geometric shapes with uniform
mass is given.
From the free body diagram, and force summation equations of the system we see that the tension in
the string will be:
โ
โ
๐
โ
๐
Inserting this for the tension, as well as the known equations for torque and angular acceleration, gives
us:
โ
๐
๐
2
โ
๐
๐
๐
2
โ
๐
This final equation gives the experimental value for the moment of inertia of the rigid body. Where here
๐ is the radius of the horizontal pulley the rigid body is attached to. (Please note that in constructing this
equation we assumed that the moment of inertia of the two pulleys are so small compared to the
moment inertia of the rigid bodyโs that they were simply ignored.)
Setup
- Using the listed equipment, construct the setup as shown.
๏ท Make sure that the step pulley is positioned on top.
๏ท To ensure that the torque equation reduces from ๐ = ๐ ๐น sin ๐ to ๐ = ๐๐น you need to
ensure that the detectable pulley is aligned to the step pulley such that the two pulleys
are perpendicular and tangential to each other, as shown in the pictures provided.
๏ท The large pulley (the bottom pulley) should already have a string attached to it. If it does
not, get a piece of string about 1.5 m in length and attach it to the large pulley.
๏ท Attach the other end of the string to a hook from the mass and hook set.
- Make sure the PASCO 850 Universal Interface is turned on and connected to the computer.
- Double click the Capstone icon to open the Capstone software.
- In the Tool Bar, on the right side of the screen, click on Hardware Setup to open the Hardware
Setup window.
- In the Hardware Setup there should be an image of
the PASCO 850 Universal Interface. If there is, skip to
set 6.
๏ท If there isnโt click on Choose Interface to
open the Choose Interface window. Select
PASPORT, then Automatically Detect, and
then click OK.
- On the image of the PASCO 850 Universal Interface
click on Digital Inputs Ch (1) to open the digital senor
list.
๏ท Scroll down the list and select Rotary Motion
Sensor. You should now see the rotary
motion sensor icon connected to digital
inputs Ch (1), and Ch (2).
๏ท Plug in the rotary motion sensor. Yellow Ch
(1), and black Ch (2).
- In the Tool Bar click on Data Summary to open up
the Data Summary window.
- In the Data Summary window click on the properties
icon right to the right of where it reads Rotary
Motion Sensor to open the properties window for
the rotary motion sensor.
๏ท In the properties window for Linear
Accessory, select Large Pulley (Groove), then
click Ok.
- At the bottom of the main screen make sure the
sample rate for the rotary motion sensor is set to 20 Hz.
- Close the Tool Bar.
- In the Display Bar, on the right side of the screen, double click the Graph icon to open up a
graph.
๏ท Click Select Measurement for the y-axis of the graph, and select Velocity (m/s).
๏ท The computer will then automatically select time (s) for the x-axis.
Analysis of Rotational Inertia Lab
Name______________________________________________ Group#________
Course/Section_______________________________________
Instructor____________________________________________
Tables (20 points)
Diameter (cm) Radius (cm) Radius (m)
Large grove pulley
Disk
Ring, inner
Ring, outer
m (g) m (kg)
Disk
Ring
Hanging mass
a (m/s
2
Disk
Disk and Ring
- Calculate the theoretical moment of inertia of the disk, and show work. (5 points)
- Calculate the theoretical moment of inertia of the ring, and show work. (5 points)
- Calculate the experimental moment of inertia of the disk, and show work. (5 points)
- Calculate the experimental moment of inertia of the ring, and show work. (5 points)
- Using the theoretical value as the accepted value, calculate the % error for the disk, and show
work. (10 points)
- Using the theoretical value as the accepted value, calculate the % error of the ring, and show
work. (10 points)
- If you repeated this experiment with a larger hanging mass, should that change the values you
would obtain for the moment of inertia of the ring and disk? Justify your answer. (10 points)
