Lab 7: Rotational Motion and Moment of Inertia- Online
version
Objective
The purpose of this experiment is to compare measured values of rotational inertia to
values calculated using the theory below.
Theory
For rotational motion we need to redefine our variables that we used to discuss
translational motion.
For position, instead of x we use the angle (in radians).
For velocity, instead of linear velocity v (m/s) we use angular velocity  (rad/s)
For acceleration, instead of linear acceleration a (m/s2) we use tangential angular
acceleration  (rad/s2).
For a particle rotating at a distance r from the axis, we have the relationship between
linear and angular variables as follows: x  r ; v  r ; a  r .
The DIRECTION is taken to be positive if the object is rotating counterclockwise, and
negative if it is rotating clockwise.
We also need to redefine our concepts of force and inertia.
For translational motion we used Newton’s second law F=ma, where F was the force
applied, m was the mass (inertia). The amount of linear acceleration depends directly on
the force applied and inversely on the mass.
For rotational motion we will use Torque () rather than force, and Moment of Inertia (I)
rather than the mass.
So our rotational version of Newton’s second law is   I
The torque is a measure of how much an applied force will cause an object to rotate, and
depends on the size of the applied force, the distance from the axis that it is applied, and
also the angle at which it is applied. The farther from the axis you applied the force (i.e.
the bigger r), the easier it is to rotate the object. For example, we have all walked up to a
glass door with a bar on it at some point in our lives, and pushed on the wrong end (by
the hinge). If you do that, you have to push really hard to open the door. We always put
the door knob at the farthest distance from the hinge as possible, to make it easy to open
the door. Similarly, if you want to open a door, you push (or pull) perpendicular to the
door to apply the greatest torque. If you pushed parallel to the door, there would be no
rotation. So our equation for the torque is:
  Fr sin  where  is the angle between the force and the vector r from the axis to
where the force is applied.
For this lab we will always apply the force perpendicular to r, so our torque is just =Fr.
The moment of inertia (I), is a measure of how difficult it is to make a certain object
rotate. It depends on the mass of the object, and how far the mass is from the center of
the object. In general, the farther the mass is from the axis, the harder it is to spin the
object. (Try spinning your textbook about its corner instead of its center).
For a particle of mass M at a distance R from the axis, I=MR2. If you have a continous
object, like your textbook, which has a lot of different atoms all at different distances
from the center, we need to integrate over the shape to get the moment of inertia. Your
textbook has a table showing the moment of inertia for a bunch of different shapes.
What we will do in this virtual lab is to apply a force to a disk by pulling on a string that
is wound around the disk. We do this by hanging a mass from the end of the string and
letting the mass drop. The applied force is then the TENSION in the string. The distance
r, is the radius of the disk, (25cm).  From these we can calculate the torque.
This torque will cause the disk to rotate, and we will measure its acceleration. From this
we can calculate the moment of inertia, I=and compare it to what it should be
theoreticallyWe will then add masses to the top of the disk and repeat the experiment.
Every mass we add will increase the moment of inertia by an amount I=MR2.
Equipment
This is the general setup of the experiment that you will find online at:
http://www.fisica.uniud.it/~deangeli/applets/Multimedia/ExplrSci/dswmedia/momofI.htm
pulley
mass hanger
and weight
m1
bar
m2
spindle,
radius r
d~1 m
floor
In this experiment, a small mass is attached to a string which rotates a
disk about its axis. By measuring the acceleration of the small mass, the
moment of inertia of the disk can be determined.
The radius, r, of the disk and the spindle are both 0.25m for the online
experiment.
Procedure
1. Go to
http://www.fisica.uniud.it/~deangeli/applets/Multimedia/ExplrSci/dswmedia/momofI.htm
2. On the right side of the screen choose the blue 100g weight and drag it up to the
holder.
3. Hit the “Release” button, and watch the disk on the left side of the screen rotate as the
mass descends. When the mass has reached the bottom, its acceleration will appear
on the right side of the screen. Record this value in table 1. When you do this
experiment in the lab, you will measur the time, t, for the weight to drop a distance, d,
and calculate a=2d/t2.
4. Using your value for a, calculate the torque with error. The force which causes the
torque is the tension in the string, FT = m(g-a), where m is the total mass on the mass
hanger. Record both FT and Frin Table 1. The radius r, of the disk in this example
is 25cm. Be sure to convert all quantities to standard SI units of kilograms and
meters!
5. Calculate the angular acceleration (=a/r). Use the torque and the angular
acceleration to calculate the moment of inertia of the disk. I, so I=. Call this
value IA, the moment of inertia of the rotational apparatus. Compare it to the value of
IA,calc = 0.3 kgm2 which we calculated for you based on the mass and radius of the
disk.
6. Trial 2: Hit the “Go again” button, which will restore the hanger to the top. At this
stage you will see 8 yellow “hot-spots” appear on the disk at the left. Four of these are
at a distance of r/2=12.5cm from the axis, and four are at a distance of r from the
center. Drag two red 250g balls to the disk, placing them at R=r/2, directly opposite
from each other as shown below. Calculate the theoretical moment of inertia, I2,calc, of
the two point masses about the axis of rotation using the appropriate formula from the
theory section.
I2,calc= m1R12+m2R22
Trial 2 layout:
7. Now repeat steps 1-5. The moment of inertia that you find in 5 is now the total
moment of inertia, IA+I2,meas. Subtract the moment of inertia of the top of the
apparatus, IA, as previously measured, to get the moment of inertia of the weights
only, I2, meas. Compare your result with the calculated value of the moment of inertia
of the weights (find percentage difference).
8. Trial 3. Repeat steps 6+7 with the two masses at the outside of the disk, i.e. at
R=25cm. Call this I3,meas.
Trial 3 layout:
9. Trial 4. Repeat steps 6+7 with four 250g masses: two masses at the outside of the
disk, and two halfway along, all in a line. Call this I4,meas. (Now I4,calc will have 4
terms in it, one for each mass)
Trial 4 layout:
10. Trial 5. Repeat steps 6+7 with four 250g masses: two masses at the outside of the
disk, and two halfway along, at right angles to each other as shown. Call this I5,meas.
Trial 5 layout:
Data
Table 1: Measurements of Moment of Inertia
Trial #
a(m/s2) FT(N)
(rad/s2)
m(g-a)
a/r
1(apparatus only)
rFT
(Nm)
Imeas
(kgm2)
2 (masses at 12.5
cm)
3 (masses at 25
cm)
4 (masses at 12.5
and 25 cm inline)
5 (masses at 12.5
and 25 cm at 90°)
Question
1. Are the moments of inertia in trial 4 and 5 the same? If so, why?
Icalc
(kgm2)
0.03
%
diff