Moment of Inertia
David Whetstone
Physics 4A
Lab Instructor: Lynda Williams
mr.fixit@mac.com
April 26, 2005
1 Objective
The objective is to experimentally determine the moments of inertia of both a solid disk
and a ring and to compare these values with those calculated using the known formulae
for moment of inertia.
2 Theory
Moment of inertia is defined simply as an object’s resistance to change in angular momentum. It is the rotational analog to mass or inertia in translational motion.
The moment of inertia (I) of a basic solid of uniform density can be
R calculated by
first deriving an appropriate formula from the general formula I = (x2 + y 2 )dm.
Formulae have already been derived for most basic shapes, making it easy to obtain
their moments of inertia. Even moments of inertia for more complicated solids can be
calculated if these solids can be decomposed into basic shapes for which the moments
of inertia are known.
But not every object lends itself
well to easy calculation of moment
of inertia.
If an object has nonuniform density or uneven weight distribution it can be difficult, if not impossible, to calculate moment of inertia.
In these cases, moment of
inertia can be experimentally determined.
In this lab, we construct an apparatus
to measure the moment of inertia of an
arbitrary object by converting a portion
of the gravitational potential energy of an
object of known mass into rotational kinetic energy of the object to be measured
(see Figure 1). The mass m is suspended
from a string at height h which (after
passing through two pulleys to change its
orientation) is wound around the axle (raFigure 1: Moment of Inertia Apparatus
dius r) of a rotating table. From the law
of conservation of energy, the gravitational potential energy of m before it drops is
equal to its translational kinetic energy and the rotational kinetic energy of the rotating
table the moment before m hits the ground.
P Eg |m → KElin |m + KErot |table
mghi = 12 mvf 2 + 21 Iωf 2
1
(1)
Solving for I results in:
I = 2mghi − mvf2
1
ωf 2
(2)
From the experiment, we are given only height, time, mass, and radius, so we
should rewrite our formula in terms of these variables. Given this data, we can calculate
the average velocity of the falling counterweight and the final angular velocity of the
table in terms of h and t, and from these, replace ωf and vf .
v=
vf − vi
hi
2hi
=
⇒ vf = 2v =
t
2
t
ωf =
2h
vf
=
r
rt
(3)
(4)
Substituting (3) and (4) into (2) gives us:
I=
2mghi − m
2hi
t
2 ! rt
2hi
2
⇒
2 2 2
2mg hi r2 t2
m22
h
i r t
−
2 2
h
22
i t
22 hi 2
(5)
Finally, we factor out mr2 and simplify to obtain our formula for moment of inertia
in terms of our four experimentally determined variables.
2
gt
2
I = mr
−1
(6)
2hi
Of course, many assumptions must be made to use this simplified calculation. The
pulleys are assumed to be both massless and frictionless. The string used is assumed
to be massless and inextensible. The rotating table is assumed to rotate without friction.
With this method, and equation (6), we can find the I for the rotating table and any
arbitrary object (within reason) placed on it. For example, to find I for our iron ring
and disk:
Iring = Iring+table − Itable
Idisk = Idisk+table − Itable
Finally, the accuracy of our calculations can be determined by comparing the moments of inertia we’ve just calculated with those calculated using known formulae for I.
Iring =
1
m(R12 + R22 )
2
Idisk =
2
1
mR2
2
3 Setup
There are three major steps to this experiment:
1. Find the moment of inertia of the empty rotating table.
2. Find the moment of inertia of the table with the iron ring.
3. Find the moment of inertia of the table with the iron disk.
The Moment of Inertia Apparatus
M ATERIALS
1 Table clamp
1 Weight hanger (mass 50g)
1 Long metal rod
1 Length of string
2 Pulleys
1 Level
2 Right angle clamps 1 Two-meter stick
1 Rotating table
1 Zero-decimal scale
1 Large iron ring
1 Stopwatch
1 Large iron disk
1 Vernier caliper
1 50g weight
Assorted paperclips
1 20g weight
Initial Setup
First, the table clamp was attached to the edge of the bench, and the long metal rod
attached vertically to it. Then the right angle clamp was used to connect one of the
pulleys to the top of the rod approximately 2 meters from the ground facing away from
the bench. The second pulley was attached to the rod roughly 8 cm from the benchtop
with the second right angle clamp. The rotating table was positioned so that the string
would enter the bottom pulley without putting lateral pressure on the pulley (necessary
to ensure low friction.) The length of string was wound around the axle of the rotating
table, and fed through the pulleys as shown in Figure 1. To ensure that the axle of the
table was vertical, the level was placed on the crossmembers and the feet were adjusted
on the table until it was level. There was a small amount of uncertainty in this step,
since the arms of the table had to be rotated to ensure that it was level in all directions.
Obtaining I
In order to overcome friction forces in the apparatus, a few paperclips were added to
the end of the string such that the table, when given a slight push, continued to rotate
at a constant angular velocity. This step was a likely source of error, as determining
constant velocity by eye is extremely inexact. The string was rewound, and a 20g
counterweight was added to the end of the string. The height of the weight above the
ground was measured with the two-meter stick. Difficulties here involved making sure
the meterstick was perpendicular to the ground, and getting an accurate measure of the
weight without disturbing its vertical position. The weight was then released, and its
descent timed until the moment it hit the ground. There is a certain amount of error
inherent in the human reaction time which would affect both the start and stop of the
timer. Both the height and the time were recorded, and the process repeated 5 times.
3
4 Data
Empty rotating table
R ADIUS OF A XLE
M ASS OF C OUNTERWEIGHT
M ASS OF PAPERCLIPS
RUN
TIME
HEIGHT
1
2
3
4
5
8.48
8.12
8.18
8.38
8.31
8.29
1.4915
1.4979
1.4820
1.4872
1.4866
1.4890
AVERAGE
0.0239m
0.0200kg
0.0002kg
I ( EXPERIMENTAL )
0.00269
0.00245
0.00252
0.00263
0.00258
0.00258
Ring
M ASS
I NNER R ADIUS
O UTER R ADIUS
M ASS OF C OUNTERWEIGHT
M ASS OF PAPERCLIPS
RUN
TIME
HEIGHT
1
2
3
4
5
14.38
14.55
14.72
14.55
14.74
14.59
1.4093
1.4012
1.4040
1.4051
1.3945
1.4028
AVERAGE
4.1720kg
0.1015m
0.1270m
0.1500kg
0.0040kg
I ( EXPERIMENTAL )
0.05919
0.06092
0.06239
0.06075
0.06291
0.06129
Disk
M ASS
R ADIUS
M ASS OF C OUNTERWEIGHT
M ASS OF PAPERCLIPS
RUN
TIME
HEIGHT
1
2
3
4
5
11.29
11.58
11.40
11.05
11.42
11.35
1.3849
1.3761
1.4161
1.3836
1.3832
1.3888
AVERAGE
4.4540kg
0.1273m
0.1500kg
0.0060kg
I ( EXPERIMENTAL )
0.03607
0.03835
0.03596
0.03448
0.03702
0.03637
4
5 Results and Discussion
The absolute possible error for the moment of inertia of the iron ring was calculated
to be 0.99%. Unfortunately, our experimental results exceeded the theoretical value by
10.97%. Since this percentage is outside the bounds of error of our measuring devices,
experimental error must be the cause.
From equation (6), the experimental value will grow if the measured values of m,
t, or r were too large, or h was too small. Since both m and r were measured once and
used throughout the experiment, they are unlikely sources of error. Measurement of h,
although problematic (as mentioned in the Setup section), is another unlikely source
of experimental error, since the method of measurement did not change throughout
the experiment, and accurate results were obtained for the iron disk. The most likely
source of error for this experiment was the measurement of time.
The human reaction delay in using the stopwatch that was mentioned in the setup
section is an unlikely culprit, again due to the fact that accurate results were obtained
for the iron disk. Also mentioned in the Setup section was the difficulty in determining
the correct number of paperclips to use to counteract the frictional forces. This is a
quite likely source of error. Another likely source is friction in some form, as it would
increase the amount of time taken for the counterweight to reach the ground. Two
potential sources of increased friction observed at various points during the experiment
were the string becoming trapped between the rotating and non rotating parts of the
table axle, and the string becoming derailed from one of the pulleys. Given sufficient
time, I would repeat this experiment to obtain usable results.
5
Addendum - Error Analysis
∆r
1×10−5
∆t
∆h
1×10−2
1×10−3
∆m
1×10−3
Partial derivatives
∂I
= r2
∂m
∂I
= 2mr
∂r
gt2
−1
2h
gt2
−1
2h
∂I
gmr2 t2
=
∂t
h
∂I
=
∂h
−
gmr2 t2
2h2
Differentials
∂I ∂m ∆m
=
=
=
∂I ∆r
∂r =
=
=
∂I ∆t
∂t =
=
=
∂I ∆h
∂h =
=
=
2
2 gt
∆m
r
−
1
2h
m
2
(0.0239m)2 (9.8 s2 )(14.59s) − 1 (1 × 10−3 kg)
2(1.4028m)
3
4.251 × 10−4 ms
2
2mr gt − 1 ∆r
2h
m
2
2(0.150kg)(0.0239m) (9.8 s2 )(14.59s) − 1 (1 × 10−5 m)
2(1.4028m)
5.329 × 10−5 kg · m · s
gmr2 t2 h ∆t
(9.8 sm2 )(0.150kg)(0.0239m)2 (14.59s)2 (1 × 10−2 s)
(1.4028m)
8.754 × 10−5 kg · m · s
−
gmr2 t2 2h2 ∆h
− (9.8 sm2 )(0.150kg)(0.0239m)2 (14.59s)2 (1 × 10−3 m)
2(1.4028m)2
4.552 × 10−5 kg · m2
6
Absolute possible error
∆I
∂I ∆m + ∂I ∆r + ∂I ∆t + ∂I ∆h
= ∂h ∂m
∂r
∂t
= 4.251 × 10−4 + 5.329 × 10−5 + 8.754 × 10−5 + 4.552 × 10−5
= 6.11 × 10−4
∆I%
=
∆I
× 100 =
Iexperimental
6.114 × 10−4
× 100
6.118 × 10−2
= 0.999%
Absolute probable error
δI
=
s
∂I
∆m
∂m
2
+
∂I
∆r
∂r
2
+
∂I
∆t
∂t
2
+
∂I
∆h
∂h
2
=
p
(4.251 × 10−4 )2 + (5.329 × 10−5 )2 + (8.754 × 10−5 )2 + (4.552 × 10−5 )2
=
4.37 × 10−4
δI% =
=
δI
Iexperimental
× 100 =
4.396 × 10−4
× 100
6.118 × 10−2
0.719%
7