2010
Dynamic Lab Report
Raniya Ali Parappil
1/1/2010
OBJECTIVE:
 Measure the dimensions of the disk and its axles and representing a scaled diagram of the disk.
 To calculate the moment of inertia of the system about its centre of mass.
 To Measure the angle of the slope
 To measure the time required for the disk to travel 0.2 m, 0.4 m, 0.6 m, 0.8 m and 1 m down the
slope and see the similarities between the theoretical and the experimental value.
 The expression for the distance travelled x as function of time t
 To sketch acceleration, velocity, distance graph as a function of time.
THEORY:
When an object at rest is set into rotation about some axis, it has a tendency to keep rotating at some
angular speed,  measured in radians/sec. This tendency is called the rotational inertia;
characterized by a physical quantity called the moment of inertia, I, of the object. Moment of
inertia is the rotational counterpart of inertial mass in linear motion. Hence the kinetic energy of a
rotating object is:
KE =
½ I2
The moment of inertia of a composite disk can be calculated by:
From the moment of inertia, an objects distance travelled, angular acceleration and velocity can be
found.
APPARATUS:
A composite disk
Callipers’
Measuring tape
Stop watch
A slope
PROCEDURE:
1. Measured the diameter of the circle and the two axels. The measurements were then noted
down.
2. The width of each axel and the disk was then measured and noted.
3. The base of the slope was measured followed by the height.
4. After measuring the composite disk and the slope, the disk was made to roll down a distance
of .2m, .4m, .8m and 1m. A stopwatch was used to record the time taken to the disk to roll
these distances. The disk was rolled thrice for each measurement.
[The figure below shows the apparatus]
RESULTS:
Calculations of m1, m2 and m3
The density of steel = 7800 kg/m^3
The total mass of the steel disk and its axles = 8.87 kg
Volumes – V1, V2 and V3
Volume = πr 2 h
V1 = π × 0.12 × 0.01325
= . 00041626102m3
V2 = π × 0.012252 × 0.017
= . 0008014399209m3
V3 = π × 0.12792 × 0.01978
= . 001016523092m3
2(m1) + 2(m2) + m3 = 8.87
Then m1 = 4.435 –
V2
m3
2
– m2
8.014399209×10−4
= 1.016523092×10−3 = 0.788412902
V3
m2
m3
1
= 0.788412902
Therefore, m2 = 1.268370923m3
Density =
m
V
equation (1)
Therefore, 7800 =
7800 =
m1
V1
+
m2
V2
+
m3
V3
m1V2V3+m2V1V3+m2V2V1
m1V2V3+m2V1V3+m2V2V1
V1V2V3
V1V2V3
= 7800
814.6821864 × 10−9 m1 + 423.1389458 × 10−9 m2 + 333.6082042 × 10−9 m3
= 7800
3.391204368 × 10−10
814.6821864 × 10−9 m1 + 423.1389458 × 10−9 m2 + 333.6082042 × 10−9 m3
= 2.645139407 × 10−6
m3
– m2) + 4.231389458 × 10−7 m2 + 3.336082042
2
× 10−7 m3 = 2.645139407 × 10−6
8.146821864 × 10−7 (4.435 –
3.613115497 × 10−6 – 4.073410932 × 10−7 m3– 8.146821864 × 10−7 m2 + 4.231389458
× 10−7 m2 + 3.336082042 × 10−7 m3 = 2.645139407 × 10−6
– 407.3410932 × 10−9 m3– 814.6821864 × 10−9 m2 + 423.1389458 × 10−9 m2
+ 333.6082042 × 10−9 m3 = −967.97609 × 10−9
−391.5432406 × 10−9 m2 − 737.2889 × 10−9 m3 = −967.97609 × 10−9
−391.5432406 × 10−9 (1.268370923m3 ) − 73.72889 × 10−9 m3 = −967.97609 × 10−9
−496.6220615 × 10−9 m3 − 73.72889 × 10−9 m3 = −967.97609 × 10−9
−570.3509515 × 10−9 m3 = −967.97609 × 10−9
m3 = 1.697158719 kg
m2 = 1.268370923m3
m2 = 1.268370923 × 1.697158719
m2 = 2.15262677 kg
V1 4.162610266 × 10−4
=
= 0.409494904
V3 1.016523092 × 10−3
m1
m3
1
= 0.409494904
m1 = 2.442032828m3
m1 = 2.442032828 × 1.697158719
m1 = 4.144517306 kg
Calculation of Moment of Inertia, I
Moment of Inertial, Ig, total =
1
2
1
1
2m1r12 + 2 2m2r22 + 2 2m3r32
Ig, total =
1
1
(2 × 4.144517306)(0.01)2 + (2 × 2.15262677)(0.01225)2
2
2
1
+ (1.697158719)(0.1297)2
2
Ig, total = 28.56145398 × 10−3 kgm3
Then Io = Ig, total + md2
Io = 28.56145398 × 10−3 + (8.87 × 0.012252 )
Io = 0.05845385438 kgm3
Calculations of Time
mgsinθ. r 2 2
x=
𝑡
2Io
Distance = 0.2 m
(8.87)(9.81)sin(6.709836808)(0.01)2 2
0.2 =
𝑡
2(0.05845385438)
t 2 = 0.2 ×
2(0.05845385438)
(8.87)(9.81)sin(6.709836808)(0.01)2
t = 4.80487368 s
Distance = 0.4 m
(8.87)(9.81)sin(6.709836808)(0.01)2 2
0.4 =
𝑡
2(0.05845385438)
t 2 = 0.4 ×
2(0.05845385438)
(8.87)(9.81)sin(6.709836808)(0.01)2
t = 6.800204009 s
Distance = 0.6 m
(8.87)(9.81)sin(6.709836808)(0.01)2 2
0.6 =
𝑡
2(0.05845385438)
t 2 = 0.6 ×
2(0.05845385438)
(8.87)(9.81)sin(6.709836808)(0.01)2
t = 8.328514985 s
Distance = 0.8 m
0.8 =
(8.87)(9.81)sin(6.709836808)(0.01)2 2
𝑡
2(0.05845385438)
t 2 = 0.8 ×
2(0.05845385438)
(8.87)(9.81)sin(6.709836808)(0.01)2
t = 9.616940737 s
Distance = 1.0 m
(8.87)(9.81)sin(6.709836808)(0.01)2 2
1.0 =
𝑡
2(0.05845385438)
t 2 = 1.0 ×
2(0.05845385438)
(8.87)(9.81)sin(6.709836808)(0.01)2
t = 10.75206661 s
1.2
1
0.8
0.6
Series1
0.4
0.2
0
0
2
4
6
8
10
12
The graph above shows the distance as an equation of time, which are the experimental value.
Distance
.2
.4
.6
.8
1
(m) Trial 1 (sec)
5.29
7.41
8.45
9.45
10.65
Trial 2 (sec)
4.98
6.92
8.75
9.96
10.33
Trial 3 (sec)
4.8
6.75
8.66
9.36
10.61
Average (sec)
5.02
7.026
8.62
9.59
10.53
12
10
8
6
Series1
4
2
0
0
0.05
0.1
0.15
0.2
The graph above shows the velocity as a function of time, showing a linear relationship.
Time
0
5.02
7.03
8.62
9.59
10.53
velocity
0
0.086846
0.121619
0.149126
0.165907
0.182169
0.024
0.018
0.012
Series1
0.006
0
0
2
4
6
8
10
12
The graph above shows the relation of accelerating as a function of time. The acceleration is
constant since the angular acceleration is a constant.
1.2
1
0.8
0.6
Series1
0.4
0.2
0
0
2
4
6
8
10
12
The graph above shows the distance as a function of time, for the values calculated theoretically.
Time distance
[calculated
theoretically]
0
0
4.8
0.2
6.8
0.4
8.3
0.6
9.6
0.8
10.7
1
DISSCUSSION:
The aim of the experiment was to increate a better understand on the kinematics of rigid bodies.
Through this experiment the following values were found:




The dimensions of the disk and its axles and representing a scaled diagram of the disk.
Calculated the moment of inertia of the system about its centre of mass.
Measured the angle of the slope
measured the time required for the disk to travel 0.2 m, 0.4 m, 0.6 m, 0.8 m and 1 m down
the slope and discussed the similarities between the theoretical and the experimental value.
The theoretical and the experimental values calculated were approximately similar. The values were
subject to numerous errors due experimental errors. Some of the errors include:
 While taking measurements, the reading where read visually, therefore it was not accurate.
 The stopwatch was operated manually, therefore was subject to late starts and finishing
depending upon the user reflexes.
 The friction between the wheel and the slope where neglected
 The values calculated were rounded off
 The disk was not exactly made to roll from the required distances; they had a change of +.1m errors.
These errors were taken into consideration and minimised. The experiment was repeated three times
and the average of the three values were taken into account.
The distance travelled and the time measured by both theoretically and experimentally was both
similar. The values differed due to factors like, friction being neglected and the reasons provided
above. When the graph for distance as a function of time “t” was plotted, the relationship was
shown to be linear. The velocity time graph was also plotted and shown to be linear. The
acceleration was a constant since angular acceleration calculated was a constant.
Another issue that was encountered during the experiment was that, the composite disk was not
symmetric. The length of the axels differed in certain area by +- .5mm. The width of the composite
disk, varied in different locations. When made roll down the slope the friction acting on the disk
and the slope where neglected. The friction acting is extremely low since the disk is made to roll
down the slope. When the error taking the values for time, an error is squared, this leads to increase
the error. This can lead to discrepancies in the theoretical and the experimental values.
CONCLUSION:
The aim of the experiment was successfully completed. There were few discrepancies in the
calculated and the experimental values, due to the errors encountered while performing the
experiment. The experiment was successful in showing how moment of inertia can be used to solve
of distance, acceleration and velocity. The relationship for acceleration, velocity and distance were
graphed as a function of time.
BIBLIOGRAPGY:







Planet
physics,
http://planetphysics.org/encyclopedia/MomentOfInertiaOfACircularDisk.html, accessed on
the 16th October 2010.
Wikipedia- moment of inertia, http://en.wikipedia.org/wiki/Moment_of_inertia, accessed on
the 16th October 2010.
Physics forum, http://www.physicsforums.com/showthread.php?t=223952, accessed on the
16th October 2010.
Hyper physics, http://hyperphysics.phy-astr.gsu.edu/hbase/tdisc.html, accessed on the 16th
October 2010.
Dj
Dunn,
Solid
Mechanics
Tutorial,
http://www.freestudy.co.uk/dynamics/moment%20of%20inertia.pdf, accessed on the 16th
October 2010.
Physics
lab,
Resource
Lesson
Rotational
Dynamics,
http://dev.physicslab.org/Document.aspx?doctype=3&filename=RotaryMotion_RotationalD
ynamicsRollingSpheres.xml, accessed on the 16th October 2010.