Object 2
In this object need to find the centre of mass of the two irregular shaped bars. To find the COM need to use at least two known masses. As the two known masses, will be used masses of the 10g and 50g.Hence, there will did 4 experiments. First of all, will be explained how each experiment was set up. Then will be shown the Table, which include the raw data during the experiment.
Part A1 (Experiment 1.1 with the 1st irregular shaped bar)
First of all, will be explained how all system were set up. In the beginning iron stand was connected to the triple beam balance. Then, using light string, bar was randomly connected to the iron beam. Then two known masses hanged to the bar and fixed in position, where whole system was balanced.
The Figure 4, which is above, will give the visual information about how equipment was set up.
Figure 4.
10g
mass
50g
mass
String
Figure 4, shows set up of equipment, but does not show the forces acting on this bar.
Figure
5.
Figure 5, which is above, is shows Free Body Diagram of this bar. Tension force of the string did not shown, because system already in equilibrium.
Where,
is the mass of the bar,
is distance between string and the beginning of the bar,
is a distance between string and
,
is a distance between string and
and
is a distance between centre of mass of the bar and the beginning of
the bar.
Part B1 (Experiment 1.2 with the 1st irregular shaped bar)
As in the Object 1, before using formula (2), experiment should be done two times. In the second experiment bar was hanged in different point. Masses also were hanged to other points. Therefore, all system must be in equilibrium. Figure 6, below, will give visual information about, how bar and masses were hanged.
Figure
6.
String
How can be clearly seen in the Figure 6, the masses are in different place. Then, to get more information and show where the forces acting to the bar, will be done Free Body Diagram. Figure 7, above, is the free body diagram, which show all forces acting on the bar.
Figure
7.
Where, is the mass of the bar, is distance between string and the beginning of the bar, is a distance between string and , is a distance between string and and is a distance between centre of mass of the bar and the beginning of the bar.
Part C1 (Raw data and calculating)
During the Experiment 1.1 and Experiment 1.2, were recording data. Data are sorted in the Table 3 for the length and Table 4 for masses.
Table 3.
Distance (cm) |
Experiment 1.1 |
Experiment 1.2 |
From 0cm to string |
41.7 |
37.3 |
From to string ( ) |
12.8 |
35.3 |
From to string ( ) |
29.7 |
22.3 |
From COM to string |
-41.7 |
-37.3 |
Table 3, shows what distance was between string and masses and COM.
Table 4.
Mass of the objects (g) |
Experiment 1.1 |
Experiment 1.2 |
Known |
50 |
50 |
Known |
10 |
10 |
Table 4, shows what kind of masses were used in Experiment 1.1 and Experiment 1.2.
After getting Free Body Diagrams and Tables with data, can be started calculating part.
Formulas.
Will be used standard formula (1.1) and then simplified for our Experiments:
(Formula
2.1)
Calculating for Experiment 1.1
For calculation in Experiment 1.1, the raw data, which in table, should be substituted into formula (2.1).
Then, need to simplify this formula:
(Formula
3.1)
Calculating for Experiment 1.2
For calculation in Experiment 1.2, the raw data, which in table, should be substituted into formula (2.1).
(Formula
3.1.1)
Then, need to simplify this formula:
(Formula
3.2)
After getting simplified formulas (3.1) and (3.2), the COM and mass of the bar could be found.
To find the mass of the bar, need to subtract from formula (3.2) formula (3.1):
Hence,
To find the centre of mass of the bar, need to substitute the mass into the formula (3.2.2):
Hence, mass and COM of bar were found. Experiment 1.1 and Experiment 1.2 is done.
Part D1 (Calculating Errors)
The
original mass of the irregular shaped bar is 232.4
and COM is 46.5cm.
Compare to experimental result, can be clearly seen that was made some errors during the Experiments. Hence, errors should be calculated in the percentage.
Error of mass:
Error of COM:
Part A2 (Experiment 2.1 with the 2nd irregular shaped bar)
First of all, will be explained how all system were set up. In the beginning iron stand was connected to the triple beam balance. Then, using light string, bar was randomly connected to the iron beam. Then two known masses hanged to the bar and fixed in position, where whole system was balanced.
The Figure 8, which is above, will give the visual information about how equipment was set up.
Figure
8.
50g
mass
10g
mass
Therefore, must to have free body diagram, as was did in the experiments before.
Figure 9.
String
Figure 9 representing the forces acting on the irregular shaped bar. Where is is the distance between string and the beginning of the bar, is the gravitational force of the bar, gravitational force of the first known mass (10g), gravitational force of the second known mass (50g), is the distance between the string and the first mass, is the distance between the string and the second mass and is the distance between the COM and the beginning of the bar.
Part B2 (Experiment 2.2 with the 2nd irregular shaped bar)
First of all, will be explained how all system were set up. In the beginning iron stand was connected to the triple beam balance. Then, using light string, bar was randomly connected to the iron beam. Then two known masses hanged to the bar and fixed in position, where whole system was balanced.
The Figure 10, which is above, will give the visual information about how equipment was set up.
Figure 10.
String
10g
mass
50g
mass
After set up equipment, were drawn Free Body Diagram of the system, including forces that action on the bar.
String
Figure 11 representing the forces acting on the irregular shaped bar. Where is is the distance between string and the beginning of the bar, is the gravitational force of the bar, gravitational force of the first known mass (10g), gravitational force of the second known mass (50g), is the distance between the string and the first mass, is the distance between the string and the second mass and is the distance between the COM and the beginning of the bar.
Part C2 (Raw data and calculating)
Raw data which were get from Experiment 2.1 and Experiment 2.2, has been recorded to the Table 5 and Table 6.
Table 5
Distance (cm) |
Experiment 2.1 |
Experiment 2.2 |
From 0cm to string |
33.9 |
30 |
From to string ( ) |
24.3 |
27.9 |
From to string ( ) |
12.1 |
20.5 |
From COM to string |
-33.9 |
-30 |
Table 5, which is above, shows the distance data. Table 6, which is bellow, representing the known masses.
Table 6
Mass of the objects (g) |
Experiment 2.1 |
Experiment 2.2 |
Known |
10 |
10 |
Known |
50 |
50 |
Formulas.
Will be used standard formula (1.1).
Calculating Experiment 2.1.
For calculating, formula (1.1), was changed for the Experiment 2.1:
(Formula
2.2)
Now substitute inside the formula (2.2) values form Table 5 and Table 6 for Experiment 2.1:
Using some simplification:
(Formula
2.2.2)
Calculating Experiment 2.2
For calculating, formula (1.1), was changed for the Experiment 2.2:
(Formula 2.1.2)
Now substitute inside the formula (2.1.2) values form Table 5 and Table 6 for Experiment 2.2:
Using some simplification:
(Formula
2.2.3)
In formulas (2.2.2) and (2.2.3), result is two unknown values. To find them should to subtract formula (2.2.2) from formula (2.2.3):
Hence,
Were found value of the mass, then by substitution value mass of the bar into the formula (2.2.3), can be found the COM:
Hence, mass and COM of bar were found. Experiment 2.1 and Experiment 2.2 is done.
Part D2 (Calculating Errors)
The original mass of the irregular shaped bar is 249 and COM is 35.5cm.
Compare to experimental result, can be clearly seen that was made some errors during the Experiments. Hence, errors should be calculated in the percentage.
Error of mass:
Error of the COM:
