Experimental Procedure

Apparatus:

Metre ruler, iron stand, triple beam balance, set of masses, light string, unknown masses and irregular shaped bars.

Object 1

In this object, originally, we need to find the mass of the one unknown object using at least one known mass and one metre ruler, which hanged on the iron stand. In my experiment I have used two unknown masses and one known mass. As unknown masses were used pen and block, as the known mas was used mass of the 50 gram. Hence, experiment has been made two times to get enough information for determination of the unknown masses. That means in different times, objects had different places on the ruler. First of all will be shown and explain each part of the Object 1. Then, will be shown Table, which includes all raw data.

Part A (Experiment 1 explanation)

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, one metre ruler was connected to the iron stand. After that, one metre was fixed in the point 48.8cm, which is the centre of mass of the ruler. When one metre ruler was in equilibrium, I started to hang objects. One known object and pen were hanged to the right side, and block was hanged to the left side. Then I found for that objects special places, where system was in equilibrium and did not shacked or moved.

The Figure 1 below is showing how experiment was set up.

Figure 1.

ß Light String

Block à Pen àß Mass

Figure 1 shows only how experiment was set up, but Figure 1 does not show the forces that acting on it. First of all, of solving problems with balanced system, should be used and shown the Free Body Diagram.

Figure 2.

0

The Figure 2 includes how the objects acting on the ruler and shows how far they placed from beginning of the ruler.Where is the gravitational force of the block, is the gravitational force of the pen and is the gravitational force of the known mass. Although, all this forces acting to the ruler.(All positions of the objects were measured from the 0cm).

Part B (Experiment 2 explanation)

Experiment 2 of the Object 1 had same set up as the Experiment 1, but the position of objects is different.

Figure 3, which is below, is a Free Body Diagram of the balanced system, where all objects has different placement.

Figure 3.

Where is the gravitational force of the block, is the gravitational force of the pen and is the gravitational force of the known mass. Although, all this forces acting to the ruler. (All positions of the objects were measured from the 0cm)

Part C (Raw data and calculating)

During the Experiment 1 and Experiment 2, were recorded raw data of the placements of the objects. Raw data was recorded into the Table. Table 1, which is below, shows raw data of the placements of the objects during the experiment.

Table 1.

Distance from 0 cm (cm)	Experiment 1	Experiment 2
COM of ruler	48.8	48.8
Pen ( )
Block ( )		38.2
Known mass ( )	93.2	95.5

The data about the mass of the objects was recorded into the Table 2, which is below.

Table 2.

Formulas.

To find the mass of unknown mass, will be used the formula of the momentum (1). Hence, our system is in the equilibrium, can be clearly seen that the sum of the forces is equal to zero. Those means, force that acting to the right side, is equal to the force that acting to the left side. In formula form, it will look like:

In our Experiments that means:

In this formula was used the point of COM of the ruler, because in formula (1) should be used distance between object and COM, but in raw data in the Table 1 said that distance took from 0 cm.

(Therefore, in calculations will not be used the gravitational force of the ruler, because it is fixed in the Centre of Mass).

Calculating for Experiment 1.

For calculation in the Experiment 1, we need substitute raw data from Table 1 and Table to the formula (2).

(3)

Now we need to simplify formula (3):

(4)

In formula (4) we did not count , because we have in both sides so we cancelled it.

Calculating for Experiment 2.

For calculation in the Experiment 1, we need substitute raw data from Table 1 and Table to the formula (2).

(5)

Now we need to simplify formula (5):

(6)

Calculating the unknown masses using the (4) and (6).

To find the unknown values of masses we need to get two simplified formulas from calculating in Experiment 1 and Experiment 2:

Now, for the easiest way of calculation, need to show in terms of

=>

Now we need to minus formula (8) from formula (7). We will get:

(9)

Hence the result which was gotten from (9), could be find the first unknown mass of the pen:

=>

=> =>

=> =>

So, mass of the pan was found. Now substitute the value of the into the formula (7), to find the mass of the second object (block):

=>

The mass of the block was found.

Part D(Calculating and Discuss Errors)

The original mass of the pen is and mass of the block is .

Compare to experimental result, can be clearly seen that was made big error during the Experiments. Hence, errors should be calculated in the percentage.

Object 2

In this object we need to find the centre of mass of the two irregular shaped bars. To find the COM we 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 A (Experiment 1.1 with the 1^st 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.

Figure 4, shows set up of equipment, but does not show the forces acting on this bar.

Figure 5, which is above, is shows Free Body Diagram of this bar. Tension force 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.

Date: 2015-01-29; view: 909