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# Circular motion lab

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Introduction

Physics Lab – Circular Motion

Candidate Name:                                                        Date:

Candidate No.:                                                                        Lab Number:

Subject Level: Physics SL                                                                Teacher’s Sign:

Aim of Experiment:

To verify the circular motion theory through data analysis and plotting relevant data graphs, thus obtaining at a value for the acceleration due to gravity.

Apparatus used:

1. String
2. Metal bob
3. Weighing scale (Least Count - ±0.01 gm)
4. Ruler (Least Count =  , 0.05 cm)
5. Weights of different weights (Least Count = ±0.01 gm) Hypothesis:

The acceleration of gravity has been proven to be equal to the ratio of:

1. Mass of bob × 4π × radius of the bob
2. Time × mass of larger body

This gives us the formula: g = 4×π×r×m ÷ t×M.

Thus using this formula, and by the process of data collection and processing, we can prove that the acceleration due to gravity on the surface of the Earth is the same as the considered value of approximately 9.8 ms-2 .

Procedure:

• A metal bob is tied to an inelastic string.
• The string is then held while the bob is rotated in a horizontal circular motion.
• The time taken for 20 revolutions is recorded using a stopwatch.
• The radius of the arc is then changed, and readings for the time are taken again.
• The procedure is repeated for 10 different radii (lengths of string) and 5 readings for the time are taken for each radius.
• The data is then used to find the average radius and average time, to find the acceleration due to gravity.

Error Propagation:

• Error Of Meter Scale = Least count of meter scale ÷ 2 (Analog Instrument)

= = 0. 05 cm

• Error of Calculated Time = Least Count ÷ 2 = 0.1 ÷ 2

0.05 s

• Small mass (m) = 18.46 ± 0.01 gms
• Big Mass (M) = 300.00 ± 0.01 gms

Middle

16.40

16.50

17.20

16.90

10.

58.00

16.60

17.60

17.90

17.80

18.10

Now, we can take all the time periods to find the Average Time Period, since

Time average = Time periods of all trials ÷ No. of Trials

For example,
T
2 (average) = (11.60 +12.00 + 11.80 +11.20 +11.0) ÷ 5     =  11.52 ± 0.05 s

 No. Length (L)[in cm]± 0.05 cm Time (T1)[in seconds]± 0.05 s Time (T2)[in seconds]± 0.05 s Time (T3)[in seconds]± 0.05 s Time (T4)[in seconds]± 0.05 s Time (T5)[in seconds]± 0.05 s Time average(in seconds)± 0.05 s 1. 49.00 10.90 11.20 11.00 11.10 11.10 11.06 2. 50.00 11.60 12.00 11.80 11.20 11.0 11.52 3. 51.00 11.60 11.10 11.20 11.00 11.30 11.24 4. 52.00 11.80 13.00 12.10 12.20 12.10 12.24 5. 53.00 13.10 13.00 12.90 13.00 13.10 13.02 6. 54.00 14.00 14.10 14.00 14.30 14.20 14.12 7. 55.00 14.90 15.20 14.50 14.90 15.10 14.92 8. 56.00 16.10 15.80 15.90 16.20 16.30 16.06 9. 57.00 17.00 16.40 16.50 17.20 16.90 16.80 10. 58.00 16.60 17.60 17.90 17.80 18.10 17.60

Next, the average time period for one oscillation as per every length is calculated by dividing the average time period of that length by a factor of 20.

Conclusion

Thus, Time period for 1 revolution= Time (average) / Number of                                                       revolutions

= 13.86 / 20

= 0.693 seconds

Average Radius = (49+50+51+52+53+54+55+56+57+58) / 10

= 53.5 cm

Thus, using the acquired values, we get the acceleration due to gravity as

g = 4×π×r×m ÷ t×M.

= 4π (53.5)2 (18.458) / (14.7)2 (300)

g= 10.25 ms-2

Conclusion:

Thus through the above experiment we find that the acceleration due to gravity is approximately 10.25 ms-2

Evaluation :

As mentioned before, the ideal acceleration due to gravity on the surface of the Earth is considered to be approximately 9.81 ms-2. However, considering the method used above and the surroundings in which it was conducted, the experiment can be called successful.

There are many reasons that would explain the deviation in the value, some of which would be:

• Wind causing improper execution of the experiment
• Ceiling fans deviating the natural path of the arc
• Possible error in the apparatus used, like the weighing scale not being completely accurate

If these factors are to be eliminated, the deviation would certainly reduce, making the value more accurate.

The percentage deviation in the experiment = (Deviation/ 9.81) × 100

= (10.25 – 9.81/ 9.81)×100

= 4.48 %

This student written piece of work is one of many that can be found in our International Baccalaureate Physics section.

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