• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Circular motion lab

Extracts from this document...

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 = image00.pngimage00.png, 0.05 cm)
  5. Weights of different weights (Least Count = ±0.01 gm)


image06.png

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)

= image00.png

= 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
...read more.

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.

...read more.

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 %

...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Physics essays

  1. Investigate the factors affecting the period of a double string pendulum

    but I would have been very hard to measure this with a 30 cm ruler possibly giving inaccurate and imprecise values. This contributes to be a major limitation to the analysis of my investigation. If I had a choice I would used a bigger metal bar perhaps of 50 cm

  2. Uniform Motion Lab. This experiment will measure the motion of a cart moving on ...

    The distance between our ticker tape dots was approximately the same, and this also shows us that the speed of the cart is approximately constant. Possible sources of error that occurred in this experiment include the ticker tape not being pulled at the same angle as the cart, which meant

  1. Suspension Bridges. this extended essay is an investigation to study the variation in tension ...

    Also, as in suspension bridges, the weight is felt by the roadway which exerts a compression force in the pillars. This in turn pulls the cables tighter and keeping this factor in mind, the second part of my study is the variation in tension when different lengths of the same string are tied between two supports.

  2. Simple Harmonic Motion Physics HL Lab Report

    Hence I plotted a graph of length (L) versus the period2 (T2) and in that graph as seen the best fit line passes through the origin. Hence we can see that the original prediction was wrong and in fact length (L)

  1. Pendulum work out the value of acceleration due to gravity (g), by using ...

    a position that they clamp the pendulum and they are in turn clamped by the clamp of the stand. 2. In order to get proper value that will be finally calculated out I have decided to take 5 readings of the oscillations of the pendulum using four different lengths of the pendulum (20cm to 90cm)

  2. This lab will test the effects of the surface area factor on acceleration due ...

    The temperature of the test environment will also remain constant, which should help maintain a consistent amount of air resistance that the paper will encounter. The vertical displacement will remain constant at 520.0cm because it will help maintain the time of descent to surface area ratio at a constant throughout

  1. Horizontal Circular Motion Lab

    Since swinging the rubber stopper with the apparatus fatigues the arm of the swinger, the three partners will take turns to swing the stopper, record the data, and time the experiment to ensure the accuracy of our data. A total of 10 trials (minimum) will be recorded for the experiment.

  2. HL Physics Revision Notes

    EM waves, resulting in the transformation of parent nuclide into daughter nuclide. Measured in Becquerel?s (Bq) transformations per second. Alpha decay: atomic nucleus emits alpha particle, equivalent to a Helium nucleus. Atomic masses and numbers balance on both sides of the equation Beta decay: atomic nucleus emits beta particle (electron or positron).

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work