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

Centripetal motion. The objective of this experiment is to verify whether the tension in a centripetal force apparatus is equal to the weight of the mass.

Extracts from this document...

Introduction

Physics Laboratory Report

Centripetal motion

Aim of experiment:

The objective of this experiment is to verify whether the tension in a centripetal force apparatus is equal to the weight of the mass.

Theory:

(Fig. 1)image01.jpg

Fig. 2 shows an object of mass m moving with constant velocity v in a circular path of radius r.

By keeping the angular speed of the rubber bung constant and considering the equilibrium of all the applied forces in the system, the theoretical value of the centripetal force F is calculated as follows:

F = mv2/r                or     F = mrω2

where v and ω are the linear and angular speeds of the object respectively.


Nevertheless, some correction should be made in this experiment. In this experiment, the following set-up is used.

image02.png(Fig.2)

As shown in Fig.3, in reality, the string is not horizontal and moves in a circle of radius r = l sinθ. The weight of the hanger with slotted mass gives the tension (T) in the string.

image03.png(Fig.3)

The horizontal component of the tension provides the net centripetal force. Therefore,

T sinθ = mrω2

T sinθ = m(l sinθ)ω2

T = mlω2


Apparatus:

Rubber bung                        x 1

Glass tube (15cm long)        x 1

Nylon thread (1.5m)        x 1

Slotted mass (50g)                x 4

Hanger (150g)                        x 1

Paper clip                        x 1

Meter rule                        x 1

Stop watch                        x 1

Adhesive tape                        x 1

Balance                                x 1

Procedures:

  1. The mass of the rubber bung (m)
...read more.

Middle

Data and data analysis:

Mass of the rubber bung (m) = 0.0211kg ± 0.00005kg

Length of the nylon thread (l) = 0.800m ± 0.0005m

Take g = 9.81ms-2

Hanger with slotted masses

Time taken for complete revolutions / s

Angular speed / rad s-1

(± 0.05 rad s-1)

Tension / N

(± 0.10055 N)

Mass (M) / kg

Weight (W= Mg) / N

30t (± 0.05s)

t

(± 0.00167s)

ω = 2π/t

T = mlω2

1st set

2nd set

3rd set

Mean

0.15

1.4715

19.35

20.40

19.80

19.85

0.662

9.49

1.52

0.20

1.9620

16.20

18.30

18.00

17.50

0.583

10.8

1.96

0.25

2.4525

16.80

16.35

16.50

16.55

0.552

11.4

2.19

0.30

2.9430

15.90

15.00

14.40

15.10

0.503

12.5

2.63

0.35

3.4335

13.80

13.95

13.80

13.85

0.462

13.6

3.12


With M = 0.15kg,

Absolute difference between W and T = 1.52 – 1.4715 = 0.0485N

        Percentage difference between W and T = 0.0485/1.4715 x 100% ≈ 3.30%

With M = 0.20kg,

Absolute difference between W and T = 1.9620 – 1.96 = 0.0020N

        Percentage difference between W and T = 0.0020/1.9620 x 100% ≈ 0.102%

With M = 0.25kg,

Absolute difference between W and T = 2.4525 – 2.19 = 0.2625N

...read more.

Conclusion


Discussion

From the results obtained, it can be easily seen that the differences between the tension of the string and the centripetal force of the circular motion of the rubber bung are 3.30%, 0.102%, 10.7%, 10.6% and 9.13%, with varying mass(M) and hence tension(T) used. It actually does not show much difference between the two values. The errors are indeed caused by the matters stated in ‘Sources of error’. Consequently, it can be concluded that the tension (T) of the string is approximately the same as the weight (W) used.

Furthermore, we cannot circle the rubber bung exactly in a horizontal plane. As shown in Fig.3, if the rubber bung is circled in a horizontal plane, the tension (T) of the string will no longer contribute a vertical force component to balance the weight (W) of the rubber bung. Hence, there will not be any force to balance the weight of the rubber bung. Consequently, the rubber bung must make an angle θ, which is more than or less than 90o, with the vertical, and thus the rubber bung cannot circle in a horizontal plane.

Conclusion

From the experiment, as the tension (T) of the string is approximately the same as the weight (W) used, it can be verified that the tension in a centripetal force apparatus is equal to the weight of the mass.

P.

...read more.

This student written piece of work is one of many that can be found in our AS and A Level Mechanics & Radioactivity 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 AS and A Level Mechanics & Radioactivity essays

  1. Flywheel experiment

    Actual value)/ Expected Value] x 100 = [3.4x10-2/ 3.43x10-2] x 100 = 99.13% 5. DISCUSSIONS/CONCLUSION Following the analysis of my results, the values of Iexperiment and Itheory differ by fairly a significant amount i.e. (a percentage error of 99.13%). The errors that led to the difference in the two values

  2. Young's Modulus of Nylon

    By calculating engineering stress, the effect of this variable can be negated. Safety Whilst the practical side of this investigation does not include any particularly dangerous apparatus or methods, there are still some safety considerations to be taken. Eye protection will have to be worn whilst the nylon has stress

  1. Force of Friction experiment

    Until limiting static friction () was reached, the block began to slide and the reading (representing ) remained basically constant. Experimental value of is much smaller than . Theoretically, should be only slightly smaller than. This may be to error. If heavier blocks are used, the difference between and might be smaller.

  2. Free essay

    CIRCULAR MOTION - revision notes and calculations

    radius r, if the road is very slippy, the only forces acting on the wheels are the normal reactions R1 and R2. The centripetal force is provided by the component of (R1 + R2) ? (R1 + R2)Sin? = mV2/r (1)

  1. The aim of this experiment is to investigate the patterns of circular motion using ...

    The radius of the orbit r is the same as the distance between the moving object and the central body. Here is a diagram that I drew explaining how the centripetal force of a object like a ball works. There are three possible factors or variables that can affect the force.

  2. Objective To measure the centripetal force for whirling a mass round a horizontal ...

    They are almost equal. Discussion T = m?2L regardless of the angle? The tension, T in the string is provided by the weight of screw nuts, Mg. i.e. T = Mg The length, L of the string above the top of glass tube is between the distance of the paper marker and the bottom of the tube.

  1. Use of technology in a hospital radiology department. The department of imaging is one ...

    The purpose of each is briefly described below. Camera collimator, this is the first place that an emitted gamma photos encounters after exiting the body. The collimator is a pattern of holes through gamma ray absorbing material, usually lead or tungsten that allows the projection of the gamma ray image onto the detector crystal.

  2. Drayton Manor Theme Park: Centrepedial Force

    The inertia in regards to circular motion is also known as centrifugal force, it is this that holds the passengers of cyclone in place.4 2nd Aspect of Physics -microgravity on the ride Apocalypse. When going on Apocalypse you have to stand on a platform and an attendant comes round a

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