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

Objective To measure the centripetal force for whirling a mass round a horizontal circle and compare the result with the theoretical value given by F= mw2r .

Extracts from this document...

Introduction

School:

Canossa College

Class:

6B

Name:

Hazel Chow Ho Ying

Class no:

3

Date:

20-10-2010

Mark:

Title

Centripetal force

Objective

To measure the centripetal force for whirling a mass round a horizontal circle and compare the result with the theoretical value given by F= mω2r .

Apparatus

  • rubber bung
  • glass tube
  • screw nuts
  • Wire hook
  • 1.5m of Nylon string
  • Small paper marker
  • metre rule
  • stop-watch

Theory

When a mass m attached to a string is whirled round a horizontal circle of radius r, the centripetal force for maintaining the circular motion is given by

F = mω2r   where ωis the angular velocity of the circular motion.

This force is provided by the tension of the string.

The formula can also be expressed in the terms of the velocity v of the mass, where ω=v/r .

Substituting ω=v/r into the formula for F , F = mv2/r

...read more.

Middle

Time 50 revolutions of the bung and calculate the angular velocity ω.Repeat several times using different lengths L of the string.

Result

Tabulate the results as follows:

        Mass of rubber bung m =   0.03491   kg

        Mass of screw nuts M =   0.13   kg

 Tension in string T = Mg =   0.13   × 9.8 N =   1.274   N

Length of string L/m

0.35

0.50

0.75

0.90

Time for 50 revolutions 50 t/s

30.85

37.04

45.00

49.65

ω = image00.png/rad s–1

10.183

8.482

6.981

6.327

mω2L/N

1.267

1.256

1.276

1.258

        Mean mω2L =   1.264   N

Conclusion

Form the results, we can find that the length of the string L is increasing, the value of angular velocity ω is decreasing. The length of the string is indirectly proportional to the value of angular velocity.

...read more.

Conclusion

ω2L .

They are almost equal.

Possible sources of errors

  1. Friction exists between the glass tube and the string.
  2. The rubber bung is not set into a horizontal circular path.
  3. The rubber bung does not move with constant speed.
  4. The length of the string beyond the upper opening is not constant.

θ increases with ω

Vertical components is T cosθ

Horizontal components is T sinθ

The system has no vertical acceleration

T cosθ = mg

The horizontal component of tension provides the centripetal acceleration

T sinθ = mrω2

Let L be the length of the string

i.e.  r = L sinθ

T sinθ = mrω2

T sinθ = m(L sinθ)ω2

T = mLω2

 mg/cosθ= mLω2

θ increases with ω

When the rubber bung is whirled around with a higher angular velocity ω , the angleθ becomes larger.

Reference

  1. Level practical physics for TAS p. 28 - 30
  2. Physics Beyond 2000 p. 40
  3. http://en.wikipedia.org/wiki/Centripetal_force
  4. http://www.greenandwhite.net/~chbut/centripetal_force.htm

...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. Young's Modulus of Nylon

    arms and feet should be kept back from below the experiment and all persons in the vicinity should be on their feet to allow them to move back quickly if required. Error Reduction The practical side of this investigation is prone to errors through a number of factors.

  2. Centripetal motion. The objective of this experiment is to verify whether the tension ...

    The hanger was first loaded with M = 150g. 3. The position of the paper clip was fixed about 1cm below the lower end of the glass tube by using adhesive tape. The rubber bung was kept about 80cm (l)

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

    Radius/cm Time taken to complete 20 revolutions / secs 100 13 125 8 150 7 175 7 200 6 225 5 I have learnt that the range I have used Is good and I will use that for the plan.

  2. Science Coursework - Investigating How Mass Influences Distance Travelled When Firing A Margarine Tub.

    I will also be doing this experiment in the corridor rather than the classroom as there is more working space which could help to improve the method. Preliminary results: Mass Of Sand (Grams) Force (Newton) Distance Travelled (cm) 30 10 167 90 10 65 150 10 50 210 10 41 Mass Of Sand (Grams)

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

    [18] 4) The company is responsibility to follow with the NHS quality system, which makes sure that all work carried, has to meet the criteria. The following quality systems are applied in UCLH hospital. o Eliminate unlawful discrimination o Promote race equality o Promote good race relation between people of different racial groups [19] 5)

  2. Drayton Manor Theme Park: Centrepedial Force

    at rest as you were when waiting for the lights to turn green, as Newton's first law of motion tells us.4 The direction of inertia is always opposite to the direction of force, as centripetal force is directed towards the centre of circular motion the inertia therefore must be directed away from the centre.

  1. The physics involved with a rollercoaster.

    "Phrases like, '...slow down..., ...speed up..., ...change speed... and change velocity...'are used to describe acceleration" (According to General)The cars are pulled by a chain to the top of the highest hill along the track. Then they are released from the chain. As the front car begins its decent, the unpowered cars have almost no speed and only a small acceleration.

  2. The Physics of an Atomic Bomb

    Since each fusion neutron directly generates 180 MeV through fission, and another 810 MeV as a result of the second generation neutrons, the energy released by fusion driven fission greatly exceeds that produced by the fusion reaction itself. Solving issues 1, 2 and 3 together is greatly complicated by the unavoidable presence of naturally occurring neutrons.

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