• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15

Centripetal Force

Extracts from this document...

Introduction

Centripetal Force

Aim:

  • To investigate the relationship between the centripetal force action on an object moving in a circle of constant radius and the frequency of revolution.

Hypothesis:

The centripetal force of an object of mass (m) moving at a constant velocity (v) and radius (r) is given by ΣF = image00.png= 4π2rmf2. where:

Fc is the centripetal force in N

Fg is the force of gravity in N

r is the radius of the string in m

ms is the mass of the rubber stopper in kg  

ml is the mass of the suspended load in kg

g is the gravitation in m s-2

From that equation, I can deduce that if the frequency of object spinning is increased, the centripetal force will increase. Also, in this experiment, the force of gravity acting on the suspended mass should be the one providing the centripetal force. This force is given by: image01.pngwhere in this case m is the mass of the suspended load and g is the gravity.

Variables:

Dependent

Independent

Controlled

The period of the revolution measured by using stopwatch, the frequency and the centripetal force

load mass (varied from 10 gr until 50 gr with interval 10 gr each)

  • The small of soft mass (rubber stopper)
  • horizontal string length

Apparatus:

Name

Quantity

Accuracy

Thin plastic tube about 15 cm long, with no sharp edges

1

±0.5cm

1.5 of fishing line

1

-

Paper clip

1

-

Small soft mass (rubber stopper)

1

-

Mass carrier and slotted masses (10 g each)

As needed

-

Stop watch

1

∆t ±0.005s

Metre ruler

1

∆l ± 0.5 cm

Electronic Scale

1

∆m ± 0.005 g

Method:

  1. Securely tie one end of the fishing line to a small, soft mass.

        (Since this is going to be twirled around your head, make sure the mass isn’t too hard!).

  1. Pass the line down through a thin plastic tube and attach a 10 g slotted mass carrier to the end as shown in the diagram. Attach a paper clip to the line to act as a marker for a measured radius of around 1 metre.
  2. Add 10 g masses to the mass carrier to make a total mass of 20 g.

image12.png

  1. Twirl the stopper in a horizontal circular path at a speed that pulls the paper clip up to, but not touching, the bottom of the tube.
  2. Get a partner to keep an eye on the position of the clip to ensure that the speed of rotation stays quite constant. Practise doing so for a while before trying any measurements.
  3. Maintain the speed of revolution and measure the time taken for 10 revolutions of the small mass.
  4. Add an extra 10 g to the mass carrier and repeat steps 2 and 3.
  5. Add another 10 g to the mass carrier and repeat steps 2 and 3. Keep adding an extra 10 g mass to the mass carrier.

Data Collection:

Radius of the string, r = 0.3 m

Mass of the rubber stopper: m=0.0187 kg

Table 3.1 Result of time taken for 10 revolutions for mass 10 g

Trial

Time Taken

t(s)

∆t = ±0.005 s

1

8.00

2

7.47

3

7.49

4

7.09

5

7.16

6

7.25

7

7.45

8

7.22

9

7.49

10

6.78

Table 3.2 Result of time taken for 10 revolutions for mass 20 g

Trial

Time Taken

t(s)

∆t = ±0.005 s

1

6.47

2

7.47

3

6.91

4

7.44

5

7.00

6

7.32

7

6.35

8

7.50

9

7.79

10

8.47

Table 3.3 Result of Time taken for 10 revolutions for mass 30 g

Trial

Time Taken

t(s)

∆t = ±0.005 s

1

7.16

2

7.25

3

6.79

4

7.72

5

6.40

6

8.04

7

7.84

8

7.41

9

6.32

10

6.53

Table 3.4 Result of time taken for 10 revolutions for mass 40 g

Trial

Time Taken

t(s)

∆t = ±0.005 s

1

7.03

2

7.40

3

6.62

4

6.87

5

7.34

6

7.12

7

6.47

8

6.60

9

5.91

10

7.72

Table 3.5 Result of time taken for 10 revolutions for mass 50 g

Trial

Time Taken

t(s)

∆t = ±0.005 s

1

5.68

2

6.03

3

7.97

4

6.34

5

6.16

6

6.90

7

7.59

8

6.81

9

7.78

10

6.67

...read more.

Middle

6.78

0.678

1.474926

2.175407

0.481308

0.068510

0.0046937

ΣFc=

4.127976

image06.png

0.0116417

Average centripetal force:image34.png

Standard deviation: image35.png

So, the centripetal force of mass 10 g is:  image36.png

Force of gravity with mass 10 g: image37.png

image02.png

image03.png

Table 4.2 Data of Centripetal Force for mass 20 g

Time for

10 revolutions (s)

Period

(T/s)

f = 1/T

(Hz)

f2

Fc

(N)

image04.png

image05.png

6.47

0.647

1.545595

2.388864

0.528535

0.101778

0.0103587

7.47

0.747

1.338688

1.792086

0.396498

-0.030259

0.0009156

6.91

0.691

1.447178

2.094324

0.463368

0.036611

0.0013404

7.44

0.744

1.344086

1.806567

0.399702

-0.027055

0.0007320

7.00

0.700

1.428571

2.040816

0.451529

0.024772

0.0006136

7.32

0.732

1.366120

1.866284

0.412914

-0.013843

0.0001916

6.35

0.635

1.574803

2.480005

0.548700

0.121943

0.0148701

7.50

0.750

1.333333

1.777778

0.393332

-0.033425

0.0011172

7.79

0.779

1.283697

1.647878

0.364592

-0.062165

0.0038645

8.47

0.847

1.180638

1.393905

0.308401

-0.118356

0.0140082

ΣFc=

4.267571

image06.png

0.0480119

Average centripetal force:

image07.png

Standard deviation

: image08.png

So, the centripetal force of mass 20 g is:  image09.png

Force of gravity with mass 20 g: image10.png

image11.png

image13.png

Table 4.3 Data of Centripetal Force for mass 30 g

Time for

10 revolutions (s)

Period

(T/s)

f = 1/T

(Hz)

f2

Fc

(N)

image04.png

image05.png

7.16

0.716

1.396648

1.950626

0.431575

-0.010600

0.0001124

7.25

0.725

1.379310

1.902497

0.420926

-0.021249

0.0004515

6.79

0.679

1.472754

2.169004

0.479891

0.037716

0.0014225

7.72

0.772

1.295337

1.677897

0.371234

-0.070941

0.0050326

6.40

0.640

1.56250

2.441406

0.540160

0.097985

0.0096011

8.04

0.804

1.243781

1.546991

0.342271

-0.099904

0.0099807

7.84

0.784

1.275510

1.626926

0.359957

-0.082218

0.0067598

7.41

0.741

1.349528

1.821225

0.402945

-0.039230

0.0015390

6.32

0.632

1.582278

2.503605

0.553921

0.111746

0.0124872

6.53

0.653

1.531394

2.345166

0.518867

0.076692

0.0058817

ΣFc=

4.421747

image06.png

0.0532685

Average centripetal force :

image14.png

Standard deviation:

image15.png

So, the centripetal force of mass 30 g is:  image16.png

...read more.

Conclusion

There will be some improvements in this experiment that I need if I’ll do this experiment again. First of all we need to practice to swing it horizontally, and when we want to record the revolution after 10 swings, we must carefully count the revolution because sometimes our eyes can’t follow the revolution. Secondly, we also need  more careful when swing because sometimes the mass can contact our head or our friends. I also need to keep the string is not moving so that the length can be measured perfectly. To get more accurate results, we can also give an alternative independent variable such as the length of the string. By getting the results from the different mass and different length of string it will make the data more accurate to prove the theory and hypothesis.

...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

    experiment as it increased I had to move my hand quicker to stop the timer.

  2. Analyzing Uniform Circular Motion

    This is a relatively high error (neglecting uncertainties). The error can be due to the fact that the length of the string was not always the radius of the circular motion; if not enough force was exerted the string will make a Pythagorean relationship, with horizontal and vertical components, where the horizontal component is the actual radius of the circle.

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

    The force applied in the form of a spherical bob was movable and thus, the readings were taken starting close to the first rigid support and moving away such that by the time I reached the second rigid support, they were 10 points at which the vertical distance was measured.

  2. Circular Motion Practical - The graphs show that there is a positive correlation between ...

    Therefore it is true that a positive correlation occurs between the force applied and the speed. But us we can see from the graph, the line thus not go through the origin, suggesting that they are not directly proportional. For test 2, force and mass is kept constant, therefore acceleration would stay the same base on Newton's second law F=ma.

  1. The Affect of Mass on the Time It Takes an Object To Fall

    Therefore, the equation that includes x to the power of -0.5 is within our uncertainty. Furthermore, when the inverse of the square root of mass was then taken, the graph became very close to linear. The positive slope of this linear line, 3.3568, shows that as the inverse of the

  2. Universal Gravitation

    12.00 0.083 0.007 0.1183 13.00 0.077 0.006 0.1020 14.00 0.071 0.005 Analysis: Graph the data and use curve straightening techniques to prepare a line graph. Derive the value for G from the straightened line graph. Determine the accuracy of G by comparing it to the theoretical value.

  1. Oscillating Mass

    Dependent: The period T of the oscillating mass; the time taken to make one full swing. This measurement will be done by using a stop watch and will rely greatly on my vision and reaction time. I could not find an apparatus that would automatically record this dependant variable.

  2. HL Physics Revision Notes

    Use A = λN and then λt½ = ln2 Topic 14: Digital Technology: Option G: Electromagnetic Waves: Outline the nature of electromagnetic (EM) waves. Oscillating electric charge produces varying electric and magnetic fields Electromagnetic waves are transverse, have the same speed Oscillating (SHM)

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