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

    Even though the data are not very accurate and prise, there are ways to improve it, which would be discussed in the next section. A few of variables are controlled during the course of the experiment to make this a fair test.

  2. Analyzing Uniform Circular Motion

    This makes sense, since having more tension in the string leads to a quicker speed, thus an increase in the frequency. The percent error for this is calculated below: Hanging Mass: Theoretical slope = 35.254; Experimental slope = 36.368 (Absolute Value)

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

    � 0.01 38.5 11 1.7821 � 0.03 17.6 1.1775 � 0.02 66.3 12.3 1.9340 � 0.03 19.5 1.2503 � 0.02 84.3 12.2 1.8447 � 0.04 19.2 1.1898 � 0.02 101 11.5 1.6092 � 0.02 17.5 1.0633 � 0.01 117 9.2 1.3181 � 0.02 14.4 0.8459 � 0.01 126 7.2 1.0349

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

    to make the centre of the bar meaning 1 cm from each side of the centre of the bar with 7.5cm on each side of the strings. Then I pulled back the left side of the metal bar with my left hand until it was 90 degrees from its original position completely perpendicular to my body facing my apparatus.

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

    Then, divide one by the square root of mass. Example: V1.5g�1.2247g0.5 1/1.224� 0.8156 1/g0.5 Inverse of the Square Root of Mass Uncertainty: First, divide the uncertainty of mass by the total mass to find what percentage of the total mass the mass uncertainty is.

  2. Oscillating Mass

    2 Planning B 2.1 Apparatus 1. 1 elastic spring 2. 1 Meter stick 3. 1 Analytical balance 4. 1 Stop watch 5. 1 Ring stand 6. 1 each of the following masses: 0.400kg, 0.600kg, 0.800kg, 1.00kg, 1.20kg, 1.40kg, 1.60kg. 2.2 Method Part I Mass m as a Variable 1.

  1. The Affect of Mass on the Period

    in the same manner, and recorded the data. A 200g mass (actual mass 200.2g) was used to collect and record data for the five trials for data point four using the exact same method of experimentation used to collect information for data point one. Finally, the five trials for data point five were conducted and the results recorded using the 250g mass (actual mass 250.3g).

  2. Finding the Spring Constant

    and the time taken for that as our example. = 0.090 + 0.21 = > 0.300 After we have done this (?Equipment + ?Reaction + ?Average uncertainty) we divide it by 10 in order to obtain ?Period. = 0.300 / 10 = 0.030 Therefore, ? Period for the first mass (0.100 kg)

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