• Our Services

Can’t find a perfect paper?

Centripetal force

This experiment aims to evaluate the following objectives:

• To experimentally measure the centripetal force in a circular motion

• To study the main features of uniform circular motion

• To verify the relationship between centripetal force, mass, and radius

Theory

All objects moving in a circular motion at a particular radius R are in acceleration and the acceleration’s radial component is called centripetal acceleration determined as ac = v2/R. The centripetal acceleration is directed towards the center of the circular path. Centripetal force is the net force which results from a moving object with a constant velocity V that is tangential to the orbit of the circle. Moreover, the force is generated due to other forces which include friction, gravity, tension, electric attraction and elasticity. Therefore, the centripetal force is directly proportional to the square if the speed of an object moving in a circular motion with mass M and at a given radius from the center R.

Data and Sample Calculations

Part 1: Dependence of centripetal force on speed of rotation (v) at constant speed

Mass of metal bob (M) = 448.1 g

Table 1. Calculations

Trial

Time for 20 rev, t (sec)

Period, T(sec)

Speed, v (m/s)

V2 (m/s)2

Hanging mass, m (kg)

Spring force, Fsp (N)

Centripetal force, Fc (N)

% difference

1

24.61

1.2305

0.9242

0.8542

0.22

2.156

2.1147

4.13%

2

19.88

0.994

1.1441

1.3090

0.265

2.54

3.2407

-70.07%

3

19.4

0.97

1.1724

1.3746

0.33

3.43

3.4031

2.69%

4

17.68

0.884

1.2865

1.6551

0.4

3.92

4.0974

-17.74%

5

16.15

0.8075

1.4084

1.9835

0.45

4.41

4.9105

-50.05%

6

16.01

0.8005

1.4207

2.0183

0.51

4.998

4.9968

0.12%

Slope from the graph of Fsp versus v2

= 2.35

Theoretical value of the slope = (2.54-2.126)/(1.3090-0.8542) = 0.84

% difference = 151.00%

Graphs

Graph 1: Fsp versus v2

Part 2: Dependence of the period of rotation (T) on the mass at constant radius

Hanging mass (m) = 0.310kg

Table 2. Calculations

Trial

Total mass on the bob, M(kg)

Time for 20 rev, t (sec)

Period, T (sec)

T2

(sec2)

1

0

20.17

1.0085

1.017

2

.100

21.87

1.0935

1.1957

3

.200

21.98

1.099

1.2078

4

Slope from the graph T2 versus M = 0.4525

Theoretical value of slope = (1.0935-1.0085)/(0.1-0) = 0.85

% difference = 39.75%

Graph 2: T2 versus M

Part 3: Dependences of the speed (v) on the radius (R) at a constant centripetal force

Mass of the bob (M) = 448.1g

Total hanging mass (m) = .490kg                   Spring Force (Fsp) = mg = 4.9N

Table 3. Calculations

Trial

Time for 20 rev, t (sec)

Period, T (sec)

Speed, v (m/s)

V2

Centripetal force, Fc (N)

(m/s)2

1

16.9

0.8450

16.50

1.2269

1.5053

4.0879

2

17

0.8500

18.50

1.3675

1.8701

4.5297

3

17.9

0.8950

18.90

1.3268

1.7605

4.1740

4

18.27

0.9135

19.90

1.3688

1.8735

4.2186

5

19.52

0.9760

20.30

1.3069

1.7079

3.7699

6

19.72

0.9860

20.60

1.3127

1.7232

3.7484

Average Fc = 4.0881N

Slope from the graph v2 versus R = 0.0497

Theoretical value of the slope = 0.1824

% difference = 13.27%

Graph 3: v2 versus R

Questions

1. How is it possible that a body moving at a constant speed is still in acceleration motion?

Ans: Acceleration is a vector quantity since it is the rate of change of velocity thus it changes unlike speed which is a scalar quantity and it is not affected by direction. Therefore, change in velocity makes the object to accelerate towards the center continuously

2. What provides the centripetal force necessary for a car to move in a circular motion at a constant speed?

Ans: The frictional force which is acting upon in the turned wheels between the tire and the road provides the centripetal force (Heard " Aravind, 2010).

3. Directions of acceleration and net effect while speed is changing in a circular motion.

Ans: The direction for acceleration is inward while that of the net force is perpendicular to the velocity vector

4. Effect on the arm hanging the bob and the pointer are not the same

Ans: If the bob is lower than the pointer, the bob will take time in hitting the pointer hence there will be a deviation in the movement of the bob from the orbit as it moves from the pointer. Additionally, if the bob is higher, the bob might never hit the pointer thus the readings will be wrong.

5. In this experiment, if there is no spring attached and the bob is rotated at a constant speed, what provides the centripetal force? Draw a diagram to explain your answer

Ans: The centripetal force for the bob is provided by the gravitational force while moving at a constant speed. The tensional forces provide the initial centripetal force.

Graph 4. Gravitational force

Conclusion

Centripetal force is a force which acts on an object moving in a circular motion at a constant speed. Moreover, the force is affected by radius of an object from the center and the period which is the time taken for one complete revolution.

Errors and obstacles

The experiment showed a large variation in the slope of the graphs with values as:

Part 1: % difference = 151.00%

Part 2: % difference = 39.75%

Part 3: % difference = 13.27%

References

Heard, J. W., " Aravind, V. R. (2010). Physics by simulation: Teaching circular motion using

applets. Latin-American Journal of Physics Education, 4(1), 35-39.