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
Radius (R) = 18.1 cm
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
Radius (R) = 18.1cm
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)
Radius, R (m)
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.
