a=v2rrf v=2rT a=v2r v=2 Centripetal Force and Centripetal Acceleration a=v2r
Introduction
Centripetal force is described as an external force necessary to make a body follow a curved path. The centripetal force causes an object to travel or accelerate to the centre of the circle and this acceleration is referred to as the centripetal acceleration. Any particular force or a combination of forces can act to provide a centripetal force. Some common examples of forces which provide centripetal forces are gravitational force, frictional forces, and magnetic forces. The centripetal force has the magnitude, Fcentripetal=mv2r . The centripetal force produces acceleration towards the center of the circle. The circular path that an object travels is determined by the magnitude of the linear velocity and the magnitude of the force acting on the object. The velocity of an object in circular motion is constantly changing in direction thus is implied that there is acceleration. In addition, throughout this experiment we will be determining the relationship between the centripetal force and the period, frequency, and the radius. Circular motion directly represents Newtons Second Law, F=ma . I f we recall, the acceleration of an object in a circular motion is towards the center of the circle. From this we can deduce that the overall force on the string is equivalent to the product of the mass and the gravitational force. In this investigation we will be using the relationship between the gravitational force,F=mg and the centripetal force to derive a general formula for centripetal force. We will to indentifying the relationship between centripetal and the period, the frequency, and the radius.
Objective
The purpose of this investigation would be to determine the relationships between centripetal force (centripetal acceleration) and radius, period, and frequency.
Materials
Rubber lab stopper Standard mass set (5-12 metal washers) 1 small paper clip 2.0 m of fishing line Glass tube ( wrapped with masking tape to prevent breakage) Metre stick Stop-watch Safety goggles Page | 1
Marker Masking tape
Uncertainty Table 1: Uncertainty Apparatus
Stop Watch Triple Beam Balance Metre Stick
L imit of Reading
0.0001 (s) 0.1 (g) 1. (m)
Uncertainty
0.05 (s) 0.05 (g) 0.05 (m)
In this investigation the margin for error is high for there was a large capacity for random and human errors. The process of receiving data is very subject for we do not know whether the paper clip is exactly 3 cm from the glass tube and this might have altered the data collected. It is important to note that the procedure was subjected to multiple errors and uncertainties, since it was not extensively reliable in collecting acceptable data. In terms of instrumental uncertainty the stop watch has an uncertainty of 0.05 (s), the triple beam balance has and uncertainty of 0.05 (g), and the metre stick used to measure the length of the string has and uncertainty of 0.05 (m). The absolute uncertainty for the stop watch is 0.15 (s) because we have to take into consideration the error associated with stopping the stop watch reflexes. A notable flaw in the data analysis is that when we are to graph the relationship between centripetal force and radius for a constant f2 we are approximating the centripetal force for a constant f2value for each corresponding radius. This is not a well processed technique in determining the centripetal force since the observer cannot directly pinpoint the values on the graph that accurately.
Data Collection Table 2: Centripetal Force (Acceleration) and Period Mass 0.05 (g)
43.70 51.47 58.20
Number of Rotations
5 5 5
Time 0.15 (s)
6.25 6.12 5.90
Period 0.03 (s)
1.25 1.22 1.18
Frequency (Hz)
0.80 2.4 % 0.82 2.4 % 0.85 2.5 %
Centripetal Force [Gravitational Force(F=mg)] 0.0005 N
0.429 0.11 % 0.505 0.10 % 0.571 0.09 %
Page | 2
64.80 72.20 76.60 85.40
5 5 5 5
5.63 5.15 4.84 4.74
1.13 1.03 0.97 0.95
0.89 2.7 % 0.97 2.9 % 1.03 3.0 % 1.06 3.2 %
0.636 0.08 % 0.708 0.07 % 0.751 0.06 % 0.838 0.06 %
Sample Calculation ( Period):
Period=TimeNumber of Rotations
Period=6.25 0.15 s5
Period=1.25 0.03 s
Note: The percentage uncertainty of the period will be equivalent to the relative uncertainty of time divided by the pure number. Sample Calculation ( Frequency): Frequency= Number of RotationsTime
Frequency= 56.25 0.15 s
Frequency= 56.25 s 2.4 %
Frequency= 0.80 2.4%
Frequency= 0.80 0.02 s-1
Note: The percentage uncertainty of frequency will be equivalent to the percentage uncertainty of time, since 5 is a pure number and does not have any uncertainty. Page | 3
Sample Calculation (Centripetal Force):
[The gravitational force] Fgravity=mg
Fgravity=Fcentripetal [Solve for Centripetal Force]
Fcentripetal= mg
Fcentripetal= 0.043760.00005 kg(9.81 ms-2)
Fcentripetal= 0.429 0.11 % N
Fcentripetal= 0.429 0.11 % N
Fcentripetal= 0.429 0.0005 N Note: The centripetal force is equal to the magnitude of the gravitational force.
Table 2: Centripetal Force and Radius Radius 0.05 (m) Number of Rotations Time 0.15 (s) Period
0.02 (s)
Frequency (Hz)
Frequency 2 ( Hz2)
Centripeta l Force 0.0005 N
0.75
10
4.34
0.43
2.30 3.5 % 2.01 3.0
5.29 7.0 % 4.00 6.0
1.96
1.00
10
4.97
0.50
1.96 Page | 4
% 1.25 10 5.50 0.55 1.82 2.7 % 1.64 2.5 %
% 3.28 5.4 % 2.69 5.0 % 1.96
1.50
10
6.09
0.61
1.96
Sample Calculation ( Period):
Period=TimeNumber of Rotations
Period=4.34 0.15 s10
Period=0.43 0.02 s
Note: The percentage uncertainty of the period will be equivalent to the relative uncertainty of time divided by the pure number. This is case since 10 is a pure and number does not have any uncertainty. Sample Calculation ( Frequency): Frequency= Number of RotationsTime
Frequency= 104.34 0.15 s
Frequency= 104.34 s 3.5 %
Frequency=2.30 010 100%+0.154.34 100%
Frequency=2.30 3.5 %
Page | 5
Frequency=2.30 0.08 s-1
Note: The percentage uncertainty of frequency will be equivalent to the percentage uncertainty of time, since 5 is a pure number and does not have any uncertainty. Sample Calculation (Frequnecy2):
Frequency2= (2.30 0.08 Hz)2
Frequency2= (2.30)2 23.5 % Hz2
Frequency2= 5.29 7.0 % Hz2
Frequency2= 5.29 0.37 Hz2
Analysis
Part A: Centripetal Force (Acceleration) and Period 1) Graph 1: Centripetal Force vs. Period
2) The relationship between centripetal force and period is that the centripetal force is inversely proportional to the period squared. It is evident that as the period is increase the frequency decreases. In addition, by looking at the graph we can determine that the graph resembles the relationship: Fcentripetal1T2
In circular motion the centripetal force is proportional to (velocity)2. Since velocity is equivalent to the circumference, C=2r, divided by the period when the period is decreased then the velocity will increase. This in relation to the centripetal force, Page | 6
Fcentripetal=mg
Fcentripetal=ma,
Where, the acceleration would be equivalent to: a=v2d Where, the velocity is equivalent to:
v=distancetime=circumferenceperiod
v= 2rT We already know that the equation, Fcentripetal=mv2r
Shows that that the centripetal force is proportional to (velocity)2 and hence, the centripetal force is inversely proportional to the (period)2:
Fcentripetal=ma
Fcentripetal=mv2r
Fcentripetal=m2rT2r
Fcentripetal=m42rT2 3) Graph 2: Centripetal Force vs. Frequency
Page | 7
4) The relationship between the centripetal force and the frequency, according the graph, is that the centripetal force is directly proportional to the (frequency)2.
Fcentripetalf2
In circular motion the centripetal force is proportional to (velocity)2. Since velocity is equivalent to the circumference, C=2r, multiplied by the frequency, when the frequency is decreased then the velocity will increase. This in relation to the centripetal force,
Fcentripetal=mg
We already know that the equation, Fcentripetal=ma
Shows that that the centripetal force is proportional to (velocity)2 and hence, the centripetal force is inversely proportional to the (period)2: Fcentripetal=mv2r
Fcentripetal=m42rT2
Fcentripetal=m42rf2 Part B: Centripetal Force and Radius 1) Graph 3: Centripetal Force vs. Frequency2
2) Table 3: Centripetal Force vs. Radius when (frequency)2 is constant
Page | 8
Radius 0.05 (m) 0.75 1.00 1.25 1.50
Centripetal Force 0.0005 N 0.74 0.98 1.20 1.46
3) Graph 4: Radius vs. Centripetal Force
4) The relationship between the centripetal force and radius is that the centripetal force is directly proportional to the radius.
Fcentripetalr Data Collection and Processing Table3: Centripetal Force, Period, and Frequency Processed Mass 0.05 (g)
43.70 51.47 58.20 64.80 72.20 76.60 85.40
Centripetal Force 0.0005 N
0.429 0.11 % 0.505 0.10 % 0.571 0.09 % 0.636 0.08 % 0.708 0.07 % 0.751 0.06 % 0.838 0.06 %
Inverse of the (Period)2 1T2
0.64 0.03 s-2 0.67 0.05 s-2 0.72 0.04 s-2 0.78 0.04 s-2 0.94 0.05 s-2 1.06 0.07s-2 1.10 0.07 s-2
Frequency2 (Hz2)
f2
0.64 0.03 Hz2 0.67 0.03 Hz2 0.72 0.04 Hz2 0.79 0.04 Hz2 0.94 0.05 Hz2 1.06 0.06 Hz2 1.12 0.07 Hz2
Page | 9
Sample Calculation ( I nverse of the (period)2):
1period2=11.25 0.03 s2
1period2=11.25 2.4 % s2
1period2=11.56 4.8 % s2
1period2=0.64 4.8 % s-2
1period2=0.64 0.03 s-2
Graph 5: Centripetal Force vs. Inverse (Period)2 [Processed]
Sample Calculation ( (frequency)2):
f2=0.80 0.02 s-12
f2=0.80 2.4% s-12
f2=0.802 20.020.80 100% Hz2
f2=0.802 22.4 % Hz2
Page | 10
f2=0.64 4.8% Hz2
f2=0.64 0.03 Hz2
Graph 6: Centripetal Force vs. (Frequency)2 [Processed]
Therefore, once we straighten the graphs it is evident that the proportionality between centripetal force and period and centripetal force and the frequency are correct. The Centripetal Force vs. Radius relation is already linear.
Discussion
1) a) The centripetal force is directly proportional to the (frequency)2.
Fcentripetal f2 Equation Fcentripetal=mg
We already know that the equation, Fcentripetal=ma,
ma=mg From this we can deduce that: Fcentripetal=mv2r
Fcentripetal=m2rf2r
Fcentripetal=m42rf2
Page | 11
The proportionality statement is true. b) The centripetal force is inversely proportional to the (period) 2. Fcentripetal 1T2
Equation Fcentripetal=mg
We already know that the equation, Fcentripetal=ma,
ma=mg From this we can deduce that: Fcentripetal=mv2r
Fcentripetal=m2rT2r
Fcentripetal=m42rT2
c) The centripetal force is inversely proportional to the radius. Fcentripetalr
Equation Fcentripetal=mg
We already know that the equation, Fcentripetal=ma,
ma=mg From this we can deduce that: Page | 12
Fcentripetal=mv2r
2) From our results, what centripetal force would be required to rotate the rubber stopper in a horizontal circle of radius 1.5 m with a frequency of 8.0 Hz?
Given: Radius = 1.5 Frequency = 8.0 Hz Deduce that the mass equals 200 g or 0.2 kg.
[Solve]
Fcentripetal=m42rf2 Fcentripetal=200 g421.5m(8.0 Hz)2] Fcentripetal=0.200 kg421.5m(8.0 Hz)2] Fcentripetal=0.200 kg421.5m(8.0 Hz)2] Fcentripetal=757.99 N Fcentripetal=7.6 102 N
Conclusion
In conclusion I would state that the theoretical equations for centripetal force are true within the experimental error. If compared to the results obtained the equations would derive results which would fall within the margin for error.
Page | 13