# The Relationship Between Centripetal Force And Velocity Environmental Sciences Essay

✅ Paper Type: Free Essay |
✅ Subject: Environmental Sciences |

✅ Wordcount: 2470 words |
✅ Published: 1st Jan 2015 |

Investigate the relationship between centripetal force and velocity in circular motion, when a stopper is swung with a string in which different weight hangers are attached to.

DATA COLLECTION AND PROCESSING

According to Mr. Isaac Newton, an object’s “natural state of motion” is to stay at rest if it’s already at rest or to continue in linear, uniform motion unless it’s subjected to a net, external force. This means that if an object is moving at constant velocity (or speed) in a straight line, it will continue to move in a straight line, at that same velocity, unless some outside force changes its motion in some way.

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Essay Writing ServiceSo in order for an object to move in a circular path, some force is needed to pull it away from the straight-line trajectory it “wants” to follow (i.e., its natural state of motion). Some force needs to pull the rotating object in at every single point along its circular path in order for it continue moving in a circular fashion (instead of allowing it to follow its natural state of motion)[1].

If an object moves in a circular path there must be centripetal Force acting on it. In this experiment we will investigate the relationship between centripetal force and velocity.

Next we will show how our raw data is going to be manipulated:

Initially we will show T for 20 revolutions, at each trial. Subsequently the average for 20 revolutions will be calculated (based on 3 trials).

Using this averaged value we will calculate the period of 1 revolution. In this process we will have to divide the uncertainty in time by 20. Details about uncertainty calculation will be added later on.

Next we are going to calculate the average linear speed, v, of the stopper for each mass of the “weight hanger”. We will include a sample calculation. We must remember that we used a fixed radius of 0.5 meters.

We have to bear in mind that we will have to add the percentage uncertainty in the radius and the percentage uncertainty on T and add them up when calculating the uncertainty for linear velocity.

Theoretically, the centripetal force should be directly proportional to the square of the speed. In order to check this, a column “v2” will be added to one of our data tables. When we do that the uncertainties on “V” must be squared.

We will also display a column indicating the centripetal force. We know that the centripetal force is equivalent to weight in this experiment; the weight, in turn, is equal to the tension on the string. When calculating the uncertainty for the resultant force (weight) we took the uncertainty on mass and multiplied it by ten which is the gravity value. In order for us to calculate centripetal force the following formulas will be used:

Fc = mv2/ r

Fc = W = mg

The uncertainties involved with the measurements which have fixed values are:

Centimeters: 1cm ± .05cm

Time: 1s ±0 .005s

Mass: 1kg ± 0.000005kg

UNCERTAINTY ON “T”

The uncertainty on T is the same of that on the stopwatch. As we start and stop the stopwatch we must, therefore, double the uncertainty: 2(±0 .005) = ±0.01

SAMPLE CALCULATION OF “T” FOR ONE REVOLUTION OF A MASS OF 0.1 KG

Average T for 20 revolutions: 15.3

Average T per revolution: 15.3 / 20 = 0.765

Uncertainty was also divided by 20: (±0.01/ 20) = ±0.0005

As the uncertainty on T was already multiplied by 2 we do not need to double it this time

CALCULATING UNCERTAINTY ON VELOCITY FOR 0.1 KG

As mentioned earlier now we will have to calculate percentage uncertainties. We will apply the following formula:

http://scidiv.bellevuecollege.edu/Physics/measure&sigfigs/Measuresigfigseq1.gif

Percentage uncertainty in radius: (0.05 / 0.5) x 100= ±1.0%

Percentage uncertainty on T for 1 revolution (calculated above): (0.00025/ 0.765) x 100 = ±0.0326%

By adding p the above uncertainties we get the percentage uncertainty for velocity which is ±1.0326% in this case.

In order to obtain the percentage uncertainty for v2 we simply square the uncertainty on “v”.

SAMPLE CALCULATION OF “V” FOR 0.1 KG

v = 2 pi r/T

V= = 4.106 ± 1.0326%

In addition we will calculate

UNCERTAINTY ON MASS

The uncertainty on mass was calculated based on the electronic scale used. The uncertainty on the scale was ±0.05 grams. Since we need the uncertainty in kg we multiply this value by 1000 and we get: ±0.00005

Table 1 – Showing magnitude of resultant force and averaged results

Mass (kg) (±0.00005)

Centripetal Force (N) (±0.000005)

T for 20 revolutions ( seconds) (±0.01)

Trial 1

Trial 2

Trial 3

0.100000

1.000000

15.50

14.81

15.56

0.150000

1.500000

13.69

13.80

13.91

0.200000

2.000000

12.31

12.76

12.43

0.250000

2.500000

11.57

11.55

11.61

0.300000

3.000000

10.40

11.20

10.80

0.350000

3.500000

10.38

10.01

10.21

In table 1 we presented the value obtained in each trial for 20 rotations. In table 2, on the other hand, we are going to present the average value of 1 rotation for each mass. By doing so we believe to have increased the accuracy of the results. In order to calculate the uncertainty for 1 oscillation we divided the uncertainties in 20 rotations by 20; as the left-most column (table 2) shows.

In spite of that there were cases where the difference between the highest and lowest value obtained were greater than the uncertainty itself. In the cases where this happened we found the differences between these values (highest and lowest) and use it as the uncertainty. Now we will show these differences between higher and lower values.

In the three trials for 0.1 Kg the difference between the highest and lowest value is: 15.56 – 14.81 =± 0.75. Hence this value will be used as the uncertainty as it is greater than the uncertainty in time.

In the three trials for 0.15 Kg the difference between the highest and lowest value is: 13.91 – 13.69 = ±0.22. Hence this value will be used as the uncertainty as it is greater than the uncertainty in time.

In the three trials for 0.2 Kg the difference between the highest and lowest value is: 12.76 – 12.31 = ±0.45. Hence this value will be used as the uncertainty as it is greater than the uncertainty in time.

In the three trials for 0.25 Kg the difference between the highest and lowest value is: 11.61 – 11.55 = ±0.06. Hence this value will be used as the uncertainty as it is greater than the uncertainty in time.

In the three trials for 0.3 Kg the difference between the highest and lowest value is: 10.80 – 10.20 = ±0.60. Hence this value will be used as the uncertainty as it is greater than the uncertainty in time.

In the three trials for 0.35 Kg the difference between the highest and lowest value is: 10.38 – 10.01 = ±0.37. Hence this value will be used as the uncertainty as it is greater than the uncertainty in time.

Table 2 – Preparing the results for graphical analysis

T for one revolution (seconds) (±0.0005)

Absolute Uncertainties (± seconds)

Percentage uncertainty on T (± %)

V (m/s)

Percentage uncertainty on V (± %)

V2 (m2/s2)

Percentage uncertainty on V2 (± %)

0.76500

0.75

0.000653

4.105

1.000653

16.859

1.001306

0.69209

0.22

0.000722

4.537

1.000722

20.857

1.001445

0.62239

0.45

0.000803

5.045

1.000803

25.457

1.001607

0.57880

0.06

0.000864

5.425

1.000864

29.430

1.001729

0.53970

0.60

0.000926

5.818

1.000926

33.846

1.001853

0.51040

0.37

0.000976

6.152

1.000976

37.850

1.001953

Now we will plot square root of centripetal force against V. We will make use of the percentage uncertainty on V to plot the horizontal error bars and the uncertainty on centripetal force to plot the vertical error bars. The uncertainty on centripetal force is ±0.000005 and, therefore must be squared to give us the uncertainty on. So we have which is equal to ±0.002236. Thus we have explained how our error bars were calculated. The graph we came across was the following:

Graph 1 – Showing correlation between and V

Not as we were expecting the graph resembles a parabola. We believe that, in order to obtain a straight line we must square both the centripetal force and velocity. This will give us the proportionality found in the formula: F = mv2/ r. We believe that, by plotting this graph we will be able to prove our prediction that the velocity squared is proportional to centripetal force. By plotting the mentioned correlation we get:

Graph 2 – Showing correlation between Fc and V2

Even though the best linear fit is not a perfect straight line there are no big discrepancies in our results (such as an outlier). The RSME value or the root means square error tells us how far the linear fit is from the plot points. The value of 0.02 is really low and suggests that the best fit is really close to the original data. Also the difference between the possible maximum value and possible minimum value of the spring is so low:

Maximum slope: 0.13

Minimum slope: 0.11

Difference: 0.02

So we have the slope of our straight line being: 0.12 ± 0.02

Because the RSME value is low, we can infer that the value obtained is realistic. In addition due to the fact that the best linear fit touches (including or not the error bars) all the data points we can infer that the graph is accurate and, consequently, so are our results.

CONCLUSION AND EVALUATION

All in all this investigation led to fairly precise results. We do however think that the experiment can be improved in several ways. The following improvements would increase the reliability of the experimental procedure.

Among the difficulties involved in the experiment we found, for instance, speed which we did not manage to keep constant when swirling the mass. Every so often the movement of our body would vary the speed at which the mass was being swung. In addition the swing rotation was not constantly a horizontal line. These factors will cause our results to become less accurate. Furthermore we faced some difficulties when swinging the stopper with constant power and speed; sometimes our hands touched the string which was not supposed to be touched during the rotations. The stopwatch delay and the human reaction time also affected our results to some extent. For example in a time of 5 seconds the human reaction time of 0.7 seconds can be very significant in the result as 1.4 seconds are involved in starting and stopping the stopwatch. Therefore these factors together are the responsible for us not obtaining a perfect parabola and consequently a perfect straight line. Moreover we found really hard to determine the initial and final point in relation to which the rotations were being counted. This probably led us to miscount the number of rotations. Therefore in some cases we might have had more or less than 20.

Many changes could have been made to the experiment to make it more accurate:

Setting up a better method of counting the rotations completed by the bung by using more advanced equipment than merely relying on human reactions.

Increase the amount of rotations to ensure greater accuracy.

· Increase the number of repeats to get a more accurate average.

· Set up computer equipment to time the experiment more accurately. This could be done using a motion sensor connected to a data logger (logger pro 3) to record the information.

· By doing the experiment outside uncontrollable factors such as wind can increase friction acting upon the bung and alter the time by small amounts which still make the experiment less accurate.

Further work as increased number of repeats could be carried out. In addition, different experiments can be done with increased number of rotations and larger radii. If one decides to investigate the effect of another variable such as radii the experiment will keep the same; the only difference will be that the weight hanger will be kept constant and the radii will vary. If we decided to increase the radii being investigated we would conclude that as the radius increases so would the time to complete 20 rotations for the bung, in a proportion directly related to the increase in distance. This is because we know that F = mv2/ r. Where F = force, m = mass, v = velocity and r = radius. So if r is increased then all the other variables increase in direct proportion to the initial increase. Newton’s First Law states that an object travels at constant velocity unless acted on by an unbalanced force. The unbalanced force is the weight in our experiment which increases the force making the speed increase because more force is being added. Therefore, this explains why the speed increases.

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View our servicesNow we will try to explain our results based on our scientific knowledge. If we draw a free body diagram of what is happening during the experiment we will come to the conclusion that the tension in the string (which is equal to the centripetal force) is being produced by the force of gravity which is acting on the load being used. From graph 2 we see that centripetal force increases in a direct proportion to the square of velocity. This relationship is further explained by the formula:

F =

Since “m” and “r” are kept constant and v is our dependent variable we see that force, in fact, should increase as our experiment suggests. Thus our experiment proves the formula for centripetal force.

Looking at the experiment we see that fairly good results were obtained. They, despite the uncertainties, allowed us to prove Hook’s Law. Due to the fact that the experiment was dynamic, a few sources of errors affected our results. We see that the curve obtained is pretty close to a straight line which is reinforced by the low RSME value. All in all this tells us that the method is reliable and lead to precise results.

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