The Relationship Between Centripetal Force And Velocity Environmental Sciences Essay

The Relationship Between Centripetal Force And Velocity Environmental Sciences Essay analyse the relationship between centripetal forte and velocity in circular motion, when a stopper is swung with a string in which different tip hangers argon attached to.DATA COLLECTION AND PROCESSINGAccording to Mr. Isaac Newton, an objects representic state of motion is to stay at rest if its already at rest or to continue in linear, uniform motion un slight its subjected to a net, external ram. This means that if an object is moving at constant velocity (or speed) in a smashing line, it bequeath continue to move in a straight line, at that same velocity, unless some outside force changes its motion in some way.So in order for an object to move in a circular runway, 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 or der for it continue moving in a circular fashion (instead of each(prenominal)owing it to follow its natural state of motion)1.If an object moves in a circular path on that point moldiness be centripetal Force performing on it. In this experiment we go out investigate the relationship between centripetal force and velocity. near we will see how our raw data is going to be manipulatedInitially we will show T for 20 revolutions, at each trial. by and by the average for 20 revolutions will be deliberate (based on 3 trials).Using this averaged appreciate we will calculate the period of 1 revolution. In this work out we will have to divide the incertitude in time by 20. Details about uncertainty calculation will be added later on.Next we argon 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 believe that we apply a fixed rundle 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 substantial of the speed. In order to check this, a column v2 will be added to one(a) of our data tables. When we do that the uncertainties on V must be featherd.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 accent 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 think of. In order for us to calculate centripetal force the following formulas will be employFc = mv2/ rFc = W = mgThe uncertainties baffling with the measurements which have fixed values areCentimeters 1cm .05cmTime 1s 0 .005sMass 1kg 0.000005kgUNCER TAINTY ON TThe 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) = 0.01SAMPLE CALCULATION OF T FOR ONE REVOLUTION OF A MASS OF 0.1 KG total T for 20 revolutions 15.3Average T per revolution 15.3 / 20 = 0.765Uncertainty was also divided by 20 (0.01/ 20) = 0.0005As the uncertainty on T was already multiplied by 2 we do not need to double it this timeCALCULATING UNCERTAINTY ON VELOCITY FOR 0.1 KGAs mentioned earlier now we will have to calculate percentage uncertainties. We will apply the following formulahttp//scidiv.bellevuecollege.edu/Physics/measuresigfigs/Measuresigfigseq1.gif plowshare uncertainty in radius (0.05 / 0.5) x 100= 1.0%pct 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 simp ly square the uncertainty on v.SAMPLE CALCULATION OF V FOR 0.1 KGv = 2 pi r/TV= = 4.106 1.0326%In addition we will calculateUNCERTAINTY ON MASSThe 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 super acid and we get 0.00005Table 1 viewing magnitude of resultant force and averaged resultsMass (kg) (0.00005)Centripetal Force (N) (0.000005)T for 20 revolutions ( seconds) (0.01) rivulet 1Trial 2Trial 30.1000001.00000015.5014.8115.560.1500001.50000013.6913.8013.910.2000002.00000012.3112.7612.430.2500002.50000011.5711.5511.610.3000003.00000010.4011.2010.800.3500003.50000010.3810.0110.21In table 1 we presented the value obtained in each trial for 20 rotations. In table 2, on the early(a) 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 th e 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 wil l 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 analysisT 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 ( %)0.765000.750.0006534.1051.00065316.8591.0013060.692090.220.0007224.5371.000 72220.8571.0014450.622390.450.0008035.0451.00080325.4571.0016070.578800.060.0008645.4251.00086429.4301.0017290.539700.600.0009265.8181.00092633.8461.0018530.510400.370.0009766.1521.00097637.8501.001953Now 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 tump over 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 followingGraph 1 display correlation between and VNot 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 getGraph 2 Showing correlation between Fc and V2Even though the best linear fit is not a perfect straight line there are no big discrepancies in our results ( much(prenominal) 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 minimal value of the spring is so lowMaximum slope 0.13Minimum slope 0.11Difference 0.02So we have the slope of our straight line universe 0.12 0.02Because 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 accurat e and, consequently, so are our results.CONCLUSION AND EVALUATIONAll in all this investigation led to evenhandedly 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. what is more we faced some difficulties when swinging the stopper with constant power and speed sometimes our hands moved(p) 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 huma n 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 unitedly are the responsible for us not obtaining a perfect parabola and consequently a perfect straight line. Moreover we found really hard to take care 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 accurateSetting up a crack 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 learn 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 through with(p) 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 incr ease in direct proportion to the initial increase. Newtons 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.Now we will try to explain our results based on our scientific knowledge. If we draw a free body draw 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 get on explained by the formulaF =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 p roves 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 maulers 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.