Jay Physics

Our First Circular Motion

(The Effect Of Radius on Speed of Circular Motion & The Effect of Force[Tension] on Speed of Circular Motion)

Question: What will happen to the velocity of a horizontal circular motion if the radius of that motion is changed and measured; or if the weight of the particle that has the motion is changed and measured? Diagram: The contraption shown in the diagram suited the experimenters’ needs to make a horizontal circular motion and change weight and radius of the system with the motion. The Diagram below shows the manner which the experimenters were able to accomplish making this contraption have a circular motion. These experimenters chose to make the motion a flat circle parallel to the ground as opposed to a cone kind of shape.

Hypothesis: (For Radius and Velocity) If the length of the radius of the circle increases, then the velocity of the stopper will increase in a linear and proportional manner, because as the equation for linear speed (velocity, in this case) is (rotations x 2Pi x radius) / Time , the variable of radius increasing would result in a larger product. (For Force and Velocity) If the weight of the system increases, then the velocity of the stopper will increase in a linear and proportional manner, because as it must swing in a constant velocity the forces must be balanced the force of tension the swinging causes must increase and therefore make it go faster to balance the force of gravity. Materials: (To make the contraption) ­Two meters of string ­A metal clip ­A rubber tube much resembling a straw, or a straw ­A rubber stopper ­Varying brass weights (enough varying to be able to go up by multiples of .02kg to .20kg) (other) ­Lab Space ­A Stopwatch Procedure: ­Create the contraption using the diagram for reference ­Measure the weight of the rubber stopper. Record data. ­Measure the weight of the brass weight(s) on the contraption and radius (the string, from the connection to the stopper to where it meets the tube)Record data. ­First measure with varying radius’ ; so have an experimenter start swinging the contraption above their head so it looks like its going about the same speed and in a circle. As that experimenter is going, have another experimenter count the rotations up to ten, starting the

stopwatch as they start counting and ending immediately after the 10th rotation. Repeat this step with a different and new radius 10 times. Record data. ­Measure with varying weights; repeat the last step but with different and new weights each time 10 times but the same radius. Record data. ­Calculate data Data: This data was measured during the trials for the experiment. Weight of rubber stopper = 0.012 kg For Radius v Velocity

Mass (kilograms) Time (seconds) Radius (meters)

0.1kg 3.82s 0.30m

0.1kg 3.92s 0.32m

0.1kg 4.10s 0.34m

0.1kg 4.13s 0.36m

0.1kg 4.20s 0.38m

0.1kg 4.50s 0.40m

0.1kg 4.57s 0.42m

0.1kg 4.63s 0.44m

0.1kg 4.71s 0.46m

0.1kg 4.74s 0.48m

For Weight v Velocity

Mass (kilograms) Time (seconds) Radius (meters)

0.20kg 3.25s 0.50m

0.18kg 3.30s 0.50m

0.16kg 3.52s 0.50m

0.14kg 3.61s 0.50m

0.12kg 3.74s 0.50m

0.10kg 3.88s 0.50m

0.08kg 3.97s 0.50m

0.06kg 4.10s 0.50m

0.04kg 4.19s 0.50m

0.02kg 4.23s 0.50m

Analysis: Data was collected and showed a parabolic function, so for radius v velocity and force v velocity there will be two sets of data; the first calculation, and then a linearization This data was measured during after the trials for the experiment using the data collected. Weight of rubber stopper = 0.012 kg For Radius v Velocity (UNLINEARIZED)

Radius (meters) From the above data. It’s just moved down to help visualize what will be translated over to the graphs

Velocity (meters / second) Calculated by using the linear speed equation with values for time and radius input from Radius v Velocity collected data (Rotations x 2Pi x Radius) / Time

0.30m (10 x 2Pi x 0.30m) / 3.82s = 4.93m/s

0.32m (10 x 2Pi x 0.32m) / 3.92s = 5.13m/s

0.34m (10 x 2Pi x 0.34m) / 4.10s = 5.21m/s

0.36m (10 x 2Pi x 0.36m) / 4.13s = 5.48m/s

0.38m (10 x 2Pi x 0.38m) / 4.20s = 5.68m/s

0.40m (10 x 2Pi x 0.40m) / 4.50s = 5.59m/s

0.42m (10 x 2Pi x 0.42m) / 4.57s = 5.77m/s

0.44m (10 x 2Pi x 0.44m) / 4.63s = 5.97m/s

0.46m (10 x 2Pi x 0.46m) / 4.71s = 6.14m/s

0.48m (10 x 2Pi x 0.48m) / 4.74s= 6.36m/s

Graph Created from above data. The Blue dots show the data while the others above and below identify uncertainty in the data.

This data was measured during after the trials for the experiment using the data collected. Weight of rubber stopper = 0.012 kg For Radius v Velocity (LINEARIZED)

Radius (meters) From the above data. It’s just moved down (again) to help visualize what will be translated over to the graphs

Velocity^2 (meters^2 / second^2) Calculated by using the above data and squaring it Velocity ^2

0.30m (4.93m/s)^2 = 24.34m/s

0.32m (5.13m/s)^2 = 26.31m/s

0.34m (5.21m/s)^2 = 27.15m/s

0.36m (5.48m/s)^2 = 30.00m/s

0.38m (5.68m/s)^2 = 32.32m/s

0.40m (5.59m/s)^2 = 31.19m/s

0.42m (5.77m/s)^2 = 33.35m/s

0.44m (5.97m/s)^2 = 35.65m/s

0.46m (6.14m/s)^2 = 37.66m/s

0.48m (6.36m/s)^2 = 40.48m/s

Graph Created from above data. The Blue dots show the data while the others above and below identify uncertainty in the data.

Mathematical Model (underlined) ­ calculated with a graphing calculator’s linear regression calculating application with above data table data. y=ax+b a=83.66 b= ­0.78 Velocity^2 = 83.66kg/s x Radius + ­0.78m^2/s^2 Can we choose a y­intercept of zero? Maximum y­value= 40.48 Calculated y­intercept= ­0.78 5% of 6.36 = 0.05 x 40.48 = 0.318 2.024 > ­0.78 A y­intercept of zero is within the level of uncertainty and was chosen as the final Y­intercept. Now we check the %error General equation: Velocity^2 = Force x Radius / Mass Velocity^2 = (Force / Mass) x Radius / Radius Velocity^2 / Radius = (0.98N / 0.012kg) (Slope from our mathematical model) = 81.67 m/s^2 83.66m/s^2 =/= 81.67m/s^2 Theoretical velocity value =Force x Radius / Mass = 81.67m/s^2 Experimental velocity value = 83.66m/s^2 %Error = abs([Theoretical Value ­ Experimental value] / Theoretical value) x100 = ([81.67m/s^2 ­ 83.66m/s^2] / 81.67) x100 = 2.4% = Percent Error ______________________________________________________________________________ For Weight v Velocity (UNLINEARIZED)

Force of Tension (Newtons) Calculated by using the equation Sum of Forces = Mass x Acceleration F(T) + F(G) = Mass x 0 F(T) + (Mass [kg]) x (­9.8m/s^2) = 0 F(T) = ­(kg x ­9.8m/s^2)

Velocity (meters / second) Calculated by using the linear speed equation with values for time and radius input from Weight v Velocity collected data (Rotations x 2Pi x Radius) / Time

­1 x ­9.8m/s^2 x 0.20kg = 1.96N (10 x 2Pi x 0.50m) / 3.25s = 9.67m/s

­1 x ­9.8m/s^2 x 0.18kg = 1.77N (10 x 2Pi x 0.50m) / 3.30s = 9.52m/s

­1 x ­9.8m/s^2 x 0.16kg = 1.57N (10 x 2Pi x 0.50m) / 3.52s = 8.93m/s

­1 x ­9.8m/s^2 x 0.14kg = 1.37N (10 x 2Pi x 0.50m) / 3.61s = 8.70m/s

­1 x ­9.8m/s^2 x 0.12kg = 1.18N (10 x 2Pi x 0.50m) / 3.74s = 8.40m/s

­1 x ­9.8m/s^2 x 0.10kg = 0.98N (10 x 2Pi x 0.50m) / 3.88s = 8.20m/s

­1 x ­9.8m/s^2 x 0.08kg = 0.79N (10 x 2Pi x 0.50m) / 3.97s = 7.91m/s

­1 x ­9.8m/s^2 x 0.06kg = 0.59 (10 x 2Pi x 0.50m) / 4.10s = 7.67m/s

­1 x ­9.8m/s^2 x 0.04kg = 0.39 (10 x 2Pi x 0.50m) / 4.19s = 7.50m/s

­1 x ­9.8m/s^2 x 0.02kg = 0.20N (10 x 2Pi x 0.50m) / 4.23s = 7.43m/s

For Weight v Velocity (LINEARIZED)

Force of Tension (Newtons) From the above data. It’s just moved down (again) to help visualize what will be translated over to the graphs

Velocity^2 (meters^2 / second^2) Calculated by using the above data and squaring it Velocity ^2

1.96N (9.67m/s) ^2 = 93.44m/s

1.77N (9.52m/s) ^2 = 90.63m/s

1.57N (8.93m/s) ^2 = 79.66m/s

1.37N (8.70m/s) ^2 = 75.74m/s

1.18N (8.40m/s) ^2 = 70.56m/s

0.98N (8.20m/s) ^2 = 65.56m/s

0.79N (7.91m/s) ^2 = 62.62m/s

0.59 (7.67m/s) ^2 = 58.71m/s

0.39 (7.50m/s) ^2 = 56.22m/s

0.20N (7.43m/s) ^2 = 55.16m/s

Mathematical Model (underlined) ­ calculated with a graphing calculator’s linear regression calculating application with above data table data. y=ax+b a=22.72 b=46.30

Velocity^2 = 22.72m/kg x Force of Tension + 46.3m^2/s^2 To actually calculate the values for this mathematical model, the data for time and position collected were input into a graphing calculator to which a linear regression calculation was applied and yielded a linear regression of y=22.72x+46.3 . What this means is that the velocity (y) is a function of 22.72 multiplied by the radius (x) with a y­intercept of 46.3 added. The y­intercept was more than 5% of the maximum y­value ­ as shown below. Maximum y­value= 93.44 Calculated y­intercept= 46.3 5% of 93.44 = 0.05 x 93.44 = 4.67 4.67 < 93.44 A y­intercept of zero is not within the level of uncertainty so this suggests that this Y­intercept is not due to human error; however, the data may be called into question when examined with the general equation. TheY­intercept represents the Velocity where the Force is 0. Realistically, this is not possible, as if the force is 0, there would be not enough force to balance the weight of the contraption it would fall over and and therefore have no velocity. When compared to the general equation is evident that there is error. General equation: Velocity^2 = Force x Radius / Mass Velocity^2 / Force = (Radius / Mass) x Force / Force (Slope from mathematical equation) = (0.5m / 0.012kg) 22.72m/kg =/= 41.67m/kg Theoretical velocity value =Force x Radius / Mass = 41.67m/kg Experimental velocity value = 22.72m/kg %Error = abs([Theoretical Value ­ Experimental value] / Theoretical value) x100 = ([41.67m/kg ­ 22.71m/kg] / 41.67) x100 = 45% = Percent Error for Force vs Velocity Conclusion: For Radius vs Velocity With a changing Radius, the relationship of radius vs velocity is a linear relation. As the radius increases, velocity increases proportionally. For the relation of radius and velocity, the hypothesis was supported in that the relation was linearly proportional. Final Mathematical Model Velocity^2 = 83.66kg/s x Radius To actually calculate the values for this mathematical model, the data for time and position

collected were input into a graphing calculator to which a linear regression calculation was applied and yielded a linear regression of y=83.66x+­0.78 . What this means is that the velocity (y) is a function of 83.66 multiplied by the radius (x) with a y­intercept of ­0.78 added. The y­intercept was less than 5% of the maximum y­value so A y­intercept of zero is within the level of uncertainty and was chosen as the final Y­intercept. The Percent Error for Radius vs Velocity is 2.4% The results are scientifically reasonable whereas the results yield less than 15% error. The experimenters believe the source of any error in this trial that one of the experimenters may not have counted to 10 and wasn’t consistent for the number of rotations that they spun the stopper in and that affected the velocity calculations by having one or two rotations more than there actually was and also the time will have been less than it actually was. For the most part this part of the experiment went well. If this experiment were to be carried out again, the experimenters recommend that when conducting this experiment, experimenters should make sure to be consistent with the number of rotations they count, knowing to start when they see a point they would like to start counting from, and then counting one each time it reaches that point after starting, and stopping the watch immediately after the tenth because it causes error. ________________________ _____________________________ For Force vs Velocity With a changing Force, the relationship of force vs velocity is a linear relation. As the force increases, velocity increases proportionally. For the relation of force and velocity, the hypothesis was supported in that the relation was linearly proportional. Mathematical Model Velocity^2 = 22.72m/kg x Force of Tension + 46.3m^2/s^2 To actually calculate the values for this mathematical model, the data for time and position collected were input into a graphing calculator to which a linear regression calculation was applied and yielded a linear regression of y=1.34x+6.94 . What this means is that the velocity (y) is a function of 22.72 multiplied by the radius (x) with a y­intercept of 46.3 added. The y­intercept was more than 5% of the maximum y­value ­ as shown below. A y­intercept of zero is not within the level of uncertainty so this suggests that this Y­intercept is not due to human error; however, the data may be called into question when examined with the general equation. TheY­intercept represents the Velocity where the Force is 0.

Realistically, this is not possible, as if the force is 0, there would be not enough force to balance the weight of the contraption it would fall over and and therefore have no velocity. The experimenters don’t know why this is the Y intercept ­ as their teacher confirmed that ideally the data should have sloped up from zero. These results for Force vs Velocity are not scientifically reasonable whereas the results yield more than 15% error. The experimenters believe the definite source of error is that one of the experimenters may not have counted to 10 and wasn’t consistent for the number of rotations that they spun the stopper in and that affected the velocity calculations by having one or two rotations more than there actually was and also the time will have been less than it actually was. The experimenters are certain this high error is not due to calculation error because the data was collected and calculated formally and flawlessly at this point after multiple revisions. If this experiment were to be carried out again, the experimenters recommend that when conducting this experiment, experimenters should make sure to be consistent with the number of rotations they count, knowing to start when they see a point they would like to start counting from, and then counting one each time it reaches that point after starting, and stopping the watch immediately after the tenth because it causes error.