Experiment 1 - Lab Report PDF

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Experiment 1: Measurements and Motion  Part 1: Distance Running  1. Based on the data shown in Table 1, can you tell which student is a stronger runner?  Based on the fact that Student 1 completed each of the six laps in a shorter period of time than Student 2, Student 1 would be considered to be the stronger runner. However, based on the fact that the coach insists on the students running at a steady pace, Student 2 would be considered a stronger runner.  2. The times in Table 1 are the accumulated lap times for every lap that each student runs. Calculate the average time per lap for each student.  Student 1 tAVG = tTOTAL  / Number of Laps tAVG = 333 seconds / 6 laps tAVG = 55.5 seconds / lap

Student 2 tAVG = tTOTAL  / Number of Laps tAVG = 393 seconds / 6 laps tAVG = 65.5 seconds / lap

 3. What is the sample standard deviation ( ) as well as the standard error (SE, ie. the uncertainty) on the average time per lap for each student? You can calculate by hand or use the appropriate functions in MS Excel. You may leave 3 decimal points in your answers.  Student 1 s = 3.271 SE = 1.335

Student 2: s = 3.332 SE = 1.360

 4. Create a new table that contains the 6 individual lap times, the average time per lap, and the standard error on the average time per lap for each student.  Table 1: Lap times, average time per lap, and standard error for two students running on an indoor track.  

Time (Seconds) Lap 1 Lap 2 Lap 3 Lap 4 Lap 5 Lap 6



Average Time Per Lap

Standard Error

Student 1

55

57

50

56

55

60

55.5

1.335

Student 2

64

64

64

68

71

62

65.5

1.360

 5. Calculate the average speed for each student as they run the 6 laps.  Let v be the speed and vAVG be the average speed. Student 1 v = Distance (m) / Time (s) v1 = 400 m / 55 s = 7.27 m/s v2 = 7.02 m/s v3 = 8 m/s v4 = 7.14 m/s v5 = 7.27 m/s v6 = 6.67 m/s vAVG = 7.228 m/s

Student 2 v = Distance (m) / Time (s) v1 = 400 m / 64 s = 6.25 m/s v2 = 6.25 m/s v3 = 6.25 m/s v4 = 5.88 m/s v5 = 5.63 m/s v6 = 6.45 m/s vAVG = 6.118 m/s

 6. Using the data from Table 1, plot a graph that shows the cumulative times vs cumulative distances that both students run. To clarify, your first data point for Student 1 would be (400 m, 55 s) and your second point would be (800 m, 112 s), etc… You should show both student’s times on the same plot. Use the linear regression tool to fit both data sets and be sure to show the uncertainties of your linear fit data. You should also show your data table in your graph file by using the “print” function in Logger Pro (and not “print graph”). Be sure to give your graph an appropriate title that describes the data shown.

  7. From the slope of a linear regression of your Graph 1, calculate the speed of each student as they run the 6 laps. 

Student 1 m (slope) = 0.1378 +/- 0.001612 m/s Speed = 1/m Speed = 7.257 m/s

Student 2 m (slope) = 0.1538 +/- 0.006416 m/s Speed = 1/m Speed = 6.502 m/s

 8. Compare the speeds obtained from your graph with the average speeds you found in Calc. 1c using a percent difference calculation. Do you feel the values of speed agree with each other?  Student 1 1 − V alue 2 % Difference =| 1/2V(Value | x 100%  alue 1 + V alue 2) % Difference = 0.4004%

Student 2 1 − V alue 2 % Difference =| 1/2V(Value | x 100%  alue 1 + V alue 2) % Difference = 6.086%

 For student 1, the values of speed agree with each other as the percent difference is less than 1 %. For student 2, the values of speed do not agree with each other as the percent difference is larger than 1 %.  Part 2: Projectile Motion  1. Find the average values of the range for each initial angle, .  Initial Angle (𝜃) Average Range (m) 25

1.581

35

1.942

45

2.065

50

2.028

60

1.787

70

1.324

 2. Using the data from Table 2, plot a graph of the average values of the maximum range vs. sin(2 ) for each initial angle. Use the linear regression tool to fit your data and be sure to show the uncertainty of your linear fit data. You should also show your data table and give your graph an appropriate title.

 3. From the slope of the linear regression data of your Graph 2, calculate the experimental value of the acceleration due to gravity, exp.  2 gexp =  9.306 m/s   4. Compare the experimental value of from Calculation 2b with an accepted value of 9.81 m/s2 using a percent error calculation. Would you say that the experimental value agrees with the accepted value?  Accepted − Experimental % error = | | x 100%  Accepted % error = |(9.81 - 9.306)/9.81| x 100% % error = 5.138%  I would say that the experimental value does not agree with the experimental value as the percent error is larger than 1%.  5. For an initial launch angle of = 60.0° and the initial speed given in the experimental details, calculate: a. The theoretical time it would take for the projectile to reach its maximum height. th = v0 sinθ / g th = (4.50 m/s)(sin60°) / (9.81 m/s2) th = 0.397 s b. The theoretical maximum height above the launch point. h = v20 sin2θ / 2g h = (4.50 m/s)2 (sin60°)2 / 2(9.81 m/s2 ) h = 0.774 m

c. The theoretical maximum range that the projectile would reach. R = v20 sin2θ / g R = (4.50 m/s)2 (2sin60°) / (9.81 m/s2 ) R = 3.575 m  Part C: Ferris Wheels  1. For each wheel, calculate its radius in meters. You can leave 3 decimal places in your final answer and you should use this precision for all other calculations involving the wheel’s radius.  Wheel 1 Diameter = 36.0 ft Radius = (18.0 ft)(1 m/3.28084 ft) Radius = 5.486 m

Wheel 2 Diameter = 28.0 ft Radius = (14.0 ft)(1 m/3.28084 ft) Radius = 4.267 m

 2. Calculate the average period of rotation (and its uncertainty) for each wheel.  Wheel 1 T = (16.0+16.5+16.0+17.0+17.0)s/5 T = 16.5 s

Wheel 2 T = (13.0+14.0+14.0+12.5+15.0)s/5 T = 13.7 s

 3. Calculate the average speed and centripetal acceleration of a rider on each wheel using the average period of rotation.  Wheel 1 v = 2πr/T v = (2π)(5.486 m)/(16.5 s) v = 2.089 m/s ac = 4π2r/T2 ac = (4π2 )(5.486 m)/(16.5 s)2  ac = 0.796 m/s2 

Wheel 2 v = 2πr/T v = (2π)(4.267 m)/(13.7 s) v = 1.957 m/s ac = (4π2)r/T2  ac = (4π2 )(4.267 m)/(13.7 s)2 ac = 0.898 m/s2 

 4. Which wheel operator was correct when they stated that their wheel provided a bigger thrill for their riders?  The second wheel operator was correct when they stated that their wheel provided a bigger thrill for their riders as the acceleration is higher than that on the first operator’s wheel.   ...