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course PHY 241
12/21/10, 11:00
Brief rotating strap and magnets#$&*
course PHY 241
12/12/10; 16:00
Brief rotating strap and magnetsUses metal strap, magnets, threaded rod with bolts and washers, die
See also Pictures related to Straps and Toy Cars
Setup A: Using the TIMER, time the strap as it rotates to rest about the threaded rod, clicking with each 180 degree rotation. Click also when the strap comes to rest, and estimate how many degrees it has rotated since its last full 180 degree rotation.
Setup B: Repeat, but this time, with a ceramic magnet on each side of the axis of rotation, halfway between the axis and the end of the strap.
Setup C: Repeat once more but with the magnets positioned at the ends of the strap.
Setup D: Same as Setup A, except that the strap is taken off the threaded rod and placed on the die.
Setup E: Same as Setup B, except that the strap is taken off the threaded rod and placed on the die.
Setup F: Same as Setup C, except that the strap is taken off the threaded rod and placed on the die.
Report your data for Setup A (simply copy and paste from the TIMER output).
****
Text1
1 0 0
2 1.171875 1.171875
3 3.445313 2.273438
Last strike is at a static position.
120+360 = 480 degrees
Second trial same setup:
Text1
1 0 0
2 .4296875 .4296875
3 .921875 .4921875
4 1.378906 .4570313
5 1.808594 .4296875
6 2.308594 .5
7 2.847656 .5390625
8 3.507813 .6601563
9 4.136719 .6289063
10 5.132813 .9960938
11 6.542969 1.410156
12 7.371094 .828125
5 rotations + 15 degrees = 1810 degrees
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Report your data for Setup B.
****
Text1
1 0 0
2 .46875 .46875
3 1.503906 1.035156
4 1.640625 .1367188
5 2.082031 .4414063
6 2.804688 .7226563
7 3.59375 .7890625
8 4.707031 1.113281
9 5.886719 1.179688
10 8.25 2.363281
4 rotations + 110 degrees = 1560 degrees
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Report your data for Setup C.
****
Text1
1 0 0
2 .59375 .59375
3 1.121094 .5273438
4 1.851563 .7304688
5 2.289063 .4375
6 3.03125 .7421875
7 3.597656 .5664063
8 4.496094 .8984375
9 5.125 .6289063
10 6.074219 .9492188
11 7.054688 .9804688
12 8.273438 1.21875
13 9.320313 1.046875
14 10.88672 1.566406
15 13.03516 2.148438
7 rotations + 95 degrees = 2615 degrees
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Report your data for Setup D.
****
1 0 0
2 1.339844 1.339844
3 2.976563 1.636719
4 5.160156 2.183594
5 7.96875 2.808594
6 10.23438 2.265625
2.5 rotatations + 30 degrees = 930 degrees
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Report your data for Setup E.
****
13 454.5 94.78906
14 455.2773 .7773438
15 456.1445 .8671875
16 457.1523 1.007813
17 458.3438 1.191406
18 459.8906 1.546875
19 462.457 2.566406
3 rotations+110 degrees= 1190 degrees
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Report your data for Setup F.
****
Text1
1 0 0
2 1.269531 1.269531
3 2.535156 1.265625
4 4.171875 1.636719
5 5.839844 1.667969
6 9.363281 3.523438
7 12.37109 3.007813
60 degrees + 3 rotations = 1140 degrees
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Analysis:
Now suppose your data for set A was
0.592
0.775
0.923
1.762
This implies four time interval, the first running from clock time 0 to clock time 0.592 sec, the second from 0.592 sec to
0.592 sec + .775 sec = 1.57 sec, the third from 1.57 sec to 1.57 sec + 0.923 sec = 2.50 sec, and the fourth from 2.50 sec to
2.50 sec + 1.762 sec = 4.26 sec.
The midpoint clock time of your second interval is halfway between 0.592 sec and 1.57 sec, at about 1.08 sec.
You could easily calculate the midpoints of all four time intervals, and should do so.
Report your midpoints below, based on your own data rather than the sample data used in the explanation:
****
A Trial 1 midpoints in time: 0.585938s, 1.722657s,
I found the midpoints my taking the last total time elapsed and adding the new total time elapsed and the divided that sum by two.
Example: t0 = 3.50781 s and the next time I hit the button for the timer = 4.13672 s summing these times and dividing by two give the mid-point of the two times= (3.50781 s +4.13672 s)/2 = 3.82266 s
A Trial 2 midpoints in time: in seconds
0.214844
0.675781
1.150391
1.59375
2.058594
2.578125
3.177735
3.822266
4.634766
5.837891
6.957032
B Trial mid points:
0.234375
0.986328
1.572266
1.861328
2.44336
3.199219
4.150391
5.296875
7.06836
C Trial mid points:
0.296875
0.857422
1.486329
2.070313
2.660157
3.314453
4.046875
4.810547
5.59961
6.564454
7.664063
8.796876
10.10352
11.96094
D Trial mid points:
0.669922
2.158204
4.06836
6.564453
9.101565
6.916018
E Trial mid-points:
454.88865
455.7109
456.6484
457.74805
459.1172
461.1738
F Trial mid-points:
0.6347655
1.9023435
3.3535155
5.0058595
7.6015625
10.8671855
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Each of your the time intervals in the sample data
0.592
0.775
0.923
1.762
corresponds to a 180 deg rotation. So for each interval you can easily find the average rate of change of angular position with respect to clock time. Find the average rates and report them below. Be sure to include a detailed calculation, with explanation, for one of your intervals.
****
Each Interval of time was recorded the intervals correspond to a certain amount of rotation [I limited myself to five mid points => at most six angular velocities being the last six points of time.] it is easy enough to convert these velocities to rad / s simply by multiplying ea. val. By pi/180. The numbers following were found by taking tf-t0 to find `dt and then using 180 degrees as the numerator and `dt and the denominator: dtheta/dt [in degrees]
Example from the last calculation of trail F:dt = 3.008 s, d(theta) = 60 degrees
60/3 = 20 degrees /s or pi * .1111111 rad/s
A 2:
333.9134 286.2113 180.7058 127.6455 18.11321
B:
249.0808 228.119 161.6842 152.5827 76.1653 26.69197
C:
189.6296 183.5856 147.6923 171.9403 114.9127 44.21813
D:
134.344 109.9761 82.43294 64.08901 13.24135
E:
207.5646 178.6069 151.0701 116.3693 42.8616
F:
141.7846 142.2222 109.9761 107.9157 51.08648 19.94808
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The moment of inertia of the strap itself is 1/12 M L^2, where M is its mass and L is its length. Assume the strap mass to be 70 grams.
The moment of inertia of a magnet attached to the strap is approximately M R^2, where M is the 50 gram mass of the magnet and R its distance from the axis of rotation (the axis of rotation is the threaded rod).
The total moment of inertia of the system is the sum of the moments of inertia of its components.
Find the moment of inertia for system A and report below:
**** I will work this one out and simply give the ans to each of the rest.
I = 1/12 * M * L^2
I believe the strap was about 24 cm [I did not record that] long.
M = .070 kg => I = about 0.000336 m * kg
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Report below a table of average rate of change of angular position vs. midpoint clock time, for the four intervals you observed for Setup A. Report also the moment of inertia:
Setup A
****
In degrees
333.9134 286.2113 180.7058 127.6455 18.11321
In radians:
5.827888 4.99533 3.153911 2.227834 0.316135
Times in seconds are vertical:
2.578125
3.822266
4.634766
5.837891
6.957032
I will work this one out and simply give the ans to each of the rest.
@& A table of one quantity vs. another shows the results in two columns, making it easy to see how one number changes with respect to the other.
I can see the trends in the information you gave, but most readers would have trouble viewing one of the quantities reported as a row, the other as a column.*@
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I = 1/12 * M * L^2
I believe the strap was about 24 cm [I did not record that] long.
M = .070 kg => I = about 0.000336 m * kg
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You are going to make the same report for the remaining systems. You can save yourself some time by using your calculator or Excel to do the necessary calculations.
Find the moment of inertia for the system in Trial B, and show your calculation in detail:
**** It = I rod + I_m1 + I_m2 = 0.000336 m * kg + 2*.003 m* kg = 0.006336 m * kg
I_m1 = .060 m * .050 kg = I_m2
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For each system report average rate of change of angular position vs. midpoint clock time, and also report the moment of inertia:
Setup B
****
249.0808 228.119 161.6842 152.5827 76.1653 26.69197
2.44336
3.550782
4.150391
5.296875
7.06836
0.006336 m * kg
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Setup C
****
189.6296 183.5856 147.6923 171.9403 114.9127 44.21813
5.59961
6.564454
7.664063
8.796876
10.10352
I = 0.095336 m * kg
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Setup D
****
134.344 109.9761 82.43294 64.08901 13.24135
Bc the die could still be considered a axis of rotation about the center => the moments of inertia are the same as above: 0.000336 m * kg
0.669922
2.158204
4.06836
6.564453
9.101565
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Setup E
****
207.5646 178.6069 151.0701 116.3693 42.8616
Linear
455.7109
456.6484
457.7481
459.1172
461.1738
I = 0.060336 m * kg
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Setup F
****
141.7846 142.2222 109.9761 107.9157 51.08648 19.94808
0.6347655
1.9023435
3.3535155
5.0058595
7.6015625
10.8671855
I = 0.095336 m * kg
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Note that average rate of change of angular position is also called angular velocity. So you have calculated a number of angular velocities.
Sketch a graph of angular velocity vs. midpoint clock time for each setup. Describe how well each graph can be 'fit' by a single straight line. In a good 'fit' the points will appear to be randomly scattered about the line.
**** I am off by 5 percent on some points but all in all it was pretty good the graph appears near some power function close to 1. Some of my data specifically trail D is perfectly linear.
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On the whole, does it seem plausible that for these systems, the angular velocity tends to decrease linearly with time?
**** yes, but the data is somewhat inconsistent from trial to trial.
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Your data look fine. I've appended instructions for analysis.
&#Good responses. Let me know if you have questions. &#
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"
@& Well done, but see my note on how a table should be reported.*@