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course Phy 231
3/5 1
Lab exercise
Using a toy truck or car and a ramp raised at one end, on a slope such that the vehicle just barely accelerates down the incline, determine by timing the vehicle as it
coasts down the ramp for a measured distance, starting from rest.
Use good timing technique and use at least three timings.
Measure the slope of the incline by measuring its rise and run.
Increase the incline slightly by adding a large washer under the raised end and repeat.
Add another washer to increase the incline and repeat again. Then add a final washer and repeat once more.
The things you have actually measured, the rises, runs and times, are your data. What we find from the data are not data and should not be reported as such.
Use your data to determine to the slope for each incline and the corresponding acceleration.
Sketch of graph of the acceleration versus the slope.
From your graph conjecture what acceleration would correspond to a slope of 1, assuming that the graph is linear (i.e., that it forms a straight line and that your
graph is not perfectly straight because of uncertainties in timing and other measurements).
Estimate the uncertainty in your various measurements, and estimate the effect of these uncertainties on your results.
As we will see later, this acceleration corresponds to the acceleration of an object falling freely under the influence of gravity.
Conduct a modified experiment by finding slopes such that the acceleration of the cart is approximately .2 m/s/s, then .4 m/s/s, then .6 m/s/s (you might extrapolate
your existing acceleration vs. slope data).
Again obtain a graph of acceleration vs. slope.
From your graph conjecture the acceleration that would correspond to slope 1, assuming that the graph is linear.
Is this modified strategy more or less accurate than the original strategy of using equal slope increments?
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1.35, 1.55, 1.75 (cm)
The rise in cm for my ramp of the three trials accurate to +-0.05 cm. The order is trial 1, Trial 2, Trial 3
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The +- .05 cm uncertainty is appropriate to these measurements.
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25.05, 25.15, 25.25 (cm)
The run in cm for my ramp of the three trials accurate to +-0,05 cm. The order is trial 1, Trial 2, Trial 3
0.0540, 0.0616, 0.0693,
Calculated slopes from the three trials to three significant figures: m = rise / run. The order is Trial 1, Trial 2, Trial3
2.885, 2.925, 2.655 (s)
Measured times 'dt in seconds of Trial 1, accurate to +-0.005 seconds.
2.822 (s)
average of the three times 'dt in trial 1 accurate to 0.005 seconds.
1.775, 1.985, 1.815 (s)
Measured times 'dt in seconds of Trial 2, accurate to +-0.005 seconds.
1.858 (s)
average of the three times 'dt in trial 2 accurate to 0.005 seconds,
1.715, 1.615, 1.665 (s)
Measured times in seconds 'dt of Trial 3, accurate to +-0.005 seconds.
1.665 (s)
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I don't believe that you can trigger the clock to a consistency of +-.005 milliseconds. The release of the ball and its behavior on the ramp also introduce some uncertainty. The variation in the times observed for the same ball on the same ramp imply an uncertainty closer to 0.1 second.
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average of the three times in trial 3 accurate to 0.005 seconds, giving a percent error = ( (1.665 - 1.670) / 1.665) *100% = 0.30%
25.1, 25.2, 25.3 (cm)
calculated slopes 'ds of the three trials using the Pythagorean theorem = ((rise^2 + run^2))^ (1/2). The order is Trial 1, Trial 2, Trial3
25.2 cm = average of three calculated slopes = 'ds
8.93, 13.6, 15.1 (cm/s)
Calculated average velocities of the three Trials in cm/s using calculation averages = 'ds / 'dt. The order is Trial1, Trial 2, Trial 3
17.86, 27.2, 30.2 (cm/s)
Calculated final velocities of the three trials in cm/s calculated by vf = vAve * 2. The order is Trial 1, Trial 2, Trial 3
6.33, 14.6, 18.1 (cm/s^2)
Calculated accelerations of three trials in cm/s^2 calculated by a = 'dv /'dt. The order is Trial 1, Trial 2, Trial 3
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From your graph conjecture what acceleration would correspond to a slope of 1, assuming that the graph is linear (i.e., that it forms a straight line and that your graph is not perfectly straight because of uncertainties in timing and other measurements).
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average of 3 accelerations = sum of three accelerations / 3 = 13.01 cm / s^2
average of three slopes = sum of three slopes / 3 = 0.06163
using the ratio of accel/slope I solve for accel below
(13.01 cm/s^2 / 0.06163) = Accel / 1
Accel = 211.1 cm/s^2
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You have apparently divided the average of your accelerations by the average of your slopes.
An average rate of change is the change in one quantity divided by the change in another, not the avearage value of one quantity divided by the change in the other.
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Conduct a modified experiment by finding slopes such that the acceleration of the cart is approximately .2 m/s/s, then .4 m/s/s, then .6 m/s/s (you might extrapolate your existing acceleration vs. slope data).
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Using my average ratio from my best line data graph, i can use my ratio to solve for the different accelerations
First i convert m/s^2 to cm/ s^2
.2 m/s^2 = 20 cm / s^2, .4 m/s^2 = 40 cm/s^2, .6 m/s^2 = 60 cm/s^2
Second I set them equal to my best line ratio of slope / accel
(0.06163 / 13.01 cm/s^2) = (slope / 20 cm/s^2), slope = 0.09474
(0.06163 / 13.01 cm/s^2) = (slope2 / 40 cm/s^2), slope2 = 0.1895
(0.06163 / 13.01 cm/s^2+ = (slope3 / 60 cm/s^2), slope3 = 0.2842
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The average of the three accelerations, and the average of the three slopes, have no important physical meaning and no bearing on the results of this experiment.
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Estimate the uncertainty in your various measurements, and estimate the effect of these uncertainties on your results.
Uncertainty in rise and run measurements was to 0.05 cm meaning that if i measured 1.1 cm, i would be 1.10 cm +- 0.05 cm of accuracy
Percent error of rise and run ((uncertainty) / actual) * 100% = (1.15 - 1.10 / 1.10) * 100% = 4.55%
Percent error of time for 1.00 sec = (uncertainty / actual) * 100% = (1.005 - 1.000 / 1.000) * 100% = 0.5%
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This is where i am unsure about percent uncertainties when there are different uncertainties and how they relate to each other, I use slope and velocity as my examples. Is this the proper method of determining final percent uncertainties for my given data?
slope = rise / run
Would the percent error in slope be +-1%
percent uncertainty = percent error rise / percent error run = 4.55% / 4.55% = 1.00%
and percent error in velocity = 'dv / 'dt = ('ds /'dt) / 'dt = (4.55 % / 0.5 %) / (0.5 %)
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It appears that you have at least a 3% uncertainty in the measurements you used to find the slope, and at least a 5% uncertainty in the measurement of the time intervals.
The percent uncertainty in measuring distances would be pretty much negligible, since the distances are much greater than the uncertainty in the measurements.
The percent uncertanty in velocity cannot be less than the largest uncertainty in the measurements used to calculate it.
The percent uncertainty in `dv / `dt is not equal to the quotient of the two uncertainties. The uncertainty in a quotient is not the quotient of the uncertainties, which could easily violate this fact.
The rules for handling uncertainties follow from the behavior of differentials, a topic you would have covered in calculus classes.
You should review the discussion of significant figures and uncertainties, which I believe is in Section 1-7 of your text.
The basic rule for propagation of errors, which applies to small uncertainties:
When you multiply or divide two quantities, the percent uncertainty of the result is the sum of the percent uncertainties in the uncertainties of the individual quantities.
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Good overall, but you do have some errors that need to be corrected.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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