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course PHY 241
10/23; 20:15
This is an at-home lab assignment. It isn't too extensive; I'm figuring an hour or two. Be as accurate as you can within that time frame, but don't let yourself get bogged down.For quick reference the TIMER can be located in the document at http://vhcc2.vhcc.edu/dsmith/forms/ph1_timer_experiment.htm . Both the .exe and Java versions are provided.
Let me know if you have questions:
Acceleration vs. Ramp Slope
Using the TIMER program, time a ball down the steel ramp when the ramp is supported by a single domino lying flat. Do five trials with this setup.
Report the median time, the distance the ball traveled from rest in this time, and the resulting acceleration.
****1.7 cm/s, 20.5 cm/s^2
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Report the rise and run between two points of the ramp and the resulting slope.
****rise = 1 cm, run = 30 cm=> 1/30 = 0.033
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Repeat with the domino lying on its long edge, so that the rise is equal to the width of the domino. Do five trials with this setup.
Report the median time, the distance the ball traveled from rest in this time, and the resulting acceleration.
****1sec, 58cm/s^2
After averaging the time I divided `dx/`dt then *2 that result giving the change in speed, after which I divided that by `dv/`dt to get a_ave.
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Report the rise and run between two points of the ramp and the resulting slope.
****2.5 cm = h; 30cm = run=> .083 for a slope
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Based on these two setups, at what rate does the acceleration of the ball appear to change with respect to ramp slope?
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2.5 * the initial slope = the second slope; 2.5 * 20cm/s^2 = the acceleration of the second slope roughly
You will need to apply the definition of rate of change to arrive at the right calculation for this question.
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Now time the toy car down the wood ramp, using two different slopes. Be sure the ramp is straight. Suggestion: Use your textbook to help. Support it at one end with something reasonably rigid, whose thickness you can measure with good accuracy (for example a couple of CD or DVD cases would be a good choice, using one for the first setup, and both for the second). Using one hand hold the wood piece flat against the book, release the car with another hand, and operate the TIMER with your third hand. If you don't have three hands, adapt the suggestions accordingly. You might also find it helpful to use the steel ramp to press the wood ramp against the book.
For the first slope:
Report the median time, the distance the ball traveled from rest in this time, and the resulting acceleration.
****2 sec, 10.4cm/s^2
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Report the rise and run between two points of the ramp and the resulting slope.
****
h=1cm, run = 28cm => a slope of .036cm_ri/cm_run
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For the second slope:
Report the median time, the distance the ball traveled from rest in this time, and the resulting acceleration.
****.87sec, 55.6cm/s^2 --I used the same method as the above stated system
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Report the rise and run between two points of the ramp and the resulting slope.
****h=2.25, run = 28 cm; slope = .08 slope
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Based on these two setups, at what rate does the acceleration of the ball appear to change with respect to ramp slope?
It is a linear relation that it a direct relationship. As above with the ramp when the angle of the slope is increase the acceleration is directly proportional to the directional slope of the object.
Again, apply the definition of rate of change.
****
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Is acceleration independent of position and velocity?
You were asked previously to design an experiment to test whether acceleration is independent of position and velocity, on a ramp with constant incline.
Do a 30-minute preliminary run, using the TIMER. Just take whatever data you can in 15 or 20 minutes, and give a brief report of your setup, your data and your results. Try to be as accurate as possible within the 15-20 minute time constraint.
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Acceleration is independent of velocity and position. But it is determined by `dv*vAve/`ds
I propose examining when the velocity is zero as a basic standard. Then when the displacement `ds is zero and when the acceleration is zero. Then we will exam a change in each velocity, `ds, and acceleration. We will change the vAve later time will not allow.
Bc when the change in velocity is 0 => `dv = 0; `dv/`dt = 0 for any time period. When the displacement is zero => `ds/`dt = 0 = > `dv/`dt = 0. So we see that the `dv and `ds can determine the acceleration. We know this to be true Bc: a * `ds = v * `dv => a = v *`dv / `ds. That is what we want to prove—we want to know whether acceleration is independent or dependent on velocity and position.
So if we start at diff point on the ramp with the ball we will see whether a change in `ds gives a proportional acceleration. The easiest way to prove this is to try to prove it wrong. With formulas it seems simpler as seen above.
After a retest of question 1 I found the `dt = .96sec, `ds=27.6 cm => v_ave = 28.7 and a = 50cm/s^2
Now we will half the `ds and then double it: hypothesis the `ds will double, the `dv will double but the acceleration will remain near constant.
For half the distance we get vAve = 19.2cm/s, `ds=12.5 cm, `dt=.65 sec => a = 59 cm/s^2
Double: `dtAve=1.55 s; `ds = 57.5 cm; vAve= 37cm/s; aAve=47.6cm/s^2 = about the same acceleration as the above which is consistent with the lab results above.
I carried out my test by acquiring two ramps and giving the whole thing an overall consistent acceleration via a constant slope. It is obvious that the double `ds will have a lower acceleration due to the dip between ramps. Off course the most inaccurate test is the shortest for it has the greatest percent error.
So we find that with a constant slope the acceleration does not vary according to only velocity or only displacement but by both in which case acceleration is dependent on both ave velocity, `dv, and displacement combined in a unique format. Namely vAve*`dv/`ds=a_ave
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you're going to have very good results once you apply the definition of rate to the two main questions
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