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course Phy 241
10/24 8
Lab-related Questions for 101013Note: Before doing the lab questions you should run through the Sketching Exercise below. That exercise starts with questions about masses pulled upward by tension and downward by gravity, much along the lines discussed in class. It continues with questions related to masses on inclines.
In lab you timed the Atwood machine (paperclips on pulley) using your bracket pendulum.
`qx001. What was the length of your pendulum? What would be the period of a pendulum of this length, based on T = .2 sqrt(L)?
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T = .2sqrt(8in) = Approx .5657sec
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`qx002. Give the time from release to first, second, third and fourth 'strikes' of the pendulum.
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.5657sec, 1.1314sec, 1.6971sec, 2.2628sec
if you counted cycles of a freely hanging pendulum this is correct
for the bracket pendulum, as discussed in class, first strike occurs after 1/4 period, subsequent strikes at intervals of 1/2 period
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`qx003. In your first set of trials there were 3 large clips on each side.
In the first line give your counts for the first set of trials, separated by commas.
In the second line give the mean of your counts.
In the third line give the time interval in seconds which is equivalent to the mean of your counts.
In the fourth line give the acceleration corresponding to the time interval just reported.
Starting in the fifth line give an explanation of the results you gave in the third and fourth lines.
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11, 14, 12, 13, 14!!!!First trail had a mistake in system,thats reason for lower count!!!
Without 1st trail mean = 13.25counts, With 1st trail included mean = 12.8counts
!!!Assuming we are to use the data without 1st trail!!! 13.25counts = approx 3.7478sec
Assuming 50cm was traveled in approx 3.7478sec, acceleration = 7.1195cm/s^2
To find the time in sec all you do is mult count/2(in this case 13.25 is our count) by .5657sec(which is the suggested period of my pendulum). We divided 13.25 because our pendulum is only completing half the period because in our pendulum system we have the bead bouncing off metal base so bead never gets a chance to complete a total oscillation. Knowing `dt and `ds we find vAve which we use to find v0(because vf =0), when we know v0 we find `dv which when divided by `dt we are given acceleration.
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qx004. In your second set of trials there were still 3 large clips on each side, but there was a small clip on the side which ascended in the first set.
In the first line give your counts for this set of trials, separated by commas.
In the second line give the mean of your counts.
In the third line give the time interval in seconds which is equivalent to the mean of your counts.
In the fourth line give the acceleration corresponding to the time interval just reported.
You don't need to include an explanation, since the procedure is identical to that of the preceding questions, which you explained in answering that question. Just make sure your results make sense.
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5, 5, 4.5, 4
Mean = 4.625counts
4.625counts = approx 1.3082sec
Assuming 50cm was traveled in approx 1.3082sec, acceleration = 58.4321cm/s^2
???That acceleration seems a lot higher than the prior, maybe to much higher for just one paper clip. Did I calculate period and acceleration correctly for both???????
the results are reasonable and the difference is about what I would expect
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`qx005. In the third set of trials a second small clip was added to each side.
In the first line give your counts for this set of trials, separated by commas.
In the second line give the mean of your counts.
In the third line give the time interval in seconds which is equivalent to the mean of your counts.
In the fourth line give the acceleration corresponding to the time interval just reported.
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2.75, 3, 3, 3
Mean = 2.9375counts
2.9375counts = approx .8309sec
Assuming 50cm was traveled in approx .8309sec, acceleration = 144.8447cm/s^2
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qx006. If there was a fourth set of trials, report as before:
In the first line give your counts for this set of trials, separated by commas.
In the second line give the mean of your counts.
In the third line give the time interval in seconds which is equivalent to the mean of your counts.
In the fourth line give the acceleration corresponding to the time interval just reported.
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??????I think we only did the three trails????????
I believe that's right
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qx007. For the trial with the greatest acceleration, sketch a force diagram showing, to scale, the tension and gravitational forces acting on the clips on the descending side of the system.
Which vector was longer?
By what percent was it longer?
What is the net force on these clips as a percent of the gravitational force?
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Weight vector was longest. Clips were accelerating at approx 1.4484m/s^2 so weight vector would have to be approx 15% greater than Tension vector so that 15% of weight vector(9.8m/s^2, or approx 10m/s^2) = approx 1.4484m/s^2. Net force is going to be between (1/6) and (1/7) the size of the gravitational force .
good, but note: the tension vector would be 15% shorter than the weight vector, which is different than the weight vector being 15% longer than the tension vector
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qx008. For the trial with the greatest acceleration, sketch a force diagram showing, to scale, the tension and gravitational forces acting on the clips on the ascending side of the system.
Which vector was longer?
By what percent was it longer?
What is the net force on these clips as a percent of the gravitational force?
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Tension vector was longest. Clips were accelerating at approx -1.4484m/s^2 so Tension vector would have to be approx 15% greater than Weight vector so that 15% of Tension vector acceleration = approx -1.4484m/s^2. Net force is going to be between (1/6) and (1/7) the size of the gravitational force .
????Considering the fact we are only taking into account tension and gravitational forces acting on the clips then if one side is being accelerated at rate of 1.4484m/s^2 by grav force then other side as well is traveling at a rate of 1.4484m/s^2 due to force of tension vector?????????
the magnitudes of the accelerations are identical
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q009. At what average rate does the acceleration of the system change with respect to the number of small paperclips?
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Approx 68.5cm/s^2 per small paperclip was average rate of change of acceleration
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q010. How much acceleration do we tend to be gaining, per added paperclip?
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Approx 68.5cm/s^2
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q011. The unbalance in the gravitational forces with each new paperclip is of course significant. It is this unbalance that causes the differences in the system's acceleration.
The total mass of the system does increase slightly with each added small paperclip, but for the moment let's assume that the resulting change in the total mass of the system isn't significant.
What percent of the acceleration of gravity do we get from each added small clip?
How is this related to the mass of a single clip as a percent of the system's total mass?
What is your conclusion about the ratio of the mass of a large clip to the mass of a small clip?
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Accel = .685m/s^2 and .685m/s^2 is approx 7% of grav force(9.8m/s^2)
Weight force = m*grav force, so knowing the weight before paper clip is added which could be solved for because grav force is constant so mass is pretty much only var to solve for on the level we are on. Then when paper clip is added you could find new mass(knowing old masses in system) by using acceleration and finding how much weight vector had increased. So everything is tied in by acceleration due to weight vector = m*grav force.
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q012. This question can be challenging. Don't let yourself get too bogged down on it:
In the preceding you drew conclusions based on the assumption that the changes in the system's total mass due to adding up to a few small paperclips was insignificant. It is perfectly possible that uncertainties in measuring the time intervals were large enough to obscure the effect of the changes in the total mass.
However refine your answers to the preceding question to take account of the change in total system mass.
(One possible approach: assume that the requested ratio is r and symbolically solve for the acceleration a in terms of the number N of added small clips, sketch a graph showing the predicted shape of your a vs. N curve, and see what value of r best matches this graph with a graph of your observed a vs. N).
Sketching exercise
The figure below is a free body diagram for a mass m on one side of an Atwood machine. The sketch is considered to be to scale.