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course Phy 241
11 pm 9/19/12
120912Do the exercise Fitting a Straight Line.
Read through Chapter 2 of your text.
`gGeneral College Physics
Giancoli: Do Text Chapter 1 Problems 16-24
Openstax: Do Text Chapter 1 Problems 29-35
Submit query_04.
In the Introductory_Problem_Sets be sure you can solve any of the problems posed in these sections, using any of the techniques demonstrated in the given solutions.
`q001. The system with two weights suspended from a pulley is called an Atwood machine.
When one domino was suspended from each side the machine was observed to accelerated from rest through 30 cm in about 3 seconds. When one paper clip was added it accelerated through the same 30 cm in about 2 seconds. With another added paper clip the time interval was about 1.7 seconds.
Find the acceleration for each trial.
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30cm, 3 s. vAve = 30/3 = 10 m/s. vf in this case is 2* vAve and aAve = vf/dt. aAve= (2*10)/3 = 6.7 m/s^2.
30 cm, 2s. vAve = 30/2 = 15 m/s. aAve = (2*15)/2 = 15 m/s^2.
30cm, 1.7s. vAve = 30/1.7 = 21.4 m/s. aAve = (2*21.4)/1.7 = 25.2 m/s^2.
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Graph acceleration vs. number of paper clips and sketch the straight line that you think best fits the trend of your results.
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0 paperclips, 6.7 m/s^2.
1 pc, 15 m/s^2.
2 pc, 25.2 m/s^2
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According to your results, what is the average rate of change of acceleration with respect to the number of clips?
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This would be the slope of the previous graph.
Thus (25.2 - 6.7)/(2-0). = 9.3 m/s^2 per paperclip.
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If the trend is in fact linear, how many clips would it take to result in an acceleration of 980 cm/s^2?
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Equation of this line is y = 9.3x + 6.7.
y = acceleration. x = paperclips.
980 = 9.3 x + 6.7.
(980 - 6.7)/9.3 = x.
x= 104.6 or 105 paperclips.
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When two dominoes were suspended from each side, data similar to that obtained for the previous system indicated accelerations of 1, 7, 10 and 15 cm/s^2 for 1, 2, 3 and 4 added clips.
At what average rate was acceleration changing with respect to the number of clips?
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(15-1)/(4-1) = 14/3 = 4.7. for average. Note this is almost exactly half the average rate of change of acceleration of the first system, which has a rate of change of 9.3 m/s^2 per paperclip.
For each interval:
(7-1)/(2-1) = 5.
(10-7)/(3-2) = 3.
(15-10)/(4-3) = 5.
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For this system, what is the slope of a linear trendline for a graph of acceleration vs. number of clips?
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(15-1)/(4-1) = 14/3 = 4.7.
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Which graph had the lesser slope?
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This graph has a lesser slope than the first.
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What was it that was different about the two systems that resulted in different slopes?
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The first system's acceleration was changing at a greater rate. This was because the second system included more dominos.
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If we conducted the same experiment for a system consisting of four dominoes, what do you think would be the slope of the graph of acceleration vs. number of clips?
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It would be half the slope of the second system (roughly 2.3) and a fourth of the slope of the first system.
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If we conducted the same experiment for a system consisting of four dominoes, what do you think would be the average rate of change of acceleration with respect to the number of added clips?
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The average rate of change of acceleration w/ respect to number of added clops IS the slope of the acceleration VS number of clips. Thus, it would be half the rate of change of the second system (roughly 2.3) and a fourth of the rate of change of the first system.
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`q002. Assume that at a length of 60 cm, a rubber band chain exerts no significant force, but that for every centimeter stretched beyond that length it exerts an addition 0.1 Newton of force. You don't need to know what a Newton is, but if you want something to relate to, a Newton is about the weight of a typical small-to-medium-sized apple grown on a typical backyard tree. I weigh about 800 Newtons. A liter of water weighs about 10 Newtons. A 20-ounce soft drink weighs about 6 newtons. A pound is about 4.4 newtons.
Now, if a given constant force was exerted on a ramp rotating on a domino, then the greater the distance through which the force is exerted, the further we would expect the ramp to coast after the force is removed. If the force was exerted through twice the distance, we might expect the ramp to rotate twice as far.
If we exerted a greater force through the same distance we would expect the ramp to coast further. If twice as much force was exerted through the same distance, we might expect the ramp to rotate twice as far.
Using these assumptions, reason out the following:
Suppose we extend the chain to a length of 68 cm, use it to set the rotating ramp in motion and find that the ramp coasts through 3600 degrees before coming to rest. If we had an additional chain identical to the first, and extended both to 68 cm, how far would we expect the same system to coast?
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Total extra length extended = (2*68) - (2*60) = 16 cm. 16 * .1 = 1.6 newtons of force.
This is twice the force exerted over twice the distance. Thus, just twice the force makes the ramp spin twice as fast, and just twice the distance causes it to spin twice as fast, these factors combined should cause the ramp to rotate four times as far.
3600*4 = 14400 degrees.
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Each rubber band is extended just 8 cm, so you would have twice the force but exerted through the same 8 cm distance.
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How much force does the chain exert at the 68 cm length?
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8cm * .1 = .8 newton.
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As the chain returns from the 68 cm length to its 60 cm length, does it exert a constant force, an increasing force or a decreasing force?
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7*.1 = .7 N.
6*.1 = .6 N.
5*.1 = .5 N.
The force is decreasing as distance decreases.
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What average force does the chain exert as its length decreases from 68 cm to 60 cm?
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.8 N, .7 N, .6 N, .5 N, .4 N, .3 N, .2 N, .1 N, 0 N. Divide by 9.
.4 N.
Also, (8-0)/2 = 4. 4*.1 = .4 N.
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Now if the chain is extended to 64 cm, how much force does it exert?
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.4 N.
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What average force does it exert as its length decreases from 64 cm to 60 cm?
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4/2 = 2. .2 N.
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If the chain produces 3600 degrees of rotation when extended to 68 cm, how much rotation does it produce when extended to 64 cm, assuming that in each case it returns to its 60 cm length before releasing the ramp?
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64 cm produces half the force over half the distance as 68 cm.
Thus we assume 1/4 the revolutions.
3600 *.25 = 900 degrees.
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If we had two identical ramps, one on top of the other, through how many degrees do you think they would rotate if the rotation was produced by a single chain extended to a 68 cm length?
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This system has twice the mass as the first system, but goes the same distance with same force, thus should, by common sense, rotate half as many times.
If the first system rotated 3600 degrees, the second system rotates 1800 degrees.
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Ultimately it's the frictional force that dissipates the energy. The frictional force is exerted at the same points as before, but since we have double the weight pushing domino and table together, we expect double the frictional force.
Same conclusion, different picture.
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If the original ramp was 2 feet long, then how far would we expect a ramp 1 foot long to coast, when the rotation is produced by a single 68 cm chain?
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This has half the mass under same conditions. Thus, would go twice as far. 7200 degrees.
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Right.
Half the frictional force would take twice the rotational distance to dissipate.
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Would you expect the 1-foot ramp to have a greater or lesser coasting acceleration than the 2-foot ramp?
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A lesser. Its deceleration would be more gradual because its going a greater distance.
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This is worth testing in the lab.
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`q003. University Physics
Give your data for rubber band chain length vs. proximity of magnets.
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Length, magnet proximity. All measurements in cm.
33, no magnet.
33.5, 10.5.
34, 8.5.
34.5, 6.5.
35.5, 5.
36.5, 3.5.
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Assuming that the force exerted by your rubber band chain is a linear function of its length, sketch a graph of force vs. proximity to the magnets. You don't know at what exact length the chain begins to exert a force, so your force axis might be subject to relabeling, but the shape of that graph will not be affected by your assumption of the zero-force length of the chain. So go ahead and make a reasonable assumption of the zero-force length and proceed accordingly.
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Zero force length = 33 cm.
.5 * .1 = .05 N. (10.5, .05).
1*.1 = .1N. (8.5, .1).
1.5*.1= .15 N. (6.5, .15).
2.5 * .1 = .25 N. (5, .25).
3.5 * .1 = .35 N. (3.5, .35).
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How does the shape of your graph compare to the shape of the 'slope graph' constructed from your previous graph of coasting distance vs. proximity of magnets?
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This data is much more consistent than the slope data. This graph is a much more even curve than the other.
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