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course Phy 241
9/25 11 pm
`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.
**** For the first trial, the machine accelerated through 30 cm in 3 seconds. The average velocity would be the change in position divided by the change in time, which would be 30 cm / 3 seconds, or 10 cm/s. Since the machine started at rest, we know the initial velocity is 0. Assuming constant acceleration, the average velocity is the average of the initial and final velocities, or vAve = (v0 + vf) / 2. If we solve this for vf, we get vf = 2 vAve - v0. The final velocity for this trial is 20 cm/s. Acceleration is the rate of change of velocity with respect to time, or a = `dv / `dt. The change in velocity is 20 cm/s and the change in time is 3 seconds, so the acceleration for this trial would be (20 cm/s) / 3 seconds, or about 6.67 cm/s^2.
Using the same reasoning for the second trial (with one paper clip), we find the average velocity is 15 cm/s, the change in velocity is 30 cm/s, and the acceleration is 10 cm/s^2.
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This trial required 2 seconds so `dv / `dt = 30 cm/s / (2 s) = 15 cm/s^2, not 10 cm/s^2.
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For the third trial (with two paper clips), the average velocity was about 17.65 cm/s, the change in velocity was about 35.29 cm/s, and the acceleration was about 11.76 cm/s^2.
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This trial only required 1.7 seconds, so the acceleration is more like 20 cm/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.
**** I used Excel to do this.
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According to your results, what is the average rate of change of acceleration with respect to the number of clips?
**** Acceleration is changing at a rate of about 2.55 cm/s^2 per clip.
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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?
**** The equation for the line is y = 2.545 x + 6.9317, where y is the acceleration and x is the number of paper clips. Using this equation, I can determine that it would take about 382 paper clips to reach an acceleration of 980 cm/s^2.
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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?
**** Averaging all three changes in acceleration together, I get the average change in acceleration to be about 4.67 cm/s^2 per paper clip.
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For this system, what is the slope of a linear trendline for a graph of acceleration vs. number of clips?
**** The slope of the trend line is 4.5.
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Which graph had the lesser slope?
**** The first one had a lesser slope.
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What was it that was different about the two systems that resulted in different slopes?
**** The first set of data was for when one domino was suspended on each side of the machine. The second set was for when two dominos were suspended on each side.
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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?
**** I’d expect the slope to be about 8. The data above implies that more dominos will increase the slope, and the second trial, which had twice as many dominos, had a little under twice the slope. The slope for the second trial was 4.5, so I’d imagine that a trial with twice the dominos would have a little under twice that slope.
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The more dominoes the less effect the paperclips would have. So the slope would be expected to decrease with an increasing number of dominoes.
Your calculations for the first set of data appear to have been incorrect, which led you to an incorrect conclusion here.
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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?
**** The rate of change of acceleration is equivalent to the slope of the graph. Using the same reasoning as stated in the previous answer, I think the rate of change of acceleration would be about 8 (cm/s^2) / clip.
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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?
**** At 60 cm, the chain exerts no significant force. At 68 cm, the chain exerts enough force to rotate the ramp 3600 degrees. By attaching another identical chain and stretching both to 68 cm, the force exerted on the ramp will double, as will the angular distance it coasts. We would expect the ramp to coast 7200 degrees.
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How much force does the chain exert at the 68 cm length?
**** If the chain exerts 1 Newton of force for every centimeter stretched beyond 60 cm, the force exerted would be 8 Newtons.
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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?
**** The stretched distance would be decreasing, so the amount of force exerted by the chain would be decreasing.
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What average force does the chain exert as its length decreases from 68 cm to 60 cm?
**** If the chain length is decreasing from 68 cm to 60 cm, the force is decreasing from 8 Newtons to 0. The initial force is 8 Newtons and the final force is 0, so assuming length of the rubber band chain is changing at a constant rate, the average force would be 4 Newtons.
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Now if the chain is extended to 64 cm, how much force does it exert?
**** At 64 cm, the chain would exert 4 Newtons of force.
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What average force does it exert as its length decreases from 64 cm to 60 cm?
**** Using the same reasoning as before, the average force, assuming constant change in chain length, would be 2 Newtons.
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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?
**** The average force exerted by the chain retracting from 64 cm is half the average force exerted by the chain retracting from 68 cm. This means the ramp should experience half the rotation, or 1800 degrees.
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At 64 cm the chain only exerts its force through 4 cm, half the distance thru which the 68 cm chain exerts its force.
This would halve the effect of the already-halved average force exerted by the 64 cm chain.
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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?
**** Two identical ramps would have twice the mass. Force = mass * acceleration, so the same force on an object of twice the mass would cause half the acceleration. The ramps would rotate half the distance, or 1800 degrees.
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The key is that the same work would be done so the two systems would gain the same amount of energy from the chain.
Since friction is presumably proportional to the weight of the supported object we would expect twice the frictional force on the system with two rods.
This would result in double the frictional torque, which would then dissipate the energy through half the rotational distance.
We could test this, and it's likely that one group or the other will do so.
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Frictional force and therefore torque being only half as great, we would expect to require twice as much rotation.
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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?
**** As stated previously, force = mass * acceleration. A ramp with half the length would have half the mass, and that means twice the acceleration. This means the ramp would rotate twice as far, or 7200 degrees
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Would you expect the 1-foot ramp to have a greater or lesser coasting acceleration than the 2-foot ramp?
**** As I said in my previous answer, the 1-foot ramp would have twice the acceleration of the 2-foot ramp.
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The analysis of this question is complex. It will turn out that the shorter ramp will have 1/8 the moment of inertia, but would experience half the coasting torque. The result would be 4 times the angular acceleration. We could test this as well.
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`q003. University Physics
Give your data for rubber band chain length vs. proximity of magnets.
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Distance between dominos Rubber band length
(n/a) 33 cm
10.5 cm 33.5 cm
8.5 cm 34 cm
6.5 cm 34.5 cm
5 cm 35.5 cm
3.5 cm 36.5 cm
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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.
**** I have the graph in Excel. As stated in the chart above, I think the zero-force length of the chain is about 33 cm. We determined this in class.
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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?
**** The points on this graph come close to forming a line, but the ‘slope graph’ from the previous experiment doesn’t seem to form anything. As I said before, this is most likely due to human error, but given what information I have, it is too chaotic to make a meaningful comparison (though, in theory, they should both be linear decreasing functions).
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&#This looks good. See my notes. Let me know if you have any questions. &#