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course Phy 241
9/16/12 11:50 pm
120910Complete and submit the following documents in the same manner as previous documents.
qa_04
qa_05
query_02
query_03
Do the experiment Using the Timer Program. Most students report that this experiment takes less than an hour.
Questions related to today's class:
`q001. For your graphs of rubber band length vs. chain length:
Find chain lengths for four different representative lengths of rubber band 1.
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Red rubber band. All measurements in cm.
Chain: 65.3 cm. Red: 13.5 cm.
Chain: 59.5. Red: 13.
Chain: 53.7: Red: 12.5
Chain: 48.5. Red: 12.5.
Unstretched. Chain: 34.5. Red: 11.5
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Find the lengths of rubber band 2 for those four chain lengths.
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Blue rubber band. All measurements in cm.
Unstretched. Chain: 34.5. Blue: 8.5.
Chain: 47. Blue: 13.
Chain: 51. Blue: 15.
Chain: 59.5. Blue: 18.5.
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Graph the lengths of rubber band 2 vs. the lengths of rubber band 1.
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on paper
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`q002. Reason out the answer to the following questions, based on the definition of average velocity as average rate of change of position with respect to clock time and average acceleration as average rate of change of velocity with respect to clock time, and the fact that for uniform acceleration the velocity vs. clock time graph is a straight line. Show explicitly and in detail how you are using the definitions and the graph.
A ball accelerates uniformly from rest, traveling 100 centimeters in 10 seconds. What are its final velocity and its acceleration?
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v0 = 0.
ave v = 100/10 = 10 cm/s. (vf + v0)/2 = 10. vf = 20.
ave a = (vf - v0)/dt. (20)/(10) = 2 cm/s.
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A ball accelerates uniformly, with its velocity decreasing from 20 cm/s to 10 cm/s as it travels 90 cm. How long does this take, and what is its acceleration?
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v0 = 20. vf= 10.
v ave = (vf + v0)/2 = ds/dt. dt = 6 seconds.
ave a = (vf - v0)/dt. (10-20)/6 = -1.6 m/s^2.
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A ball accelerates uniformly through a displacement of 60 centimeters, during which its average velocity is 20 cm/second and its final velocity is 30 cm/second. How long does it take and what is its acceleration?
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dt = ds/ (ave v). dt = 60/20 = 3 seconds.
(30 + v0)/2 = 20. v0 = 10 m/s.
a ave = (30 - 10)/3. a ave = 20/3 cm/s^2 = 6.7 cm/s^2.
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A ball accelerates from 10 cm/s to 30 cm/s, accelerating at a uniform 6 cm / s^2. How long does this take and how far does the ball travel?
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dt = (vf - v0)/ave a.
dt = (30-10)/6 = 20/6 s = 3.33 s.
ds = v ave * dt. (10 + 30)/2 * 3.33 = 66.6 cm.
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`q003. Give your data for the rotating strap, along with a brief explanation of how you obtained your data.
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We had a rotating arm balanced on a domino and die. This arm (15 cm long) had a magnet taped to one end in such a way that when we held another magnet, opposite sides facing, close to it, the repulsive force would propel the arm in a circular motion. This was timed by myself counting pendulum oscillations. Pendulum length 42cm. Chris Schram observed the angles swept out.
We believe our error to be within +- 10 degrees and +- .5 oscillations.
(Angle in degrees swept out by rotating arm, time)
(315, 3)
450, 5
720, 7
300, 3.5
150, 2
270, 4
190, 2.5
330, 4
1000, 11.5
270, 4
345, 4.5
160, 2
280, 3
1470, 16
210, 2.5
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Find the average angular velocity of the strap for each trial.
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Omega (angular velocity) = change in angle over change in time. d theta/dt.
(315, 3). omega = 315/3 = 105 degrees/oscillation. All omega units hereafter in degrees/oscillation.
450, 5. omega = 90.
720, 7. omega = 102.8.
300, 3.5. omega = 85.7.
150, 2. omega = 75.
270, 4. omega = 67.5.
190, 2.5. omega = 76.
330, 4. omega = 82.5.
1000, 11.5. omega = 87.
270, 4. omega = 67.5.
345, 4.5. omega = 76.7.
160, 2. omega = 80.
280, 3. omega = 93.
1470, 16. omega = 91.8.
210, 2.5. omega = 84.
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Find the angular acceleration of the strap for each trial, assuming that the angular acceleration is constant.
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aAve = change in omega/dt. change in omega = (vf-v0). vf? (vf+v0)/2 = vAve, which we have. Since all our trials start from rest, vf = 2* vAve and aAve = vf/dt.
All aAve units in degrees/oscillations^2
450, 5. omega = 90. aAve = (90*2)/5 = 36.
720, 7. omega = 102.8. aAve = 29.4.
300, 3.5. omega = 85.7.aAve = 49.
150, 2. omega = 75. aAve = 75.
270, 4. omega = 67.5. aAve = 33.8.
190, 2.5. omega = 76. aAve = 60.8.
330, 4. omega = 82.5. aAve = 41.3.
1000, 11.5. omega = 87. aAve = 15.
270, 4. omega = 67.5. aAve = 33.8.
345, 4.5. omega = 76.7. aAve = 34.1.
160, 2. omega = 80. aAve = 80.
280, 3. omega = 93. aAve = 62.
1470, 16. omega = 91.8. aAve = 11.5.
210, 2.5. omega = 84. aAve = 67.2.
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Which appears to have the greater angular acceleration, the unloaded strap or the strap loaded with magnets? (University Physics students might not have observed the strap loaded with magnets).
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N/A
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For the unloaded strap, construct a graph of angular acceleration vs. average angular velocity, and describe any trend of this graph.
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N/A
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`q004. Once more consider the domino on the balanced metal track. If you balance the system with the domino near the end of the track, you find that you can move it some distance without upsetting the equilibrium. If you then move the domino twice as close to center of the track and rebalnce it, do you think you would be able to move the domino further without upsetting the equilibrium, or the same distance, or would a lesser distance suffice to upset the equilibrium?
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A lesser distance I would think because the domino is not traveling as far seeing as the arm length is shorter. Also, less momentum, so it would probably stop spinning sooner.
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Try this next time you get a chance.
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`q005. University Physics Students: Give your data for the angular displacement of the coasting strap vs. magnet proximity. Describe how your results were obtained.
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Setup is the same as the angle vs time scenario.
We measured distance away from the two magnets using a meter stick. We started with the big magnet 10 cm away and moved 1 cm closer each trial. 3 cm was as close as we could get without upsetting the balance of the arm. All angle measurements in degrees.
To get data, we held the arm in place with a wooden pencil while holding the large magnet in place. We then pulled the pencil away and let the repulsive forces set the arm in motion.
We believe our error to be within +- 10 degrees and +- .5 oscillations.
(distance, separation)
10, 60.
9, 90.
8, 120.
7, 225.
6, 360.
5, 405.
4, 530.
3, 600.
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Sketch a graph of the angular displacement of the coasting strap vs. magnet proximity, and sketch a smooth curve that follows the trend you believe is indicated by your data.
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on paper
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This assumption might or might not prove to be reasonable, but for the moment we will assume that the amount of energy required to rotate the strap through each degree is the same. So we can invent a unit we will call the degree_energy, which is the energy required to rotate the strap through one degree, in opposition to the frictional torque which opposes its motion. Our assumption is equivalent to the assumption that frictional torque for this system is constant, independent of the strap's position and angular velocity. With this assumption, each degree of rotation corresponds to one degree_energy, so that your graph can now be interpreted as a graph of energy vs. proximity.
Divide your graph into intervals, each interval corresponding to 1 centimeter of proximity, and find the slope for each of these intervals. For each interval list the slope and the midpoint of the interval:
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10-9. Slope = -30. MP = 9.5.
9-8. Slope = -30. MP 8.5.
8-7. Slope = -105. MP 7.5.
7-6. Slope = -135. MP 6.5.
6-5. Slope = -45. MP 5.5.
5-4. Slope = -125. MP 4.5.
4-3. Slope = -70. MP 3.5.
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Coasting acceleration is pretty erratic, as revealed by your previous results, so it would take a larger number of repetitions to get a valid trend. I would guess that about 11 repetitions at each separation, using the median result, would provide us with a fairly reasonable curve.
Another alternative is to use something with a more consistent angular acceleration.
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Sketch a graph of slope vs. midpoint, and describe your sketch:
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Well... Our data is a bit odd, so the graph is sort've all over the place. However, this is because we're graphing point to point. If we made a graph that fit our data to a function, it would look something like at 3.5 the slope is very negative. You have a negative y value. The further you progress along the x axis, the less negative your y values become.
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Explain why your last graph depicts rate of change of energy with respect to position vs. proximity.
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Because this is the change in energy over a given time interval. Thus a rate.
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Explain why, since work is force * distance and energy is equivalent to work, this last graph therefore depicts the magnetic force vs. proximity.
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Force = w/d. This is work divided by distance, or energy divided by distance. In our problem, this is a rate because our initial work or energy and initial distance are 0.
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`q006. University Physics Students: We could set up a rotating system similar to that used with the magnets, but relying on electrostatic forces, which might exert an inverse-square force of the form F = k / r^2, where r is the proximity of our two charges.
For the moment let's leave everything unitless and just look at the numerical implications of a force function F = 144 / r^2.
How much force will be exerted at proximity r = 3, r = 6, r = 9 and r = 12?
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r = 3. F = 16.
r = 6. F= 4.
r = 9. F = 16/9= 1.8.
r = 12. F = 1.
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What is the approximate average force for the interval from r = 3 to r = 6? Answer also for the intervals between r = 6 and r = 9, and between r = 9 and r = 12.
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3-6. (16+4)/2 = 10.
6-9. (4 + 1.77)/2 = 2.9.
9-12. (1.8 + 1)/2 = 1.4.
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If the charges are brought to the r = 3 separation and released, then what would be the product of the approximate average force and displacement from r = 3 to r = 6? What would be the same result for the intervals between r = 6 and r = 9, and between r = 9 and r = 12?
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3-6. (Approximate average force) * (r2 - r1). 10*3 = 30.
6-9. 2.9 * 3 = 8.7.
9-12. 1.4 * 3 = 4.2.
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We could divide the interval from r = 3 to r = 12 into more than three subintervals. We could divide this interval into 10, or 100, or a million subintervals, and perform exactly the same type of analysis (though I wouldn't recommend it for the million-interval case). The approximation errors would pretty quickly become insignificant, so that with increasing numbers of subintervals our results would quickly approach a limiting value. What would this limiting value be?
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If you take the limit of 144/r^2 as r approaches 0 because the intervals are getting smaller and smaller, this limit approaches infinity.
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The r interval remains [3, 12]. r doesn't approach zero.
The lengths of the subintervals approach zero.
In your previous approximation, for example, you averaged 16 and 4 to get 10 and treated that as the average force on the interval [3, 6]. However the midpoint of that interval is 4.5, at which point the force would be about 7. So the midpoint value and the average of initial and final values on the interval are different. And the average force on that interval would also be different (how would you get the average force on that interval?).
If our subintervals were much shorter the midpoint value and average of endpoint values would be much closer together (the endpoint values themselves would be much closer together and our two results couldn't differ by more that the endpoint values). So approximation errors approach zero as subinterval width approaches zero.
The question remains, then, what would be get in the limit?
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Good, but check my notes, and especially see if you can reconcile that last question.
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