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course Phy 241
9/23 11 pm
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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Rubber band length Chain length
8.5 38.1
13.2 49.0
13.8 50.2
19.2 64.3
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Find the lengths of rubber band 2 for those four chain lengths.
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Rubber band length Chain length
10.7 40.9
12.1 51.0
12.5 58.1
13.4 65.8
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Graph the lengths of rubber band 2 vs. the lengths of rubber band 1.
**** The points form a line with a slope of about 11/3.
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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?
**** The ball traveled 100 cm in 10 seconds. Average velocity is the rate of change of position with respect to clock time, so the average velocity is 100cm / 10 s, or 10 cm/s. If acceleration is constant, then the average velocity is the average of the initial velocity and the final velocity. Since the initial velocity is 0, the final velocity is 20 cm/s. Velocity changes by 20 cm/s in 10 seconds. Average acceleration is the average rate of change of velocity with respect to clock time, so the average acceleration is (20 cm/s) / 10 s, or 2 cm/ s^2.
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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?
**** Since acceleration is constant, the average velocity is the average of the initial and final velocities, so the average velocity for the ball is 15 cm/s. The ball traveled 90 cm at an average of 15 cm/s. The change in clock time is 90 cm / (15 cm/s), or 6 s. The velocity changes -10 cm/s in 6 seconds. The average acceleration is (-10 cm/s) / 6 s, or about -1.67 cm/ 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?
**** The ball traveled 60 cm at an average rate of 20 cm/s. The change in clock time is 3 seconds. Because acceleration is uniform, the average velocity is the average of the initial and final velocities. The initial velocity is 10 cm/s. The change in velocity is 20 cm/s in 3 seconds. Acceleration is (20 cm/s) / 3 s, or about 6.67 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?
**** The change in velocity is 20 cm/s at a rate of 6 cm/ s^2. The change in clock time is (20 cm/s) / (6 cm/ s^2), or about 3.33 seconds. Average velocity is the average of the initial and final velocities. The average velocity is (10 cm/s + 30 cm/s) / 2, or 20 cm/s. The ball traveled at a rate of 20 cm/s for about 3.33 seconds. The total distance travelled is (20 cm/s) * 33.3 s, or 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.
**** We took a metal strap, marked on one end, and set it on a die. We would spin the strap and measure how many rotations it went through before coming to rest. We used a pendulum to time how long it took for the strap to come to rest.
Trial Angular displacement (measured in degrees) Cycles
1 315 3
2 450 5
3 720 7
4 300 3.5
5 150 2
6 270 4
7 190 2.5
8 330 4
9 1000 11.5
10 270 4
11 345 4.5
12 160 2
13 270 3
14 1470 16
15 210 2.5
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Find the average angular velocity of the strap for each trial.
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Trial Angular velocity (degrees/cycle)
1 105
2 90
3 102.9
4 85.71
5 75
6 67.5
7 76
8 82.5
9 86.96
10 67.5
11 76.67
12 80
13 90
14 91.88
15 84
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Find the angular acceleration of the strap for each trial, assuming that the angular acceleration is constant.
**** Assuming acceleration is constant, average velocity is the average of the initial and final velocities. The final velocities are 0, so the initial velocities will be twice the average.
Trial Change in velocity (degrees / cycle) Cycles Angular acceleration (degrees / cycle^2)
1 -210 3 -70
2 -180 5 -36
3 -205.8 7 -29.4
4 -171.42 3.5 -48.98
5 -150 2 -75
6 -135 4 -33.75
7 -152 2.5 -60.8
8 -165 4 -41.25
9 -173.92 11.5 -15.12
10 -135 4 -33.75
11 -153.34 4.5 -34.08
12 -160 2 -80
13 -190 3 -63.33
14 -183.76 16 -11.49
15 -168 2.5 -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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For the unloaded strap, construct a graph of angular acceleration vs. average angular velocity, and describe any trend of this graph.
**** The points are scattered but seem to be grouped near the line x (velocity) = 85.
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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 rebalance 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?
**** If the domino is closer to the center of the track it should be able to move further without upsetting the equilibrium. This is based on my memory of levers. The closer a force is to the fulcrum, the less effect it has on the lever.
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Also, the radius of the circle on which the domino travels is less than that of the circular path of the domino near the end. Both complete a cycle in the same time. Since the outer domino travels further in equal time, it's moving faster.
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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.
**** To find this data, we held a strap with a magnet attached to it in place and positioned another magnet next to it so the two would repel. We then released the strap and measured its angular displacement.
Distance between magnets Angular displacement (degrees)
10 cm 60
9 cm 90
8 cm 120
7 cm 225
6 cm 360
5 cm 405
4 cm 530
3 cm 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.
**** I have it 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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Slope Midpoint
-70 3.5
-125 4.5
-45 5.5
-135 6.5
-105 7.5
-30 8.5
-30 9.5
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Your data are pretty scattered, due to the sporadic nature of friction.
However a smooth curve through your points would decrease at a decreasing rate (i.e., would be decreasing and concave up). Such a curve would yield steadily decreasing slopes. Slopes would likely decrease in magnitude, changing from perhaps -150 deg / cm to around -30 deg / cm.
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Sketch a graph of slope vs. midpoint, and describe your sketch:
**** The graph is chaotic, rising, falling, and rising again without any noticeable trend. This is most likely because of human error in collecting the data. The general shape of the graph is an ‘M’.
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If you based your slopes on a smooth curve, your slope graph would have a clear trend.
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Explain why your last graph depicts rate of change of energy with respect to position vs. proximity.
**** The slope of a graph is equivalent to the rate at which it is changing. By taking the slope of an energy vs. proximity graph, I find the rate of change of energy. Therefore, the slope vs. midpoint graph is a graph of the rate of change of energy with respect to position vs. proximity.
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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.
**** We are discussing a graph of the rate of change of energy with respect to position vs. proximity. Since energy is work, and work = force * distance, change in energy with respect to position = (force * distance) / distance = force.
`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.
**** In this function, the higher the r value, the smaller the force will be. It makes sense; if the magnets are far apart, the force they exert on each other will be small, but if they are close together, the force will be much greater.
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How much force will be exerted at proximity r = 3, r = 6, r = 9 and r = 12?
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r = 3; F = 144 / (3)^2 = 16
r = 6; F = 144 / (6)^2 = 4
r = 9; F = 144 / (9)^2 = 1.78
r = 12; F = 144 / (12)^2 = 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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From r = 3 to r = 6, the average force is 10.
From r = 6 to r = 9, the average force is 2.89.
From r = 9 to r = 12, the average force is 1.39.
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Good.
Do note that due to the upward concavity of the curve, these averages overestimate the magnitudes of the force.
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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?
**** The displacement for all three intervals is 3. The product of the approximate average force and displacement for each interval is respectively:
30
8.67
4.17
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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?
**** This would be the integral of the force function on the interval 3 to 12.
F = 144 / r^2
The integral of this function is -144 / r | from 3 to 12.
(-144 / 12) - (-144 / 3) = 36
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Good.
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&#This looks good. See my notes. Let me know if you have any questions. &#