course Phy 121 Nȕ˝y̰Nx{{assignment #007
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12:48:17 `q001. We obtain an estimate of the acceleration of gravity by determining the slope of an acceleration vs. ramp slope graph for an object gliding down an incline. Sample data for an object gliding down a 50-cm incline indicate that the object glides down the incline in 5 seconds when the raised end of the incline is .5 cm higher than the lower end; the time required from rest is 3 seconds when the raised end is 1 cm higher than the lower end; and the time from rest is 2 seconds when the raised end is 1.5 cm higher than the lower end. What is the acceleration for each trial?
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RESPONSE --> v0=0, so 50cm/5sec=10cm/s for average velocity. vf=20cm/s because 10cm/s is the average. so then 20cm/s/5s=4cm/s^2. v0=0, use 'ds=v0 'dt+.5 a 'dt^2. plug in the numbers to get 'ds=0'dt+.5 a'dt^2. with the 0 at the front we are left with 'ds=.5 a 'dt^2, then to get acceleration put 'a' on one side by itself to get a= 'ds*2/dt^2. then plug in the numbers to get a=50cm*2/3sec^2 which gives us around 11cm/s^2. for the one with 2 seconds we do the same. v0=0, use 'ds=v0 'dt+.5 a 'dt^2. plug in the numbers to get 'ds=0'dt+.5 a'dt^2. with the 0 at the front we are left with 'ds=.5 a 'dt^2, then to get acceleration put 'a' on one side by itself to get a= 'ds*2/dt^2. also plug in the numbers to get a=50 cm*2 / 2 sec^2 which gives us 25 cm/s^2 confidence assessment: 1
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12:48:27 We can find the accelerations either using equations or direct reasoning. To directly reason the acceleration for the five-second case, we note that the average velocity in this case must be 50 cm/(5 seconds) = 10 cm/s. Since the initial velocity was 0, assuming uniform acceleration we see that the final velocity must be 20 cm/second, since 0 cm/s and 20 cm/s average out to 10 cm/s. This implies a velocity change of 20 cm/second a time interval of 5 seconds, or a uniform acceleration of 20 cm/s / (5 s) = 4 cm/s^2. The acceleration in the 3-second case could also be directly reasoned, but instead we will note that in this case we have the initial velocity v0 = 0, the time interval `dt = 3 sec, and the displacement `ds = 50 cm. We can therefore find the acceleration from the equation `ds = v0 `dt + .5 a `dt^2. Noting first that since v0 = 0 the term v0 `dt must also be 0,we see that in this case the equation reduces to `ds = .5 a `dt^2. We easily solve for the acceleration, obtaining a = 2 `ds / `dt^2. In this case we have a = 2 * (50 cm) / (3 sec)^2 = 11 cm/s^2 (rounded to nearest cm/s^2). For the 2-second case we can use the same formula, obtaining a = 2 * (50 cm) / (2 sec)^2 = 25 cm/s^2.
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RESPONSE --> ok self critique assessment: 3
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12:54:04 `q002. What are the ramp slopes associated with these accelerations?
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RESPONSE --> slope is rise/run. one with 5 seconds-0.5cm/50cm=0.01 cm one with 3 seconds-1cm/50cm=0.02cm one with 2 seconds-1.5cm/50cm=0.03cm confidence assessment: 2
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12:54:15 For the 5-second trial, where acceleration was 4 cm/s^2, the 'rise' of the ramp was .5 cm and the 'run' was nearly equal to the 50-cm length of the ramp so the slope was very close to .5 cm / (50 cm) = .01. For the 3-second trial, where acceleration was 11 cm/s^2, the 'rise' of the ramp was 1 cm and the 'run' was very close to the 50-cm length, so the slope was very close to 1 cm / (50 cm) = .02. For the 2-second trial, where the acceleration was 25 cm/s^2, the slope is similarly found to be very close to 1.5 cm / (50 cm) = .03.
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RESPONSE --> ok self critique assessment: 2
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13:01:55 `q003. Sketch a reasonably accurate graph of acceleration vs. ramp slope and give a good description and interpretation of the graph. Be sure to include in your description how the graph points seem to lie with respect to the straight line that comes as close as possible, on the average, to the three points.
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RESPONSE --> take the acceleration of each time and pair them with the slopes. (0.01cm, 4cm/s^2), (0.02cm, 11cm/s^2), and (0.03cm, 25cm/s^2). put them on the graph, with slope on the x-axis and the acceleration on the y-axis. the graph has a positive slope. confidence assessment: 2
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13:02:22 The graph will have acceleration in cm/s^2 on the vertical axis (the traditional y-axis) and ramp slope on the horizontal axis (the traditional x-axis). The graph points will be (.01, 4 cm/s^2), (.02, 11.1 cm/s^2), (.03, 25 cm/s^2). The second point lies somewhat lower than a line connecting the first and third points, so the best possible line will probably be lower than the first and third points but higher than the second. The graph indicates that acceleration increases with increasing slope, which should be no surprise. It is not clear from the graph whether a straight line is in fact the most appropriate model for the data. If timing wasn't particularly accurate, these lines could easily be interpreted as being scattered from the actual linear behavior due to experimental errors. Or the graph could indicate acceleration vs. ramp slope behavior that is increasing at an increasing rate.
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RESPONSE --> didnt talk about the line and the points and close they were to the line self critique assessment: 2
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13:46:02 `q004. Carefully done experiments show that for small slopes (up to a slope of about .1) the graph appears to be linear or very nearly so. This agrees with theoretical predictions of what should happen. Sketch a vertical line at x = .05. Then extend the straight line you sketched previously until it intersects the y axis and until it reaches past the vertical line at x = .05. What are the coordinates of the points where this line intersects the y-axis, and where it intersects the x =.05 line? What are the rise and the run between these points, and what therefore is the slope of your straight line?
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RESPONSE --> with the x=0.05, i got about 38cm/s^2, so (0.05, 38) and with the y-axis its 0 and about -8cm/s^2, so (0, -8). the rise is 38-(-8)=46 and the run is 0.05-0=0.05 therefore the slope is 46cm/s^2/0.05=920cm/s^2. confidence assessment: 3
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13:46:12 A pretty good straight line goes through the points (0, -6 cm/s^2) and (.05, 42 cm/s^2). Your y coordinates might differ by a few cm/s^2 either way. For the coordinates given here, the rise is from -6 cm/s^2 to 42 cm/s^2, a rise of 48 cm/s^2. The run is from 0 to .05, a run of .05. The slope of the straight line is approximately 48 cm/s^2 / .05 = 960 cm/s^2. Note that this is pretty close to the accepted value, 980 cm/second^2, of gravity. Carefully done, this experiment will give us a very good estimate of the acceleration of gravity.
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RESPONSE --> ok self critique assessment: 3
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13:50:43 `q005. The most accurate way to measure the acceleration of gravity is to use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) for the period of a pendulum. Use your washer pendulum and time 100 complete back-and-forth cycles of a pendulum of length 30 cm. Be sure to count carefully and don't let the pendulum swing out to a position more than 10 degrees from vertical. How long did it take, and how long did each cycle therefore last?
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RESPONSE --> i got 105 so 105sec/100cycles=1.05sec/cycle confidence assessment: 2
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13:51:19 100 cycles of a pendulum of this length should require approximately 108 seconds. This would be 108 seconds per 100 cycles, or 108 sec / (100 cycles) = 1.08 sec / cycle. If you didn't count very carefully or didn't time very accurately, you might differ significantly from this result; differences of up to a couple of cycles are to be expected.
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RESPONSE --> mine was a few seconds less, therefore the seconds for each cycle was less self critique assessment: 3
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13:55:36 `q006. You now have values for the period T and the length L, so you can use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) to find the acceleration g of gravity. Solve the equation for g and then use your values for T and L to determine the acceleration of gravity.
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RESPONSE --> g=4 `pi^2 L / T^2. with T=1.05 and L=30 we can plug in to get g=4'pi^2 (30)/1.05^2 to get g=1074cm/s^2 confidence assessment: 2
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13:55:46 Solving T = 2 `pi / `sqrt(g) * `sqrt(L) for g, we can first multiply both sides by `sqrt(g) to obtain T * `sqrt(g) = 2 `pi `sqrt(L). Then dividing both sides by T we obtain `sqrt(g) = 2 `pi `sqrt(L) / T. Squaring both sides we finally obtain {}g = 4 `pi^2 L / T^2. Plugging in the values given here, L = 30 cm and T = 1.08 sec, we obtain g = 4 `pi^2 * 30 cm / (1.08 sec)^2 = 1030 cm/s^2. You should check these calculations for accuracy, since they were mentally approximated.
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RESPONSE --> ok self critique assessment: 3
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