course Phy 231 6/15 2:45 007. Acceleration of Gravity Question: `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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Given Solution: 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. Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `q002. What are the ramp slopes associated with these accelerations? Your solution: Acceleration: 4cm/s^2, rise .5cm, run 50cm, slope: .5cm/50cm = .01 Acceleration: 11cm/s^2, rise 1cm, run 50cm, slope: 1cm/50cm = .02 Acceleration: 25cm/s^2, rise 1.5cm, run 50cm, slope: 1.5cm/50cm = .03 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `q003. Sketch a reasonably accurate graph of acceleration vs. ramp slope and give a good description and interpretation of the graph. Be sure to include in your description how the graph points seem to lie with respect to the best-fit straight line (i.e., the straight line that comes as close as possible, on the average, to the three points). Your solution: On the axes, the acceleration is on the y axis, the rise, and the slope is on the x axis, the run. The graph of acceleration versus slope increases from left to right. The three points that I have cluster around the best fit line with the second data point (.02, 11cm/s^2) being a little bit lower than the others. This would make the best fit line a little bit lower on the axes. The slope of the line is constant, it seems, because all of the points lie in a relatively straight line. It is also a steeps slope because the rate at which the acceleration increases is large compared to the change in the slope. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. • Alternatively the graph could that acceleration vs. ramp slope is increasing at an increasing rate. • The best-fit straight line for this data wouldn't go through any of the graph points; some points will lie above the line and some below. • So the points would to an extent be scattered about the line. • As long as the degree of scattering can be explained by the uncertainties in our measurements, then it is possible that the straight-line, or linear, model is consistent with the data. Self-critique (if necessary): It is uncertain whether the graph will be linear or curved. You would need more data points to be able to tell for sure. Self-critique rating #$&* 3 ********************************************* Question: `q004. Carefully done experiments show that for small slopes (up to a slope of about .1) the graph appears to be linear or very nearly so. This agrees with theoretical predictions of what should happen. Sketch a vertical line at x = .05. Then extend the straight line you sketched previously until it intersects the y axis and until it reaches past the vertical line at x = .05. What are the coordinates of the points where this line intersects the y-axis, and where it intersects the x =.05 line? What are the rise and the run between these points, and what therefore is the slope of your straight line? Your solution: y-intercept: (0, -8cm/s^2) vertical line intercept: (.05, 44cm/s^2) Rise: 44cm/s^2 - -8cm/s^2 = 52cm/s^2 Run: .05 – 0 = .05 Slope = 52cm/s^2 / .05 = 1040cm/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. There is in fact only one best-fit line for the data, but since we're estimating the best-fit line it is expected that our estimates of its slope and location will differ somewhat. Certainly it would be possible for one person to get an estimate of 840 cm/s^2, while another might estimate 960 cm/s^2 and another 1060 cm/s^2. If all three people used the same three data points, and actually found the unique line that best fits those three points, then all three would get identical results. Self-critique (if necessary): I did the calculations for the slope and everything correctly, but the slope for my line was steeper than the actual acceleration of gravity. This is because my line of best fit was different than the actual line. Self-critique rating #$&* 3 ********************************************* Question: `q005. The most accurate way to measure the acceleration of gravity is to use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) for the period of a pendulum. Use your washer pendulum and time 100 complete back-and-forth cycles of a pendulum of length 30 cm. Be sure to count carefully and don't let the pendulum swing out to a position more than 10 degrees from vertical. How long did it take, and how long did each cycle therefore last? Your solution: It took 108.7 seconds to do 100 complete cycles. Each cycle lasted 1.087 seconds. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `q006. You now have values for the period T and the length L, so you can use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) to find the acceleration g of gravity. Solve the equation for g and then use your values for T and L to determine the acceleration of gravity. Your solution: T = 2pi/sqrt g *sqrt L T/sqrt L = 2pi/sqrt g T sqrt g = 2pi sqrt L Sqrt g = (2pi sqrt L) / T G = ((2pi sqrt L) / T)^2 = 4 pi^2 L / T^2 G = 4 pi^2 30cm / (1.087 sec) ^2 = 1184.35cm / 1.18sec^2 = 1002 cm/s^2 Confidence Assessment: 2
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Given Solution: Solving T = 2 `pi / `sqrt(g) * `sqrt(L) for g, we can first multiply both sides by `sqrt(g) to obtain • T * `sqrt(g) = 2 `pi `sqrt(L). Then dividing both sides by T we obtain • `sqrt(g) = 2 `pi `sqrt(L) / T. Squaring both sides we finally obtain • g = 4 `pi^2 L / T^2. Plugging in the values given here, L = 30 cm and T = 1.08 sec, we obtain g = 4 `pi^2 * 30 cm / (1.08 sec)^2 = 1030 cm/s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The number I got was pretty close to the correct acceleration. Self-critique rating #$&* 3 "