#$&* course Phy 121 9/18 4:22 pm 007. Acceleration of Gravity
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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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The 50cm ramp is the hypotenus of a right triangle in each example a^2 + b^2 = c^2. We are given a as the height of the ramp at the high end (this will be our rise). c is the 50 cm ramp, so we solve for b( which will be our run). First subtract a^2 from both sides: b^2 = c^2 - a^2 Take sqrt of both sides b = sqrt (c^2 - a^2 ) in this case we will only use the positive square root since it is a length. To find slope we divide rise/run or in this case a/b When the raised end of the incline is .5 cm higher than the lower end: b = sqrt (c^2 - a^2 ) b = sqrt [(50cm)^2 -( .5cm)^2] b= sqrt (2500cm^2 - .25cm^2) b = sqrt 2499.75 b = 50cm (approx) slope = .5cm/50cm = .01 When the raised end of the incline is 1 cm higher than the lower end b = sqrt (c^2 - a^2 ) b = sqrt [(50cm)^2 - 1cm)^2] b= sqrt (2500cm^2 - 1cm^2) b = sqrt 2499 b = 50cm (approx) slope = 1 cm/50cm = .02 When the raised end of the incline is 1.5 cm higher than the lower end b = sqrt (c^2 - a^2 ) b = sqrt [(50cm)^2 - 1.5 cm)^2] b= sqrt (2500cm^2 - 2.25cm^2) b = sqrt 2498 b = 50cm (approx) slope = 1.5 cm/50cm = .03 The a^2+ b^2=c^2 stuff really didn't make a difference here, but I'm sure it would for a steeper slope. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph is labeled Ramp Slope on the x axis and Acceleration on the y-axis. There are three data points on the graph. The best fit straight line lies about .001 units to the right of points (.01, 4) and (.03, 25) and about .002 units to the left of of point (.02, 11). This line shows acceleration increasing at a constant rate with regards to ramp slope. However, a curved line might actually present a better picture of what is really happening. The curve line would be concave up which would mean accleration is increasing at an increasing rate with respect to ramp slope. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. STUDENT COMMENT I am a little unclear about why the lines could be considered scattered from an actual linear behavior, unless it is because the line is not exactly straight and seems to have a curve. INSTRUCTOR RESPONSE The graph points don't line along a perfectly straight line. Based on these three points, you could possibly infer that the graph curves. A curved graph would be consistent with the data. If experimental uncertainties are large enough, the graph could also be consistent with a best-fit straight line. 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):Ok - I did not include the units cm/s^2 in my decsription. This was an oversight, I usually do. ------------------------------------------------ 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My graph was not big enough to accomodate the intersection at .05, but I was able to extend the y xis to accomodate the y -intercept. The y intercept is (0, -7.6cm/s^2). I can calculate slope from this and one other distant point (.03, 24.6cm/^2) The rise is 23.6cm/^2 - - 7.6cm/s^2 = 31.2cm/s^2 The run is .03 - 0 = .03 The slope is 31.2 c/s^2/.03 = 1040 cm/s^2 per unit of slope. 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. STUDENT QUESTION Once I graphed the points using the correct coordinates I was able to see the correct graph, and its slope of 48cm/s^2/ .05 =960cm/s^2. ??????????My question here is what if on my graph the points were at around 42cm/s^2 instead of the 48cm/s^2, and my answer was 840 or so. Is there room for the variation in how you plot the graph, if I started with different number or something, or my graph is not as precise.????????????????????????? ?????????????Also I am still a little unclear, seeing how I had my point graphed wrong in the first place, about why we would use the slope of the graph .5/ 50cm to get .01....I really cannot seem to rationalize this. ????????????????????????? INSTRUCTOR RESPONSE Your slope is based on two points; it can't be calculated based on a single point because you need two points to get the rise and run. If the two points on your straight line were, say, (0, -10 cm/s^2) and (.05, 48 cm/s^2) then the rise would be 48 cm/s^2 - 10 cm/s^2 = 58 cm/s^2, your run would be .05 and your slope would be 58 cm/s^2 / .05 = 1060 cm/s^2. If the rise between the two points on your straight line was 42 cm/s^2, with the run still .05, then your slope would be 42 cm/s^2 / .05 = 840 cm/s^2, as you say. 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):OK. I didn't get the same result but it was within the parameters of the instructor response. I understand the concepts. ------------------------------------------------ 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It took 104.4 seconds to complete 100 cycles. Each cycle therefor took 104.4s/100cycles = 1.044 s. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 100 cycles of a pendulum of this length should require approximately 108 seconds. This would be 108 seconds per 100 cycles, or 108 sec / (100 cycles) = 1.08 sec / cycle. If you didn't count very carefully or didn't time very accurately, you might differ significantly from this result; differences of up to a couple of cycles are to be expected. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Okay - I think I'm close here. ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q006. You now have values for the period T and the length L, so you can use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) to find the acceleration g of gravity. Solve the equation for g and then use your values for T and L to determine the acceleration of gravity. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T = 2 `pi / `sqrt(g) * `sqrt(L) divide each side by `sqrt(L) T/`sqrt(L) = 2`pi/`sqrt(g) multiply each side by `sqrt(g) sqrt(g) * T/`sqrt(L) = 2pi divide each side by T/`sqrt(L) sqrt (g) = 2pi/(T/`sqrt(L)) take sqrt of each side g = sqrt[2pi/(T/`sqrt(L))] substitute g = sqrt[ 6.28/(1.04s/ 5.48cm^1/2)] g = sqrt (6.28/.19) At this point the units have become too muddled for me g = sqrt (33.05 units) g = 5.74 m/s^2 I just know it should be m/s squared by why I don't know. 5.74 seems way off. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. You should check these calculations for accuracy, since they were mentally approximated. STUDENT QUESTION I do not understand how if you divide both sides by the sqrt (g), the you end up with t* sqrt(g) = 2' pi ' sqrt (L), when sqrt(L) was divided by 2 'pi.......why wouldnt it be t * sqrt (g) = 2'pi / sqrt(L)....what happens to the division sign there????? INSTRUCTOR RESPONSE The division sign is between pi and sqrt(g). It doesn't apply to sqrt(L). Multiplications and divisions are done in order, as stated in the order of operations. This means that you multiply 2 by pi, then divide the result by sqrt(g), then multiply the result by sqrt(L). If the sqrt(L) was to be part of the denominator, the expression would have to be written 2 pi / ( sqrt(g) * sqrt(L) ). In that case you would begin by multiplying sqrt(g) by sqrt(L), then you would do multiplications and divisions in order, multiplying 2 by pi then dividing by the product (sqrt(g) sqrt(L)). The correct expression: The expression 2 pi / sqrt(g) * sqrt(L) is represented in a computer algebra system as This representation is consistent with the order of operations. The common misinterpretation of the expression: The expression 2 pi / sqrt(g) * sqrt(L) is often misinterpreted as However the above represents 2 pi / (sqrt(g) * sqrt(L) ), not our original expression 2 pi / sqrt(g) * sqrt(L) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I set up the equation slopplily and then confused squaring with square root. Had I done this properly it would have looked like this. g = 4 `pi^2 L / T^2 g = 39.4 * 30cm/ (1.04s)^2 = 1182cm/ 1.08s^2 = 1095cm/s^2 ------------------------------------------------ Self-critique rating:3" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: