course PHY 201
.................................................
......!!!!!!!!...................................
18:40:09 `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?
......!!!!!!!!...................................
RESPONSE --> 'ds = 50cm 'dt = 5s,3s,2s because intial velocity is 0m/s it can be assumed that 'ds = v0 'dt + .5 a * 'dt^2 50cm = .5 a * (5s)^2 50cm = .5 a * 25s^2 50cm/25s^2 = .5 a 2cm/s^2 * 2 = a .5cm, 4cm/s^2 1cm, 11.11cm/s^2 1.5 cm, 25cm/s^2
.................................................
......!!!!!!!!...................................
18:42:22 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.
......!!!!!!!!...................................
RESPONSE --> check
.................................................
......!!!!!!!!...................................
18:44:45 `q002. What are the ramp slopes associated with these accelerations?
......!!!!!!!!...................................
RESPONSE --> lets see, it was 50cm long (run) and it was raised .5cm, 1cm, 1.5cm (rise) .01 .2 .03
.................................................
......!!!!!!!!...................................
18:47:10 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.
......!!!!!!!!...................................
RESPONSE --> check, not sure why I had .2
.................................................
......!!!!!!!!...................................
18:53:15 `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.
......!!!!!!!!...................................
RESPONSE --> as ramp slope increased, so did acceleration. The acceleration rises about 10 (approx) for each .01 slope increase. Though I'm sure that further analysis would indicate greater variation.
.................................................
......!!!!!!!!...................................
18:55:45 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.
......!!!!!!!!...................................
RESPONSE --> okay, so I needed to go into more detail such as what was labeled X and Y. I believe a curve of increasing at an increasing rate is appropriate. That does coincide with linear discription as well.
.................................................
......!!!!!!!!...................................
19:18:01 `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?
......!!!!!!!!...................................
RESPONSE --> .05, 35.5 0, it seems it is 0 acceleration. But this would make sense because the force of gravity is perpindicular to the surface of the slope. The slope seems to be 710
.................................................
......!!!!!!!!...................................
19:23:51 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.
......!!!!!!!!...................................
RESPONSE --> okay, but X = slope if I'm not mistaking, and Y is acceleration. well if X = 0 that would mean you would have a flat surface, and the force of gravity would be perpendicular. Should the line not reflect this fact? Is this even a fact? Initially, around -5.6 might have been considered, but because of this, I brought my line into question.
.................................................
......!!!!!!!!...................................
20:00:46 `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?
......!!!!!!!!...................................
RESPONSE --> it took 109.125 / 100 1.09
.................................................
......!!!!!!!!...................................
20:01:55 100 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.
......!!!!!!!!...................................
RESPONSE --> 1.09, that's not far off.
.................................................
......!!!!!!!!...................................
20:30:18 `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.
......!!!!!!!!...................................
RESPONSE --> 1.08 = 2pi / sqrt(g) * sqrt(30) 1.08 / sqrt(30) = 2pi / sqrt(g) sqrt(g) * 1.08 / sqrt(30) = 2pi 2pi / (1.08 / sqrt(30)) = sqrt(g) 1015cm/s^2
.................................................
......!!!!!!!!...................................
20:47:57 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.
......!!!!!!!!...................................
RESPONSE --> check 1015cm/s^2 is more accurate.
.................................................
"