course Phy 231 ????????r????assignment #007007. Acceleration of Gravity
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06:40:35 `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 --> Displacement ('ds) for all trials is 50 cm. Time interval ('dt) for the first trial is 5 s. For the second trial, it is 3 s. It is 2 s for the third trial. v0 for all trials is 0 cm/s. For the first trial (0.5 cm higher on one end), since v0 is zero, the vf is equal to vAve * 2, which is 'ds / 'dt * 2 = 50 cm / 5 s * 2 = 20 cm/s. We can now set up an equation for finding acceleration: ( vf - v0 ) / 'dt = ( 20 cm/s - 0 cm/s ) / 5 s = 4 cm/s/s For the second trial (1 cm higher on one end), since v0 is zero, the vf is equal to vAve * 2, which is 'ds / 'dt * 2 = 50 cm / 3 s * 2 = 33.3 cm/s. We can now set up an equation for finding acceleration: ( vf - v0 ) / 'dt = ( 33.3 cm/s - 0 cm/s ) / 3 s = 11.1 cm/s/s For the third trial (1.5 cm higher on one end), since v0 is zero, the vf is equal to vAve * 2, which is 'ds / 'dt * 2 = 50 cm / 2 s * 2 = 50 cm/s. We can now set up an equation for finding acceleration: ( vf - v0 ) / 'dt = ( 50 cm/s - 0 cm/s ) / 2 s = 25 cm/s/s confidence assessment: 3
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06:41:20 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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06:44:59 `q002. What are the ramp slopes associated with these accelerations?
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RESPONSE --> For the 0.5cm incline, the slope would be 0.5 cm / 50 cm, which = 1/100 For the 1.0 cm incline, the slope would be 1.0 cm / 50 cm, which = 1/50 For the 1.5 cm incline, the slope would be 1.5 cm / 50 cm, which = 3/100 confidence assessment: 3
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06:45:08 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: 3
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06:48:59 `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 --> The line-of-best-fit slopes upward, and the plotted points seem to indicate that the acceleration is increasing at an increasing rate, as the slope of the ramp becomes greater. The first point (.01, 4 cm/s/s) lines slightly above the line, as does the third point (.03, 25 cm/s/s). The second point (.02, 11.1 cm/s/s) lies slightly below the line. confidence assessment: 3
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06:50:11 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 --> OK I forgot to explain which axis was which - the acceleration is the vertical axis, and the slope is the horizontal axis. self critique assessment: 3
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06:55:27 `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 --> The line intersects the y-axis at (0, -8) and intersects the vertical line at 0.5, at (0.05, 38) The rise = 38cm/s^2 + 8 cm/s^2 = 46 cm/s^2 The run = 0.05 - 0 = 0.05 Slope = Rise / Run = 46 cm/s^2/ 0.05 = 920 cm/s^2 confidence assessment: 3
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06:55:43 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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07:03:48 `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 --> To complete 100 cycles, it took 110 seconds. Each cycle therefore lasted 1.1 s confidence assessment: 3
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07:04:08 100 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 --> OK self critique assessment: 3
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07:14:34 `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 --> 1.1/s * sqrt(g) = 2 pi * sqrt (L) 1.1/s * sqrt(g) = 2 pi * sqrt (30 cm) 1.1/s * sqrt(g) = 2 pi * 5.5 sqrt(cm) sqrt(g) = 34.6 sqrt(cm) / (1.1/s) sqrt(g) = 31.4 sqrt(cm)/s g = (31.4 sqrt(cm/s)^2 g = 986 cm/s^2 Not sure how to get the units correct on this one - I added a "" /s "" on the end of the cycles value. Not sure if this was correct or not, but seems to make sense. confidence assessment: 2
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07:15:26 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 I see - it would have been 1.1 s, not 1.1/s. Now I understand. self critique assessment: 2
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?????????????assignment #008 008. Using the Acceleration of Gravity Physics I 03-01-2009
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20:53:16 `q001. The accepted value of the acceleration of gravity is approximately 980 cm/s^2 or 9.8 m/s^2. This will be the acceleration, accurate at most places within 1 cm/s^2, of any object which falls freely, that is without the interference of any other force, near the surface of the Earth. If you were to step off of a table and were to fall 1 meter without hitting anything, you would very nearly approximate a freely falling object. How fast would you be traveling when you reached the ground?
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RESPONSE --> v0 = 0 cm/s, 'ds = 1000 cm, a = 980 cm/s^2 vf^2 = v0^2 + 2 * a * 'ds vf^2 = 0 + 2 * 980 cm/s^2 * 1000 cm vf^2 = 1,960,000 cm^2/s^2 sqrt(vf^2) = (1,960,000 cm^2/s^2) vf = 1400 cm/s confidence assessment: 3
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20:55:58 You would have an initial vertical velocity of 0, and would accelerate at 9.8 m/s^2 in the same direction as your 1 meter vertical displacement. You would also have a slight horizontal velocity (you don't step off of a table without moving a bit in the horizontal direction, and you would very likely maintain a small horizontal velocity as you fell), but this would have no effect on your vertical motion. So your vertical velocity is a uniform acceleration with v0 = 0, `ds = 1 meter and a = 9.8 m/s^2. The equation vf^2 = v0^2 + 2 a `ds contains the three known variables and can therefore be used to find the desired final velocity. We obtain vf = +- `sqrt( v0^2 + 2 a `ds) = +- `sqrt ( 0^2 + 2 * 9.8 m/s^2 * 1 m)= +- `sqrt ( 19.6 m^2 / s^2) = +- 4.4 m/s, approx. Since the acceleration and displacement were in the direction chosen as positive, we conclude that the final velocity will be in the same direction and we choose the solution vf = +4.4 m/s.
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RESPONSE --> OK I see my error. I equated 1 m = 1000 cm, which is clearly wrong. 1m = 100 cm. I was off by a factor of 10. Had I done this correctly, I would have done the problem correctly. Everything else seems to make sense. I would have come up with a final answer of 442.72 cm/s, which is close to your approx of 4.4 m/s self critique assessment: 2
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21:00:27 `q002. If you jump vertically upward, leaving the ground with a vertical velocity of 3 m/s, how high will you be at the highest point of your jump? Note that as soon as you leave the ground, you are under the influence of only the gravitational force. All the forces that you exerted with your legs and other parts of your body to attain the 3 m/s velocity have done their work and are no longer acting on you. All you have to show for it is that 3 m/s velocity. So as soon as you leave the ground, you begin experiencing an acceleration of 9.8 m/s^2 in the downward direction. Now again, how high will you be at the highest point of your jump?
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RESPONSE --> at the highest point, the vf will equal 0 m/s, therefore: v0 = 3 m/s, vf = 0 m/s, a = -9.8 m/s^2 first, we must find 'dt: 'dt = (vf - v0) / a = (0 m/s - 3 m/s) / -9.8 m/s^2 = 0.31 s now we can find 'ds: 'ds = (vf + v0) / 2 * 'dt 'ds = (0 m/s + 3 m/s) / 2 * 0.31s = 0.465 m confidence assessment: 3
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21:00:56 From the instant the leave the ground until the instant you reach your highest point, you have an acceleration of 9.8 m/s^2 in the downward direction. Since you are jumping upward, and since we can take our choice of whether upward or downward is the positive direction, we choose the upward direction as positive. You might have chosen the downward direction, and we will see in a moment how you should have proceeded after doing so. For now, using the upward direction as positive, we see that you have an initial velocity of v0 = + 3 m/s and an acceleration of a = -9.8 m/s^2. In order to use any of the equations of motion, each of which involves four variables, you should have the values of three variables. So far you only have two, v0 and a. {}What other variable might you know? If you think about it, you will notice that when objects tossed in the air reach their highest point they stop for an instant before falling back down. That is precisely what will happen to you. At the highest point your velocity will be 0. Since the highest point is the last point we are considering, we see that for your motion from the ground to the highest point, vf = 0. Therefore we are modeling a uniform acceleration situation with v0 = +3 m/s, a = -9.8 m/s^2 and vf = 0. We wish to find the displacement `ds. Unfortunately none of the equations of uniformly accelerated motion contain the four variables v0, a, vf and `ds. This situation can be easily reasoned out from an understanding of the basic quantities. We can find the change in velocity to be -3 meters/second; since the acceleration is equal to the change in velocity divided by the time interval we quickly determine that the time interval is equal to the change in velocity divided by the acceleration, which is `dt = -3 m/s / (-9.8 m/s^2) = .3 sec, approx.; then we multiply the .3 second time interval by the 1.5 m/s average velocity to obtain `ds = .45 meters. However if we wish to use the equations, we can begin with the equation vf = v0 + a `dt and solve to find `dt = (vf - v0) / a = (0 - 3 m/s) / (-9.8 m/s^2) = .3 sec. We can then use the equation `ds = (vf + v0) / 2 * `dt = (3 m/s + 0 m/s) / 2 * .3 sec = .45 m. This solution closely parallels and is completely equivalent to the direct reasoning process, and shows that and initial velocity of 3 meters/second should carry a jumper to a vertical height of .45 meters, approximately 18 inches. This is a fairly average vertical jump. If the negative direction had been chosen as positive then we would have a = +9.8 m/s^2, v0 = -3 m/s^2 (v0 is be in the direction opposite the acceleration so if acceleration is positive then initial velocity is negative) and again vf = 0 m/s (0 m/s is the same whether going up or down). The steps of the solution will be the same and the same result will be obtained, except that `ds will be -.45 m--a negative displacement, but where the positive direction is down. That is we move .45 m in the direction opposite to positive, meaning we move .45 meters upward.
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RESPONSE --> OK! self critique assessment: 3
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21:14:56 `q003. If you roll a ball along a horizontal table so that it rolls off the edge of the table at a velocity of 3 m/s, the ball will continue traveling in the horizontal direction without changing its velocity appreciably, and at the same time will fall to the floor in the same time as it would had it been simply dropped from the edge of the table. If the vertical distance from the edge of the table to the floor is .9 meters, then how far will the ball travel in the horizontal direction as it falls?
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RESPONSE --> First, we must find the vf in the vertical direction: vf^2 = v0^2 + 2 * a * 'ds vf^2 = 0 + 2 * 9.8 m/s^2 * 0.9m = 17.64 m^2/s^2 sqrt(vf^2) = sqrt(17.64 m^2/s^2) vf = 4.2 m/s Now we can find the 'dt in the vertical direction: 'dt = (vf - v0) / a = (4.2 m/s - 0 m/s) / 9.8 m/s^2 'dt = 0.43 s Now that we know the time to fall, we can find the distance in the horizontal direction: v0 = 3 m/s, 'dt = 0.43 s, a = 9.8 m/s^2 first we need to find vf in the horizontal direction: vf = v0 + a * 'dt = 3 m/s + 9.8 m/s^2 * 0.43s vf = 7.21 m/s now, we can find the 'ds: 'ds = (vf + v0) / 2 * 'dt 'ds = (7.21 m/s + 3 m/s) / 2 * 0.43s 'ds = 5.11 m/s * 0.43s = 2.2 m confidence assessment: 3
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21:16:43 A ball dropped from rest at a height of .9 meters will fall to the ground with a uniform vertical acceleration of 9.8 m/s^2 downward. Selecting the downward direction as positive we have `ds = .9 meters, a = 9.8 m/s^2 and v0 = 0. Using the equation `ds = v0 `dt + .5 a `dt^2 we see that v0 = 0 simplifies the equation to `ds = .5 a `dt^2, so `dt = `sqrt( 2 `ds / a) = `sqrt(2 * .9 m / (9.8 m/s^2) ) = .42 sec, approx.. Since the ball rolls off the edge of the table with only a horizontal velocity, its initial vertical velocity is still zero and it still falls to the floor in .42 seconds. Since its horizontal velocity remains at 3 m/s, it travels through a displacement of 3 m/s * .42 sec = 1.26 meters in this time.
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RESPONSE --> OK Looks like the mistake I made was in trying to calculate the vAve in the horizontal direction, when I should have understood that the 3 m/s was the vAve in the horizontal direction. Had I done that correctly, I would have come up with the correct answer.
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