Phy 202
Your 'flow experiment' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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6/6 130pm
** #$&* Is flow rate increasing, decreasing, etc.? **
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The picture below shows a graduated cylinder containing water, with dark coloring (actually a soft drink). Water is flowing out of the cylinder through a short thin tube in the side of the cylinder. The dark stream is not obvious but it can be seen against the brick background.
You will use a similar graduated cylinder, which is included in your lab kit, in this experiment. If you do not yet have the kit, then you may substitute a soft-drink bottle. Click here for instructions for using the soft-drink bottle.
In this experiment we will observe how the depth of water changes with clock time.
In the three pictures below the stream is shown at approximately equal time intervals. The stream is most easily found by looking for a series of droplets, with the sidewalk as background.
Based on your knowledge of physics, answer the following, and do your best to justify your answers with physical reasoning and insight:
As water flows from the cylinder, would you expect the rate of flow to increase, decrease or remain the same as water flows from the cylinder?
Based on what I have learned about physics so far, I would expect the rate of flow to decrease as the liquid flows from the cylinder. I can tell this is true based on the pictures above. The difference between the flows in the first and second pictures is about half an inch, while the difference between the flows in the second and third pictures is about a quarter of an inch.
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As water flows out of the cylinder, an imaginary buoy floating on the water surface in the cylinder would descend.
Would you expect the velocity of the water surface and hence of the buoy to increase, decrease or remain the same?
I would expect the velocity of the water surface and the buoy to decrease, because it is directly related to the rate of flow of water, which is decreasing.
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How would the velocity of the water surface, the velocity of the exiting water, the diameter of the cylinder and the diameter of the hole be interrelated? More specifically how could you determine the velocity of the water surface from the values of the other quantities?
The diameter of both the cylinder and the hole play into how far the water level drops and how far the flow flies. Taken over subsequent time intervals, the difference in these distances divided by the time intervals gives us the velocities of both the water surface and the exiting water.
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The water exiting the hole has been accelerated, since its exit velocity is clearly different than the velocity it had in the cylinder.
Explain how we know that a change in velocity implies the action of a force?
Acceleration by definition is a change in velocity over a time interval, and Newtons second law of motion states that the acceleration of an object is directly proportional to the net force acting on it.
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What do you think is the nature of the force that accelerates the water from inside the cylinder to the outside of the outflow hole?
The nature of the force accelerating the water is gravity. As the mass of the liquid in the cylinder escapes, there is less of a mass for gravity to act upon, so as the mass decreases, the total acceleration due to gravity upon the mass will decrease.
Your argument is well-conceived, but the application of Newton's Second Law is more subtle. Newton's Second Law applies only to the net force, as you stated previously. The water in the container is subject to forces in addition to that of gravity (e.g., the container holds the water up (and in)), so the net force on the water in the container is actually quite small. Only the water that exits the hole speeds up, and the net force on that water turns out to be due to pressure.
We'll see the details later.
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From the pictures, answer the following and justify your answers, or explain in detail how you might answer the questions if the pictures were clearer:
Does the depth seem to be changing at a regular rate, at a faster and faster rate, or at a slower and slower rate?
The depth appears to be changing at a slower and slower rate, because the difference between the first and second pictures is about Ό, while the difference between the second and third pictures is about 1/8.
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What do you think a graph of depth vs. time would look like?
A graph of depth vs. time would look like an exponential graph, with a steep left side that gradually levels off as it proceeds down and to the right.
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Does the horizontal distance (the distance to the right, ignoring the up and down distance) traveled by the stream increase or decrease as time goes on?
The horizontal distance traveled by the stream decreases as time goes on. This can be noted in the pictures, because the stream in the third picture is much closer than the stream in the first or second pictures.
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Does this distance change at an increasing, decreasing or steady rate?
The distance is changing at a decreasing rate, because the difference between the flows in the first and second pictures is about ½, while the difference between the flows in the second and third pictures is about Ό.
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What do you think a graph of this horizontal distance vs. time would look like? Describe in the language of the Describing Graphs exercise.
The graph of the horizontal distance vs. time would look like an exponential graph, decreasing at a decreasing rate (with a decreasing slope).
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You can easily perform this experiment in a few minutes using the graduated cylinder that came with your kit. If you don't yet have the lab materials, see the end of this document for instructions an alternative setup using a soft-drink bottle instead of the graduated cylinder.
Setup of the experiment is easy. You will need to set it up near your computer, so you can use a timing program that runs on the computer. The cylinder will be set on the edge of a desk or tabletop, and you will need a container (e.g., a bucket or trash can) to catch the water that flows out of the cylinder. You might also want to use a couple of towels to prevent damage to furniture, because the cylinder will leak a little bit around the holes into which the tubes are inserted.
Your kit included pieces of 1/4-inch and 1/8-inch tubing. The 1/8-inch tubing fits inside the 1/4-inch tubing, which in turn fits inside the two holes drilled into the sides of the graduated cylinder.
Fit a short piece of 1/8-inch tubing inside a short piece of 1/4-inch tubing, and insert this combination into the lower of the two holes in the cylinder. If the only pieces of 1/4-inch tubing you have available are sealed, you can cut off a short section of the unsealed part and use it; however don't cut off more than about half of the unsealed part--be sure the sealed piece that remains has enough unsealed length left to insert and securely 'cap off' a piece of 1/4-inch tubing.
Your kit also includes two pieces of 1/8-inch tubing inside pieces of 1/4-inch tubing, with one end of the 1/8-inch tubing sealed. Place one of these pieces inside the upper hole in the side of the cylinder, to seal it.
While holding a finger against the lower tube to prevent water from flowing out, fill the cylinder to the top mark (this will be the 250 milliliter mark).
Remove your thumb from the tube at the same instant you click the mouse to trigger the TIMER program.
The cylinder is marked at small intervals of 2 milliliters, and also at larger intervals of 20 milliliters. Each time the water surface in the cylinder passes one of the 'large-interval' marks, click the TIMER.
When the water surface reaches the level of the outflow hole, water will start dripping rather than flowing continuously through the tube. The first time the water drips, click the TIMER. This will be your final clock time.
We will use 'clock time' to refer to the time since the first click, when you released your thumb from the tube and allowed the water to begin flowing.
The clock time at which you removed your thumb will therefore be t = 0.
Run the experiment, and copy and paste the contents of the TIMER program below:
1 34.1569999999992 34.1569999999992
2 36.3210000000036 2.16400000000431
3 38.5610000000015 2.23999999999796
4 40.5450000000055 1.98400000000402
5 42.8810000000012 2.33599999999569
6 45.3130000000019 2.4320000000007
7 48.1280000000042 2.81500000000233
8 51.096000000005 2.96800000000076
9 54.4720000000016 3.37599999999657
10 58.5280000000057 4.05600000000413
11 63.5679999999993 5.0399999999936
12 71.8559999999998 8.28800000000047
13 74.648000000001 2.79200000000128
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Measure the large marks on the side of the cylinder, relative to the height of the outflow tube. Put the vertical distance from the center of the outflow tube to each large mark in the box below, from smallest to largest distance. Put one distance on each line.
0.95
2.96
4.91
6.90
8.81
10.71
12.61
14.45
16.30
18.15
20.02
21.88
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Now make a table of the position of the water surface vs. clock time. The water surface positions will be the positions of the large marks on the cylinder relative to the outflow position (i.e., the distances you measured in the preceding question) and the clock times will as specified above (the clock time at the first position will be 0). Enter 1 line for each event, and put clock time first, position second, with a comma between.
For example, if the first mark is 25.4 cm above the outflow position and the second is 22.1 cm above that position, and water reached the second mark 2.45 seconds after release, then the first two lines of your data table will be
0, 25.4
2.45, 22.1
If it took another 3.05 seconds to reach the third mark at 19.0 cm then the third line of your data table would be
5.50, 19.0
Note that it would NOT be 3.05, 19.0. 3.05 seconds is a time interval, not a clock time. Again, be sure that you understand that clock times represent the times that would show on a running clock.
The second column of your TIMER output gives clock times (though that clock probably doesn't read zero on your first click), the third column gives time intervals. The clock times requested here are those for a clock which starts at 0 at the instant the water begins to flow; this requires an easy and obvious modification of your TIMER's clock times.
For example if your TIMER reported clock times of 223, 225.45, 228.50 these would be converted to 0, 2.45 and 5.50 (just subtract the initial 223 from each), and these would be the times on a clock which reads 0 at the instant of the first event.
Do not make the common error of reporting the time intervals (third column of the TIMER output) as clock times. Time intervals are the intervals between clicks; these are not clock times.
0.00 , 21.88
2.16 , 20.02
4.40 , 18.15
6.39 , 16.30
8.72 , 14.45
11.16 , 12.61
13.97 , 10.71
16.94 , 8.81
20.32 , 6.90
24.37 , 4.91
29.41 , 2.96
37.69 , 0.95
40.49 , 0.00
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You data could be put into the following format:
clock time (in seconds, measured from first reading) Depth of water (in centimeters, measured from the hole)
0 14
10 10
20 7
etc. etc.
Your numbers will of course differ from those on the table.
The following questions were posed above. Do your data support or contradict the answers you gave above?
Is the depth changing at a regular rate, at a faster and faster rate, or at a slower and slower rate?
The depth is changing at a slower and slower rate, because the water level dropped 20.02 18.15 = 1.87 cm in 4.40 2.16 = 2.24 seconds, or 0.835 cm/sec; later it dropped 12.61 10.71 = 1.90 cm in 13.97 11.16 = 2.81 seconds, or 0.676 cm/sec; and towards the end, it dropped 2.96 0.95 = 2.01 cm in 37.69 29.41 = 8.28 seconds, or 0.243 cm/sec. I believe my data supports the answers I gave above.
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Sketch a graph of depth vs. clock time (remember that the convention is y vs. x; the quantity in front of the 'vs.' goes on the vertical axis, the quantity after the 'vs.' on the horizontal axis). You may if you wish print out and use the grid below.
Describe your graph in the language of the Describing Graphs exercise.
The graph is decreasing at a decreasing rate, similar to an exponential graph, with an actual y-intercept at 21.88 cm.
For reasons you don't really need to know, the graph is actually parabolic, not exponential. An exponential would have a consistent 'half-life' and a horizontal asymptote; this graph actually has a vertex, which indicates that the flow, even theoretically, will cease.
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caution: Be sure you didn't make the common mistake of putting time intervals into the first column; you should put in clock times. If you made that error you still have time to correct it. If you aren't sure you are welcome to submit your work to this point in order to verify that you really have clock times and not time intervals
Now analyze the motion of the water surface:
For each time interval, find the average velocity of the water surface.
Explain how you obtained your average velocities, and list them:
For each line starting with the second, I subtracted the previous clock time from the current clock time to arrive at the time interval, which matches my TIMER information. I then subtracted the previous distance from the current distance to determine the change in depth, which I divided by the time interval to arrive at the average velocities for each time interval.
-0.861
-0.835
-0.930
-0.794
-0.754
-0.676
-0.640
-0.565
-0.491
-0.387
-0.243
-0.339
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Assume that this average velocity occurs at the midpoint of the corresponding time interval.
What are the clock times at the midpoints of your time intervals, and how did you obtain them? (Give one midpoint for each time interval; note that it is midpoint clock time that is being requested, not just half of the time interval. The midpoint clock time is what the clock would read halfway through the interval. Again be sure you haven't confused clock times with time intervals. Do not make the common mistake of reporting half of the time interval, i.e., half the number in the third column of the TIMER's output):
To arrive at the midpoints of the time intervals, I took the time intervals obtained in the previous question, divided each by half, then added the result to the previous clock time.
1.08
3.28
5.40
7.56
9.94
12.57
15.46
18.63
22.35
26.89
33.55
39.09
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Make a table of average velocity vs. clock time. The clock time on your table should be the midpoint clock time calculated above.
Give your table below, giving one average velocity and one clock time in each line. You will have a line for each time interval, with clock time first, followed by a comma, then the average velocity.
1.08 , -0.861
3.28 , -0.835
5.40 , -0.930
7.56 , -0.794
9.94 , -0.754
12.57 , -0.676
15.46 , -0.640
18.63 , -0.565
22.35 , -0.491
26.89 , -0.387
33.55 , -0.243
39.09 , -0.339
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Sketch a graph of average velocity vs. clock time. Describe your graph, using the language of the Describing Graphs exercise.
The graph for the most part appears to be a linear graph, increasing at a constant rate. The third and last points appear to be anomalies when compared to the other points
surface tension and other factor do start to affect the data near the end
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For each time interval of your average velocity vs. clock time table determine the average acceleration of the water surface. Explain how you obtained your acceleration values.
I determined the average acceleration by taking the difference of the midpoint values for each line and the difference of the average velocities for each midpoint, then dividing the velocity difference by the midpoint difference.
-0.797
0.012
-0.045
0.063
0.017
0.030
0.013
0.024
0.020
0.023
0.022
-0.017
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Make a table of average acceleration vs. clock time, using the clock time at the midpoint of each time interval with the corresponding acceleration.
Give your table in the box below, giving on each line a midpoint clock time followed by a comma followed by acceleration.
1.08 , -0.797
3.28 , 0.0119
5.40 , -0.0448
7.56 , 0.0628
9.94 , 0.0167
12.57 , 0.0297
15.46 , 0.0126
18.63 , 0.0235
22.35 , 0.0198
26.89 , 0.0230
33.55 , 0.0216
39.09 , -0.0174
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Answer two questions below:
Do your data indicate that the acceleration of the water surface is constant, increasing or decreasing, or are your results inconclusive on this question?
Do you think the acceleration of the water surface is actually constant, increasing or decreasing?
My data indicate that the acceleration of the water surface is constant, because the majority of the points hover around the 0.02 horizontal line.
I think the acceleration of the water is actually constant, because the velocity at which the water level drops appears to be steadily slowing down.
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Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
It took approximately an hour and a half to complete the experiment.
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Excellent work. See my notes.