Phy 232
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.
** Your initial message (if any): **
** Is flow rate increasing, decreasing, etc.? **
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?
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I would expect the rate of flow to decrease.
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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?
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If water is flowing out of the cylinder, then the pressure is decreasing, so the velocity 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?
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I'm pretty sure that it's v1A1 = v2A2, or (velocity of water surface) * (area of surface of cylinder as determined by diameter) = (velocity of exiting water) * (area of hole as determined by diameter of hole).
So it's a directly proportional relationship for v1 and A1 or for v2 and A2, but an inverse relationship for v1 and v2, v1 and A2, A1 and v2, and A1 and A2.
So it seems that if the areas are the same, and if the velocity of the exiting water is going down, then to balance the equation, the velocity of the water surface would have to be increasing. So I revise my answer to the previous question.
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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?
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Well, F = ma, which can be written F = m * (dv/dt), so assuming m and dt are the same, there must be a change in F.
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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?
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I'm not entirely sure, but I assume it's the force of gravity pushing down and the mass of the water? Maybe some air pressure getting in on the action?
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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?
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It seems to be changing (decreasing) at a slower rate (decreasing rate). However, I predicted that it should be decreasing at an increasing rate. Well, we'll see.
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* What do you think a graph of depth vs. time would look like?
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If it's decreasing at a decreasing rate, the slope would be getting less steep as time went on. If it's decreasing at an increasing rate, the slope would be getting steeper as time went on.
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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?
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It decreases.
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* Does this distance change at an increasing, decreasing or steady rate?
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From what I can tell, it decreases at a decreasing rate.
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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.
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If it's decreasing at a decreasing rate, it would be a curve downward, and the slope would be getting less steep as time went on.
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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:
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1 1518.246 1518.246
2 1520.477 2.230469
3 1523.035 2.558594
4 1525.543 2.507813
5 1528.137 2.59375
6 1531.27 3.132813
7 1534.328 3.058594
8 1537.934 3.605469
9 1542.035 4.101563
10 1546.98 4.945313
11 1553.531 6.550781
12 1565.293 11.76172
13 1568.445 3.152344
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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.
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0.85cm
2.8cm
4.7cm
6.6cm
8.5cm
10.3cm
12.1cm
13.8cm
15.5cm
17.2cm
18.9cm
20.5cm
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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.
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0, 20.5cm
2.23, 18.9cm
4.79, 17.2cm
7.3, 15.5cm
9.89, 13.8cm
13.02, 12.1cm
16.08, 10.3cm
19.68, 8.5cm
23.78, 6.6cm
28.73, 4.7cm
35.28, 2.8cm
47.04, 0.85cm
50.19, 0cm
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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?
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It's decreasing at a slightly slower and slower rate, so it contradicts my prediction that it would decrease at an increasing rate.
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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.
image110.gif (4103 bytes)
Describe your graph in the language of the Describing Graphs exercise.
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The graph is slightly decreasing at a decreasing rate, meaning the slope is getting slightly less steep as you move from the right to the left of the graph.
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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:
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-0.72 cm/s
-0.66 cm/s
-0.68 cm/s
-0.66 cm/s
-0.54 cm/s
-0.59 cm/s
-0.50 cm/s
-0.46 cm/s
-0.38 cm/s
-0.29 cm/s
-0.17 cm/s
-0.27 cm/s
I just took all of the slope intervals from the position graph (i.e., (20.5cm - 18.9cm)/(0s - 2.23s) =~ -0.72 cm/s.
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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):
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1.12
3.51
6.05
8.60
11.46
14.55
17.88
21.73
26.26
32.01
41.16
48.62
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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.
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1.12, -0.72 cm/s
3.51, -0.66 cm/s
6.05, -0.68 cm/s
8.60, -0.66 cm/s
11.46, -0.54 cm/s
14.55, -0.59 cm/s
17.88, -0.50 cm/s
21.73, -0.46 cm/s
26.26, -0.38 cm/s
32.01, -0.29 cm/s
41.16, -0.17 cm/s
48.62, -0.27 cm/s
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* Sketch a graph of average velocity vs. clock time. Describe your graph, using the language of the Describing Graphs exercise.
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Well, my graph is fairly squiqqly and definitely ascending, but from what I've drawn, I'd say a best-fit line (the slope being the acceleration) would be close to linear (velocity increasing at a constant rate).
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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.
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0.025 cm/s/s
-.008 cm/s/s
.008 cm/s/s
0.042 cm/s/s
-.016 cm/s/s
0.027 cm/s/s
0.010 cm/s/s
0.018 cm/s/s
0.016 cm/s/s
0.013 cm/s/s
-0.013 cm/s/s
The acceleration values are just the slope intervals from the velocity graph (i.e., (-.66cm/s - (-.72cm/s))/(3.51s - 1.12s) =~ 0.025cm/s/s).
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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.
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2.32, 0.025 cm/s/s
4.78, -.008 cm/s/s
7.33, .008 cm/s/s
10.03, 0.042 cm/s/s
13.01, -.016 cm/s/s
16.22, 0.027 cm/s/s
19.81, 0.010 cm/s/s
24.00, 0.018 cm/s/s
29.14, 0.016 cm/s/s
36.59, 0.013 cm/s/s
44.89, -0.013 cm/s/s
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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?
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Well, my best-fit line was nearly linear, so that would imply that the acceleration of the water surface is constant. The results/graph are a little iffy, though, so maybe it was just an inconclusive result. I have a feeling there's some atmospheric/air pressure dealio going on that I'm missing.
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`gr5
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