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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.
** Flow Experiment_labelMessages **
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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?
Your answer (start in the next line):
I would expect the rate of flow to decrease as the water flows from the cylinder.
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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?
Your answer (start in the next line):
I would expect the velocity of the water surface to decrease as water flows out of the cylinder.
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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?
Your answer (start in the next line):
The diameter of the cylinder and the diameter of the hole would affect the velocity of the water surface. The larger
the diameter of the cylinder and diameter of hole the greater the velocity of the water surface and exiting water
will be.
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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?
Your answer (start in the next line):
The water would not accelerate without a force being exerted on it. If there was no action by a force the velocity
would be the same inside and outside the cylinder.
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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?
Your answer (start in the next line):
Gravity and pressure are acting against the water inside and outsid the cylinder.
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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?
Your answer (start in the next line):
The depth seems to be changing at a slower and slower rate.
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What do you think a graph of depth vs. time would look like?
Your answer (start in the next line):
As time increased the depth would decrease. Slope would be negative and get less negative 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?
Your answer (start in the next line):
The horizontal distance traveled by the stream decreases as time goes on.
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Does this distance change at an increasing, decreasing or steady rate?
Your answer (start in the next line):
The distance changes 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.
Your answer (start in the next line):
As the value of time increased the value of horizontal distance would decrease at a decreasing rate.
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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. If you will be using that alternative, read all the
instructions, then at the end you will see instructions for modifying the procedure to use a soft drink bottle.
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:
Your answer (start in the next line):
1 207.5195 207.5195
2 209.4453 1.925781
3 210.3398 .8945313
4 211.8867 1.546875
5 213.3359 1.449219
6 215.0898 1.753906
7 217.082 1.992188
8 220.1016 3.019531
9 222.5273 2.425781
10 223.7773 1.25
11 226.6055 2.828125
12 228.2514 1.645874
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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.
Your answer (start in the next line):
0cm
1.8cm
3.6cm
5.4cm
7.2cm
9cm
10.8cm
12.6cm
14.4cm
16.2cm
18cm
19.8cm
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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.
Your answer (start in the next line):
0cm, 0
1.8cm, 1.925781
3.6cm, 2.8203123
5.4cm, 4.3671873
7.2cm, 5.8164063
9cm, 7.5703123
10.8cm, 9.5625003
12.6cm, 12.5820313
14.4cm, 15.0078123
16.2cm, 16.2578123
18cm, 19.0859373
19.8cm, 20.7318113
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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?
Your answer (start in the next line):
The rate sped up and slowed down unlike my prediction. This contradicts my hypothesis.
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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.
Your answer (start in the next line):
The graph increases as the slope increases.
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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:
Your answer (start in the next line):
I divided the distance by the amount of time spent covering that distance.
0.93468572
1.276454384
1.236493795
1.23787776
1.188854521
1.129411729
1.001428124
0.959500273
0.99644403
0.943102752
0.955054033
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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):
Your answer (start in the next line):
0.9628905
2.37304665
3.5937498
5.0917968
6.6933593
8.5664063
11.0722658
13.7949218
15.6328123
17.6718748
19.9088743
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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.
Your answer (start in the next line):
0.9628905, 0.93468572
2.37304665, 1.276454384
3.5937498, 1.236493795
5.0917968, 1.23787776
6.6933593, 1.188854521
8.5664063, 1.129411729
11.0722658, 1.001428124
13.7949218, 0.959500273
15.6328123, 0.99644403
17.6718748, 0.943102752
19.9088743, 0.955054033
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Sketch a graph of average velocity vs. clock time. Describe your graph, using the language of the Describing Graphs
exercise.
Your answer (start in the next line):
For the most part the slope of the graphe increases as x-values increase.
@&
It appears that the overall trend of the graph is to decrease.
*@
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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.
Your answer (start in the next line):
I divided each individual time interval by its corresponding time interval.
0
0.485354108
1.426953293
0.799349524
0.854168873
0.677832518
0.566920255
0.331650221
0.395542826
0.797155224
0.333472796
0.580271657
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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.
Your answer (start in the next line):
0.9628905, 0.485354108
2.37304665, 1.426953293
3.5937498, 0.799349524
5.0917968, 0.854168873
6.6933593, 0.677832518
8.5664063, 0.566920255
11.0722658, 0.331650221
13.7949218, 0.395542826
15.6328123, 0.797155224
17.6718748, 0.333472796
19.9088743, 0.580271657
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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?
Your answer (start in the next line):
Overall acceleration decreases, but there are some point where it increases.
I think the acceleration of the water surface is actually decreasing.
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Go back to your graph of average velocity vs. midpoint clock time. Fit the best straight line you can to your data.
What is the slope of your straight line, and what does this slope represent? Give the slope in the first line, your
interpretation of the slope in the second.
How well do you think your straight line represents the actual behavior of the system? Answer this question and
explain your answer.
Is your average velocity vs. midpoint clock time graph more consistent with constant, increasing or decreasing
acceleration? Answer this question and explain your answer.
Your answer (start in the next line):
slope = 0.0075
A slope of 0.0075 means that the line is positive and increasing. The slope implies that average velocity is increasing.
y=0.0075x+0.9228
I think the straight line does a poor job of representing the actual behavior of the system.
The actual graph is mor consisten with a decreasing acceleration. If you take the first point out of the data the slope of the graph is generally decreasing.
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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?
Your answer (start in the next line):
1.5 hours
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You may add any further comments, questions, etc. below:
Your answer (start in the next line):
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*#&!
&#Your work on this lab exercise looks good. Let me know if you have any questions. &#