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?
One would expect the rate of flow to decrease because there is less force pushing the fluid out of the cylinder as the fluid level decreases.
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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?
One would expect the velocity of the water surface to decrease because the water will flow out of the cylinder at a slower and slower rate. Therefore, the surface of the water will be traveling less distance with respect to time.
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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 velocity of the water surface could be determined by first finding the cross sectional area of the hole. This value would be pi*radius of the hole^2. This area would then be multiplied by the velocity of the exiting water to find the volume of water leaving the cylinder over a given amount of time. Finally, the cross sectional area of the cylinder would be found with pi*radius of cylinder^2 and the volume of water would be divided by this value.
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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?
According to Newtons 1st Law of Motion, an object will remain in its same state of motion or rest unless an external force is applied. Therefore, the water changing its velocity signifies that an external force is acting on the system.
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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 is the reduction in volume as the water enters the hole. This reduction in volume is what ultimately forces the water out with a greater velocity.
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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?
It is difficult to tell from the pictures; however, it appears that the depth is changing at a slower and slower rate.
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What do you think a graph of depth vs. time would look like?
The graph of depth vs. time would look like the downward half of a quadratic function. In other words, it would decrease sharply at first and level off over time.
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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 over time as shown by the pictures.
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Does this distance change at an increasing, decreasing or steady rate?
This 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.
Like the depth vs. time graph, this graph would look like the half of a quadratic function with a negative slope. Because the distance is changing at a decreasing rate, the graph would slope sharply downward at first and then level off as time progresses.
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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 analternative 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 47.23438 47.23438
2 49.11719 1.882813
3 51.4375 2.320313
4 53.75 2.3125
5 56.27344 2.523438
6 59.01563 2.742188
7 62.02344 3.007813
8 65.42188 3.398438
9 69.28125 3.859375
10 73.95313 4.671875
11 80.57813 6.625
12 96.75781 16.17969
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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.
.6 cm
2.2 cm
3.8 cm
5.4 cm
6.9 cm
8.5 cm
9.9 cm
11.4 cm
12.9 cm
14.4 cm
15.9 cm
17.4 cm
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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.
0,17.4
1.88281,15.9
4.20312,14.4
6.51562,12.9
9.03906,11.4
11.78125,9.9
14.78906,8.5
18.1875,6.9
22.04687,5.4
26.71875,3.8
33.34375,2.2
49.52343,.6
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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 data shows that the depth is changing at a slower and slower 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.
Describe your graph in the language of the Describing Graphs exercise.
The graph looks like a quadratic function. It initially slopes down sharply; however, the slope becomes less negative over time.
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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:
The average velocities were found by dividing the change in distance from the hole over a given time interval by the length of the time interval. These velocities are listed in order starting from the beginning of the experiment: -.797 cm/s, -.646 cm/s, -.649 cm/s, -.594 cm/s, -.547 cm/s, -.465 cm/s, -.471 cm/s, -.389 cm/s, -.342 cm/s, -.242 cm/s, -.099 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):
These values were obtained by dividing each time interval by 2 and then adding this value to the time at which the interval started.
.941405
3.042965
5.35937
7.77734
10.410155
13.285155
16.48828
20.117185
24.38281
30.03125
41.43359
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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.
.941405,-.797
3.042965,-.646
5.35937,-.649
7.77734,-.594
10.410155,-.547
13.285155,-.465
16.48828,-.471
20.117185,-.389
24.38281,-.342
30.03125,-.242
41.43359,-.099
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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 appears to be a straight line with a positive slope.
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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.
The acceleration is simply the slope between each point on the average velocity graph. Therefore, these values were found using the formula for slope.
0.071851
-0.0013
0.022746
0.017852
0.028522
-0.00187
0.022596
0.011018
0.017704
0.012541
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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.
1.992185,0.071851
4.2011675,-.0013
6.568355,.022746
9.0937475,.017852
11.847655,.028522
14.8867175,-.00187
18.3027325,.022596
22.2499975,.011018
27.20703,.017704
35.73242,.012541
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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?
As shown by the deterioration of difference quotients experiment, the acceleration data has magnified the uncertainty in the original data. Therefore, solely based on the acceleration data, one would have to say the results are inconclusive. However, it is obvious that the acceleration should be constant due to the nature of the experimental system.
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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?
1.5 hours
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Very well done.
Further Notes:
Based on the apparent linearity of the v vs. t graph, you could conclude uniform acceleration.
This in turn would be consistent with the apparent parabolic nature of the graph, which you mentioned.
The random fluctuations in the accelerations are likely due only to the degree of uncertainty in your original measurement, which is magnified as you first calculate average velocities, then use these results to get the average accelerations.
In terms of calculus, the acceleration is the derivative of the velocity function, which in turn is the derivative of the position function. Thus uniform acceleration implies a linear velocity function, which in turn implies a quadratic acceleration function.
As you will see soon, and energy analysis would predict uniform acceleration.
As you will also see soon, analysis of the pressure and the forces that accelerate the water out of the hole would predict a quadratic depth vs. clock time function.
The force analysis and energy analysis therefore yield compatible results.