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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.
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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?
Your answer (start in the next line):
As water flows from the cylinder, the rate would decrease because the amount of water in the cylinder decreases causing less pressure to push the water out below it. The water flow rate continues to decrease as the water 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?
Your answer (start in the next line):
The velocity of the water surface and buoy would decrease because as the water level decreases, the amount of water above the exit point also decreases. This causes there to be lesser water pushing down on the exiting water as time goes on. The more water in the cylinder, the faster it flows out of the bottom, and as time goes on, the pressure becomes less and less causing the velocity to slow down.
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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):
All of these factors are interrelated because the assigned value of one determines the other. The velocity of the exiting water determines how fast the water in the cylinder goes down. And the size of the exiting hole determines the exiting speed of the water. Also, the size of the cylinder determines how much water it can hold which relates to how fast the water runs of the cylinder at the bottom.
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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):
If the velocity in the system changes, we know that an external force must have been added. The velocity in the water in the cylinder differing from the exiting waters velocity is a sign of this force being added. The velocity of the water at the two points is the same without any external forces being added to the system, but this is not the case here.
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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):
For the exiting velocity to be increased, we know that pressure must have been applied to the water in the cylinder. This is because we know that added pressure to the water in the cylinder causes the exiting waters velocity to increase because more pressure from above has nowhere to exit but from the hole at the bottom. Therefore there must be an external force to the water in 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 of the water inside the cylinder is moving at a slower and slower rate. This is because as more water escapes through the exiting hole, there is less water in the cylinder. With less water in the cylinder, there is less pressure on the water exiting causing its speed to be decreased. This means that as the water decreases, it flows out slower and slower causing the depth to decrease more slowly as time passes.
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What do you think a graph of depth vs. time would look like?
Your answer (start in the next line):
A depth vs. time graph would be very similar to an exponential graph because the depth doesnt decrease in a uniform rate, meaning it isnt a straight line going downward. The graph does have a negative slope because the water is decreasing but because it is decreasing at a decreasing rate, it forms a curved line, just like an exponential graph.
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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 of the water stream exiting the cylinder decreases with time. This occurs because there continues to be less and less water in the cylinder, meaning that there is less water pushing out the water at the bottom of the cylinder. This causes the water to gradually exit the cylinder at a slower rate and therefore the water doesnt shoot as far out of the hole because it is moving at a slower rate as time passes.
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Does this distance change at an increasing, decreasing or steady rate?
Your answer (start in the next line):
This distance changes at a decreasing rate because it is connected to the rate at which the water flows out of the hole in the cylinder. We know that the water speed and level decreases at a decreasing rate, and that same rate is applied to the horizontal distance it travels. How fast the water shoots out of the hole determines the distance at which the water travels.
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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):
This graph would be relatively the same as the other graph with the water level on it. Since both are decreasing at a decreasing rate, they both would have the same shape to them. This graph would have a negative slope because the distance is decreasing as time passes.
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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 22.58594 3.40625
2 26.29688 3.710938
3 32.63281 6.335938
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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):
4.75 inches
2.375 inches
1.1875 inches
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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):
Clock times(seconds) Depth of water(inches from hole)
0 4.75
3.4 2.375
9.73 1.1875
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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):
They support my hypothesis above because the water is decreasing at a decreasing rate. The numbers and a graph of these numbers show an exponential behavior of the graph and indicate that the rate isnt constant through the water draining from the cylinder.
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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.
Your answer (start in the next line):
This graph has a negative slope because the water level is decreasing. Its shape is similar to an exponential function, and that is because the water is exiting the cylinder a decreasing rate, causing the slope to be less and less steep as time goes on. This graph shows those features and indicates how the water level behaves in this scenario. Decreasing at a decreasing rate is the best way to describe this 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:
Your answer (start in the next line):
1st interval: total distance / total time = velocity. 4.75 inches - 2.375 inches = 2.375 inches total distance traveled. 3.4 seconds is the total time elapsed. This means that 2.375 / 3.4 = .6985 inches per second.
2nd interval: total distance / total time = velocity. 2.375 inches - 1.1875 inches = 1.1875 inches total distance traveled. 6.33 seconds is the total time elapsed. This means that 1.1875 / 6.33 = .1876 inches per second.
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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):
1st interval: 3.4 seconds + 0 seconds = 3.4 seconds. This divided by 2 gives us our midpoint time for this interval, which equals 1.7 seconds.
2nd interval: 3.4 seconds + 6.33 seconds = 9.73 seconds. This divided by 2 gives us our midpoint time for this interval, which equals 4.865 seconds.
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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):
Clock times(seconds) Average velocity(inches per second)
1.7 .6985
4.865 .1876
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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):
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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):
The average acceleration value for the time interval listed above would be .8861 inches per second / 3.2885 seconds = .2695 inches / seconds^2. The acceleration is the derivative of velocity and after calculating the numbers shown in the graph above, the acceleration is determined.
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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):
Clock times(seconds) Acceleration
3.2885 .2695
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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):
The acceleration of the water surface is decreasing because the velocity is not constant, therefore there is acceleration, and the water surface is truly slowing down as time goes on. If the velocity of the water is decreasing then so will its acceleration. These graphs show that the water surface is slowing down at a slowing down rate just as predicted early on.
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Decreasing velocity doesn't mean decreasing acceleration, it just means negative acceleration.
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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):
The slope is -.1648. This comes from rise = .1876 - .6985 = -.5109 inches. Run = 4.8 - 1.7 = 3.1. Rise over run gives you the slope of the graph.
The slope is the steepness of the graph and shows how it behaves at each point in order to determine its direction.
I think my line represents the data fairly well considering outside forms of error and bias will always be present in experiments. Miscalculations and improper timings can be targets for bad data used for analyzing.
My average velocity vs. midpoint shows a decreasing acceleration because the velocity of the water inside of the cylinder is decreasing as time goes on and as more water flows out of the cylinder. The velocity will slow down and because there is less water above pushing it downward.
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See my previous note.
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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 hour and 10 minutes
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You may add any further comments, questions, etc. below:
Your answer (start in the next line):
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&#This lab submittion looks good. See my notes. Let me know if you have any questions.
Revision isn't requested, but if you do choose to submit revisions, clarifications or questions, please insert them into a copy of this document, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
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