Some of your work might not have been posted. This is usually due to student error, but in this case it was a clerical error on my part. Let me know if there's other work you have submitted that does not appear on your access site.
Your work on flow experiment has been received (and is excellent). Scroll down through the document to see my comment at the end.
Your initial message (if any):
I just used the first part of the video experiment on the dvd. My kit didn't come with a graduated cylinder, and I am not sure if it was supposed to or not. We are working on getting that all straightened out.
Is flow rate increasing, decreasing, etc.?
From the picture it appears as though the rate of flow is decreasing because the stream of liquid is shooting out farther at the beginning of the experiment, but as time goes on and fluid drains out of the cylinder the stream of fluid is draining with less pressure and therefore at a slower rate.
Is the velocity of the water surface increasing, decreasing, etc.?
According to Bournulli's principle, pressure is low where velocity is high, and pressure is high where velocity is low. I think the pressure at the water surface is probably lower than the pressure at the bottom, but since the water at the top cannot go anywhere until the water at the bottom leaves the hole it will move slower. So, since the pressure at the bottom of the cylinder is so high, the velocity will be low, and since the water at the top of the cylinder cannot move anywhere until the water exits the hole, the water surface will move at a decreasing speed.
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?
Velocity is displacement/time, or (final position - initial position)/(elapsed time). Area = (pi)(r^2) so you can use the diameter to find the area of the hole and the area of the cylinder. You can find out how much water is leaving the hole per second. Then you can figure out how much water is leaving the cylinder and you can find the displacement by subtracting the amount that is leaving the cylinder from the initial amount and then divide is by the number of seconds and it will give you the velocity of the water surface.
Explain how we know that a change in velocity implies the action of a force:
Newton's first law states that every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it. That means, that in order for velocity or direction to change, a net force must act on the object. The water has been accelerated due to the pressure forcing the water out of the small hole. Also gravity is pulling the water toward the ground.
Does the depth seem to be changing at a regular rate, at a faster and faster rate, or at a slower and slower rate
We do not know if the pictures were taken at exactly the same distance away from the cylinder, so we can't really take measurements by taking the height of the liquid from the screen, but assuming the camera was exactly the same distance away from the cylinder in each picture, it looks as though the depth is changing at an decreasing rate. The water surface line is going down by less with each equal time interval.
What do you think a graph of depth vs. time would look like?
The Y-axis would be depth and the X-axis would be time, with zero at the point where the two axes meet. The line would start at zero seconds but the depth would be very high because the cylinder is filled to the top at the beginning and gradually flows out and becomes less. As seconds go by the line will be on a steady decline, with the slope getting gradually less steep with time. It is getting less steep because the outflow is slowing down which will also slow down the displacement of the water surface.
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?
You can tell from the picture that the horizontal distance traveled by the stream decreases as time goes on, and this is because there is less pressure pushing the water out of the hole.
Does this distance change at an increasing, decreasing or steady rate?
This distance also appears to be a decreasing rate, because it looks like the horizontal distance is decreasing by less and less with each equal interval.
What do you think a graph of this horizontal distance vs. time would look like?
The horizontal distance would be on the Y-axis and the time would be on the X-axis, with zero being where the two axes meet. The line would start at zero seconds but the horizontal distance would be very great because of all the pressure pushing the water out of the hole. The line would be decreasing very steadily at first with a fairly steep slope, but the slope would get gradually less and less because the horizontal distance is decreasing less and less with each time interval.
The contents of TIMER program as you submitted them:
1 516.0625 516.0625
2 527.207 11.14453
3 541.1563 13.94922
4 543.2813 2.125
5 545.3828 2.101563
6 547.7773 2.394531
7 550.1914 2.414063
8 552.3945 2.203125
The vertical positions of the large marks as you reported them, relative to the center of the outflow hole
5 cm
6 cm
7 cm
8 cm
9 cm
10 cm
20 cm
30 cm
Your table for depth (in cm) vs clock time (in seconds)
0,30
11.14, 20
13.95, 10
2.12, 9
2.10, 8
2.39, 7
2.41, 6
2.20, 5
Is the depth changing at a regular rate, at a faster and faster rate, or at a slower and slower rate?
In support of my answer given in the earlier question, the depth is changing at a slower and slower rate because the water surface level is decreasing by less and less with each passing second.
Your description of your depth vs. t graph:
Water depth is along the Y-axis, and Clock time is along the X-axis.
From the distance of 30 cm above the hole to 10 cm above the hole, the water surface level is dropping at a very fast rate; the slope of the line is very steep during the 30 to 10 cm decline. After the water surface passes below the mark that is 10 cm above the hole, the slope of the line gets much less steep than before. This supported my earlier answer.
Your explanation and list of average average velocities:
Velocity = (final position - initial position)/(time elapsed)
So you would just find the displacement of the water surface by subtracting its final position from its initial position, then dividing that number by the time elapsed, and since velocity is a vector, the sign on the answer you calculate tells you which direction the water surface is traveling. Since we know the water surface is traveling downward along the y-axis (if we drew an imaginary x- and y- axis through the cylinder), the velocities we calculate should have a negative sign telling us the water is moving downward.
Example Calculation:
Velocity = (final - initial)/ time elapsed
Velocity = (20cm - 30cm)/11.15sec
Velocity = -.90 cm/sec OR .9 cm/sec downward
Velocities for each time interval:
.90 cm/sec downward
.72 cm/sec downward
.47 cm/sec downward
.48 cm/sec downward
.42 cm/sec downward
.41 cm/sec downward
.45 cm/sec downward
The midpoints of your time intervals and how you obtained them:
To get the midpoint I subtracted the second recorded time from the first recorded time. Then I divided their difference by two, and added that number back on to the the first recorded time and that gave me the midpoint of the time interval.
Example Calculation:
527.21 - 516.06 = 11.15
11.15 / 2 = 5.575
5.575 + 516.06 = 521.635 (the midpoint between the first two recorded times)
Your table of average velocity of water surface vs. clock time:
521.635, .90 downward
534.185, .72 downward
542.22, .47 downward
544.33, .48 downward
546.58, .42 downward
548.985, .41 downward
551.29, .45 downward
*I'm not sure if I am supposed to leave the negatives in front of the velocity. Am I supposed to leave the negatives in front of the velocities when graphing and finding the acceleration too?
Your description of your graph of average velocity vs clock time:
Average Velocity is on the y-axis and clock time is on the x-axis. The clock time was actually the midpoints between each time interval. The line is decreasing at an increasing rate from left to right and the slope is very steep. Then the line levels out and slope is not so steep. The line kind of goes up and down just a little bit but the general curve is decreasing at a decreasing rate. There is an exception of two points that go back up, instead of down like the other points, but that could have been human error in measuring the time. I may have gotten a little excited and clicked the mouse too early.
Your explanation of how acceleration values were obtained:
Average Acceleration = (final velocity - initial velocity)/(elapsed time)
So you would just find the difference between the final and initial velocity and then divide that number by the time that has elapsed. Acceleration is a vector, so the sign on the solution tells you which direction the water surface is accelerating.
Example Calculation:
Acceleration = (final velocity - initial velocity)/elapsed midpoint time
Acceleration = [(-.72cm) - (-.90cm)]/(534.185sec - 521.635sec)
Acceleration = .014 cm/sec
Acceleration Values:
.014 cm/sec
.031 cm/sec
-.005 cm/sec
.027 cm/sec
.004 cm/sec
.017 cm/sec
*Again, I don't know if I was supposed to carry the negatives through my calculations of acceleration, so if I WAS NOT supposed to, then my calculations are going to be backwards.
Your acceleration vs clock time table:
527.91, .014
538.20, .031
543.27, -.005
545.45, .027
547.78, .004
550.14, .017
* I am a little confused here because the acceleration is between each midpoint clock time, so there are not as many acceleration calculations as there are midpoints. I found the midpoints of the midpoints but that is pretty much just the same as the times I started with? Is that what I was supposed to do?
According to the evidence here, is acceleration increasing, decreasing, staying the same or is in not possible to tell?
My data indicates that the acceleration is increasing through the first two points, but then it decreases drastically at the third point of -.005. After decreasing so much it increases again at .027 and decreases again to .004 then increases to .017. So I would say that my results are inconclusive because the acceleration just increases and decreases back and forth over the chart.
I think the acceleration of the water should be decreasing. I think it should be increasing toward the top of the cylinder, but after reaching a certain point, probably around the 10 or 9 cm mark, it should be decreasing.
06-02-2006 18:39:22
Excellent work on the flow experiment. You clearly have a very good foundation in first-semester physics--very good technical and conceptual skills. You obviously had a very good course, and my guess is that you were also a very good student in that course.
After the due date we will be discussing this experiment further via an online forum.