#$&*
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 **
Interesting experiment. As I mention in my comments below, working with running water and computers simultaneously is not the safest thing in the world for one's data. Then again, I don't have any better idea about how to go about data collection in a scenario where one can't divide labor among lab partners, so I'm not sure how constructive this criticism is.
@&
The water flow isn't that fast and the TIMER can be operated at a safe distance with one hand, or by the mouse at an even greater distance, so safety isn't a big concern. Convenience, however, could be another matter.
*@
** **
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):
- RESPONSE: I would expect the rate of flow to decrease. This is the nature of Torricelli's Law (for the situation in question, assuming that the diameter and thus the radius and cross-sectional area of the graduated cylinder >> the diameter and thus the radius and cross-sectional area of the hole at the bottom, the velocity of the out-flowing water is equal to sqrt[2 * gravity * height of water column]).
@&
This depends, of course, on the velocity of the descending water in the cylinder being insignificant compared to that of the exiting water, a condition equivalent to your stated condition on cross-sectional areas.
You are probably aware that this velocity is the same as that of a freely falling object dropped from rest at the height of the water column, and that both results are the result of conversion of gravitational PE to KE.
The result also depends on the absence of dissipative force. Because of the viscosity of water and its adhesion to the edges of the hole, this ideal is not achieved.
*@
Because the velocity of the water flowing through an aperture of a given size determines the volume flow (and, for a noncompressible fluid, the mass flow) per unit time of that water, the change in the amount of water remaining and thus the change in height of that remaining water are both proportional to the velocity and therefore to sqrt(2gh). Sqrt(2gh) = sqrt(2g) * sqrt(h) = sqrt(2g) * (h^0.5), so the change in the change of the height of the water (i.e., the second derivative of the height equation) = the first derivative of the velocity equation = d/dt(k * (sqrt(2g)) * (h^0.5)) = k(sqrt(2g) * 0.5h)dh = k([0.5(sqrt[2g])]h)dh. Therefore, as height decreases, velocity decreases proportionally.
- - Note that here, the constant k actually signifies the area of the aperture (because volume flow per unit time = volume of water per unit time * area of aperture), which = [pi]([radius of aperture]^2).
#$&*
* 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):
- RESPONSE: I would expect the magnitude of the velocity of the water surface and hence the magnitude of the buoy's velocity to decrease. (Because the direction of these velocities is negative and because these velocities get closer and closer to zero, some sticklers for sign placement would say that the velocities become less negative and therefore increase, but those people would be missing the point about the physical behavior of the system.) Because the volume of a cylinder is directly proportional to its height (= [pi][radius^2][height]), an event having a given effect on the volume of a cylindrical water column with fixed radius will have an effect on that water column's height that is directly proportional to the effect on its volume. Therefore, when volume flow decreases over time, so will the magnitude of the rate of change in the water's height (this statement holds for either direction of flow, but its application here involves outflow of the water and negative change in volume and height).
#$&*
@&
The positive direction is arbitrary, and can certainly be chosen as downward. This would in fact be the natural choice in this case.
Changing the word 'volume' to 'speed' would also reduce the need for qualification.
*@
* How would the velocity of the water surface, the velocity of the exiting water, the diameter of the cylinder and the diameter of the hole be interrelated? More specifically how could you determine the velocity of the water surface from the values of the other quantities?
Your answer (start in the next line):
- RESPONSE: The velocity of the water surface would vary a) directly with (for these purposes, in linear proportion to) each of i) the velocity of the exiting water and ii) the square of the diameter of the hole and b) inversely with the square of the diameter of the cylinder.
- a) 1) Change in volume per unit time = volume flow per unit time = velocity of exiting water * area of hole = Velocity:ExitingWater * [pi]([hole radius]^2) = Velocity:ExitingWater * [pi]([.5(Diameter:Hole)]^2) = Velocity:ExitingWater * .25pi([(Diameter:Hole)^2]).
- -- 2) Velocity of water surface = rate of change of cylinder height = ([rate of change of volume of cylinder contents] / [area of cylinder]) = ([RateOfChange:Volume:Cylinder]/[(pi)([Radius:Cylinder]^2)]) = ([RateOfChange:Volume:Cylinder]/[(pi)([.5(Diameter:Cylinder)]^2)]) = ([RateOfChange:Volume:Cylinder]/[.25(pi)([Diameter:Cylinder]^2)]).
- b) Therefore, substituting the formula derived in 1) for RateOfChange:Volume:Cylinder in 2), velocity of water surface = ([Velocity:ExitingWater * .25(pi)([(Diameter:Hole)^2])]/[.25(pi)([Diameter:Cylinder]^2)]) = ([Velocity:ExitingWater * ([Diameter:Hole]^2)]/[(Diameter:Cylinder)^2]).
- c) This derivation passes the (necessary but not sufficient) check of evaluation in terms of dimensional analysis: Velocity = (L^1)(T^-1); diameter = (L^1); (L^1)(T^-1) ?= ([(L^1)(T^-1) * (L^2)]/[L^2]) Yes= (L^1)(T^-1).
#$&*
* 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):
- RESPONSE: Newton's First and Second Laws of Motion (bodies at rest [...] and force = mass * acceleration [with corollary 'acceleration = (force/mass)'], respectively)
#$&*
* What do you think is the nature of the force that accelerates the water from inside the cylinder to the outside of the outflow hole?
Your answer (start in the next line):
- RESPONSE: Both a) the force accelerating the water at hole level from inside the cylinder through the outflow hole to the surrounding environment and b) the force providing downward acceleration to the molecule-thin layers of water above that water are the sum of the weight of the water molecules at a given layer and the weight of the water molecules in all the layers above that layer. This weight is equal to those molecules' mass * the force of gravity, which equals gravity * the density of water * the volume of the portion of the water column at or above a given level (hole level for water going through; a graduated-cylinder measurement for water still above hole level). This equals gravity * water's density * the area of the cylinder * the height of the water above the layer in question, i.e., gravity * water's density * pi([cylinder radius]^2) * water height above the layer in question. Assuming water molecules to be infinitesimally small, the factor affecting a given water molecule is equal to this force / the area of the layer, which corresponds to the conceptual definition of pressure. (Gravity * water's density * cylinder's area * height of water at and above layer or molecule)/(cylinder's area) = gravity * water's density (generalizable to fluid's, or at least ideal fluid's, density) * height of water (generalizable to [at least ideal] fluid's height), or density * gravity * height for short, which is how Pascal (I think it was Pascal) formalized the application to fluids of this conceptual definition of pressure.
#$&*
@&
Very good.
The analysis is possible at the molecular level, but requires multivariable calculus and differential equations.
It's simpler here to consider a 'plug' of water being forced out of the cylinder, but this model while useful is a bit of an oversimplification.
*@
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):
- RESPONSE: The depth seems to be changing at a slower and slower rate. See my discussion in the first through third Responses; velocity of water surface = rate of change of depth, and because a) velocity of water surface is directly proportional to velocity of outflowing water and b) the rate of change of the velocity of outflowing water is directly proportional to constant * water height (= constant * depth), the rate of change of depth is directly proportional to (constant * depth), meaning that as depth decreases (and time increases), the rate of change of depth will likewise decrease.
#$&*
@&
The net force on the 'plug' depends on the cross-sectional area of the 'plug'; the distance over which it is applied depends on the length of the 'plug' and is hence proportional to the mass of the 'plug' as well as to the work done on it. The work done on it is equal to the change in its kinetic energy, so the work done per unit of mass is half the change in squared velocity. Just a couple of simple steps to 1/2 rho v^2.
*@
* What do you think a graph of depth vs. time would look like?
Your answer (start in the next line):
- RESPONSE: A graph of depth vs. time would look like the graph of ([initial volume of water / cross-sectional area of cylinder] - [volume of water lost so far / cross-sectional area of cylinder]) = ([initial volume of water - volume of water lost so far] / cross-sectional area of cylinder). The volume of water lost so far at time t is equivalent to the integral from 0 to t of the volume-flow-per-unit-time function, and the volume-flow-per-unit-time function is equal to the aperture's area (= [pi][(aperture radius)^2]) * the velocity of the outflowing water, i.e., ([pi][(aperture radius)^2]sqrt[2*gravity*height]). The velocity and its integral (= volume lost so far) will therefore depend on the initial height h, and I forget (and don't have time to figure out) how to integrate recursively defined functions, so I can't work the whole thing out, but here's the point: The equation will be the equation of a positive initial value minus a value that increases at a decreasing rate over time, so its graph will start at t = 0 with a positive y-intercept and then decrease at a decreasing rate (and thus be concave upward, because it involves a constant minus a value that changes in a concave-downward manner), approaching a horizontal asymptote at 0.
#$&*
@&
Good.
FYI, the velocity of the outflowing water is proportional the sqrt(y), where y is water depth relative to the hole.
Thus
dy/dt = k sqrt(y)
so that
dy / sqrt(y) = k dt.
Integrating both sides we get
2 sqrt(y) = k t + constant
so that
y = (k t + constant) ^2
where the right-hand proportionality constant k and the integration constant have 'absorbed' the factor 1/2.
This shows that in the idea case, water depth is a quadratic function of t. The integration constant turns out to be negative, resulting in a negative multiple of t to go with the positive coefficient k^2 of t^2. The result is a decreasing parabola that opens upward.
*@
* 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):
- RESPONSE: The distance decreases as time goes on. Ignoring the up and down distance is something of a misnomer because that distance determines the (non-varying) amount of time (= sqrt[2(Distance:ApertureToGround)/g]: Distance:ApertureToGround = .5g(t^2), so t^2 = [Distance:ApertureToGround]/[.5g]) for which the water can travel at whatever horizontal velocity it has. Because the water's velocity decreases as time goes on, the distance it travels (= the velocity * sqrt[2h/g]) decreases as well.
#$&*
@&
Good.
Note also that due to the result mentioned in the preceding note the water depth being a quadratic function of t, it changes at a linear rate, meaning that the range of the horizontal stream is a linear function of time.
*@
* Does this distance change at an increasing, decreasing or steady rate?
Your answer (start in the next line):
- RESPONSE: The distance changes (specifically, decreases) at a decreasing rate because i) the travel time is constant per the Response above and ii) the velocity decreases at a decreasing rate as described in the first Response of this assignment.
#$&*
* 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):
- RESPONSE: The graph of horizontal distance vs. time would start at a positive y-intercept (= time * sqrt(2g[Height:Initial]) = [2(Distance:ApertureToGround)/g]sqrt(2g[Height:Column:Initial]) = ([2sqrt2 * Distance:ApertureToGround * sqrt(Height:Column:Initial)]/sqrt[g]) = (2[Distance:ApertureToGround] * sqrt[(2 * Height:Column:Initial)/g]) and then decrease at a decreasing rate (making the graph concave upward) as it would approach a horizontal asymptote of 0.
#$&*
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. ???? I'm assuming that you mean for me to insert the end with the (1/8)-inch tube inside the (1/4)-inch tube into the bottom hole. I'm doing so for two reasons: First, although the .25- and .125-inch tubes once nested could qualify as a combination regardless of which end of the assembly went in, inserting the end with the nested/combined portion seems like a more pure application of the instructions. Second, putting the .125-inch tube on the other end would cause retention of water in the .25-inch tube and mess up the data. Please let me know if I'm wrong about which end to insert. ???? 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):
- RESPONSE:
- - 00 336.4922 336.4922
- - 01 338.4375 1.945313
- - 02 340.3906 1.953125
- - 03 342.4219 2.03125
- - 04 344.6016 2.179688
- - 05 346.6953 2.09375
- - 06 349.2344 2.539063
- - 07 351.75 2.515625
- - 08 354.5547 2.804688
- - 09 357.6016 3.046875
- - 10 360.9375 3.335938
- - 11 364.8828 3.945313
#$&*
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):
- RESPONSE: 1 cm = 10 mm = 12 to 12.[12]rptg. mL; 1 mm = 1.2 to 1.[21]rptg. mL; 1 mL = 0.825 mm to (almost exactly same)
- - 30 mL, 4.5 mm
- - 50 mL, 24.25 mm
- - 70 mL, 43.5 mm
- - 90 mL, 62.5 mm
- - 110 mL, 81.5 mm
- - 130 mL, 99 mm
- - 150 mL, 117.25 mm
- - 170 mL, 134.75 mm
- - 190 mL, 151.75 mm
- - 210 mL, 168.5 mm
- - 230 mL, 185 mm
- - 250 mL, 201.5 mm
#$&*
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):
- RESPONSE: To make calculations a bit more obvious and in units that are easier to work with, I've formatted the table as follows: (reading number, cylinder reading in L at gradation mark, [TIMER reading - 300 sec (I had it idling for a while at first while I moved everything around)], clock time [= time elapsed = (TIMER reading - TIMER reading at reading #0)], water column height in m at gradation mark [readings are rounded to nearest 0.25 mm = 0.00025 m]).
- - 00, 0.250, 36.4922, 00.0000, 0.20150
- - 01, 0.230, 38.4375, 01.9453, 0.18500
- - 02, 0.210, 40.3906, 03.8984, 0.16850
- - 03, 0.190, 42.4219, 05.9297, 0.15175
- - 04, 0.170, 44.6016, 08.1094, 0.13475
- - 05, 0.150, 46.6953, 10.2031, 0.11725
- - 06, 0.130, 49.2344, 12.7422, 0.09900
- - 07, 0.110, 51.7500, 15.2578, 0.08150
- - 08, 0.090, 54.5547, 18.0625, 0.06250
- - 09, 0.070, 57.6016, 21.1094, 0.04350
- - 10, 0.050, 60.9375, 24.4453, 0.02425
- - 11, 0.030, 64.8828, 28.3906, 0.00450
#$&*
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):
- RESPONSE: For the most part, the depth is changing at a slower and slower rate, as shown by the following table of intervals indicating a (usually) longer and longer time required to accomplish each water-level drop of a 20-mL magnitude.
- - Note: a) This table places reading intervals (corresponding to column-height intervals) on the independent-variable axis and time required, which for fixed changes in volume corresponds to mass flow rate, on the dependent-variable axis.
- - ----- b) This table plots differences in time vs. what amount to equally spaced differences in distance. As such, its plot is analogous to the graph of the first derivative of the (time required)-vs.-height function; therefore, its slope (an approximation of its first derivative) approximates the second derivative of the (time required)-vs.-(column height) function.
- - Reading interval Time interval (= Time:X - Time:[X - 1])
- - 00 to 01 1.945313
- - 01 to 02 1.953125
- - 02 to 03 2.031250
- - 03 to 04 2.179688
- - 04 to 05 2.093750
- - 05 to 06 2.539063
- - 06 to 07 2.515625
- - 07 to 08 2.804688
- - 08 to 09 3.046875
- - 09 to 10 3.335938
- - 10 to 11 3.945313
- - The only anomalous intervals are the 04-to-05 and 06-to-07 interval, which take less time than one might expect; the deviation, however, is within the range attributable to human error (clicking too late when ending the previous interval and starting the one in question and/or too early when ending it and starting the next one).
#$&*
* Sketch a graph of depth vs. clock time (remember that the convention is y vs. x; the quantity in front of the 'vs.' goes on the vertical axis, the quantity after the 'vs.' on the horizontal axis). You may if you wish print out and use the grid below.
[image110.gif (4103 bytes)]
Describe your graph in the language of the Describing Graphs exercise.
Your answer (start in the next line):
- RESPONSE: Because my earlier tables/plots put column height on the x-axis and time on the y-axis, I'm redoing my table here to put clock time on the x-axis per your instructions. Also, I'm including time and distance intervals as a third and fourth column because they're very instructive (see explanation below) and because I get the feeling that I will or at least should be using them soon anyway.
- - (ClkTim, Depth:m, TimeIntvl:s, DistIntvl:m)
- - 00.0000, 0.20150, -------- -------
- - 01.9453, 0.18500, 1.945313, -0.0165
- - 03.8984, 0.16850, 1.953125, -0.0165
- - 05.9297, 0.15175, 2.031250, -0.01675
- - 08.1094, 0.13475, 2.179688, -0.017
- - 10.2031, 0.11725, 2.093750, -0.0175
- - 12.7422, 0.09900, 2.539063, -0.01825
- - 15.2578, 0.08150, 2.515625, -0.0175
- - 18.0625, 0.06250, 2.804688, -0.019
- - 21.1094, 0.04350, 3.046875, -0.019
- - 24.4453, 0.02425, 3.335938, -0.01925
- - 28.3906, 0.00450, 3.945313, -0.01975
- - Note that the distance intervals are all fairly close (as one would expect from a well-calibrated graduated cylinder and ruler measuring that cylinder). The point, as discussed above with the variables on the opposite axes, is that more time (here, x-value) is required for each increment of a similar level, making the slope of the position-vs.-time graph and hence its velocity less and less each time (with the exception of at least one of the two anomalies).
#$&*
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):
- RESPONSE: Average velocity = distance interval / time interval.
- - (TimeInt#, TimeIntvl:s, DistIntvl:m, AvgVeloc:[m(s^-1)])
- - 00 to 01 1.945313, -0.0165, -0.00848
- - 01 to 02 1.953125, -0.0165, -0.00845
- - 02 to 03 2.031250, -0.01675, -0.00825
- - 03 to 04 2.179688, -0.017, -0.0078
- - 04 to 05 2.093750, -0.0175, -0.00836
- - 05 to 06 2.539063, -0.01825, -0.00719
- - 06 to 07 2.515625, -0.0175, -0.00696
- - 07 to 08 2.804688, -0.019, -0.00677
- - 08 to 09 3.046875, -0.019, -0.0062
- - 09 to 10 3.335938, -0.01925, -0.00577
- - 10 to 11 3.945313, -0.01975, -0.00501
#$&*
* Assume that this average velocity occurs at the midpoint of the corresponding time interval.
What are the clock times at the midpoints of your time intervals, and how did you obtain them? (Give one midpoint for each time interval; note that it is midpoint clock time that is being requested, not just half of the time interval. The midpoint clock time is what the clock would read halfway through the interval. Again be sure you haven't confused clock times with time intervals. Do not make the common mistake of reporting half of the time interval, i.e., half the number in the third column of the TIMER's output):
Your answer (start in the next line):
- RESPONSE: Midpoint of time interval = ([Time:IntervalX + Time:Interval(X - 1)]/2); midpoints are listed in center column, with time interval numbers to left and average velocities to right.
- - (TimeInt#, TimeMidpt:s, AvgVeloc:[m(s^-1)])
- - 00 to 01, 0.9727, -0.00848
- - 01 to 02, 2.9219, -0.00845
- - 02 to 03, 4.9141, -0.00825
- - 03 to 04, 7.0196, -0.0078
- - 04 to 05, 9.1563, -0.00836
- - 05 to 06, 11.4727, -0.00719
- - 06 to 07, 14.0000, -0.00696
- - 07 to 08, 16.6602, -0.00677
- - 08 to 09, 19.5860, -0.0062
- - 09 to 10, 22.7774, -0.00577
- - 10 to 11, 26.4180, -0.00501
#$&*
* 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):
- RESPONSE: See table above -- I guess I instinctively foresaw this request and consolidated it with the previous one.
#$&*
* Sketch a graph of average velocity vs. clock time. Describe your graph, using the language of the Describing Graphs exercise.
Your answer (start in the next line):
- RESPONSE: The graph consists of negative values best fit by a curve having an always-positive tangent line showing that the series of values is decreasing in absolute value (and increasing in numerical value) over time; if we had defined the downward direction as positive, the values would be positive and best fit by a curve whose tangent line always has a negative slope. The absolute value decreases at a decreasing rate, matching what we think of as deceleration of an object traveling in the negative direction, although the decrease in rate is masked by a) reduction of the number of time data points by one and b) a side drawback of using midpoints, namely that the time increase occurring within each original interval is split between the pair of half-intervals before it and the pair of half-intervals after it, with the effect that the increase in distance between x-axis data points isn't as obvious as it was in the original graph.
#$&*
* 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):
- RESPONSE: Average acceleration = (AverageVelocity:[(X - 1) to X] - AverageVelocity:[(X - 2) to (X - 1)])/(TimeMidpoint:[(X - 1) to X] - TimeMidpoint:[(X - 2) to (X - 1)])
- - (#:IntBtwTimeInts, MidptBtwTimeMidpts:sec, TimeElapsedBtwMidpts:sec, DiffBtwAvgVelocs:m[sec^-1], AvgAccel:m[sec^-2])
- - (00 to 01) to (01 to 02) 1.9473 1.9492 +0.00003 +0.00015
- - (01 to 02) to (02 to 03) 3.9180 1.9922 +0.00020 +0.00010
- - (02 to 03) to (03 to 04) 5.9669 2.1055 +0.00045 +0.00021
- - (03 to 04) to (04 to 05) 8.0880 2.1367 -0.00056 -0.00026
- - (04 to 05) to (05 to 06) 10.3145 2.3164 +0.00117 +0.00051
- - (05 to 06) to (06 to 07) 12.7364 2.5273 +0.00023 +0.00009
- - (06 to 07) to (07 to 08) 15.3310 2.6602 +0.00019 +0.00007
- - (07 to 08) to (08 to 09) 18.1231 2.9258 +0.00057 +0.00019
- - (08 to 09) to (09 to 10) 22.1817 3.1914 +0.00043 +0.00013
- - (09 to 10) to (10 to 11) 24.5977 3.6406 +0.00076 +0.00021
#$&*
* Make a table of average acceleration vs. clock time, using the clock time at the midpoint of each time interval with the corresponding acceleration.
Give your table in the box below, giving on each line a midpoint clock time followed by a comma followed by acceleration.
Your answer (start in the next line):
- RESPONSE: See second and fifth columns of above table (again, I guess I instinctively incorporated these data into the table).
#$&*
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):
- RESPONSE:
- a) My data: My calculations of average acceleration are themselves inconclusive, but drawing a line of best fit to the position data shows a conclusive relationship; cf. the lesson from the deterioration of difference quotients exercise/lab that it's better to infer relationships directly from concrete data than to abstract from those data and then use the abstracted calculations in a likely-vain attempt to reconstruct something approximating a mathematical expression of the variables' relationship.
- b) Hypothesis re: actual behavior: The acceleration of the water surface is actually decreasing in numerical as well as absolute value (numerical because the acceleration is positive given that the water's velocity is in the negative direction; absolute because the acceleration's magnitude is getting smaller); hence the at a decreasing rate portion of the statement the velocity is decreasing at a decreasing rate.
#$&*
Go back to your graph of average velocity vs. midpoint clock time. Fit the best straight line you can to your data.
* What is the slope of your straight line, and what does this slope represent? Give the slope in the first line, your interpretation of the slope in the second.
* How well do you think your straight line represents the actual behavior of the system? Answer this question and explain your answer.
* Is your average velocity vs. midpoint clock time graph more consistent with constant, increasing or decreasing acceleration? Answer this question and explain your answer.
Your answer (start in the next line):
- RESPONSE:
- - a1) The slope is about +0.00009.
- - a2) This slope shows the rate of change in the first-order/linear function that best fits the average-velocity-vs.-time relationship.
- - a3) I think that neither this nor any other straight line represents the actual behavior of the system. See my first through third and sixth Responses in this lab writeup. Again, the rate of change in velocity itself decreases over time, such that velocity approaches an asymptote at zero; linear functions like this line of best fit don't do that.
#$&*
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):
- RESPONSE: I spent about 5.5 to 6 hours on this lab, but that factor includes some time spent multitasking, 30 to 45 minutes trying to keep track of small pieces of equipment (see comments below), and an hour and a half or so in which I got really absorbed in the calculus in the preliminary questions, trying to see how much I could figure out and/or derive a priori.
#$&*
You may add any further comments, questions, etc. below:
Your answer (start in the next line):
- RESPONSE: Comments:
- a) I'm not sure it's a good idea to use one hand to fill, hold shut, and manipulate a tall graduated cylinder full of tap (i.e., non-[distilled and deionized]) water while using one's other hand to carry around and operate one's laptop computer as the use of the TIMER program requires. If data were heat, the reaction between tap water and laptops would be extremely exothermic ;-)
- b) The small pieces of rubber tube are hard to handle: Especially because they're clear and springy/bouncy, they are easy to drop, easy to lose track of, and difficult to find once they're lost. (Once lost, they can also pose a choking hazard for pets who may find and chew on them.) Maybe next year use tubes of a bright color?
#$&*
*#&!
@&
Clearly water and computers don't mix. With reasonable precautions that wouldn't be a problem, but it's probably worth a warning.
Colored tubing of precise dimensions isn't available, but smaller pieces embedded in larger could make them easier to handle. This is worth some thought. Obviously these experiments will sometimes be done in the presence of children and pets.
*@
@&
I've inserted some brief comments pointing to deeper analysis. You're already thinking beyond the level of this course. I don't expect you to take the time to pursue them now, given the time pressures of this course, but feel free to follow up later if you wish.
*@