#$&*
PHY 201
Your 'energy conversion 1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
#$&* Your optional message or comment: **
Energy Conversion I Lab submitted 13 Mar 2011 around 6:46PM.
#$&* How far and through what angle did the block displace on a single trial, with rubber band tension equal to the weight of two dominoes? **
Energy Conversion I Lab
Note that the data program is in a continual state of revision and should be downloaded with every lab.
Most students report completion times between 2 and 3 hours, with some as short as 1 hour and some as long as 5 hours.
For part of this experiment you will use the calibrated rubber band you used in the preceding experiment 'Force vs. Displacement 1', as well as the results you noted for that experiment.
• For this experiment you will need to use at least one rubber band in such a way as to make it useless for subsequent experiments. DO NOT USE ONE OF YOUR CALIBRATED RUBBER BANDS. Also note that you will use four of the thin rubber bands in a subsequent experiment, so DO NOT USE THOSE RUBBER BANDS HERE.
• If your kit has extra rubber bands in addition to these, you may use one of them.
You are going to use the rubber band to bind three of your dominoes into a block. If you don't have extra rubber bands, you could use some of the thread that came with your kit, but rubber bands are easier to use.
• The idea of binding the dominoes is very simple. Just set one domino on a tabletop so that it lies on one of its long edges. Then set another right next to it, so the faces of the two dominoes (the flat sides with the dots) are touching. Set a third domino in the same way, so you have a 'block' of three dominoes.
• Bind the three dominoes together into a 'block' using a rubber band or several loops of thread, wrapping horizontally around the middle of the 'block', oriented in such a way that the block remains in contact with the table. The figure below shows three dominoes bound in this manner, resting on a tabletop.
Now place a piece of paper flat on the table, and place the block on the paper, with the block at one end of the paper.
• Give the block a little push, hard enough that it slides about half the length of the paper.
• Give it a harder push, so that it slides about the length of the paper, but not quite.
• Give it a push that's hard enough to send it past the other end of the paper.
You might need to slide the block a little further than the length of one sheet, so add a second sheet of paper:
• Place another piece of paper end-to-end with your first sheet.
• Tuck the edge of one sheet slightly under the other, so that if the block slides across the first sheet it can slide smoothly onto the second.
You are going to use a calibrated rubber band to accelerate the blocks and make them slide across the table.
• Tie two pieces of thread through to the rubber bands holding the blocks, at the two ends of the block, so that if you wanted you could pull the block along with the threads. One thread should be a couple feet long--long enough that if the block is at one edge of one paper, the other end of the thread extends beyond the edge of the other paper. The other thread needs to be only long enough that you can grasp it and pull the block back against a small resistance.
• At the free end of the longer thread, tie a hook made from a paper clip.
• Use the rubber band you used in the preceding experiment (the 'first rubber band' from your kit, the one for which you obtained the average force * distance results). Hook that rubber band to the hook at the free end of the longer thread.
• Make another hook, and put it through the other end of the rubber band loop, so that when you pull on this hook the rubber band stretches slightly, the string becomes taut and the block slides across the tabletop.
You will need something to which to attach the last hook:
• Now place on the tabletop some object, heavy enough and of appropriate shape, so that the last hook can in one way or another be fixed to that object, and the object is heavy enough to remain in place if the rubber band is stretched within its limits. That is, the object should be able so remain stationary if a few Newtons of force is applied. Any rigid object weighing, or being weighted by, about 5-10 pounds ought to be sufficient.
• Your goal is to end up with a moderately massive object, to which the last hook is tied or attached, with the rubber band extending from the hook to another hook, a thread from that hook to the block (with a shorter thread trailing from the other end of the block)
• With a slight tension in the system the block should be a few centimeters from the 'far' edge of the paper which is furthest from the massive object.
• If the block is pulled back a little ways (not so much that the rubber band exceeds its maximum tolerated length) the rubber band will stretch but the last hook will remain in place, and if the block is then released the rubber band will snap back and pull the block across the tabletop until the rubber band goes slack and the block then coasts to rest.
• The figure below shows the block resting on the paper, with the thread running from a hook to the rubber band at the far end, which is in turn hooked to the base of a flatscreen monitor.
At the far end the rubber band is ready to be stretched between two hooks. A measuring device is shown next to the rubber band; to get accurate measurements of rubber band length it is recommended that a piece of paper be placed beneath the rubber band, and two points carefully marked on the paper to indicate the positions of the ends. The separation of the points can later be measured. Alternatively the two points can be marked in advance at the desired separation and the system stretched accordingly.
Consult your previous results and determine the rubber band length required to support the weight of two dominoes. Pulling by the shorter piece of thread (the 'tail' of thread), pull the block back until the rubber band reaches this length, and on the paper mark the position of the center of the block (there might well be a mark at the center of the domino; if not, make one, being sure it is within 1 millimeter of the center, and mark the paper according to this mark). Release the thread and see whether or not the block moves. If it does, mark the position where it comes to rest as follows:
• Make a mark on the paper where the center mark comes to rest by drawing a short line segment, perhaps 3 mm long, starting from the center mark and running perpendicular to the length of the block.
• Make another mark about twice the length of the first, along the edge of the block centered at the center mark.
• This will result in a mark that looks something like the following, with the longer line indicating the direction of the block and the two lines coming together at the center mark: __|__. In the first figure below the lowest two marks represent the positions of the center of the dominoes at initial point and at the pullback point. The mark next to the domino is the horizontal part of a mark that looks something like |- ; the vertical part of that mark is obscured by the blocks, and the mark it also tilted a bit to coincide with the slightly rotated orientation of the block. In the second figure most of the |- mark can be seen.
You will make a similar mark for the final position for each trial of the experiment, and from these marks you will later be able to tell where the center mark ended up for each trial, and the approximate orientation of the block at the end of each trial.
• Based on this first mark, how far, in cm, did the block travel after being released, and through approximately how many degrees did it rotate before coming to rest?
• If the block didn't move, your answers to both of these questions will be 0.
Answer in comma-delimited format in the first line below. Give a brief explanation of the meaning of your numbers starting in the second line.
Your answer (start in the next line):
10 cm, 30 deg
The 10 cm is the distance the block traveled after it was released from the first mark.
#$&* _ 2 rb tension how far and thru what angle
Tape the paper to the tabletop, or otherwise ensure that it doesn't move during subsequent trials.
• Repeat the previous instruction until you have completed five trials with the rubber band at same length as before.
Report your results in the same format as before, in 5 lines. Starting in the sixth line give a brief description of the meaning of your numbers and how they were obtained:
Your answer (start in the next line):
9.8 cm, 35 deg
10 cm, 35 deg
10 cm, 33 deg
9.9 cm, 30 deg
10 cm, 30 deg
#$&* _ trials on paper
Now, without making any marks, pull back a bit further and release.
• Make sure the length of the rubber band doesn't exceed its original length by more than 30%, with within that restriction what rubber band length will cause the block to slide a total of 5 cm, then 10 cm, then 15 cm.
• You don't need to measure anything with great precision, and you don't need to record more than one trial for each sliding distance, but for the trials you record:
• The block should rotate as little as possible, through no more than about 30 degrees of total rotation, and
• it should slide the whole distance, without skipping or bouncing along.
• You can adjust the position of the rubber band that holds the block together, the angle at which you hold the 'tail', etc., to eliminate skipping and bouncing, and keep rotation to a minimum.
Indicate in the first comma-delimited line the rubber band lengths that resulted in 5 cm, 10 cm and 15 cm slides. If some of these distances were not possible within the 30% restriction on the stretch of the rubber band, indicate this in the second line. Starting in the third line give a brief description of the meaning of these numbers.
Your answer (start in the next line):
12.5 cm, 32 deg
14 cm, 31 deg
17 cm, 30 deg
#$&* _ rb lengths for 5, 10, 15 cm slides
Now record 5 trials, but this time with the rubber band tension equal to that observed (in the preceding experiment) when supporting 4 dominoes. Mark and report only trials in which the block rotated through less than 30 degrees, and in which the block remained in sliding contact with the paper throughout.
Report your distance and rotation in the same format as before, in 5 lines. Briefly describe what your results mean, starting in the sixth line:
Your answer (start in the next line):
12.8 cm, 31 deg
13.0 cm, 30 deg
12.9 cm, 33 deg
13.0 cm, 30 deg
13.0 cm, 30 deg
#$&* _ 5 trials 4 domino length
Repeat with the rubber band tension equal to that observed when supporting 6 dominoes and report in the same format below, with a brief description starting in the sixth line:
Your answer (start in the next line):
19 cm, 32 deg
20 cm, 30 deg
20 cm, 31 deg
19.9 cm, 30 deg
20 cm, 30 deg
#$&* _ 5 trials for 6 domino length
Repeat with the rubber band tension equal to that observed when supporting 8 dominoes and report in the same format below, including a brief description starting in the sixth line:
Your answer (start in the next line):
21 cm, 31 deg
21 cm, 30 deg
20.5 cm, 31 deg
20.9 cm, 30 deg
21 cm, 30 deg
#$&* _ 5 trials for 8 domino length
Repeat with the rubber band tension equal to that observed when supporting 10 dominoes and report in the same format below, including your brief description as before:
Your answer (start in the next line):
22 cm, 31 deg
21.5 cm, 30 deg
22 cm, 31 deg
22 cm, 30 deg
21.5 cm, 30 deg
#$&* _ 5 trials for 10 domino length
In the preceding experiment you calculated the energy associated with each of the stretches used in this experiment.
The question we wish to answer here is how that energy is related to the resulting sliding distance.
• For each set of 5 trials, find the mean and standard deviation of the 5 distances. You may use the data analysis program or any other means you might prefer.
• In the space below, report in five comma-delimited lines, one for each set of trials, the length of the rubber band, the number of dominoes supported at this length, the mean and the standard deviation of the sliding distance in cm, and the energy associated with the stretch.
• You might choose to report energy here in Joules, in ergs, in Newton * cm or in Newton * mm. Any of these choices is acceptable.
• Starting in the sixth line specify the units of your reported energy and a brief description of how your results were obtained. Include your detailed calculations and specific explanation for the third interval. Be sure to give a good description of how you obtained the energy associated with each stretch:
Your answer (start in the next line):
9.8 cm, 10 cm, 10 cm, 9.9 cm, 10 cm, RB - 9.5 cm, 3 dom, Mean = 9.94, STDV = 0.08944.
12.8 cm, 13.0 cm, 12.9 cm, 13.0 cm, 13.0 cm, RB - 10 cm, 4 dom, Mean = 12.94, STDV = 0.08944.
19 cm, 20 cm, 20 cm, 19.9 cm, 20 cm, RB - 12 cm, 6 dom, Mean = 19.78, STDV = 0.4382.
21 cm, 21 cm, 20.5 cm, 20.9 cm, 21 cm, RB - 12 cm, 8 dom, Mean = 20.88, STDV = 0.2168.
22 cm, 21.5 cm, 22 cm, 22 cm, 21.5 cm, RB - 12 cm, 10 dom, Mean = 21.8, STDV = 0.2739.
@& You don't appear to have reported the energies.*@
#$&* _ for each set of trials length, # dom, mean, std of sliding dist, energy _ describe how results obtained esp energy calculations
Sketch a graph of sliding distance vs. energy, as reported in the preceding space.
• Fit the best possible straight line to your graph, and give in the first comma-delimited line the slope and vertical intercept of your line.
• In the second line specify the units of the slope and the vertical intercept.
• Starting in the third line describe how closely your data points cluster about the line, and whether the data points seem to indicate a straight-line relationship or whether they appear to indicate some sort of curvature.
• If curvature is indicated, describe whether the curvature appears to indicate upward concavity (for this increasing graph, increasing at an increasing rate) or downward concavity (for this increasing graph, increasing at a decreasing rate).
Your answer (start in the next line):
#$&* _ sliding dist vs. energy slope, vert intercept of st line, how close to line, describe curvature if any
@& Did you graph sliding distance vs. energy?*@
Now repeat the entire procedure and analysis, but add a second rubber band to the system, in series with the first.
• For each trial, stretch until the first rubber band is at the length corresponding to the specified number of dominoes, then measure the second rubber band and record this length with your results.
• When graphing mean sliding distance vs. energy, assume for now that the second rubber band contributes an amount of energy equal to that of the first. You will therefore use double the energy you did previously.
• When you have completed the entire procedure report your results in the space es below, as indicated:
Report in comma-delimited format the length of the first rubber band when supporting the specified number of dominoes, and the length you measured in this experiment for second band. You will have a pair of lengths corresponding to two dominoes, four dominoes, ..., ten dominoes. Report in 5 lines:
Your answer (start in the next line):
#$&* _ lengths of 1st and 2d rbs in series each of 5 trials
Report for each set of 5 trials your mean sliding distance and the corresponding standard deviation; you did five sets of 5 trials so you will report five lines of data, with two numbers in each line:
Your answer (start in the next line):
#$&* _ sliding dist and std dev each tension
Give the information from your graph:
• Give in the first comma-delimited line the slope and vertical intercept of your line.
• In the second line specify the units of the slope and the vertical intercept.
• Starting in the third line describe how closely your data points cluster about the line, and whether the data points seem to indicate a straight-line relationship or whether they appear to indicate some sort of curvature.
• If curvature is indicated, describe whether the curvature appears to indicate upward concavity (for this increasing graph, increasing at an increasing rate) or downward concavity (for this increasing graph, increasing at a decreasing rate).
Your answer (start in the next line):
#$&* _ slope, vert intercept, describe curvature
In the space below, report in the first line, in comma-delimited format, the sliding distance with 1 rubber band under 2-domino tension, then the sliding distance with 2 rubber bands under the same 2-domino tension.
Then in the subsequent lines report the same information for 4-, 6-, 8- and 10-domino tensions.
You will have five lines with two numbers in each line:
Your answer (start in the next line):
#$&* _ 5 lines comparing 1 rb to 2 rb trials
Your preceding answers constitute a table of 2-rubber-band sliding distances vs. 1-rubber-band sliding distances.
Sketch a graph of this information, fit a straight line and determine its y-intercept, its slope, and other characteristics as specified:
• Give in the first comma-delimited line the slope and vertical intercept of your line.
• In the second line specify the units of the slope and the vertical intercept.
• Starting in the third line describe how closely your data points cluster about the line, and whether the data points seem to indicate a straight-line relationship or whether they appear to indicate some sort of curvature.
• If curvature is indicated, describe whether the curvature appears to indicate upward concavity (for this increasing graph, increasing at an increasing rate) or downward concavity (for this increasing graph, increasing at a decreasing rate).
Your answer (start in the next line):
#$&* _ graph 2 rb dist vs 1 rb dist _ slope and intercept _ describe any curvature
To what extent do you believe this experiment supports the following hypotheses:
The sliding distance is directly proportional to the amount of energy required to stretch the rubber band. If two rubber bands are used the sliding distance is determined by the total amount of energy required to stretch them.
Your answer (start in the next line):
#$&* _to what extend is hypothesis of sliding dist prop stretching energy supported _ to what extent for 2 rb
Copy this document, from this point down, into a word processor or text editor.
* Follow the instructions, fill in your data and the results of your analysis in the given format.
* Regularly save your document to your computer as you work.
* When you have completed your work:
Copy the document into a text editor (e.g., Notepad; but NOT into a word processor or html editor, e.g., NOT into Word or
FrontPage).
Highlight the contents of the text editor, and copy and paste those contents into the indicated box at the end of this form.
Click the Submit button and save your form confirmation.
This lab exercise is based on the observations you previously made of a ball rolling down ramps of various slopes. We further investigate
the relationship between ramp slope and acceleration.
The mean time reported to complete this exercise is 2 hours. The most frequently reported times range from 1 hour to 3 hours, with some
reports of shorter or longer times.
Note that there are a number of repetitive calculations in this exercise. You are encouraged to use a spreadsheet as appropriate to save
you time, but be sure your results check out with a handwritten analysis of at least a few representative trials.
Document your data
For ramps supported by 1, 2 and 3 dominoes, in a previous exercise you reported time intervals for 5 trials of the ball rolling from right
to left down a single ramp, and 5 trials for the ball rolling from left to right.
If in that experiment you were not instructed to take data for all three setups in both directions, report only the data you were
instructed to obtain.
(Note: If you did the experiment using the short ramp and coins, specify which type of coin you used. In the instructions below you
would substitute the word 'coins' for 'dominoes').
Go to your original data or to the 'readable' version that should have been posted to your access page, and copy your data as indicated in
the boxes below:
Copy the 10 trials for the 1-domino setups, which you should have entered into your original lab submission in the format specified by the
instruction
'In the box below, give the time interval for each trial, rounded to the nearest .001 second. Give 1 trial on each line, and give the 5
trials for the first system, then the 5 trials for the second system. You will therefore give 10 numbers on 10 lines.'
In the 'readable' posted version this data will follow the boldfaced heading
'5 trials each way 1 domino'
Enter your 10 numbers on 10 lines below, and on the first subsequent line briefly indicate the meaning of the data:
------>>>>>> ten trials for 1-domino setups
Your answer (start in the next line):
1.984
2.047
1.953
1.859
1.750
2.031
1.813
1.828
1.734
1.734
The first five intervals are the first 5 trials rolling the ball down the ramp. The second set of 5 are the intervals from rotating the
ramp 180 degrees. The intervals were timed from the release of the ball, till the point where the ball hit the bracket. The ball traveled a
total of 27.4cm during each trial. As the ball rolled down the ramp it's velocity was increasing.
#$&*
Enter your data for the 2-domino setups in the same format, being sure to include your brief explanation:
On the 'readable' posted version this data will follow the boldfaced heading
'5 trials each way 2 dominoes'
------>>>>>> 2 domino results
Your answer (start in the next line):
1.203
1.234
1.188
1.266
1.266
1.234
1.188
1.250
1.250
1.328
#$&*
Enter your data for the 3-domino setups in the same format, including brief explanation.
On the 'readable' posted version this data will follow the boldfaced heading
'5 trials each way 3 dominoes'
------>>>>>> 3 domino results
Your answer (start in the next line):
1.063
0.953
0.953
1.031
0.984
1.109
0.984
0.953
1.000
1.109
#$&*
Calculate mean time down ramp for each setup
In the previous hypothesis testing exercise, you calculated and reported the mean and standard deviation of times down each of the two
1-domino setups, one running right-left and the other left-right.
You may use any results obtained from that analysis (provided you are confident that your results follow correctly from your data), or you
may simply recalculate this information, which can be done very quickly and easily using the Data Analysis Program at
http://www.vhcc.edu/dsmith/genInfo/labrynth_created_fall_05/levl1_15\levl2_51/dataProgram.exe\
In any case, calculate as needed and enter the following information, in the order requested, giving one mean and standard deviation per
line in comma-delimited format:
* Mean and standard deviation of times down ramp for 1 domino, right-to-left.
* Mean and standard deviation of times down ramp for 1 domino, left-to-right.
* Mean and standard deviation of times down ramp for 2 dominoes, right-to-left.
* Mean and standard deviation of times down ramp for 2 dominoes, left-to-right.
* Mean and standard deviation of times down ramp for 3 dominoes, right-to-left.
* Mean and standard deviation fof times down ramp or 3 dominoes, left-to-right.
On the first subsequent line briefly indicate the meaning of your results and how they were obtained:
------>>>>>> mean, std dev each setup each direction
Your answer (start in the next line):
1.919, .1161
1.828, .1216
1.231, .03568
1.042, .5123
.9968, .04886
1.031, .07318
The first column contains the mean's of the data, the second column containes the stadard deviations. The numbers are organized as listed
right above.
#$&*
Calculate average ball velocity for each setup
Assuming that the ball traveled 28 cm from release until the time it struck the bracket, determine each of the following, using the mean
time required for the ball to travel down the ramp:
* Average ball velocity for 1 domino, right-to-left.
* Average ball velocity for 1 domino, left-to-right.
* Average ball velocity for 2 dominoes, right-to-left.
* Average ball velocity for 2 dominoes, left-to-right.
* Average ball velocity for 3 dominoes, right-to-left.
* Average ball velocity for 3 dominoes, left-to-right.
Report your six results in the box below, one result per line, in the order requested above.
Starting in the seventh line explain how you obtained your results, giving the details of how you obtained at least one of your results.
These details should include the definition of the average velocity, and should explain how you used the mean time and the distance down
the ramp to arrive at your result, and should show the numbers used and the numbers obtained in each step.
------>>>>>> ave velocities each of six setups
Your answer (start in the next line):
14.6
15.3
22.7
26.9
28.1
27.2
I got these velocities from dividing the mean times from 28cm. 28.0cm/1.919sec= 14.6cm/s vAve= `ds/`dt
#$&*
Calculate average ball acceleration for each setup
Assuming that the velocity of the ball changed at a constant rate in each trial, use the mean time interval and the 28 cm distance to
determine the average rate of change of velocity with respect to clock time. You will determine your results in the following order:
* Average rate of change of ball velocity with respect to clock time for 1 domino, right-to-left.
* Average rate of change of ball velocity with respect to clock time for 1 domino, left-to-right.
* Average rate of change of ball velocity with respect to clock time for 2 dominoes, right-to-left.
* Average rate of change of ball velocity with respect to clock time for 2 dominoes, left-to-right.
* Average rate of change of ball velocity with respect to clock time for 3 dominoes, right-to-left.
* Average rate of change of ball velocity with respect to clock time for 3 dominoes, left-to-right.
Report your six results in the box below, one result per line, in the order requested above.
Starting in the seventh line explain how you obtained your results, giving the details of how you obtained at least one of your results.
These details should include the definition of the average rate of change of velocity with respect to clock time and should explain, step
by step, how you used the mean time and the distance down the ramp to arrive at your result, and should show the numbers used and the
numbers obtained in each step.
------>>>>>> ave roc of vel each of six setups
Your answer (start in the next line):
15.2
16.7
36.9
51.6
56.4
52.8
These results are in the units cm/s^2. I got these by taking the average velocities and multiplying by 2 to get vf. All the v0 values were
equal to 0. 14.6cm/s* 2= 29.2 for Vf a= `dv/`dt (29.2cm/s-0cm/s)/(1.919sec)= 15.2cm/s^2
#$&*
Average left-right and right-left velocities for each slope
For the 1-domino system you have obtained two values for the average rate of change of velocity with respect to clock time, one for the
right-left setup and one for the left-right. Average those two values and note your result.
For the 2-domino system you have also obtained two values for the average rate of change of velocity with respect to clock time. Average
those two values and note your result.
For the 3-domino system you have also obtained two values for the average rate of change of velocity with respect to clock time. Average
those two values and note your result.
Report your results in the box below, giving one average rate of change of velocity with respect to clock time per line, in the order
requested. Starting the the first subsequent line, briefly indicate how you obtained your results and what you think they mean.
------>>>>>> ave of right-left, left-right each slope
Your answer (start in the next line):
15.95
44.25
54.60
These are the averages of my right-left, left-right accelerations for 1, 2, 3 domino trials. The units are cm/s/s
#$&*
Find acceleration for each slope based on average of left-right and right-left times
Average the mean time required for the right-to-left run with the mean time for the left-to-right run.
These are the average velocities
Using this average mean time, recalculate your average rate of velocity change with respect to clock time for the 1-domino trials
Do the same for the 2-domino results, and for the 3-domino results.
Report your results in the box below, giving one average rate of change of velocity with respect to clock time per line, in the order
requested. In the subsequent line explain how you obtained your results and what you think they mean.
------>>>>>> left-right, right-left each setup, ave mean times and give ave accel
Your answer (start in the next line):
14.95cm/s
24.81cm/s
27.62cm/s
These are the average velocities for 1,2,3 dominos
1.8735sec
1.1365sec
1.0139sec
These are the means of the time intervals for 1,2,3 dominos
14.96*2= vf of 29.9/(1.8735) = 15.95 cm/s^2 I used this method to get all the accelerations here.
15.95cm/s^2
43.66cm/s^2
54.48cm/s^2
These values are the means of the rate of change in velocity with respect to clock time(average of the clock times) of the
1domino,2domino,and 3domino trials. These values represent the average accelerations.
#$&*
Compare acceleration results for the two different methods
You obtained data for three basic setups, each with a different slope. Each basic setup was done with a right-left and a left-right
version.
* You previously calculated a single average rate of change of velocity with respect to clock time for each slope, by averaging the
right-left rate with the left-right rate.
* You have now calculated a single average rate of change of velocity with respect to clock time for each slope, but this time by using
the average of the mean times for the right-left and left-right versions.
Answer the following questions in the box below:
Since both methods give a single average rate of change of velocity with respect to clock time, would you therefore expect these two
results to be the same for each slope?
Are the results you reported here, based on the average of the two mean times, the same as those you obtained previously by average the two
rates? Are they nearly the same?
Why would you expect that they would be the same or nearly the same?
If they are not exactly the same, can you explain why?
------>>>>>> ave of mean vel, ave based on mean of `dt same, different, why
Your answer (start in the next line):
average of mean vel: 15.95, 44.25, 54.60 ave based on mean of `dt: 15.95, 43.66, 54.48
These values are pretty much the same if rounded up. They difference is very small. I would assume they are different because not only were
they calculated a little different, but human error may have played a role. If I rounded some of the values differently when calculating
them it may have made this difference.
#$&*
Associate acceleration with ramp slope
Your results will clearly indicate that, as expected, acceleration increases when ramp slope increases. We want to look further at just
how the acceleration changes with ramp slope.
If you set up the ramps according to instructions, then the ramp slopes for 1-, 2- and 3-domino systems should have been approximately
equal to .03, .06 and .09 (if you used coins and the 15 cm ramp instead of dominoes and the 30-cm ramp, your ramp slopes will be different;
each dime will correspond to a ramp slope of about .007, each penny to a slope of about .010, each quarter to a slope of about .013).
For each slope you have obtained two values for the average rate of change of velocity with respect to clock time on that slope. You may
use below the values obtained in the preceding box, or the values you obtained in the box preceding that one. Use the one in which you have
more faith.
In the box below, report in the first line the ramp slope and the average rate of change of velocity with respect to clock time for the
1-domino system. Use comma-delimited format.
Using the same format report your results for the 2-domino system in the second line, and for the 3-domino system in the third.
In your fourth line specify the units of these quantities. Ramp slope is a unitless quantity; be sure you report this. Also briefly
explain how you got your results and what they tell you about this system:
------>>>>>> ramp slope ave roc of vel each system
Your answer (start in the next line):
.03, 15.95
.06, 44.25
.09, 54.60
The units are cm/s^2 for the second column. The ramp slope is in the first column which is unitless. I got the ramp slopes from the slopes
for 1,2,3 dominos. The accelerations were obtained by dividing average clock time from mean of velocity changes.
#$&*
Graph acceleration vs. ramp slope
A graph of acceleration vs. ramp slope will contain three data points. The graph will visually represent the way acceleration changes with
ramp slope. A straight line through your three data points will have a slope and a y-intercept, each of which has a very significant
meaning.
Your results constitute a table with three rows and two columns, representing rate of velocity change vs. ramp slope.
* Sketch in your lab notebook a graph of the table you have just entered. The graph will be of rate of change of velocity with respect
to clock time vs. ramp slope. Be sure to follow the y vs. x convention to put the right quantities on the horizontal and vertical axes (if
it's y vs. x, then y is on the vertical, x on the horizontal axis).
Your graph might look something like the following. Note, however, that this graph is a little too long for its height. On a good
graph the region occupied by the data points should be about as high as it is wide. To save space on the page, graphs depicted here are
often not high enough for their width
* Sketch the best possible straight line through your 3 data points. Unless the points lie perfectly along a straight line, which due
to experimental uncertainty is very unlikely, the best possible line will not actually pass through any of these points. The best-fit line
can be constructed reasonably well by sketching the line which passes as close as possible, on the average, to the 3 points.
For reference, other examples of 3-point graphs and best-fit lines are shown below.
Describe your best-fit line by giving the following:
* On the first line, the horizontal intercept of your best-fit line. The horizontal intercept will be specified here by a single
number, which will be the coordinate at which the line passes through the horizontal axis of your graph.
* On the second line, the vertical intercept of your best-fit line. The horizontal intercept will be specified here by a single number,
which will be the coordinate at which the line passes through the vertical axis of your graph.
* On the third line, give the units of your horizontal intercept and the meaning of that intercept.
* On the fourth line, give the units of your vertical intercept and the meaning of that intercept.
Starting in the fifth line, give a brief written description of your graph and an explanation of what you think it might tell you about the
system:
------>>>>>> horiz int, vert int, units and meaning of horiz, then vert int
Your answer (start in the next line):
-0.0015
0.5
ramp slope has no units, -0.0015 is were my line of best fit crosses my x-axis.
cm/s^2, 0.5 is were the line crosses the y-axis.
The graph tells me their is a direct relationship between ramp slope and acceleration. The graph is increasing from left to right at an
increasing rate.
#$&*
Mark the point on your best-fit line which would correspond to a ramp slope of .10. Determine as accurately as you can the rate of velocity
change that goes with this point, so that you have both the horizontal and vertical coordinates of the point.
Report the horizontal and vertical coordinates of that point on the first line below, in the specified order, in comma-delimited format.
Starting at the second line, explain how you made your estimate and how accurate you think it might have been. Explain, briefly, what your
numbers mean and how you got them.
------>>>>>> mark and report best fit line coord for ramp slope .10
Your answer (start in the next line):
0.1, 67
The first number is were the ramp slope was at .10. By looking at my line of best fit the vertical point corresponding would be 67cm/s^2.
These measurements are not extremely accurate. They would be more accurate if generated from microsoft excel.
#$&*
Determine the slope of the best-fit line
We defined rise, run and slope between graph points:
* The 'run' from one graph point to another is the change in the horizontal coordinate, from the first point to the second.
* The 'rise' from one graph point to another is the change in the vertical coordinate, from the first point to the second.
* The slope between the two graph points is the rise-to-run ratio, calculated as slope = rise / run.
As our first point we will use the horizontal intercept of your best-fit line, the point where that line goes through the horizontal axis.
As our second point we will use the point on that line corresponding to ramp slope .10.
* In the box below give on the first line the run from the first point to the second.
* On the second line give the rise from the first point to the second.
* On the third line give the slope of your best-fit straight line.
* Starting in the fourth line, give a brief explanation and an indication of what you think the slope might tell you about the system.
------>>>>>> slope of graph based on horiz int, ramp slope .10 point
Your answer (start in the next line):
run= 0.10- (-0.0015)= 0.1015
rise= 67-0= 67
slope= 660.1
The slope tells me that the system is increasing at an increasing rate. Their is a direct relationship between the x and y values.
#$&*
Assess the uncertainties in your result
The rest of this exercise is optional for Phy 121 and Phy 201 students whose goal is a C grade
Calculate average of mean times and average of standard deviations for 1-domino ramp
Since there is uncertainty in the timing data on which the velocities and rates of velocity change calculated in this experiment have been
based, there is uncertainty in the velocities and rates of velocity change.
We first estimate this uncertainty for the 1-domino case.
In the box below, report in the first line the right-to-left mean time, the left-to-right mean time and the average of these two mean times
on the 1-domino ramp. This third number, which you also calculated previously, will be called 'the average of the mean times'.
In the second line report the standard deviation of right-to-left times, the standard deviation of left-to-right times and the average of
these standard deviations for the 1-domino ramp. This third number will be called 'the average of the standard deviations'.
Starting in the next line give a brief explanation and speculate on the significance of these results.
------>>>>>> 1 dom ramp mean rt-left and left-rt, then std def of both
Your answer (start in the next line):
:
#$&*
Use average time and standard deviation to estimate minimum and maximum possible velocity and acceleration for first ramp
We will use the average of the mean times and the average of the standard deviations to estimate our error in the average velocity and in
the acceleration on the 1-domino ramp.
We will assume that the actual time down the ramp is within in the interval defined by mean +- std dev, where 'mean' is in this case the
average of the mean times, and 'std dev' is the average of the standard deviations.
Using these values for mean and std dev:
* Sketch a number line and sketch the interval from mean - std dev to mean + std dev. The interval will be centered at the average of
the mean times as you reported it in the previous box, and will extend a distance equal to the average of the standard deviations (as also
reported in the previous box) on either side.
* So for example if the average of the mean times was 1.93 seconds and the standard deviation .11 second, the interval would extend
from 1.93 sec - .11 sec = 1.82 sec to 1.93 sec + .11 sec to 2.04 sec.. This interval would be bounded on the left by 1.82 sec and on the
right by 2.04 sec..
Report in the first line of the box below the left and right boundaries of your interval. Starting in the second line explain briefly, in
your own words, what these numbers represent.
------>>>>>> boundaries of intervals rt-left, left-rt
Your answer (start in the next line):
:
#$&*
Instead of 'rate of velocity change with respect to clock time' we will now begin to use the word 'acceleration'. So 'average
acceleration' means exactly the same thing as 'average rate of velocity change with respect to clock time', and vice versa.
Since we are assuming here that acceleration is constant on a straight ramp, in this context we can simply say 'acceleration' rather than
'average acceleration'.
Using this terminology:
* If the time down the ramp is equal to that of the left-hand boundary of the interval you just sketched, then what would be the
average velocity and the acceleration of the ball? Report in comma-delimited format on the first line below.
* Find the same quantities for the right-hand boundary of your interval, and report in similar format on the second line.
* In the third line report the resulting minimum and maximum possible values of acceleration on this interval, using comma-delimited
format. Your results will just be a repeat of the results you just obtained.
* Starting on the fourth line, explain what your numbers represent and why it is likely that the actual acceleration of the ball on a
1-domino ramp, if set up carefully so that right-left symmetry is assured, would be between the two results you have given.
------>>>>>> 1-dom vel and accel left boundary of interval, rt boundary, min and max possible accel
Your answer (start in the next line):
:
#$&*
Repeat for 2- and 3-domino ramps
Do the same for the 2-domino data, and report in identical format, including explanations:
------>>>>>> 2-dom vel and accel left boundary of interval, rt boundary, min and max possible accel
Your answer (start in the next line):
:
#$&*
Do the same for the 3-domino data, and report in identical format, including explanations:
------>>>>>> 3-dom vel and accel left boundary of interval, rt boundary, min and max possible accel
Your answer (start in the next line):
:
#$&*
Now make a table of your results, as follows.
You will recall that slopes of .03, .06 and .09 correspond to the 1-, 2- and 3-domino ramps.
* In the first line report the slope and the lower limit on acceleration for the 1-domino ramp.
* In the second line report the slope and the lower limit on acceleration for the 2-domino ramp.
* In the third line report the slope and the lower limit on acceleration for the 3-domino ramp.
* In the fourth line report the slope and the upper limit on acceleration for the 1-domino ramp.
* In the fifth line report the slope and the upper limit on acceleration for the 2-domino ramp.
* In the sixth line report the slope and the upper limit on acceleration for the 3-domino ramp.
Starting in the seventh line give a brief explanation, in your own words, of what these numbers mean and what they tell you about the
system:
------>>>>>> slope and lower limit 1, 2, 3 dom; slope and upper limit 1, 2, 3 dom
Your answer (start in the next line):
:
#$&*
Plot acceleration vs. ramp slope using vertical segments to represent velocity ranges
On your graph of acceleration vs. ramp slope, plot the points specified by this table.
When you are done you will have three points lying directly above the .03 label of your horizontal axis. Connect these three points with a
line segment running vertically from the lowest to the highest.
You will also have three points above the .06 label, which you will similarly connect with a segment, and three points above the .09 label,
which you will also connect.
Your graph will now contain the best-fit straight line you made earlier, and the three short vertical line segments you have just drawn.
Your graph will look something like the one below, though your short vertical line segments will probably be a little thinner than the ones
shown here, and unlike yours the graph shown here does not contain the best-fit line. And of course your points won't be the same as those
used in constructing this graph:
Does your graph fit this description?
Does your best-fit straight line pass through the three short vertical segments?
Give your answer and be sure to include a couple of sentences of explanation.
------>>>>>>
best-fit line thru error bars?
Your answer (start in the next line):
:
#$&*
Determine max and min possible slopes of acceleration vs. ramp slope graph
It should be possible to draw a number of straight lines which pass through all three vertical segments. Some of these lines will have
greater slopes than others. For example note that the figures below show two lines which pass through all three vertical segments, with
the line in the second graph being steeper than the line in the first.
Draw the steepest possible straight line which passes through all three vertical segments on your graph.
* Using the x-intercept of this line and the point on this line corresponding to ramp slope .10, determine the slope of the line.
* In the first line below report the rise, run and slope of your new line. Use comma-delimited format.
* Starting in the second line give a brief statement of what your numbers mean, including an explanation of how you obtained your
slope. Be sure to include the coordinates of the two points you used and the resulting rise and run.
------>>>>>>
max possible graph slope
Your answer (start in the next line):
:
#$&*
Now draw the least-steep possible straight line which passes through all three vertical segments.
Follow the same instructions as before, and report your results for this line in the same way, including a brief explanation:
------>>>>>> min possible graph slope
Your answer (start in the next line):
:
#$&*
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):
:
#$&*
*#&!
@& You left some items blank, but the ones you answered were excellent. You have very good data.
You do need to work out the rubber band energies and the graph in the first of these experiments.*@