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Phy201+L
Your 'rubber band calibration' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
#$&* Your initial comment (if any): **
#$&* first line ruler markings, distance in actual cm between ends, how obtained: **
1 hr.
#$&* The basis for your uncertainty estimate: **
In this experiment you 'calibrate' six rubber bands by measuring their lengths when stretched by varying forces. You will obtain for each rubber band a table of force vs. length, and you will construct force vs. length graphs for four of the six bands. These rubber bands will be used in subsequent experiments.
Most students report that this experiment takes between 2 and 3 hours; some report times of less than 1 hour, some report times in excess of 4 hours. This version of the experiment defers analysis of two of the six bands and should require about 15% less time than the version on which these reports are based.
Taking Data for Calibration:
Note: You should not stretch any of the marked rubber bands more than 35% beyonds its maximum unstretched length. If you stretch a rubber band beyond this length you will permanently distort it. This means, for example, that if a rubber band is 8 cm long you should not stretch it by more than 2.8 cm, to a maximum length of 10.8 cm.
Important: Throughout the course you will be using the rubber bands and the calibration graphs you make here, so be sure you keep the rubber bands and the graphs in a place where you can locate them, and be sure the graphs are clearly labeled so you know which one goes with which rubber band.
For this experiment you will use one of the plastic bags that came with your lab materials and the dominoes from the packet, along with a ruler, paper clips and marked rubber bands.
You have a bundle of thin rubber bands and a pack of over 100 thicker rubber bands. You will use rubber bands from the pack.
Pick at random six of these rubber bands from your lab kit. If any of the selected rubber bands have obvious flaws, discard then and replace with other randomly selected bands. Preferably using a permanent marker, put 1, 2, 3, 4, 5 and 6 marks on the respective rubber bands, so you can easily identify them later.
Using paperclips bent into the shape of hooks, form a 'chain' of all six of your marked rubber bands (a chain of two rubber bands is shown below). Be sure you observe which is which, and when you record data make sure that the individual rubber bands are clearly identified by the number of marks.
Hang the plastic bag from the chain.
Place one domino in the bag.
Measure as accurately as possible the length of the topmost of your rubber bands. Be sure you keep track of which is which.
• Measure from one end of each rubber band to the other. You will therefore be recording the positions of both ends of each rubber band. Be sure you measure the end-to-end distance, from the point where one end of the rubber band ceases and the air beyond the end begins, to the similar point at the other end.
• You should not attempt to align the end of your measuring device with either of the positions you are recording. Rather align one of the markings (e.g., the 10.0 cm marking) on your measuring device with one end of the rubber band, see what marking corresponds to the other end, and record both markings.
• To get the most precise measurement possible you should use a reduced copy of a ruler. To make sure the measurement is also accurate, you should take into account any tendency toward distortion in the corresponding part of that copy. You can choose whichever level of reduction you think will give you the most accurate and precise measurement.
In the box below, indicate in the first line the ruler markings of both ends of the first rubber band, entering two numbers in comma-delimited format.
In the second line indicate the distance in actual centimeters between the ends, to an estimated precision of .01 cm..
In the third line explain how you obtained the numbers in the second line, and what the meaning of those numbers is. Also indicate how this rubber band is marked, and the limits within which you think your measurement is accurate (e.g., +- .03 cm, indicating that you believe the actual measurement to be between .03 cm less and .03 cm greater than the reported result).
Your answer (start in the next line):
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Explain the basis for your estimate of the uncertainty of the length of the first rubber band.
Your answer (start in the next line):
The uncertainty in my estimate is about a tenth of a centimeter.
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Measure as accurately as possible the lengths of the remaining rubber bands. Be sure you keep track of which is which. You may move your measuring device from one rubber band to the next.
In the space below enter the ruler markings of the ends of the first rubber band, delimited by commas, in the first line (this will be the same information you entered in the first line of the last space ), the ruler markings of the ends of the second rubber band on the second line, etc., until you have a comma-delimited line for each rubber band.
Then put the word 'End' in the very next line.
Follow this in the very next line by a comma-delimited line containing the numerical distances in cm, each estimated to within .01 cm, of the rubber bands in your chain.
Follow this by a line indicating the markings on the rubber bands.
Finally indicate the uncertainty in your measurements, which should probably be the same as the uncertainty as that given in the preceding space .
Your answer (start in the next line):
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Add another domino to the bag and repeat your measurements. The positions of the ends should be recorded in your lab book, and should be backed up electronically in a way you can easily interpret at any future date (a comma-delimited text file or a spreadsheet file would be good; a tab-delimited file would also work but tabs can be variable and invisible so if you are going to use a text file, a comma-delimited is probably the better choice).
You won't enter the endpoint information here, but as cautioned above be sure you have it so if the information reported here has any anomalies, you can go back to your raw data and correct them.
Determine the distances in centimeters between the ends of each rubber band, and enter them in the space below, in the same order you entered them in the preceding space . Use one line and use comma-delimited format.
In the second line indicate that these results were from the weight of two dominoes.
Your answer (start in the next line):
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Continue adding dominoes and measuring until one of the rubber bands exceeds its original length by 30%, or until you run out of dominoes, then stop. To keep the time demands of this experiment within reason, you should beginning at this point adding two dominoes at a time. So you will take measurements for 4, 6, 8, ... dominoes until the 'weakest' of your rubber bands is about to stretch by more than 30% of its original length, or until you run out of dominoes.
If one rubber band reaches its limit while the rest are not all that close to theirs, remove this rubber band from the experiment and modify your previous responses to eliminate reference to the data from this band. However, keep the band and keep your copy of its behavior to this point.
In the space below, enter on the first line the actual lengths in cm of your rubber bands when supporting four dominoes, in comma-delimited format. Enter in the same order you used previously.
On the second line enter the number 4 to indicate that this result is for four dominoes.
On the third line enter in comma-delimited format the lengths in cm when supporting 6 dominoes.
On the fourth line enter the number 6 to indicate the six dominoes being supported.
Continue in this manner until you have entered all your lengths and numbers of dominoes.
Then on the next line enter 'End'.
You may then enter any brief identifying information or commentary you wish. However since the nature of the information has been defined by previous spaces, this is optional.
If you have reason to believe the uncertainty in your measurements has changed, indicate this also. Otherwise it will be assumed that your previous uncertainty estimates apply.
Your answer (start in the next line):
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Compiling and Graphing your Data
Each domino is pulled downward by the Earth's gravitational field. Each rubber band resists this force by stretching out, which creates a tension equal and opposite to the force exerted by the Earth (each rubber band also supports the rubber bands below it, but the rubber bands don't weigh much so we neglect that weight). The force exerted by the Earth on each domino is about .19 Newtons.
Make a table of the force exerted by each of the first four rubber bands vs. the length of the rubber band. You do not need to do this with all six, but you should retain the last two rubber bands and your data for those two, in case you have need of them in later experiments.
Make a force vs. length table for each of these four bands. The length will go in the first column, the force in the second. Your graph will be of the type shown below, but you probably won't have quite as many data points; your forces will also differ from the forces indicated by this graph.
There is a tendency for students at the beginning of a physics course to connect graphs point-to-point. This is a usually a very bad idea in physics, since there are experimental uncertainties in our data and we learn nothing by following those uncertainties around. The graph below is an example of this Bad Idea.
Note also the REALLY bad idea, which is to treat the 'origin' as if it is a data point. In this example, we never measured the force at the 8 cm length, and there is no justification at all for using the 'origin' as a data point (actually the point where the axes come together in this graph is not the origin, it's the point (8 cm, 0); the origin would be (0 cm, 0) and is well off the scale of this graph ).
It is a good idea to add a smooth curve to the data. This is because we expect that force will change smoothly with rubber band length. However we acknowledge that errors might occur in our data, so we never attempt to make the smooth curve pass through the actual data points, though we don't try to avoid them either.
In the example below the curve wobbles around from point to point instead of smoothly following the trend of the points.
In the next example the curve doesn't try to 'hit' each data point, but rather to follow the pattern of the actual force vs. length. It passes among the data points, remaining as smooth as possible and coming as close as possible to the data points without making unsightly 'wobbles' in an attempt to pass through specific data points.
In the space below give your table in a series of lines.
The first line will contain, in the previous order, the lengths the rubber bands supporting 1 domino, separated by commas, followed by the downward force exerted by gravity on 1 domino ( i.e., the number, indicating .19 Newtons). You can copy most of this information (all except the .19) from a previous space .
The second line will contain, in the previous order, the lengths the rubber bands supporting 2 dominoes, separated by commas, followed by the downward force exerted by gravity on 2 dominoes. Again you can copy most of this from a previous space .
Continue in this manner until you have all the lengths and downward forces, in the same comma-delimited syntax described above.
Follow your data with a line containing the word 'End'.
In subsequent lines specify the meaning of each column of your table, the units and the quantity measured in each.
Your answer (start in the next line):
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If you haven't already done so, construct a graph for each rubber band and fit a smooth curve that you think best depicts the actual behavior of that rubber band.
In the space below describe the shape of the curve you drew to approximate the force vs. length behavior of first rubber band. The curve in the last figure above could be described as 'increasing at a decreasing rate, then increasing at an increasing rate'. Other possible descriptions might be 'increasing at an increasing rate throughout', 'increasing at a decreasing rate throughout', 'increasing at an increasing rate then increasing at a decreasing rate', etc.).
Then describe the shapes of all six rubber bands. Follow your last description by a line containing the word 'End'. You may if you wish add comments starting on the next line.
Your answer (start in the next line):
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Estimating Forces
We can now use our curve to estimate the force at a given length, or to estimate the length that will give us a specified force.
In the figure below we estimate the force for the 9.5 cm length.
• From the data point it might appear that the force corresponding to 9.5 cm is about 1.5 Newtons. However we're going to put our trust in the curve.
• We project a line from the L = 9.5 point on the horizontal axis, straight up to the curve, then straight over to the F axis.
• Reading the point on the y axis as F = 2.6 or maybe F = 2.7 we see that the curve gives us a force between 2.6 and 2.7 Newtons.
• If our curve has been drawn carefully and if it appears to make good sense then we believe that the curve is more reliable than our data points, and we will tend to believe this estimate more than our data point.
Similarly we use the curve to estimate the length that gives us a force of 2 Newtons.
• We project a horizontal line from the F = 2 point on the vertical axis to the curve, then from this point we project vertically downward to the horizontal axis.
• We read a length of about 10.4 cm. Again we use the curve, which 'averages out' the characteristics of several data points, to estimate the required length.
If you haven't already done so, include in your report a table of your data for force vs. length for each of the four selected rubber bands.
Now for the first rubber band, sketch your best smooth curve, the one you believe best shows the real force vs. length behavior of a rubber band. Describe your curve and describe your thinking about how to construct the curve.
Use your curve for the first rubber band (the one with 1 mark) to do the following:
• Estimate the force in Newtons corresponding to a length of 9.8 cm and report the number in the first line of the space below.
Your answer (start in the next line):
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• Estimate the length in cm of a rubber band that gives a force of 1.4 Newtons and report the number in the second line.
Your answer (start in the next line):
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• From the curve estimate the force in Newtons corresponding to each of the lengths you actually observed. For example, if you observed lengths of 8.7, 8.9, 9.3, 9.8, 10.1 cm with 1, 2, 4, 6 and 8 dominoes, what forces would be predicted by the curve for each of these lengths? Give your estimates in the first line, using comma-delimited format. In the second line indicate by how much the estimate of the curve differs from the actual weight supported.
Your answer (start in the next line):
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• From the curve estimate, using or your first graph, report in comma-delimited format, in the first line, the length corresponding to each of the forces .19 N, .38 N, .76 N, 1.14 N, etc.. In the second line indicate in comma-delimited format by how much each of these lengths differs from the length you actually observed when the rubber band was resisting this force.
Your answer (start in the next line):
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• Which do you have more faith in, the values from the curve you just created or the values you reported in your table, and why?
• If you were to estimate a force for a given length using one of your graphs, what do you think would be the uncertainty in that force (e.g., +- .12 N, or +- .03 N, etc.) and what is your evidence for this estimate?
Your answer (start in the next line):
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• If you were to estimate a length for a given force using one of your graphs, what do you think would be the uncertainty in that length (e.g., +- .05 cm, or +- .13 cm, etc.) and what is your evidence for this estimate?
Your answer (start in the next line):
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*#&!
@& I don't see any content (i.e., I see no data and no answers) in this submission.*@
@& Please submit once more and include your results and your analysis.*@
Determine the mean difference in accelerations for the first set of 5 trials and the standard deviation of these differences. Give your results as 2 number is the first line, delimited by commas. Give the mean and standard deviation of the acceleration differences for the second set of 5 trials in the second line.
In the space below, explain why you think the magnitude of the acceleration up the incline is greater than the the magnitude of the acceleration down, or vice versa, and in the process consider the following questions:
• What force is it that is responsible for the difference?
• Is the work done on the ball by this force positive or negative as the ball travels up the incline?
• Is the work done on the ball by this force positive or negative as the ball travels down the incline?
• Does your answer depend on whether the incline is sloped to the right or to the left?
----->>>> mean diff in accel first 5 trials, std dev of diff; same for 2d set of 5 trials; force responsible, work on ball pos or neg up, work pos or neg down, does answer depend on orientation of incline
Your answer (start in the next line):
:2.202 cm/s/s, 4.437 cm/s/s
2.116 cm/s/s, 1.292 cm/s/s
For the 5 trials left to right increasing ramp, three of my trials had magnitude of the acceleration down the ramp greater than magnitude up, one trial vice versa and one trial actually had the same acceleration up and back.
For the 5 trials of the right to left increasing ramp, all of the trials had magnitudes of acceleration down the ramp greater than the magnitudes up the ramp.
There are several forces acting on the steel ball, the force I give it to start it in motion, the force of gravity from its weight, and any friction between the ball and the metal ramp.
As the steel ball rolls up the ramp, the force of rolling friction is in the direction opposite the motion. The force the ball exerts against friction is in the direction of is motion so it does positive work. The work done by the ball against gravity is in the form of PE. PE will increase as it goes up the incline until KE =0 as the ball also does work against gravity.
In the slope from left to right, I feel the results varied do the amount of “push” I was giving the steel ball to start it rolling. As I went through more trials I got better at a gently nudging the ball and starting and stopping the timer. I still feel the nudge in the second set of trials played a part in the ball having a greater magnitude going up than down. Because considering the fact that the steel ball ends up where it started, and in the process does positive work against friction (or alternatively that friction does negative work on it) the steel ball should be moving more slowly at the end.
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@& Your comments are right-on. The average velocity when descending the ramp should be less than when ascending, and this will be associated with greater magnitude of acceleration when ascending than when descending.
It's difficult to get really accurate results here, for the reasons you mention.*@
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You may if you wish do the remainder of this experiment using your data from the experiment 'Uniformity of Acceleration for a Ball on a Ramp' (that experiment is in turn a continuation of the experiment Ball and Ramp Projectile Behavior, which preceded it). In those experiments you set up '2-, 3- and 4-domino ramps' and carefully observed the range of the ball as it rolls down and off the ramp. Based on the ranges and the vertical drop, you used the data analysis program to determine the speed of the ball at the end of the ramp.
• If you have only the 15-cm ramp then you would have used half the specified number of dominoes (i.e., 1 domino instead of the specified 2, and 2 dominoes instead of the specified 4), and your 10, 20 and 30 cm rolls were probably 5, 10 and 15 cm).
• The data program requests the number of dominoes, from which it calculates a slope based on a 30-cm ramp. A 15-cm ramp has only half the 'run' of a 30-cm ramp, so if you used a 15-cm ramp you would enter double the number of dominoes you actually used. That is, if you use 1 domino on a 15-cm ramp, tell the program you used 2; if you used 2 dominoes tell the program you used 4. If you happened to use 3 or 4 dominoes, you would tell the program you used 6 or 8, respectively.
If not, you should review the instructions for those experiments, set up ramps with 2 and 4 dominoes (1 and 2 dominoes if using a 15-cm ramp). Using 5 trials you should obtain horizontal ranges for the ball rolling distances of 15 cm and 30 cm along each ramp (use 7.5 and 15 cm if you have only the 15-cm ramp), and calculate the mean and standard deviation of ranges for each set of 5 trials. You will have a total of 4 five-trial means and standard deviations to report.
If you use the data from the experiments you will have information for 10, 20 and 30 cm rolls on 2- and 4-domino ramps, for a total of 6 five-trial means and standard deviations, which you will have reported once already (you should report once more).
In the space below report in the first line the number of dominoes, the distance of the roll down the ramp, and the mean and standard deviation of the horizontal range for your first setup. In the second line report the same information for your second setup, etc., until you have reported your results for all of your setups (4 setups if you obtained new data for the experiment, 6 setups if you used your old data). In the first subsequent line, give the distance the ball fell after leaving the end of the ramp.
-------->>>> 10, 20, 30 cm rolls 2 and 4 dom; report # dom, dist, mean and std dev of horiz range first ramp; same reversed ramp; then 2d, then if present 3d; also vert dist of fall
Your answer (start in the next line):
:2 dominoes, 10 cm, 12.57 cm, .1155 cm
2 dominoes, 20 cm, 14.27 cm, .05774 cm
2 dominoes, 30 cm, 15.17 cm, .05774 cm
4 dominoes, 10 cm, 11.4 cm, .0624 cm
4 dominoes, 20 cm, 15.9 cm, .0133 cm
4 dominoes, 30 cm, 20.5 cm, .0952 cm
73.1 cm vertical distance fall
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You might already have your data and your results from the previous experiment. However if you are using data obtained specifically for this experiment, you will need to use your mean horizontal ranges to find the corresponding velocities. The instructions for doing so were given in the previous experiments, and are repeated here for easy access:
Using the 'Experiment-Specific Calculations' button, select 1, as you did in the preceding experiment, and respond with the information necessary to calculate the speed of the ball at the end of the ramp, based on the mean distance observed for your first set of 5 trials.
The program is the simplest way to get your results, and the majority of students follow instructions and use it successfully. If you can't get that program to give you reasonable results, you can follow the alternative procedure below. This will take a little longer but the process is the same for every trial, and should quickly become routine.
• Using the acceleration of gravity and the distance to the floor, figure out the time `dt it takes the ball to fall from rest from the end of the ramp. For typical distances to the floor the time of fall will be in the neighborhood of .4 seconds.
• Using your value of `dt and the horizontal displacement of the ball after leaving the end of the ramp, calculate the resulting average horizontal velocity. This is a fairly good approximation of the velocity at the end of the ramp.
• Multiply your horizontal velocity by the slope of the ramp. This gives you a pretty good approximation of the vertical component of the ball's speed as it leaves the end of the ramp. Note that the slope of your ramp (its rise / run) will probably be between .05 and .12.
• Using the ball's vertical velocity when it leaves the end of the ramp, and the acceleration of gravity, find the corrected time of fall, which we will call `dt_corr. Be sure you are careful about the signs of your quantities.
• Using `dt_corr and your horizontal displacement, find the corrected horizontal velocity.
• Your corrected horizontal velocity still won't be exactly right (you would have to continue the above process and infinite number of times to refine it to an exact value), but the difference is unlikely exceed experimental uncertainty.
• The difference between the horizontal component of the ball's and the magnitude of its velocity at the end of the ramp will also be insignificant.
In the space below, report in the first line the number of dominoes, the distance of the roll and the final velocity on the ramp for the first trial. In subsequent lines report the same information for subsequent trials: If you got your results by calculating rather than by using the program, include a sample calculation for one of the trials.
------>>>>> # dom, dist of roll, final vel 1st trial; then subsequent trials
Your answer (start in the next line):
2 dominoes, 10 cm, 32.7 cm/s
2 dominoes, 20 cm, 37.15 cm/s
2 dominoes, 30 cm, 39.5 cm/s
4 dominoes, 10 cm, 30.1 cm/s
4 dominoes, 20 cm, 42.16 cm/s
4 dominoes, 30 cm, 54.60 cm/s
I used the program provided by the lab, but for curiosity sake I calculated the velocity for the 10 cm, 4 domino trial: 11.4 cm/.386 sec = 29.53 cm/s, only a .57 cm/s difference. (.386 sec was calculated in the previous experiment using vf^2 = v0^2 + 2a`ds and `dt = `ds/vAve) I also had to set up the trial for the 4 dominoes because in the previous experiment we were not required to do this setup if our access number was even, and mine was.
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Measure the height of a stack of 4 dominoes, as accurately as you can. In the space below give in the first line the height per domino. In the second line describe how you made your measurement and how you then determined the height per domino.
------>>>>> height per domino
Your answer (start in the next line):
.825 cm/domino
I stacked up 4 dominoes and measure to the nearest millimeter the height and found it to be 3.3 cm. I then divided 3.3 cm/4 which resulted in .825 cm. If you measure the domino individually the edge falls in between the 8th and 9th millimeter as expected from the calculation.
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For the 2-domino ramp, the ramp descends through a distance equal to the height of two dominoes while rolling a distance of about 30 cm. In the space below, give in the first line the total descent of the ball for the entire 30 cm of roll. In the second line give the amount of descent per cm of roll. Starting in the third line explain how you obtained your two results.
Note that the word 'descent' is clearly defined in the above paragraph being equal to the height of two dominoes; 'descent' refers to the vertical distance through which the ball descends. If the ball rolls 30 cm, it does not descend 30 cm. 30 cm would be the distance of roll for the trial in which the domino rolled the entire length of the ramp. Its descent for the entire 30-cm roll will in this case be about equal to the height of the two dominoes (perhaps a little more, depending on exactly where the dominoes were positioned).
------>>>>> dist of descent, descent per cm, 2 dom
Your answer (start in the next line):
1.65 cm, .055 cm/1 cm of roll
The height of one domino was found to be .825 cm. For the 2 domino ramp, the vertical descent would be 2*.825 cm = 1.65 cm. Finally, to find the descent per cm of roll, 1.65cm/30 cm = .055 cm/cm of roll.
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For the 2-domino ramp, use the preceding results to determine the amount of descent that would correspond to rolls of 10 cm, 15 cm, 20 cm and 30 cm. Give these results as 4 numbers in the first line, delimited by commas. In the second line explain how you obtained your results.
To check whether your results make sense, note that the more centimeters the ball rolls, the further it will descend. You just got done calculating the amount of descent per cm of roll, which provides a basis for the present calculation.
------->>>>>>> descent for 10, 15, 20 and 30 cm on 2 dom ramp
Your answer (start in the next line):
.55cm, .825 cm, 1.1 cm, 1.65 cm
Yes, these make sense - after rolling 15 cm the steel ball has descended the height of one domino, .825 cm and after 30 cm the descended the total of 1.65 cm or 2 dominoes.
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Using a similar strategy find the descent corresponding to rolls of 10 cm, 15 cm, 20 cm and 30 cm on a 4-domino ramp and report in similar syntax in the space below:
----->>>>>> descent for 10, 15, 20 and 30 cm four-dom ramp
Your answer (start in the next line):
1.1 cm, 1.65 cm, 2.2 cm, 3.3 cm
Total height of 4 dominoes is 4*.825 cm = 3.3 cm which for a 30 cm ramp corresponds to a 3.3 cm/30 cm of roll = .11 cm/cm of roll. The results are reasonable because after rolling 15 cm, the steel ball has descended a height of two dominoes.
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For your remaining calculations, assume that the mass of the ball is 100 grams. This isn't completely accurate, but the mass of the ball isn't critical and if necessary results could later be adjusted very easily for the accurate mass.
Determine how much work it would take to raise the mass of the ball, against the downward gravitational pull, through each of the distances of descent you have calculated. In the space below indicate in the first line the first distance of descent, then the work. In subsequent lines give the remaining distances of descent and corresponding amounts of work. In the first line following your data lines, specify the units of the quantities you have given, and explain how you obtained your results. Include a set of sample calculations for one of your lines.
----->>>> work to raise 100 g thru each dist of descent, each line dist of descent and work to raise
Your answer (start in the next line):
.55 cm, .00539 J
.825 cm, .008085 J
1.1 cm, .01078 J
1.65 cm, .01617 J
2.2 cm, .02156 J
3.3 cm, .03234 J
First, I converted on my units in centimeters to meters and the mass to a unit of kilograms. `dw = mass*acceleration due to gravity*descending height: 0.0055m * 9.8 m/s/s * .1kg = .00539 Joules
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The differences in acceleration observed at the beginning of this experiment, between the ball traveling up the ramp and the ball traveling down the ramp, are due to the change in the direction of the frictional force.
• At any point on the ramp, the gravitational force has a component along the ramp, which is the same whether the ball is moving up or down.
• However the frictional force is always in the direction opposite the motion, so when the ball is moving up the ramp the frictional force is in the same direction as the gravitational force component along the ramp, whereas when the ball is moving down the ramp the frictional force is in the opposite direction.
• It follows that the difference in net force while the ball is traveling up the ramp, and the net force while the ball is traveling down the ramp, is double the force of friction.
Based on the above:
• What was the average of the differences previously observed between the accelerations up the ramp and the acceleration down the ramp? Answer in the first line of the space below.
• On a 100-gram ball, how much difference in force would be required, to result in this much acceleration? Answer in the second line of the space below.
• What therefore is the force of friction on the ball? Answer in the third line of the space below.
• Starting in the fourth line, explain how you obtained your results. Include a set of sample calculuations.
------>>>>>> ave of differences between accel, force diff required, frictional force
Your answer (start in the next line):
2.202 cm/s/s, 2.116 cm/s/s
2.202*10^-4 N, 2.116*10^-4 N
.0004404 N, .0004232 N
The differences in the mean acceleration of the steel ball going up the ramp vs. coming down the ramp were found earlier and given in the first line. I then found the force for each acceleration, F = m*a = .1kg * .002202 m/s/s = 2.202*10^-4 for the 2 domino ramp and .1kg * 2.116*10^-4 = 2.116*10^-4 N for the 4 domino ramp.
The force of friction on the ball almost turned out the same for both, which ideally it should. As mentioned in the lab, it follows that the difference in net force while the ball is traveling up the ramp, and the net force while the ball is traveling down the ramp, is double the force of friction. Therefore I doubled the net force and found the force for friction.
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For each trial with the ball rolling up and then down the ramp, how much work did the frictional force exert on the ball as it moves up the ramp then back down?
In the space below report distance of roll and work against friction, in comma-delimited format, reporting the results for one distance per line. In the first line after your data report, indicate how you obtained your results.
-------->>>>>>>> dist of roll, work against friction, each up-and-down trial
Your answer (start in the next line):
.10 m, 4.404*10^-5
.20 m, 8.808*10^-5
.30 m, 1.3212*10-4
.10 m, 4.232*10^-5
.20 m, 8.464*10^-5
.30 m, 1.27*10^-4
The work done by the frictional force is found by the product of the force through a distance. I took the frictional force found for the 2 domino ramp and multiplied the force by each distance in the trial. I did the same for the 4 domino ramp. The first three in the list are for the 2 domino ramp and the last three; the 4 domino ramp. Work = .0004404 N * .10 m = 4.404*10^-5 Joules.
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The translational kinetic energy of the ball is 1/2 m v^2, where m is the mass of the ball and v its velocity.
For each 30-cm setup, based on the average of the velocities observed for that setup, report the number of dominoes, the distance of roll, and the translational kinetic energy attained by the ball, one setup per line in comma-delimited format. Then indicate the units of your results, and how you obtained your results.
-------->>>>>> each 30 cm setup # dom, dist of roll, transl KE
Your answer (start in the next line):
2 dominoes, 30 cm, 0.00780125 J
4 dominoes, 30 cm, 0.0149058 J
The velocity was previously calculated using the data program provided with the labs. Kinetic energy is defined as .5*mass*velocity^2. First convert mass to kilograms and velocity from cm/s to m/s, then KE = .5*.1kg*.395 m/s^2 = .000780125 Joules.
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The potential energy (PE) loss of the ball as it descends on the ramp is equal to the work required to raise it against gravity through the distance of descent.
If the ball was rolling, without slipping, on a smooth and ungrooved ramp its rotational KE would be 2/5 of its translational KE.
Assuming this to be the case (it isn't, but we'll use it as a first approximation), report in the first line of the space below the number of dominoes, distance of roll, PE loss, work done against friction, translational KE and rotational KE, for the first 30-cm setup. You will report 6 numbers in comma-delimited format. In subsequent lines indicate the same information for each remaining 30-cm setup. In the first remaining line, indicate the units of PE loss, work against friction, translational KE and rotational KE.
----->>>>> each 30-cm setup # dom, dist of roll, PE loss, work against friction, transl KE, rot KE
Your answer (start in the next line):
2 dominoes, 30 cm, .01617 J, 1.3213*10^-5, .00780125 J, 0.0031205 J
4 dominoes, 30 cm, .03234J, 1.2696*10^-5, .0149058 J, .00596232 J
The unit of PE loss, work against friction, translational KE and rotational KE are all in Joules, which is a Newton*meter.
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We need to compare PE loss, work against friction, translational KE and rotational KE. Are your units for these quantities all the same?
----->>>>> are units compatible?
Your answer (start in the next line):
Yes, they are all in Joules and therefore compatiable.
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Standard units of work/energy are:
• Joules, which are the same as N * m, or kg * m^2 / s^2, and
• ergs, which are the same as dyne * cm, or g * cm^2 / s^2.
We expect that PE loss should be equal to total KE gain + work done against friction, where total KE gain is translational KE gain + rotational KE gain.
Use the results you previously reported, but when necessary convert these quantities to consistent units so they can be compared. Report in the first line of the space below the PE loss for your first 30-cm setup, KE gain + work done against friction for that setup, and the second quantity as a percent of the first (i.e., KE gain + work done against friction as a percent of PE loss); you will report three numbers in comma-delimited format. In subsequent lines report the same information for the remaining setups. Starting in the first remaining line, give the units of your results, and explain how you obtained your results.
----->>>>>>> 30 cm setups PE loss, KE gain + `dW_frict, 2d as % of 1st
Your answer (start in the next line):
.01617 J, .014 J, 86%
.03234 J, .021 J, 65%
PE units as well as KE units are in Joules. PE was calculated previously using the height of the dominoes and the corresponding cm/cm of roll. KE gain = KE tranaslational + KE rotational, both were also calculated earlier along with the work done against friction. KE gain (2dom) = 1.3213*10^-5 J + .00780125 J + .0031205 J = .014 J
The percent were found by dividing (.014 J/.01617)*100 = 86%
All the energy should have been conserved converting from KE to PE going up the ramp and then PE back to KE going down the ramp. An 86% and 65% success rate is not perfect results, but with all the uncertainties in calculations and human error involved with measurement and timing, the results are within reason.
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Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
• Approximately how long did it take you to complete this experiment?
Your answer (start in the next line):
4.5 hours
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@& Your analysis is good; the results don't show the anticipated behavior due to experimental uncertainties.
We expect a greater magnitude of acceleration when ascending, since in that case the frictional force and the parallel component of the gravitational force are in the same direction and hence reinforce.*@