initial timing experiment

Your 'initial timing experiment' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

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Initial Timing Experiment


You may enter any message or comment you wish in the box below:

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In this experiment you will use the TIMER program, a hardcover book, a cylinder or some other object that will roll along the book in a relatively straight line, and a ruler or the equivalent (if you don't have one, note the RULERS link on the Assignments page).

Place the book on a flat level tabletop.  You will prop one end of the book up a little bit, so that when it is released the object will roll without your assistance, gradually speeding up, from the propped-up end to the lower end.  However don't prop the end up too much.  It should take at least two seconds for the ball to roll down the length of the book when it is released from rest.

Then reverse the direction of the book on the tabletop, rotating the book and its prop 180 degrees so that the ball will roll in exactly the opposite direction.  Repeat your measurements.

In the box below describe your setup, being as specific as possible about the book used (title, ISBN) and the object being used (e.g., a solid glass marble, a small can of tomato paste (full or empty?), a ball-point pen), and what you used to prop the object up (be as specific as possible).   Also describe how well the object rolled--did it roll smoothly, did it speed up and slow down, did it roll in a straight line or did its direction change somewhat?

Note:  Don't trust this form.  Compose your answer in Notepad or a word processor, saving it every few minutes, then copy and paste it into the box.  Power could surge, your computer could malfunction, in any of a number of ways the work you put into this form could be lost.  Compose it elsewhere and keep a copy.

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The book used was University Physics, (ISBN # -201603225). The object being used was a red ball point pen without the lid. The object used to prop up the book was a Bostitch Stapler. The object rolled smoothly and with quite abit of speed. It rolled in a steady, smooth, straight line each time.

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In the box below report your data.  State exactly what was measured, how it was measured, how accurately you believe it was measured and of course what the measurements were.  Try to organize your report so the reader can easily scan your data and see any patterns that might occur.

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The first set of data is involving all the necessary materials such as the book, pen and stapler. The book is angled to the right with the incline beginning at the left. The book laying flat on the surface of the table is 4cm and with the incline support of the stapler is now at 10cm. The data for the first five tries is as follows:

1 5440.82 5440.82

2 5443.043 2.222656

3 5445.707 2.664063

4 5448.922 3.214844

5 5452.188 3.265625

Now for the second attempt. THe book/items were rotated 180 degrees and the experiment was repeated. Here are the data collections.

1 5783.008 5783.008

2 5785.672 2.664063

3 5787.855 2.183594

4 5789.258 1.402344

5 5791.383 2.125

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Using your data determine how fast the object was moving, on the average, as it rolled down the incline.  Estimate how accurately you believe you were able to determine the object's average speed, and give the best reasons you can for your estimate of the accuracy.

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I took each of the ten time intervals and averaged them and came with the conclusion of 1124.357019

Then, I seperately took the average time of the first five and got an answer of 1090.437438 and then second set and received an average time of 1158.2766.

5783.008 seconds is not a time interval which corresponds to any of your trials and shouldn't be included in your averaging. The other intervals seem reasonable, though the 1.4 seconds might be a little short compared to the others.

Otherwise your procedure is good.

To further elaborate on my answers. I took each time interval from the first set and divided it by 26.5 which was the distance the pen traveled. My answers are as follows.

26.5/ 5440.82 = 0.00487

26.5/ 2.222656= 11.92267

26.5/ 2.664063= 9.947212

26.5/ 3.214844= 8.243012

26.5/ 3.265625= 8.114832

The average of these times is 7.6465192 which is the average of the pen the first five times.

Your calculation divides the change in position (length of the roll) by the change in the clock time, and is consistent with the definition vAve = `ds / `dt in the following discussion.

A common definition reads v = d / t. Many students use this definition to calculate the average velocity here, and it works fine for this example. However this formula has some shortcomings and won't be used in this course. Read on:

In this course t will generally stand for clock time and `dt for time interval. The symbol `d represents the capital Greek delta (the triangle symbol) and denotes 'change in ...', so `dt stands for 'change in t'.

Since d is used in `d to stand for 'change in', we generally won't use d to stand for any actual quantity (it would be confusing if we wanted to represent 'change in d' by '`dd').

We will use s or x to stand for position, so that `dx or `ds stands for the change in position.

What the equation v = d / t would calculate is actually the average speed or average velocity. We'll worry later about the difference between speed and velocity. Using `d for 'change in', t for clock time and s for position we use here the definition

vAve = `ds / `dt.

So for example if the ball travels 24 cm while the clock time changes from, say, 28.1 sec to 32.9 sec, we would observe the following:

`ds is the change in the position of the ball, which in this example is 24 cm.

`dt is the change in the clock time; if the TIMER program is triggered at clock times 28.1 sec and 31.7 sec, then the corresponding time interval will be the difference 31.7 sec - 28.1 sec = 3.6 sec.

So here we would have

vAve = `ds / `dt = 24 cm / 3.6 sec = 6.7 cm / sec.

Note that your distance is probably accurate only to the nearest cm; your time interval is indicated to a much higher precision, but as you saw in the

TIMER program, the precision of the program itself is limited to around .01 sec. So even if your own triggering of the mouse was flawless, the TIMER couldn 't give you as much precision as used in this calculation.

As a rule of thumb, when the precision of one of your observed quantities is limited to 2 significant figures (as is the case here), the precision of the result cannot exceed two significant figures. In this experiment the limits of precision of the distance measurement usually dictates that the velocity is good to only two significant figures.

So for example if the distance is 24 cm and the TIMER reports an interval of 1.8071 seconds, we wouldn't report an average velocity of 24 cm / (1.8071 sec) = 13.28094737 cm / sec. The ... 071 in the time interval is pretty meaningless, since the TIMER itself, and also the user, do not have that sort of precision. So most of the figures in that result are meaningless. The best we can report is probably that the average velocity is 13 cm/s. If we think the measurements are really close we might report 13.3 cm/s, but we would probably mention that the .3 part is questionable.

Be sure you use units in your calculations. For example if the ball rolls 24 cm in 3.6 seconds, the calculation is 24 cm / (3.6 seconds) = 6.7 cm/second, not just 24 / 3.6.

The second set of data is as follows:

26.5/5783.008= .00458239

26.5/2.664063= 11.92267

26.5/2.183594= 12.135955

26.5/1.402344= 18.896932

26.5/2.125 = 12.470588

The average of these times is 10.69105388 which is the average time the pen had the second time this was conducted.

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Devise and concuct an experiment to determine whether or not the object is speeding up as it rolls down the incline.  If you have set the experiment up as indicated, it should seem pretty obvious that the object is in fact speeding up.  But figure out a way to use actual measurements to support your belief.

Explain how you designed and conducted your experiment, give your data and explain how your data support your conclusions.

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During this expirament I wanted to see if there was a difference in using a hard bound book as opposed to a paper back book. Whereas, in the first expirament a hard back book was used and the results are previously mentioned. This time around the same expirament will be conducted and a paper bound book (a Quill business supply ordering book) will be used. All other materials, the stapler and the pen and the amount of times and measurements will be the same as before.

set 1 data

1 8700.223 8700.223 26.5/8700.223= .003045898

2 8700.883 .6601563 26.5/.660156= 40.14202703

3 8702.047 1.164063 26.5/1.164063= 22.76509089

4 8702.715 .6679688 26.5/.6679688= 39.67625116

5 8703.355 .640625 26.5/.640625= 41.3658536

Avg. Speed = 28.7897058

Set 2 data book rotated 180'

book laying flat on table is 4cm high

1 8725.98 8725.98 26.5/8725.98= .003036908

2 8727.141 1.160156 26.5/1.160156=22.841755

3 8728.102 .9609375 26.5/.9609375=27.5772357

4 8728.535 .4335938 26.5/.4335938=61.117110069

5 8729.734 1.199219 26.5/1.199219=22.09771526

Avg speed. 26.7273705

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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: 

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2 hours

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You may add optional comments and/or questions in the box below.

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I've inserted some notes, mainly to reinforce the ideas you have already used in your work. Be sure to read my notes carefully and completely. There is no need to respond unless you have questions, in which case those questions are more than welcome.