#$&* course PHY 231 8/26 7:06 002. `ph1 query 2
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Given Solution: Average velocity is defined as the average rate of change of position with respect to clock time. The average rate of change of A with respect to B is (change in A) / (change in B). Thus the average rate of change of position with respect to clock time is • ave rate = (change in position) / (change in clock time). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: Why can it not be said that average velocity = position / clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because this statement is false. Average velocity is measuring the rate of change of one variable (position) relative to the change in another variable (time). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The definition of average rate involves the change in one quantity, and the change in another. Both position and clock time are measured with respect to some reference value. For example, position might be measured relative to the starting line for a race, or it might be measured relative to the entrance to the stadium. Clock time might be measure relative to the sound of the starting gun, or it might be measured relative to noon. So position / clock time might, at some point of a short race, be 500 meters / 4 hours (e.g., 500 meters from the entrance to the stadium and 4 hours past noon). The quantity (position / clock time) tells you nothing about the race. There is a big difference between (position) / (clock time) and (change in position) / (change in clock time). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: Explain in your own words the process of fitting a straight line to a graph of y vs. x data, and briefly discuss the nature of the uncertainties encountered in the process. For example, you might address the question of how two different people, given the same graph, might obtain different results for the slope and the vertical intercept. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The process of fitting a straight line to the graph was based on an eye test involving human error in which the y-int and slope appear to position the line to where the sum of the deviations above and below line are equal for the samples measured. Two different people can see the slope very differently, which also affects where the Y-Intercept would be placed. Example: You could take a slope one person has found, make it steeper, and move the line down to form a new solution that appears to pass the eye test as well. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: (Principles of Physics students are invited but not required to submit a solution) Give your solution to the following, which should be in your notes: Find the approximate uncertainty in the area of a circle given that its radius is 2.8 * 10^4 cm. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A = pi r^2 R = 2.8 * 10^4cm +- .05 * 10^4cm Normal Area A = pi (2.8*10^4 cm)^2 =24.6 * 10^8 cm^2 =2.46 * 10 ^9 cm2 Lower bound A = pi (2.75*10^4 cm)^2 =23.8 * 10^8 cm^2 =2.38 * 10 ^9 cm^2 Difference in area is .08 * 10^9cm^2 ~ 3.3% error in area. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm. We regard 2.75 *10^4 cm as the lower bound and 2.85 *10^4 cm as the upper bound on the radius. 2.75 is .05 less than 2.8, and 2.85 is .05 greater than 2.8, so we say that the actual number is 2.8 +- .05. • Thus we express the actual radius as (2.8 +- .05) * 10^4 cm, and we call .05 * 10^4 cm the uncertainty in the measurement. The area of a circle is pi r^2, with which you should be familiar (if for no reason other than that you used it and wrote it down in the orientation exercises).. With this uncertainty estimate, we find that the area is between a lower area estimate of pi * (2.75 * 10^4 cm)^2 = 2.376 * 10^9 cm^2 and and upper area estimate of pi * (2.85 * 10^4 cm)^2 = 2.552 * 10^9 cm^2. • The difference between the lower and upper estimate is .176 * 10^9 cm^2 = 1.76 * 10^8 cm^2. • The area we would get from the given radius is about halfway between these estimates, so the uncertainty in the area is about half of the difference. • We therefore say that the uncertainty in area is about 1/2 * 1.76 * 10^8 cm^2, or about .88 * 10^8 cm^2. Note that the .05 * 10^4 cm uncertainty in radius is about 2% of the radius, while the .88 * 10^8 cm uncertainty in area is about 4% of the area. • The area of a circle is proportional to the squared radius. • A small percent uncertainty in the radius gives very nearly double the percent uncertainty in the squared radius. ** STUDENT COMMENT: I don't recall seeing any problems like this in any of our readings or assignments to this point INSTRUCTOR RESPONSE: The idea of percent uncertainty is presented in Chapter 1 of your text. The formula for the area of a circle should be familiar. Of course it isn't a trivial matter to put these ideas together. STUDENT COMMENT: I don't understand the solution. How does .176 * 10^9 become 1.76 * 10^8? I understand that there is a margin of error because of the significant figure difference, but don't see how this was calculated. INSTRUCTOR RESPONSE: .176 = 1.76 * .1, or 1.76 * 10^-1. So .176 * 10^9 = 1.76 * 10^-1 * 10^9. Since 10^-1 * 10^9 = 10^(9 - 1) =10^8, we have • .176 * 10^9 = 1.76 * 10^8. The key thing to understand is the first statement of the given solution: • Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm. This is because any number between 2.75 and 2.85 rounds to 2.8. A number which rounds to 2.8 can therefore lie anywhere between 2.75 and 2.85. The rest of the solution simply calculates the areas corresponding to these lower and upper bounds on the number 2.8, then calculates the percent difference of the results. STUDENT COMMENT: I understand how squaring the problem increases uncertainty and I understand the concept of a range of uncertainty but I am having trouble figuring out how the range of 2.75 * 10^4 and 2.85*10^4 were established for the initial uncertainties in radius. INSTRUCTOR RESPONSE: The key is the first sentence of the given solution: 'Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm.' You know this because you know that any number which is at least 2.75, and less than 2.85, rounds to 2.8. Ignoring the 10^4 for the moment, and concentrating only on the 2.8: Since the given number is 2.8, with only two significant figures, all you know is that when rounded to two significant figures the quantity is 2.8. So all you know is that it's between 2.75 and 2.85. STUDENT QUESTION I honestly didn't consider the fact of uncertainty at all. I misread the problem and thought I was simply solving for area. I'm still not really sure how to determine the degree of uncertainty. INSTRUCTOR RESPONSE Response to Physics 121 student: This topic isn't something critical to your success in the course, but the topic will come up. You're doing excellent work so far, and it might be worth a little time for you to try to reconcile this idea. Consider the given solution, the first part of which is repeated below, with some questions (actually the same question repeated too many times). I'm sure you have limited time so don't try to answer the question for every statement in the given solution, but try to answer at least a few. Then submit a copy of this part of the document. Note that a Physics 201 or 231 student should understand this solution very well, and should seriously consider submitting the following if unsure. This is an example of how to take a solution phrase by phrase and self-critique in the prescribed manner. ##&* ** Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. Understood completely. ##&* We regard 2.75 *10^4 cm as the lower bound and 2.85 *10^4 cm as the upper bound on the radius. 2.75 is .05 less than 2.8, and 2.85 is .05 greater than 2.8, so we say that the actual number is 2.8 +- .05. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. Understood completely. ##&* Thus we express the actual radius as (2.8 +- .05) * 10^4 cm, and we call .05 * 10^4 cm the uncertainty in the measurement. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. Understood completely. ##&* The area of a circle is pi r^2, with which you should be familiar (if for no reason other than that you used it and wrote it down in the orientation exercises). With this uncertainty estimate, we find that the area is between a lower area estimate of pi * (2.75 * 10^4 cm)^2 = 2.376 * 10^9 cm^2 and and upper area estimate of pi * (2.85 * 10^4 cm)^2 = 2.552 * 10^9 cm^2. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. Understood completely. ##&* The difference between the lower and upper estimate is .176 * 10^9 cm^2 = 1.76 * 10^8 cm^2. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. Understood completely. ##&* The area we would get from the given radius is about halfway between these estimates, so the uncertainty in the area is about half of the difference. We therefore say that the uncertainty in area is about 1/2 * 1.76 * 10^8 cm^2, or about .88 * 10^8 cm^2. Note that the .05 * 10^4 cm uncertainty in radius is about 2% of the radius, while the .88 * 10^8 cm uncertainty in area is about 4% of the area. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. Understood completely. ##&* The area of a circle is proportional to the squared radius. A small percent uncertainty in the radius gives very nearly double the percent uncertainty in the squared radius. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. Understood completely. ##&* STUDENT QUESTION I said the uncertainty was .1, which gives me .1 / 2.8 = .4. INSTRUCTOR RESPONSE A measurement of 2.8 can be taken to imply a number between 2.75 and 2.85, which means that the number is 2.8 +- .05 and the uncertainty is .05. This is the convention used in the given solution. (The alternative convention is that 2.8 means a number between 2.7 and 2.9; when in doubt the alternative convention is usually the better choice. This is the convention used in the text. It should be easy to adapt the solution given here to the alternative convention, which yields an uncertainty in area of about 8% as opposed to the 4% obtained here). Using the latter convention, where the uncertainty is estimated to be .1: The uncertainty you calculated would indeed be .04 (.1 / 2.8 is .04, not .4), or 4%. However this would be the percent uncertainty in the radius. The question asked for the uncertainty in the area. Since the calculation of the area involves squaring the radius, the percent uncertainty in area is double the percent uncertainty in radius. This gives us a result of .08 or 8%. The reasons are explained in the given solution. NOTE FOR UNIVERSITY PHYSICS STUDENTS (calculus-based answer): Note the following: A = pi r^2, so the derivative of area with respect to radius is dA/dr = 2 pi r. The differential is therefore dA = 2 pi r dr. Thus an uncertainty `dr in r implies uncertainty `dA = 2 pi r `dr, so that `dA / `dr = 2 pi r `dr / (pi r^2) = 2 `dr / r. `dr / r is the proportional uncertainty in r. We conclude that the uncertainty in A is 2 `dr / r, i.e., double the uncertainty in r. STUDENT QUESTION I looked at this, and not sure if I calculated the uncertainty correctly, as the radius squared yields double the uncertainty. I know where this is in the textbook, and do ok with uncertainty, but this one had me confused a bit. INSTRUCTOR RESPONSE: In terms of calculus, since you are also enrolled in a second-semester calculus class: A = pi r^2 The derivative r^2 with respect to r is 2 r, so the derivative of the area with respect to r is dA / dr = pi * (2 r). If you change r by a small amount `dr, the change in the area is dA / dr * `dr, i.e., rate of change of area with respect to r multiplied by the change in r, which is a good commonsense notion. Thus the change in the area is pi * (2 r) `dr. As a proportion of the original area this is pi ( 2 r) `dr / (pi r^2) = 2 `dr / r. The change in the radius itself was just `dr. As a proportion of the initial radius this is `dr / r. The proportional change in area is 2 `dr / r, compared to the proportional change in radius `dr / r. That is the proportional change in area is double the proportional change in radius. STUDENT COMMENT I used +-.1 instead of using +-.05. I understand why your solution used .05 and will use this method in the future. INSTRUCTOR RESPONSE Either way is OK, depending on your assumptions. When it's possible to assume accurate rounding, then the given solution works. If you aren't sure the rounding is accurate, the method you used is appropriate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK, this one took a bit longer and I had a very basic error. I started out with my uncertainty as .1, but quickly thought that out in my head and realized it could be a number all the way down to 2.75. After this the problem was very easy as I have already done all the practice problems in chapter 1 of the book. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: What is your own height in meters and what is your own mass in kg (if you feel this question is too personal then estimate these quantities for someone you know)? Explain how you determined these. What are your uncertainty estimates for these quantities, and on what did you base these estimates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My Height 67 inches * (2.54 cm/1 inch) = 170 cm * ( 1 m / 100cm) = 1.7 m Uncertainty is 1% found by (.5 inch/67 inches) My Weight 155 lbs * (1 kg/2.2 lb) = 70.5 kg Uncertainty is .5 lb/155lb = .03% confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Presumably you know your height in feet and inches, and have an idea of your ideal weight in pounds. Presumably also, you can convert your height in feet and inches to inches. To get your height in meters, you would first convert your height in inches to cm, using the fact that 1 inch = 2.54 cm. Dividing both sides of 1 in = 2.54 cm by either 1 in or 2.54 cm tells us that 1 = 1 in / 2.54 cm or that 1 = 2.54 cm / 1 in, so any quantity can be multiplied by 1 in / (2.54 cm) or by 2.54 cm / (1 in) without changing its value. Thus if you multiply your height in inches by 2.54 cm / (1 in), you will get your height in cm. For example if your height is 69 in, your height in cm will be 69 in * 2.54 cm / (1 in) = 175 in * cm / in. in * cm / in = (in / in) * cm = 1 * cm = cm, so our calculation comes out 175 cm. STUDENT SOLUTION 5 feet times 12 inches in a feet plus six inches = 66 inches. 66inches * 2.54 cm/inch = 168.64 cm. 168.64 cm * .01m/cm = 1.6764 meters. INSTRUCTOR COMMENT: Good, but note that 66 inches indicates any height between 65.5 and 66.5 inches, with a resulting uncertainty of about .7%. 168.64 implies an uncertainty of about .007%. It's not possible to increase precision by converting units. STUDENT SOLUTION AND QUESTIONS My height in meters is - 5’5” = 65inches* 2.54cm/1in = 165cm*1m/100cm = 1.7m. My weight is 140lbs* 1kg/2.2lbs = 63.6kg. Since 5’5” could be anything between 5’4.5 and 5’5.5, the uncertainty in height is ???? The uncertainty in weight, since 140 can be between 139.5 and 140.5, is ?????? INSTRUCTOR RESPONSE Your height would be 5' 5"" +- .5""; this is the same as 65"" +- .5"". .5"" / 65"" = .008, approximately, or .8%. So the uncertainty in your height is +-0.5"", which is +-0.8%. Similarly you report a weight of 140 lb +- .5 lb. .5 lb is .5 lb / (140 lb) = .004, or 0.4%. So the uncertainty is +-0.5 lb, or +- 0.4%. STUDENT QUESTION I am a little confused. In the example from another student her height was 66 inches and you said that her height could be between 65.5 and 66.5 inches. but if you take the difference of those two number you get 1, so why do you divide by .5 when the difference is 1 INSTRUCTOR RESPONSE If you regard 66 inches as being a correct roundoff of the height, then the height is between 65.5 inches and 66.5 inches. This makes the height 66 inches, plus or minus .5 inches. This is written as 66 in +- .5 in and the percent uncertainty would be .5 / 66 = .007, about .7%. If you regard 66 inches having been measured only accurately enough to ensure that the height is between 65 inches and 67 inches, then your result would be 66 in +- 1 in and the percent uncertainty would be 1 / 66 = .015 or about 1.5%. STUDENT QUESTION If a doctor were to say his inch marker measured to the nearest 1/4 inch, would that be the uncertainty? Meaning, would I only have to multiply that by .0254 to find the uncertainty in meters, dividing that by my height to find the percent uncertainty? INSTRUCTOR RESPONSE That's pretty much the case, though you do have to be a little bit careful about how the rounding and the uncertainty articulate. For example I'm 72 inches tall. That comes out to 182.88 cm. It wouldn't make a lot of sense to say that I'm 182.88 cm tall, +- .64 cm. A number like 182.88 has a ridiculously high number of significant figures. It wouldn't quite be correct to just round up and say that I'm 183 cm tall +- .64 cm. We might be able to say that I'm 183 cm tall, +- .76 cm, but that .76 cm again implies more precision than is present. We would probably end up saying that I""m 183 cm tall, +- 1 cm. Better to overestimate the uncertainty than to underestimate it. As far as the percent uncertainty goes, we wouldn't need to convert the units at all. In my case we would just divide 1/4 in. by 72 in., getting about .034 or 3.4%. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up additional speed before rolling off the end of that book. Suppose you know all the following information: • How far the ball rolled along each book. • The time interval the ball requires to roll from one end of each book to the other. • How fast the ball is moving at each end of each book. How would you use your information to calculate the ball's average velocity on each book? How would you use your information to calculate how quickly the ball's speed was changing on each book? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I would find the distance traveled along each book and divide by the time spent rolling on each book. This will provide the individual average velocities as vAve = change in position/ change in time. We could compared the three average velocities to determine they are getting faster and faster; yet, I believe there would problems with this such as the time/angle of the drop (lost or gained speed in the air)? The truest way to evaluate this would be to evaluate two points at the beginning of each book and two points of equal position or time at the end of each book. If the position is equal we would want to see less time due to a higher velocity. If the time was equal we would want to see a greater position change due to the higher velocity. confidence rating #$&*: 2 (Only because I was unsure if you can compare the average velocities of the individual books. Thinking this would not be accounting for the drops (gain/loss in speed). Or can this be overlooked as gravity is vertical, air friction is minimal, and there is no surface friction during the drop? ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up additional speed before rolling off the end of that book. Suppose you know all the following information: • How far the ball rolled along each book. • The time interval the ball requires to roll from one end of each book to the other. • How fast the ball is moving at each end of each book. How would you use your information to calculate the ball's average velocity on each book? How would you use your information to calculate how quickly the ball's speed was changing on each book? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I would find the distance traveled along each book and divide by the time spent rolling on each book. This will provide the individual average velocities as vAve = change in position/ change in time. We could compared the three average velocities to determine they are getting faster and faster; yet, I believe there would problems with this such as the time/angle of the drop (lost or gained speed in the air)? The truest way to evaluate this would be to evaluate two points at the beginning of each book and two points of equal position or time at the end of each book. If the position is equal we would want to see less time due to a higher velocity. If the time was equal we would want to see a greater position change due to the higher velocity. confidence rating #$&*: 2 (Only because I was unsure if you can compare the average velocities of the individual books. Thinking this would not be accounting for the drops (gain/loss in speed). Or can this be overlooked as gravity is vertical, air friction is minimal, and there is no surface friction during the drop? ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!