course PHY 201 June 15 around 1:20 pm 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 #$&* 3 ********************************************* Question: Why can it not be said that average velocity = position / clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The position and clock time can be either digit, so this is no correct. It is actually the “change” in position and the “change” in clock time, which has a huge difference on the equation of average velocity. It is also missing one change with respect to another change, which is very important to the definition of average velocity or rate change. confidence rating #$&* 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 #$&* 3 ********************************************* Question: Explain in your own words the process of fitting a straight line to a graph of y vs. x data, and briefly discuss the nature of the uncertainties encountered in the process. For example, you might address the question of how two different people, given the same graph, might obtain different results for the slope and the vertical intercept. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When you are trying to fit a straight line to a graph of y vs. x data, you are always going to have uncertainty because you will not get the line at its perfect location. First, you try your best to get the line in the middle of the data points and draw that line. Next, you find two points on the line according to the x and y axis numbers, and estimate where your line passes through the y axis. Then you subtract the difference between the two coordinates and, then divide them, and you get your slope of your line. Then you use the equation y = m x + b, and substitute your slope and one of the points and solve for b, the y-intercept. Then, you compare the difference between your y-intercept and the actual best-fit y-intercept, and see how far you were off. There are a lot of uncertainties in this situation, from guessing where the line passes through the y-intercept too the coordinates you picked out on your graph. If you do the estimation over, I bet you will get different answer than you got before, because of the uncertainty in this process. confidence rating #$&* 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: First you get the area of a circle by using the formula A = pi * r^2, and substitute what you know into the equation: A = pi * (2.8 * 10^4) ^2 = 87964.5943. Next, I used the percent error equation to solve for the uncertainty of the problem: .0001/87964.5943 * 100 = .000000114% confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. I understand. ##&* 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. I understand. ##&* 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. I understand ##&* 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. I understand ##&* 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. I understand. ##&* 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. I do not understand why the area we would get from the given radius is about halfway between these estimates. Is it halfway between 2.75 and 2.85 or is it something different? ##&* 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. I understand. ##&* 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 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): I understand the method of getting the uncertainty now and will apply this knowledge in future problems. ------------------------------------------------ Self-critique rating #$&* 2 ********************************************* 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: 71 inches = 1.80 meters 156 lbs = 70.91 kg’s For the height, I converted units: if there are 12 inches in 1 foot, 3.281 feet in 1 meter, then I multiplied and divided to get my answer. For the weight, I converted units: if there are 2.20 lbs in 1 kg, then I multiplied and divided and got my answer. For the height: 1.8 %, I based this estimate on the book’s way of doing uncertainty, which is .01 or how many significant digits there were in the original number. For the weight: .0141 %, I based this estimate on the book’s way of doing uncertainty as well, which is .01. confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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%. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Now I understand the way to get the percent uncertainty, but I am not sure what the book is trying to do. I understand your way, so I am going to stick with that method. ------------------------------------------------ Self-critique rating #$&* 2 ********************************************* 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: If you knew how far it went and the time it took you can determine the ball’s average velocity, because is the change in position divided by the change in clock time. You would need the time interval the ball requires to roll from one end of each book to the other, and how fast the ball is moving at each end of each book, and you could get the ball’s average speed each book, because the average rate of the how fast it was going with respect time change would equal to the change in speed divided by the change in time. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ "