course PHY 121

6/15 6:06pm

ph1 query 0Most queries in this course will ask you questions about class notes, readings, text problems and experiments. Since the first two assignments have been lab-related, the first two queries are related to the those exercises. While the remaining queries in this course are in question-answer format, the first two will be in the form of open-ended questions. Interpret these questions and answer them as best you can.

Different first-semester courses address the issues of experimental precision, experimental error, reporting of results and analysis in different ways and at different levels. One purpose of these initial lab exercises is to familiarize your instructor with your work and you with the instructor 's expectations.

Comment on your experience with the three lab exercises you encountered in this assignment or in recent assignments.

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Question: This question, related to the use of the TIMER program in an experimental situation, is posed in terms of a familiar first-semester system.

Suppose you use a computer timer to time a steel ball 1 inch in diameter rolling down a straight wooden incline about 50 cm long. If the computer timer indicates that on five trials the times of an object down an incline are 2.42sec, 2.56 sec, 2.38 sec, 2.47 sec and 2.31 sec, then to what extent do you think the discrepancies could be explained by each of the following:

• The lack of precision of the TIMER program.

To what extent to you think the discrepancies are explained by this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

There is error using the timer program at the .01 mark there is less accuracy anything faster than a tenth of a second

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• The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)

To what extent to you think the discrepancies are explained by this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

The factor of human triggering causing uncertainty in the precision is very likely because it is not possible to click the exact moment on the computer by eyeing a moving object and doing so repeatedly, you have to assume there will be error

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• Actual differences in the time required for the object to travel the same distance.

To what extent to you think the discrepancies are explained by this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

The object may take different times to travel the same distance due to many factors such as it wasn’t released at the same point each time, or the distance was measured differently each time, or the timer was not started at an accurate time or stopped at an accurate time

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• Differences in positioning the object prior to release.

To what extent to you think the discrepancies are explained by this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Discrepancies are cause also by releasing the object at different points each time, just by eyeing the release position there is going to be error and even the slightest position off will change the outcome and time it takes for the object to move

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• Human uncertainty in observing exactly when the object reached the end of the incline.

To what extent to you think the discrepancies are explained by this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

There is definite uncertainty when observing when the object reached the end of the line especially when there are not sensors monitoring the exact moment the object reaches the end only by sight then no 2 people would see the same ending point for multiple readings

#$&*

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Question: How much uncertainty do you think each of the following would actually contribute to the uncertainty in timing a number of trials for the ball-down-an-incline lab?

• The lack of precision of the TIMER program.

To what extent to you think this factor would contribute to the uncertainty?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

There is error using the timer program at the .01 mark there is less accuracy anything faster than a tenth of a second

#$&*

• The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)

To what extent to you think this factor would contribute to the uncertainty?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

There is uncertainty in precision due to not pressing the start button at the exact moment the ball starts rolling down the incline

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• Actual differences in the time required for the object to travel the same distance.

To what extent to you think this factor would contribute to the uncertainty?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

#$&*

• Differences in positioning the object prior to release.

To what extent to you think this factor would contribute to the uncertainty?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

#$&*

• Human uncertainty in observing exactly when the object reached the end of the incline.

To what extent to you think this factor would contribute to the uncertainty?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

#$&*

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Question: What, if anything, could you do about the uncertainty due to each of the following? Address each specifically.

• The lack of precision of the TIMER program.

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Unless the TIMER program was changed then no matter how accurate you press the button the timer itself is only going to be precise to the .01 mark

#$&*

• The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Practicing runs would increase accuracy, as long as there is a human pressing a button there is going to be some error, the only way to prevent error would be electronically using sensors to monitor the beginning and end of the run

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• Actual differences in the time required for the object to travel the same distance.

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Be sure and measure the distance, accurately, release from the exact spot each time, have a sensor device measure the time required from start to finish

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• Differences in positioning the object prior to release.

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Instead of just starting by letting the ball go with your hand set up a device when you push a button or lever the ball is released to make sure the ball is starting at the same point each time

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• Human uncertainty in observing exactly when the object reached the end of the incline.

What do you think you could do about the uncertainty due to this factor?

your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

The mark the moment the object reaches the “finish line” there could be a computer based sensor set up to mark the exact moment the object reached the end of the incline so the time would be correct and also the position would be correct each time as well

#$&*

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Question: If, as in the object-down-an-incline experiment, you know the distance an object rolls down an incline and the time required, explain how you will use this information to find the object 's average speed on the incline.

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Your solution:

Avg. speed will be found if you know the distance and time required by using the formula

Speed=distance/time

confidence rating #$&* 3

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Question: If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer is connected to your experience.

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Your solution:

40cm distance

5 seconds time

Avg. velocity=change in distance/change in time

Avg. velocity=40cm/5sec=8cm/sec

When rolling an object down an incline avg velocity was calculated during the incline experiement using change in distance and change in time

confidence rating #$&* 3

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Question: If the same object requires 3 second to reach the halfway point, what is its average velocity on the first half of the incline and what is its average velocity on the second half?

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Your solution:

First half incline:

Distance is 20cm time is 3 sec

So avg. velocity 20cm/3sec=6.67 cm/sec

Second half:

Distance 20 cm time 2 sec

Avg. velocity=20cm/2sec=10cm/sec

confidence rating #$&* 3

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Question: `qAccording to the results of your introductory pendulum experiment, do you think doubling the length of the pendulum will result in half the frequency (frequency can be thought of as the number of cycles per minute), more than half or less than half?

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Your solution:

According to the results doubling the length of the pendulum did not result in half th frequency it was less than half

confidence rating #$&* 3

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Question: `qNote that for a graph of y vs. x, a point on the x axis has y coordinate zero and a point on the y axis has x coordinate zero. In your own words explain why this is so.

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Your solution:

The point on the x axis that has a y coordinate zero is the x intercept and the point on the y axis has a x coordinate zero is the y intercept this is the point where the line crosses the x axis and y axis

confidence rating #$&* 3

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Question: `qOn a graph of frequency vs. pendulum length (where frequency is on the vertical axis and length on the horizontal), what would it mean for the graph to intersect the vertical axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the vertical axis)? What would this tell you about the length and frequency of the pendulum?

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Your solution:

If the line representing frequency vs. length goes through the vertical axis then this tells you that the length of the pendulum is 0 because the only way for the line to go through the vertical axis would be if the length of the pendulum was zero and if the length was zero the frequency would be zero

confidence rating #$&* 2

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Question: `qOn a graph of frequency vs. pendulum length, what would it mean for the graph to intersect the horizontal axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the horizontal axis)? What would this tell you about the length and frequency of the pendulum?

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Your solution:

For the graph to intersect the horizontal axis on the frequency vs. pendulum length it would mean the frequency would have to be zero and if there was not frequency then the pendulum was either at rest or had a length of zero

confidence rating #$&* 2

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Question: `qIf a ball rolls down between two points with an average velocity of 6 cm / sec, and if it takes 5 sec between the points, then how far apart are the points?

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Your solution:

Avg. velocity 6cm/sec

Time 5 sec

Distance?

Avg. velocity=change in distance/change in time

6cm/sec=change in distance/5sec

5sec*6cm/sec=change in distance

Distance=30cm

confidence rating #$&*3

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Given Solution:

`aOn the average the ball moves 6 centimeters every second, so in 5 seconds it will move 30 cm.

The formal calculation goes like this:

• We know that vAve = `ds / `dt, where vAve is ave velocity, `ds is displacement and `dt is the time interval.

• It follows by algebraic rearrangement that `ds = vAve * `dt.

• We are told that vAve = 6 cm / sec and `dt = 5 sec. It therefore follows that

• `ds = 6 cm / sec * 5 sec = 30 (cm / sec) * sec = 30 cm.

The details of the algebraic rearrangement are as follows:

• vAve = `ds / `dt. We multiply both sides of the equation by `dt:

• vAve * `dt = `ds / `dt * `dt. We simplify to obtain

• vAve * `dt = `ds, which we then write as{}`ds = vAve *`dt

Be sure to address anything you do not fully understand in your self-critique.

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Your solution:

Avg. velocity 6cm/sec

Time 5 sec

Distance?

Avg. velocity=change in distance/change in time

6cm/sec=change in distance/5sec

5sec*6cm/sec=change in distance

Distance=30cm

confidence rating #$&* 3

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Question: `qYou were asked to read the text and some of the problems at the end of the section. Tell your instructor about something in the text you understood up to a point but didn't understand fully. Explain what you did understand, and ask the best question you can about what you didn't understand.

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Your solution:

I WAS CONFUSED ABOUT UNCERTAINTY BUT I READ THE SECTION AND LOOKED OVER THE EXAMPLE PROBLEM AND IT EXPLAINED THE CONCEPT BETTER

I UNDERSTOOD HOW TO CALCULATE PERCENT UNCERTAINTY BUT I WAS UNSURE ON HOW THEY WERE GETTING THE UNCERTAINTY TO START WITH

HOW DO YOU DETERMINE THE UNCERTAINTY WHEN IT IS NOT GIVEN IN THE PROBLEM?

You will typically assess the accuracy and precision of your measuring instruments.

#$&*

STUDENT QUESTION

I understand that we cannot measure to exact precision, but when we are dealing with estimated uncertainty, do we always

just increment our lowest unit by one and that is our uncertainty? Is there a standard that is used to figure out this?

INSTRUCTOR RESPONSE

The standard answer is that we assume an uncertainty of +- 1 of our smallest unit of precision. However, depending on how well we can 'see' that smallest unit, we can get pretty close to +- 1/2 of a unit.

A more sophisticated answer can be given in terms of the statistics of the normal distribution, but in this course we're not going to go into a whole lot of depth with that. A calculus background would be just about required to understand the analysis well enough to apply it meaningfully.

STUDENT QUESTION

I fully understand how to calculate uncertainty, but what if the uncertainty isn’t given? For example,

problem 6 asks us for the uncertainty of 1.67. Do we just use .01 as the uncertainty?

INSTRUCTOR RESPONSE

Depending on the nature of the instrument and the observation, +- .01 might be necessary, but we could go to +-.005 if can regard 1.67 as an accurate roundoff.

Without very good reason, though, +-.01 would be the safer assumption.

STUDENT QUESTION: I understood the portion discussing the nature of science and felt familiar with much of the measurement. What I did not fully understand was how do you know when to write an answer using the powers of 10 or to leave it alone? Several of the tables had values in powers of 10 for metric prefixes such as centi and mili.

INSTRUCTOR RESPONSE

Whether you use scientific notation or not depends a lot on the context of the situation.

As a rule of thumb, I would recommend going to scientific notation for numbers greater than a million (10^6) and less than a millionth (10^-6). When numbers outside this range are involved in an analysis it's a good idea to put everything into scientific notation.

And when you know that scientific notation is or is not expected by your audience, write your numbers accordingly.

QUESTION RELATED TO UNIVERSITY PHYSICS (relevant only to University Physics students)

I don’t fully understand the dot product rule

INSTRUCTOR RESPONSE

The dot product of vectors A = a_1 i + a_2 j + a_3 k and B = b_1 i + b_2 j + b_3 k is a_1 * b_1 + a_2 * b_2 + a_3 * b_3. The dot product is simply a number.

The magnitude of A is | A | = sqrt( a_1 ^ 2 + a^2 ^ 2 + a_3 ^ 2); the magnitude of B is found in a similar manner.

The dot product is equal to | A | * | B | * cos(theta), where theta is the angle between the two vectors.

If you have the coefficients of the i, j and k vectors, it is easy to calculate the dot product, and it's easy to calculate the magnitudes of the two vectors. Setting the two expressions for the dot product equal to one another, we can easily solve for cos(theta), which we can then use to find theta.

More importantly for physics, we can find the projection of one vector on another. The projection of A on B is just the component of A in the direction of B, equal to | A | cos(theta). The projection of one vector on another is important in a number of situations (e.g., the projection of the force vector on the displacement, multiplied by the displacement, is the work done by the force on the interval corresponding to the displacement).

Dot products are a standard precalculus concept. Check the documents at the links below for an introduction to vectors and dot products. You are welcome to complete these documents, in whole or in part, and submit your work. If you aren't familiar with dot products, it is recommended you do so.

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/pc2/pc2_qa_09.htm

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/pc2/pc2_qa_10.htm

confidence rating #$&* 3 seems like uncertainty is problem for a lot of people at first

TYpically there are lots of questions about that.

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Question: `qTell your instructor about something in the problems you understand up to a point but don't fully understand. Explain what you did understand, and ask the best question you can about what you didn't understand.

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Your solution:

The question was this

“what, approximately, is the percent uncertainty for the measurement given as 1.57 m^2 ?

So I am assuming the uncertainty is +-.01 , I think this because the number is 1.57 m^2 and it goes to the hundredth place so the uncertainty is plus or minus one to the hundredth place

So to calculate percent uncertainty for the measurement .01/1.57=(approx).6364

So what I was unsure about is do I round up and give you whole number % like the answer is 1% or do I do sig figs and the answer is .636%

That depends on how confident you think you can be in your estimated percent. Rounding to 1% is 'safe', for a situation in which you want to be careful not to overestimate the accuracy of your answer. In some cases, though, you could regard .636% as 'about half a percent'.

#$&*

SOME COMMON QUESTIONS:

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QUESTION: I didn’t understand how to calculate uncertainty for a number such as 1.34. When given examples we had problems such as 1.34 ±0.5 and with that we had a formula (0.5/1.34)*100. So I do not understand how to compute uncertainty when no estimated uncertainty is given.

INSTRUCTOR RESPONSE:

The +- number is the uncertainty in the measurement.

The percent uncertainty is the uncertainty, expressed as a percent of the number being observed.

So the question in this case is simply, 'what percent of 1.34 is 0.5?'.

• 0.5 / 1.34 = .037, approximately. So 0.5 is .037 of 1.34.

• .037 is the same as 3.7%.

I recommend understanding the principles of ratio, proportion and percent as opposed to using a formula. These principles are part of the standard school curriculum, though it does not appear that these concepts have been well mastered by the majority of students who have completed the curriculum. However most students who have the prerequisites for this course do fine with these ideas, after a little review. It will in the long run save you time to do so.

There are numerous Web resources available for understanding these concepts. You should check out these resources and let me know if you have questions.

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QUESTION: I understood the main points of changing the different units, but I’m not sure when in the problem I should change the number to 10 raised to a certain power. In example 1-8 I did not understand why they changed 70 beats/min to 2 x 10^9 s.

2 * 10^9 is about the number of seconds in 70 years.
70 beats / min were not changed to 2 * 10^9 seconds; in changing the beats / minute to beats in a lifetime, there was a step where it was necessary to multiply by 2 * 10^9 seconds.
The example actually used 80 beats / min as a basis for the solution. This was converted to beats / second by the calculation
80 beats / min * 1 minute / (60 seconds), which would yield about 1.33 beats / second.
This was then multiplied by 2 * 10^9 seconds to get the number of beats in a lifetime:
2 * 10^9 seconds * 1.33 beats / second = 3 * 10^9 beats.
In the given solution 80 beats / min * 1 minute / (60 seconds) was not actually calculated; instead 80 beats / min * 1 minute / (60 seconds) was multiplied by 2 * 10^9 seconds in one step
80 beats / min * 1 minute / (60 seconds) * 2 * 10^9 seconds = 3 * 10^9 beats.
In your instructor's opinion the unit 'beats' should have been left in the result; the text expressed the result simply as 3 * 10^9, apparently ignoring the fact that the unit 'beats' was included in the quantities on the left-hand side.
Also the text identified this number as 3 trillion. In the British terminology this would be correct; in American terminology this number would be 3 billion, not 3 trillion.

COMMENT:

I thought that these problems were pretty basic and felt that I understood them well. However, when I got to questions 14 (determine your own mass in kg) and 15 (determining how many meters away the Sun is from the Earth), I did not understand how to complete these. I know my weight in pounds, but how can that be converted to mass in kilograms? I can look up how to convert miles to meters, but is this something I should already know?

INSTRUCTOR RESPONSE:

Both of these questions could be answered knowing that an object with a mass of 1 kg has a weight of 2.2 lb, and that an inch is 2.54 centimeters. This assumes that you know how many feet in a mile, and that the Sun is 93 million miles away. All these things should be common knowledge, but it doesn't appear to be so.

For my own weight I would reason as follows:

I weigh 170 lb and every kg of my mass weighs 2.2 lb. I'll have fewer kg of mass than I will pounds of weight, so it's reasonable to conclude that my mass is 170 / 2.2 kg, or about 78 kg.

More formally 170 lb * (1 kg / (2.2 lb) ) = 170 / 2.2 kg = 78 kg, approx.. (technical point: this isn't really right because pounds and kilograms don't measure the same thing--pounds measure force and kg measure mass--but we'll worry about that later in the course).

Converting 93 million miles to kilometers:

93 million miles * (5280 feet / mile) * (12 inches / foot) * (2.54 cm / inch) * (1 meter / (100 cm) ) = 160 billion meters (approx.) or 160 million kilometers.

QUESTION

What proved to be most tricky in the problems portion was the scientific notation. I am somewhat familiar with this from

past math classes, but had trouble when dealing with using the powers of 10. I had trouble dealing with which way to move my decimal according to the problems that were written as 10^-3 versus 10^3. Which way do you move the decimal when dealing with negative or positive powers of 10?

INSTRUCTOR RESPONSE

Using your numbers, 10^3 means 10 * 10 * 10 = 1000.

• When you multiply a number by 1000 you move the decimal accordingly. For example 3.5 * 1000 = 3500.

10^-3 means 1 / 10^3 = 1 / (10 * 10 * 10) = 1 / 1000.

• When you multiply by 10^-3 you are therefore multiplying by 1 / 1000, which is the same as dividing by 1000, or multiplying by .001.

• For example 3.5 * 10^-3 = 3.5 * .001 = .0035.

• As another example 5 700 000 * 10^-3 would be 5 700 000 * (1 / 1000) = 5 700.

• From these examples you should be able to infer how the decimal point moves.

You can also search the Web under 'laws of exponents', 'arithmetic in scientific notation', and other keywords.

There isn't a single site I can recommend, and if I did find a good one its URL might change by the time you try to locate it. In any case it's best to let you judge the available materials yourself.

When searching under 'arithmetic in scientific notation' using Google, the following appear as additional suggested search phrases:

scientific notation

exponents

scientific notation metric prefixes

significant digits

multiply with scientific notation

scientific notation decimal

scientific notation lessons

addition and subtraction with scientific notation

scientific notation metric system

'scientific notation lessons' might be a good place to look.

QUESTIONS AND RESPONSES

1)In the text question five asks for the percent uncertainty of a measurement given 1.57 m^2

I think that we figure this by an uncertainty of .01/1.57m^2 = .6369 or approximately one. ??????Am I correct in how I

calculate this??????? Can I asuume that if the number given was 1.579 then we would calculate it by .001/1.57 = .1 % approximately or am I incorrect?????

You're on the right track.

There are two ways to look at this.

1.57 m^2 represents a quantity which rounds off to 1.57, so presumably lies between 1.565 and 1.575.

This means that the quantity is within .005 of 1.57.

.005 / 1.57 = .003, approx., so the uncertainty is .003 of 1.57, which is the same as 0.3%, of 1.57.

Another way to look at it:

1.57 could be interpreted to mean a number between 1.56 and 1.58. The uncertainty would then be .01, which is .01 / 1.57 = .006, or 6%, of 1.57.

2)In the text question number 11 the book asks what is the percent uncertainty in the volume of a sphere whose radius is

r=2.86 plus or minus .09.

I know that the Volume of a sphere is 4/3 pi r^3, so I calculated the volume to be 4/3 pi (2.86)^3 = 97.99 and to get the

percent uncertainty I tried to divide 0.09/97.99 * 100 =.091846, but the book answer is 9% ??????I am not sure what i am doing wrong here?????????????????

Again there are two ways to approach this.

I believe the book tells you that the uncertainty in the square of a number is double the uncertainty in the number, and the uncertainty in the cube of the number is trip the uncertainty in the number.

An uncertainty of .09 in a measurement of 2.86 is .09 / 2.86 = .03, approx., or about 3%. As you state, you cube the radius to find the volume. When 2.86 is cubed, the resulting number has three times the uncertainty, or about 9%.

Another approach:

Calculate the volume for r = 2.86.

Then calculate the volume for r = 2.86 - .09 = 2.77.

You will find that the resulting volumes differ by about 9%.

You could just as well have calculated the volume for r = 2.86 + .09 = 2.95. Again you would find that the volume differs from the r = 2.86 volume by about 9%.

QUESTION RELATED TO UNIVERSITY PHYSICS (relevant only to University Physics students)

I understand everything but the part on measuring the individual i j k vectors by using cosine.

INSTRUCTOR RESPONSE

It's not completely clear what you are asking, but I suspect it has to do with direction cosines.

The vector A = a_1 i + a_2 j + a_3 k makes angles with the directions of the x axis, the y axis and the z axis.

Let's consider first the x axis.

The direction of the x axis is the same as the direction of the unit vector i.

The projection of A on the x direction is just a_1. This is obvious, but it can also be found by projecting the A vector on the i vector.

This projection is just | A | cos(alpha), where alpha is the angle between A and the x direction.

Now A dot i = A = (a_1 i + a_2 j + a_3 k) dot i = A = a_1 i dot i + a_2 j dot i + a_3 k dot i = a_1 * 1 + a_2 * 0 + a_3 * 0 = a_1.

It's also the case that A dot i = | A | | i | cos(alpha). Since | i | = 1, it follows that A dot i = | A | cos(alpha), so that

• cos(alpha) = A dot i / | A | = a_1 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ).

Making the convention that alpha is the angle made by the vector with the x direction, we say that cos(alpha) is the direction cosine of the vector with the x axis.

If beta and gamma are, respectively, the angles with the y and z axes, reasoning similar to the above tells us that

• cos(beta) = a_2 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ) and

• cos(gamma) = a_3 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ).

cos(alpha), cos(beta) and cos(gamma) are called the 'direction cosines of the vector A' with respect to the three coordinate axes.

Recall that alpha, beta and gamma are the angles made the the vector with the three respective coordinate axes.

If we know the direction cosines and the magnitude of the vector, we can among other things find its projection on any of the coordinate axes.

Please feel free to include additional comments or questions:

&#Good responses. See my notes and let me know if you have questions. &#

course PHY 121

6/15

002. `ph1 query 2

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Question: Explain how velocity is defined in terms of rates of change.

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Your solution:

Velocity is defined in terms of rates of change of distance with respect to clock time

So the average rate of change in velocity is equal to the rate of change for quantity A divided by the rate of change for quantity B

Where vAve=change of A/change of B

confidence rating #$&*3

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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).

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Self-critique (if necessary): 3

Rate of change of position not rate of change of distance

And I need to memorize this word for word

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Self-critique rating #$&*3

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Question: Why can it not be said that average velocity = position / clock time?

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Your solution:

This is not an accurate statement because vAve= change in position/change in clock 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).

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Self-critique (if necessary): ok

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Self-critique rating #$&*ok

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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.

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Your solution:

First to fit a straight line to a graph a lot has to do with your personal opinion when you are trying to decide the best place for a line that is in reasonable position to all data points, then if you are estimating where the line will cross the y intercept then there is going to be differences in people’s choice of the y intercept and therefore differences when finding the slope

confidence rating #$&* 3

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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.

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Your solution:

confidence rating #$&*

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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.

#$&*

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 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.

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Self-critique (if necessary): 3

I have read this very carefully, I too was having difficulty understanding uncertainty but I think this has cleared up the problem

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Self-critique rating #$&* 3

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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?

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Your solution:

I determined my height by using the height I had the last time I went to the dr. and my weight by kitchen scales

Height in inches= 65 inches

First inches to centimeters 1 inch=2.54 cm

So 65in/1 * 2.54cm/1in= 165.1cm

Then cm to meters

165.1cm/1 * 1m/100cm = 1.651 meters

Mass in pounds= 130lbs

1 pound = 453.6 grams

So 130lbs/1 * 453.6 grams/1 lb = 58968grams

Grams to kilograms

58968grams/1 * 1kg/1000grams = 58.97 kg

Uncertainty estimates of the height is 1.651 meters +-.001 so then %uncertainty is .001/1.651 *100=.061%

Uncertainty estimates of the weight is 58.97 kg +-.01 then the % uncertainty is .01/58.97 *100= .17%

These are the uncertainties because they are the either +-.001 or +-.01 away from the actual number and uncertainty is generally assumed to be one or a few units in the last digit specified

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%.

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Self-critique (if necessary): ok

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Self-critique rating #$&* ok

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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?

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Your solution:

To calculate the balls vAve on each book

You would use the distance of each book and the time it takes to roll from each book individually

Ex. vAve while crossing book 1 would be vAve=distance book 1/time to roll across book 1

And we can solve for the final velocity to fill into the acceleration formula knowing that

vAve=(vf + v0)/2

To calculate the how quickly the ball’s speed was changing or the acceleration you would use the formula aAve=(vf-v0)/time

confidence rating #$&* 3

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You will typically assess the accuracy and precision of your measuring instruments.

TYpically there are lots of questions about that.

That depends on how confident you think you can be in your estimated percent. Rounding to 1% is 'safe', for a situation in which you want to be careful not to overestimate the accuracy of your answer. In some cases, though, you could regard .636% as 'about half a percent'.

&#Good responses. See my notes and let me know if you have questions. &#

&#Good work. Let me know if you have questions. &#