Physics query 2

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course PHY 241

9/23 10 pm

002. `ph1 query 2#$&* delim

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

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Your solution:

Velocity is the rate of change of position with respect to clock time.

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): I was not as specific with my answer, but I understand the definition.

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Self-critique Rating:

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

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Your solution:

Velocity is the rate of change of position with respect to clock time. The word missing from the above stated definition is change. Position / clock time might describe an object with constant velocity, but if the velocity is changing, it will give the wrong answer.

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):

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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:

Fitting a straight line to a scatter graph is choosing the line that best represents the trend in the data. In other words, the line of best fit is the line that best follows the path of the data points. Unless this is done with a computer, there is significant room for error. Two people may find two different lines that fit the same data. Because the lines were estimated, they are subject to human error, so most likely neither is truly the line of best fit.

confidence rating #$&*: 3

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Question:

(Principles of Physics and General College Physics students) What is the range of speeds of a car whose speedometer has an uncertainty of 5%, if it reads 90 km / hour? What is the range of speeds in miles / hour?

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Your solution:

If the speedometer has an uncertainty of 5%, then that means the actual velocity can vary by as much as 5% in either direction. 5% of 90 is 4.5, so the actual velocity can range between 85.5 km/hour and 94.5 km/hour. To find what the range of speeds is in miles/hour, I need to convert my units. There are 1000 meters in a kilometer, and there are 100 cm in a meter, so there are 100,000 cm in a kilometer. This means the speedometer reads the equivalent of 9,000,000 cm/hour. There are 2.54 cm in an inch, 12 inches in a foot, and 5280 feet in a mile, so a similar speedometer marked in miles/hour reads about 55.92 miles/hour. 5% of 55.92 is 2.796, so the range of speeds in miles/hour is between 53.124 miles/hour and 58.716 miles/hour.

confidence rating #$&*: 3

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Given Solution: 5% of 90 km / hour is .05 * 90 km / hour = 4.5 km / hour. So the actual speed of the car might be as low as 90 km / hour - 4.5 km / hour = 85.5 km / hour, or as great as 90 km / hour + 4.5 km / hour = 94.5 km / hour.

To convert 90 km / hour to miles / hour we use the fact, which you should always know, that 1 inch = 2.54 centimeters. This is easy to remember, and it is sufficient to convert between SI units and British non-metric units.

Using this fact, we know that 90 km = 90 000 meters, and since 1 meter = 100 centimeter this can be written as 90 000 * (100 cm) = 9 000 000 cm, or 9 * 10^6 cm.

Now since 1 inch = 2.54 cm, it follows that 1 cm = (1 / 2.54) inches so that 9 000 000 cm = 9 000 000 * (1/2.54) inches, or roughly 3 600 000 inches (it is left to you to provide the accurate result; as you will see results in given solutions are understood to often be very approximate, intended as guidelines rather than accurate solutions). In scientific notation, the calculation would be 9 * 10^6 * (1/2.54) inches = 3.6 * 10^6 inches.

Since there are 12 inches in a foot, an inch is 1/12 foot so our result is now 3 600 000 *(1/12 foot) = 300 000 feet (3.6 * 10^6 * (1/12 foot) = 3 * 10^5 feet).

Since there are 5280 feet in a mile, a foot is 1/5280 mile so our result is 300 000 * (1/5280 mile) = 58 miles, again very approximately.

So 90 km is very roughly 58 miles (remember this is a rough approximation; you should have found the accurate result).

Now 90 km / hour means 90 km in an hour, and since 90 km is roughly 58 miles our 90 km/hour is about 58 miles / hour.

A more formal way of doing the calculation uses 'conversion factors' rather than common sense. Common sense can be misleading, and a formal calculation can provide a good check to a commonsense solution:

We need to go from km to miles. We use the facts that 1 km = 1000 meters, 1 meter = 100 cm, 1 cm = 1 / 2.54 inches, 1 inch = 1/12 foot and 1 foot = 1 / 5280 mile to get the conversion factors (1000 m / km), (100 cm / m), (1/2.54 in / cm), (1/12 foot / in) and (1/5280 mile / ft) and string together our calculation:

90 km / hr * (1000 m / km) * (100 cm / m) * (1/2.54 in / cm) * (1/12 foot / in) * (1/5280 mile / ft) = 58 mi / hr (again not totally accurate).

Note how the km divides out in the first multiplication, the m in the second, the cm in the third, the inches in the fourth, the feet in the fifth, leaving us with miles / hour.

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Self-critique (if necessary):

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Question: (Principles of Physics students are invited but not required to submit a solution) Give your solution to the following: 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:

The area of a circle with the given radius is pi * (2.8 * 10^4 cm)^2, or pi * (7.84 *10^8) cm^2. The approximate uncertainty of the radius of the circle is 50 cm. This means the radius could be anywhere between 2.75 * 10^4 cm and 2.85 * 10^4 cm. Thus, the area of the circle could be anywhere between pi * (7.5625 *10^8) cm^2 and pi * (8.1225 *10^8) cm^2. The area could be off by as much as pi * (2.825 * 10^7) cm^2. The percentage uncertainty of the area of the circle is about 3.6%.

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 break down a solution phrase by phrase and self-critique in the prescribed manner.

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** 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.

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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.

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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.

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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.

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The difference between the lower and upper estimate is .176 * 10^9 cm^2 = 1.76 * 10^8 cm^2.

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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.

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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.

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If you wish you can submit the above series of questions in the usual manner.

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.

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Self-critique (if necessary): The way I worked the problem is slightly different, so my answer is slightly different, but I understand what to do. One possible problem in my calculations is that I kept pi separate as much as possible, which was not difficult to do but it might have led to a mistake along the way. Another possibility is how instead of dividing the difference between the two extreme areas by 2, I found which one was more extreme (further away from the area found with the given radius) and used the difference between it and the expected area (which, in theory, should have made my percent greater than yours instead of less). Despite any mistakes I might have made, after reading the given solution I now understand how to find the uncertainty.

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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 am about 6 feet tall. There are 12 inches in a foot, 2.54 cm in an inch, and 100 cm in a meter, so I am about 1.83 meters tall. I weigh about 150 lb. There are about 2.2 pounds in a kilogram (though technically kilogram is a unit of mass and pound is a unit of force so they are not really convertible, but in this problem I don’t think it will be an issue). Therefore, I weigh about 68.18 kg. I’d guess the uncertainty for my height is about 2 inches, or 5.08 cm. This is because the yardstick I used to measure myself, which is not the most accurate, said I was about 5’ 10”. However, I remember that I am about an inch shorter than a cousin of mine, and he is close to 6’ 3”. This would make the percentage uncertainty for my height about 2.78%. For my weight, I’d estimate the uncertainty to be about 5 pounds, or 2.27 kg. I haven’t weighed myself in a while and the scale I used to verify my weight for this question is at least 10 years old, so it might be slightly off. This would make the percentage uncertainty for my weight about 3.33%.

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%.

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Self-critique (if necessary):

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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:

Since we know how far the ball rolled along each book and how long it took for the ball to roll that far, the ball’s average velocity could be found by dividing the change in position by the change in time. We also know how fast the ball was moving at the end of each book. When the ball fell from one book to the other, it would not have gained any velocity in the horizontal direction, so the initial velocity for each book would be equal to the final velocity of the previous book. We then take the change in velocity divided by the change in time, and we have the average acceleration.

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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:

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Physics query 2

&#Very good work. Let me know if you have questions. &#