Query 2

course Phy 201

6/7 10

002. `ph1 query 2 Question: Explain how velocity is defined in terms of rates of change.

Your solution:

Velocity is the change in position divided by the change in clock time. It is defined by using the change in two other quantities, one with respect to the other.

confidence rating #$&*: 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 #$&* OK

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

Your solution:

Velocity has to be defined in terms of changing quantities. It has to be **change in** position / **change in** clock time. You cannot only have position and time because the change in them is the important part to finding velocity.

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 #$&* OK

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:

To fit a straight line to a group of points to take a string and put the string down in a line that seems to fit the data. You have to take into account the vertical distance between the string and the points because they all have to be about the same distance away. It is hard to eyeball the line, especially if the points are very spread out. The uncertainty comes because, if the distances between the points and the string are small, people may get 2 completely different lines that look correct for the data, but are not exactly right. It is difficult to find the line of best fit by just looking at the graph. It is also hard to find the line if you have to determine the coordinates of the points, because people my get different coordinates just by looking at it, which then throws off the calculations and, hence, the equation of the line.

confidence rating #$&*: 2

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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:

2.8*10^4 cm uncertainty = .1*10^4 (taking the last decimal place, right?????), so +/- .05

Pi(2.8*10^4)^2 = 2.463*10^9 sq cm this is ~halfway between the other 2 areas

Area = pi r^2 pi (2.85*10^4)^2 = 2.553*10^9 sq cm

Pi (2.75*10^4)^2 = 2.376*10^9 sq cm

Difference between the estimates (the uncertainty)= 1.77*10^8 sq cm

Because the area was about half, the uncertain should be as well, so the uncertainty = .5(1.77*10^8) = 8.85*10^7 sq cm.

confidence rating #$&*: 1

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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.

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 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. **

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.

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.

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Self-critique (if necessary): So what you do if find the total uncertainty and use it to find the upper and lower bounds of the measurement.

Then you find the area with both bounds and find the difference, which gives you the uncertainty for the area. Then find the area with the given measurement and see where it falls. Since it was in the middle, the uncertainty is half of the distance as well.

You can also find that the uncertainty in the radius was 4%, then multiply that by 2 (because you are going from radius to area) and get that the uncertainty would be 8%, so you take 8% of the area and find the uncertainty.

???Does it matter if I put the uncertainty in scientific notation with 10^7 as opposed to 10^8???

Your solution was fine.

Self-critique rating #$&* 3

Question: What is your own height in meters and what is your own mass in kg (if you feel this question is too personal then estimate these quantities for someone you know)?

Explain how you determined these.

What are your uncertainty estimates for these quantities, and on what did you base these estimates?

Your solution:

Height: 1.73 meters Weight: 297 kilograms

Convert the height from 5.67 feet (12 inches / 1ft) (2.54cm / 1in) (1m / 100cm) = 5.67 *12*2.54 / 100 = 1.7282 meters

Convert weight from 135 lbs (1kg/ 2.2 lb) = 135/2.2 = 61.4 kilograms

Uncertainty for height +/- .5 in = .5 / 68 = +/- .735%

Uncertainty for weight +/- .5 lb = .5 / 135 = +/-.37%

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.

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%.

Self-critique (if necessary): To do this you find the uncertainty of the measurements. Then you divide the uncertainty by the measurement and multiply by 100 to get the percent uncertainty of the measurements and the conversions.

Self-critique rating #$&* 3

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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 required 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 find the average velocity you would divide the total distance travelled by the ball: the length of each book. Then you divide each respective distance by the time it took the ball to travel down each book. This would give you the average velocity.

If you wanted to calculate the rate at which the speed was changing on each book, you find the initial speed and the final speed of the ball on each book. You find the difference in the speeds and divide it by the time it took the ball to roll down the book. You have to be sure that speeds used from the first book are used for the rate of change of the speed on the first, not use it also on the second rate of change because that would mess up the results.

confidence rating #$&* 3

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

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