Gas station without pumps

2012 October 2

Chapter 14 done

Filed under: home school — gasstationwithoutpumps @ 16:21
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We’re a week behind already, because we did not get any of the Chapter 14 homework done during the trip to Boulder, but we did finally get Chapter 14 done today. The twelve problems I assigned for Chapter 14 (plus one program) were too many, because they were almost all just plug-in-numbers-and-turn-the-crank exercises (despite having been given “P” for problem rather than “X” for exercise). I was rather tired last night and this morning when I did the problems, and I made a huge number of copying errors: copying the charge of an electron wrong (and using it for all the exercises that needed charge of proton or electron), copying the wrong line from a problem, copying numbers wrong from my calculator to the paper, … .  I don’t think I’ve ever had so many wrong answers. On only one problem when my son and I compared answers did we get different results that stemmed from his error rather than mine (and that was also a clerical error, not a conceptual one).

There was one problem that wasn’t just an exercise, though my son treated it as one.  That problem is 14P40 part c:

The electric field at a location C points north, and the magnitude is 1×106 N/C.  Give numerical answers to the following questions:

(c) where should you place a proton and an electron, at equal distances from C, to produce this field.

Locus of positions for electron with C at origin. The proton would be symmetrically located at (x,-y) for an electron located at (x,y). Locus plotted by Wolfram Alpha command “plot y/(x^2+y^2)^(3/2)=347.2E12″

The trivial answer places an electron to the north and a proton to the south, and students just compute the distance. But there are an infinite number of solutions and the locus of the solutions is an interesting one.  I don’t know whether the authors were thinking of this infinite set of solutions, or if they had only considered the trivial solution.

Since C is equidistant from the proton and electron, it must be on the perpendicular bisector of the line segment between them. Since the field points north, and we’re on the perpendicular bisector, the electron must be due north of the proton. If we do a 2D plot, putting C at the origin and the electron at (x,y), we get a formula of the form $y/(x^2+y^2)^{3/2}=347.2E12$, which we can ask Wolfram Alpha to plot for us. Note: I’m deliberately not providing the derivation for the number, so that students have to do some work before copying this answer!

The trivial solution is the one on the y-axis (not the origin). I had not expected to see the lower part of the curve, where the locus approaches the origin, but it makes sense. If we look at the formula in polar coordinates, we get $r = \sqrt{\sin(\theta)/347.2E12}$, where r is the distance from C to the electron (and to the proton), and θ is the angle from the horizontal axis (angle north of east). This parameterization also makes it easier to find where the x value is maximized by taking the derivative of $\cos(\theta)\sqrt{\sin(\theta)/347.2E12}$ with respect to θ, and setting it to 0. I get the maximal value for x as about 3.33E-08 m, at about 35.26°.

For very small angles, the electron and proton need to be very close together, though the solutions with them too close together are bogus, because classical electrostatics breaks down once quantum effects become significant.

The polar plot made by giving Wolfram Alpha “polar plot r=sqrt(sin(theta) / 347.2e12)” is cleaner and faster than solving for the x,y values directly.

For next week, we’ll have to read Chapter 15. I should have some problems selected before this weekend.

2012 September 17

Chapter 14 homework

Filed under: home school — gasstationwithoutpumps @ 16:05
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We’re already almost a week behind in physics, having intended to finish Chapter 13 last week, and instead finishing up the homework for tomorrow. We’ve both read Chapter 14, which is a short chapter on electric fields, so we should be able to get back on schedule, finish Chapter 14 by next week (except that my son and I will be visiting my Dad in Boulder Saturday through Tuesday, which will cut into our physics time, so it is highly probable that the schedule will slip again).

Chapter 14 homework—due 2012 Sept 25

14P23, 14P24, 14P40, 14P43, 14P47, 14P51, 14P52, 14P53, 14P70, 14P73, 14P75, 14P76
Computational problem: 14P78

2012 August 31

Physics C: E&M curriculum for year and Chapter 13 homework

Filed under: home school — gasstationwithoutpumps @ 14:28
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I’ll be homeschooling my son in calculus-based physics again this year, using the Matter and Interactions book.  The school  year has started for him, so now I have to put together a schedule for finishing the book this year. I want him prepared  for the AP Physics C: Electricity and Magnetism test in May. We never got around to Chapter 13 last year on thermodynamics, so I’ll tack that onto the beginning of this year’s schedule.

It looks like I have 34 Tuesdays to cover 568 pages, which is about 17 pages a week.  Assuming that each chapter’s length is roughly proportional to how much time it takes to cover it, I’m planning on

Chapter pages weeks Finish by
13 41 2 Sept 11
14 31 2 Sept 25
15 44 3 Oct 16
16 36 2 Oct 30
17 46 3 Nov 20
18 41 2 Dec 4
19 42 3 Jan 1
20 42 2 Jan 15
21 63 4 Feb 12
22 50 3 Feb 26
23 38 2 Mar 12
24 55 4 Apr 9
25 40 2 Apr 23

Chapter 13 homework—due 2012 Sept 11

13P19, 13P20, 13P22, 13P26, 13P27, 13P28, 13P32

We should probably go over the first three problems on 2012 Sept 4 and do a gas pressure lab of some sort (probably pressure and temperature measurements).  I bought some physics toys last spring that may be suitable for the lab.

2012 June 8

Reporting bugs in books and web sites

Filed under: home school — gasstationwithoutpumps @ 08:07
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Yesterday, while working problems in Chapter 12 of Matter and Interactions (about entropy and temperature),my son and I each found a bug.  The one I found was in the book, the one he found was in Wolfram Alpha.  I reported both of them to the respective sources.

Here is the exchange with Bruce Sherwood, one of the authors of Matter and Interactions.  Last night I wrote

I was doing the entropy calculations for 12.P.40 and decided to see if the number of states values were consistent.  I found that they were consistent with having 170 oscillators, which made the choice of 100 atoms seem strange to me.  What arrangement of atoms gives only 170 degrees of freedom for 100 atoms?

In just over an hour, Prof. Sherwood responded

Thanks! It’s clearly an error. And I note that with 170/3 = about 57 atoms, the per-atom heat capacity is about 2.8e-23 J/K, still less than the classical 3k_B limit, as is expected for the low temperature (about 113 K).

The simplest way to “fix” this would seem to be to say in part (c) that there are 57 atoms in this nanoparticle. Would you agree? You might be the only person in the world who has gone to the trouble to check whether the “100 atoms” made sense, though I would like to think we did a reality check on this in making up the problem but made a mistake.

While I’m pleased by how quickly he responds to my comments and that I apparently worked the problem correctly (all  this quantum mechanics is new to me), I’m a little disturbed by the assumption that I’m the only one checking my physics homework problems for consistency.  (Though since no one else has previously reported the problem to Prof. Sherwood, his assumption may well be correct.)

The problem my son found was in doing 12.P.48, which gives a mechanics problem that involves forces, rotating objects, and a brake.  The problem ends with a computation of how much the iron masses heat up as a result of the energy dissipated by the brake.  We  both used the $3 k_B$ high-temperature approximation of the atomic specific heat of Fe, getting a molar specific heat of about 25 J/(K mol), which should be fairly universal for simple solids at high temperature.  He looked up the specific heat of iron on Wolfram Alpha (his go-to source for numeric and mathematical information) and found that they had the molar specific heat as 251 J/(K mol).  He came to me for help, because he couldn’t find an error in his computation.  I suggested looking at some other references on the web.  They all had values around 25 J/(K mol), so I reported the apparent data entry error to Wolfram Alpha, suggesting that they should have done consistency checks on their molar specific heats.  I’m amazed that we seem to be the first ones to have caught this error—has no one ever used this entry in Wolfram Alpha’s database? Does no one do consistency checks on the data they find on the web? (I know that consistency checks on data entry are often done, but things slip through even well-designed data screens.)

So far all we have from Wolfram Alpha is the robomessage:

We appreciate your feedback regarding Wolfram|Alpha. The issue you reported has been passed along to our development team for review. Thank you for helping us improve Wolfram|Alpha.

But we can hardly expect commercial companies to work the sorts of hours that professors do, nor to admit to errors without piles of internal bureaucracy.

Of course, Wolfram Alpha is not the first web source whose information I’ve attempted to correct.  I have frequently sent Google Maps corrections on the bicycle routing information in Santa Cruz.  They usually respond in between 2 weeks and 2 months, and usually agree that I’m correct and claim to have fixed the problem.  About half the time they have fixed the problem, but one remains persistent—they keep sending bicyclists up the downhill-only leg of the bike path on the UCSC campus, though they have twice claimed to have fixed the problem.  The bike path is on a steep hill and is divided like a freeway, with the uphill path going around the hill and the downhill path going over the top and straight down.  Because the path is fairly narrow and downhill speeds exceed  35 mph, it is extremely dangerous to be routing bikes or pedestrians up the downhill path. The path is dangerous enough for downhill riders (I lost my spleen in a solo crash coming down that path in March 2000) without oncoming traffic.

I suspect that Google Maps has a database representation problem, lacking the ability to code a route as both one-way and bikes-only, so the fix may require more development work than the QA staff is authorized to do—that’s why I always check to see if it has really been fixed as they claim.  Their repeated reporting of the problem as fixed when it clearly is not indicates a serious management problem among Google Maps QA staff, who probably are evaluated on how many problems they clear, with no check for whether the problems are really resolved properly.

2012 June 5

Physics homework chapter 12

Filed under: home school — gasstationwithoutpumps @ 09:33
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Chapter 12 of Matter and Interactions is entropy and temperature.

I’ve not come up with any labs I really want us to do for Chapter 12, and with class cancelled next week (June 15)  I’m not sure when we’ll get to do an entropy lab.  The only lab I’ve thought of is to build an electric calorimeter and measure the specific heat of water, but that sounds like more effort than it is worth. The microwave and styrofoam cup exercise in 12.P.45 doesn’t seem much better than just looking up the specific heat of water. Does anyone have a more interesting lab we could do?

I would like us to continue on the rocket experiment, now that the simulation seems to be working.  Launching next to a tall building and making a video recording may be our best bet for measuring height with some semblance of accuracy, and using a superpulley and thread may be our best bet for measuring initial launch speed.

Anyway, Chapter 12 homework, which we’ll try to have at least half done by Fri June 8, our next meeting: 12.P.40, 12.P.43, 12.P.48, 12.P.62, 12.P.63, 12.P.66, 12.P.68.

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