Final exam layout

The final exam will be broken up into 5 sections. In each section, you will have a few required multiple choice questions and a choice of problems (4 problems given, solve any 2). The multiple choice questions will require some calculations, but will be much shorter than the problems. Here are the details of each section, with the corresponding sections of the book noted.

Overall, you need to do 10 problems and 12 multiple choice questions. An extensive formula sheet will be given, and you will be allowed to bring in two sheets of paper (front and back) with your own notes.

Section 1: Kinematics

• 2.2-10 (1D motion)
• 4.2-7 (2D motion)
• 5.2-9, 6.2-5 (forces)

2 required multiple choice questions, choose 2 of 4 problems

Section 2: momentum and energy

• 7.2-8 (K, work)
• 8.2-8 (PE, consv. E)
• 9.2-10 (p)

4 required multiple choice questions, choose 2 of 4 problems

Section 3: rotation and gravitation

• 10.2-10 (rotation)
• 11.2-4, 6-11 (torque, angular momentum)
• 13.2-8 (gravitation)

1 required multiple choice question, choose 2 of 4 problems

Section 4: oscillations and waves

• 15.2-7 (oscillations)
• 16.2-10, 17.2-5 (waves)

2 required multiple choice questions, choose 2 of 3 problems

Section 5: fluids & thermodynamics

• 14.2-10 (fluids)
• 18.6-11 (temperature, heat)
• 19.3-5, 11 (kinetic theory of gasses)

3 required multiple choice questions, choose 2 of 4 problems

Homework typo

I just realized that there is a typo in the due dates. Problems 3 & 4 are due today, but problems 5-10 are not due until Wednesday, 19 Feb (not today). 

Snowpocalypse 2014

I'm going to make a wild leap and assume classes will be held tomorrow.

On the outside chance that they are not, the HW due date and exam will shift by one class. Specifically, if there are no classes tomorrow, the HW will be due on Friday and the exam will move to Monday. Again, so long as classes are still on tomorrow, the HW is due Wednesday and the exam will be Friday.

Right. Better get back to shoveling my car out of the snow drifts so I can stock up on bread and milk.

Exam duration / syllabus confusion

The syllabus is not totally clear as to how long the exams are. The in-class exams will all be 50 minutes long, no matter what day they occur on. 

There is some confusion because my original (internal) schedule had all the exams on M or W, presupposing rather long exams. After rethinking it a bit, the exams didn't end up coming at natural logical breaks in the course material, and of course long exams just suck. At that point, I moved the exam dates around, putting most of the exams on Friday instead (so long as there was a natural conceptual break in the material), but forgot to update the syllabus. (One exam remains on a Monday because the preceding and following Fridays would make for awkward exam coverage.)

What the syllabus should say is that all four "hour" exams will designed to take 50 minutes, independent of what day they are on. For the one exam scheduled on Monday, you will still have a 50 minute exam, not a 110 minute exam. We will probably use the remainder of the 110 minute class period to review or do a short lab.

Webassign / Textbook

Some of you have been wondering, and I wasn't crystal clear about it - you don't need a WebAssign code with your textbook. Your homework is purely pencil & paper type stuff.

If you already got a WebAssign code with your book, the chances are extremely high that someone in one of the PH105 sections could use it, or you could just try to exchange it.  

Homework due at what time?

So I have said that the homework is due at the end of the day formally, i.e., 11:59pm on the day in question. That is still true. There are 'grey areas,' however.

Less formally, it will not be considered "late" if I am still awake after that time. Tonight that means you have about another hour or maybe 90 minutes. In general, 1-2am is a reasonable bet. If you're staying up until 2am doing homework, that is hardcore, and I'll give you some leeway ;-)

Minimal points are taken off for submissions arriving before I begin grading the homework, usually a day or two after the due date. Things happen, you get busy. Don't make it a habit.

Submissions after I have already posted the solutions online are taken on a case-by-case basis, but will receive some credit for at least bothering to copy the solutions. Learning by osmosis can work sometimes.

The message here is again: always turn it it. It is never given zero credit.

Problem 9 / HW10

There is a mistake in the beginning of the hint ... the first expression for velocity v(y) is inverted.

I will fix that and clean it up a bit in the next hour. The proper expression for velocity is

v(y) = (const)*sqrt[y/(R+y)]

where "const" involves G's, M's, and R's. Possibly even a 2.

Keep an eye out here for the revision soon.

Constraints vs. real forces

It is confusing sometimes to realize what is a real, live force, and what is not. For instance, circular motion. You draw a free-body diagram, and sum up the forces. This is one side of the equation. The other side is your constraint on the motion, namely, that for a circular path a net force of mv^2/r is required. Sometimes, we call this "centripetal force," and that is highly misleading. It is not a real force, but a fictitious force, which basically means that if you impose the requirement that the motion is circular, an specific constraint is placed on your force balance.

When you think about it, your force balance always has a constraint, namely, that the net force has to result in mass times the acceleration required to produce the observed path. For a straight line path, zero acceleration, the force balance is zero: equilibrium. For a circular path, the required acceleration is v^2/r, so if circular motion is observed then the net force divided by mass must give v^2/r - if not, then you don't have circular motion. For generic paths, it is more complex: an observed path constrains the force balance, but it depends on the speed along the path and the local radius of curvature.

The confusion is, in my opinion, largely an unfortunate artifact of history and terminology. There is no centripetal force, it is just a boundary condition on your force balance that enforces circular motion.

Anyway: here's what I wrote to one of you earlier. We'll touch on this again in class tomorrow.

In the middle of page 336, in Equation 13-12 (F_n - m*a_g = m(-(omega^2)R)), why do we have a negative sign on the right side. I feel like gravitational acceleration (weight) and centripetal acceleration, both being directed toward the center of the earth, should have the same sign. Here, if you put them on the same side, they have opposite signs.

It is a little bit confusing because the centripetal force is not a true force in its own right, but only the *result* of all other forces. The language in the text is confusing in this regard.

What it is saying is that there are two real forces acting: the normal force acting upward, and the gravitational force acting inward. Since the box follows circular motion by virtue of being on the earth's (rotating) surface, we know that the *constraint* on the force balance in magnitude is that it must sum to mv^2/r. The constraint on direction is that it must be directed radially inward to result in a circular path.

Saying that "centripetal force points toward ..." or "the centripetal acceleration points ..." is misleading. The left side of the force balance contains all the real forces; the right side contains the constraints on the motion from the known path:

(1) straight line: constraint is that forces sum to zero
(2) circle: constraint is that forces give mv^2/r to be consistent with the path, pointing radially inward.
(3) general: what we derived a while back; sum of forces relates to the rate of change of speed and the local radius of curvature of the path.

So it is OK that they have the same sign - it says that the net force balance comes out in favor of gravity, and the net force (which we call the centripetal force, misleadingly) points in the same direction ... we stay on the surface and don't fly off.

Tuesday's class / Gravitation

For better or worse, there will be much mathness tomorrow. Two things that can help either before or after the fact:

• Skim Ch. 13 of your text (gravitation)
• Remind yourself about motion on generic curved paths.

The idea is that we're going to "derive" Kepler's laws of planetary motion and Newton's law of universal gravitation from some basic principles and one astronomical observation (viz., the existence of orbits described by conic sections).

In the end, you will not be responsible for the derivations, only the main results. You will see the derivations again in PH301 or PH302 (and possibly MA227). The hope is to show you that the main points of Ch. 13 can be derived with a bit more work, which will ideally help you appreciate them a bit better. Also, all that scary math at the beginning of the semester will pay off again, which is nice.

Once that is out of the way, we'll work on some of the homework problems. Thursday, we will make use of our shiny new results, and be able to show that the remainder of Ch. 13 is a bunch of special cases.

Anyway: tomorrow will be a lot of 'I derive stuff not in the book and you watch' than usual, but not without good reason.

HW Clarification

A question from one of you:

Is a symbolic solution necessary for the simple parts of 8.41
such as part a? Also, our b part should serve as a sketch shouldn't it?

Correct on both accounts - if the problem just asks for a number in one part, no symbolic solution is necessary. Since the problem asks for U(x), that is sufficient for a sketch.

Exam details, part one

You have an exam this comingTuesday. Here are some details:

• There will be 8 problems, you can do any 4 - your choice
• Heavy partial credit, no multiple choice
  
• Covers chapters 1-6
• You can bring in 1 sheet of 8.5x11 inch paper with whatever you want on it
• I'll provide a basic formula sheet with all end-of-chapter formulas and constants

Why are there so many bastard problems?

So, I don't think this will make me seem like *less* of an ogre, but I thought I'd put down some of my reasoning behind your homework sets.

The homework is really, really difficult. I won't pretend otherwise. In fact, it is comparable to what physics majors at places like MIT would see - meaning if you are getting through it, or at least understanding the solutions after the fact, you are learning physics as well as anyone anywhere.

(Your exam problems will not be as hard. Not even close. I'll post some sample problems tomorrow to give you an idea.)

Out of every (say) three problems I give you, I usually expect one of them to be pretty easy or very similar to an example problem in the book or online. A second one usually has a painful but straightforward brute-force solution, and a more mathematically sophisticated but shorter solution. Then, there is usually the bastard problem, which is just going to suck.

So why the bastard problem? One, I can't really see how much you have learned, and whether you can apply it in unfamiliar situations, if you ace all the problems. In that regard, it is OK, and expected, if you don't complete the whole thing. On the toughest problems, you will be well-rewarded for just trying something sensible, I tend to grade the 'bastard problems' pretty leniently. If you don't get those problems, you are still going to end up just fine - you will not need to solve all of them to get an "A." With the partial credit scheme based on the homework template, even just setting up the bastard problem, without solving it completely, can get you 75% partial credit.

Two, it is still very valuable if you try the problem and get stumped. There is heavy partial credit for trying, but that is not entirely the point either. If I just solve the really tough problems in class for you, without you having tried them, they will either be hard to follow or seem deceptively easy. So the theory goes, at least.

Three, for most of you it is going to 'click' at some point, particularly once we learn some new techniques in the coming weeks. Most of this week's homework can be solved in a few lines with the benefit of hindsight we'll have at that point. It is just hard in the beginning, period.

Four, you are free to (and encouraged to) collaborate. This is not a justification for making the homework difficult, but a reminder that you should help each other get through it when you can ...

Finally, part of the point is that our students often get to 300 level physics classes and get creamed. I'd rather it be reversed, at least for honors students - by the time you get to PH301, you should manhandle it.

So how do I calibrate the homework? I solved all of this week's problems on Friday night and just made a second pass to double check. If I can't do the whole homework set in about an hour, using only the same things we covered in class, it is too much.

As an aside: part of this is about time management as well. Is it OK to spend time x and get 90% credit, or should you spend time 2x to get 95%? Multiply that by 4 or 5 courses and life outside classes ... don't spend three extra hours on physics for 5% more if you can spend it on calculus for 25% more.

Hints on PS3, cont.

The hint file now contains a free-body diagram for number 6.

Remember, the centripetal force must be the result of forces in the radial direction for circular motion. It is not a separate force acting on a body, but a constraint that must be obeyed to satisfy circular motion.

Random info on PS3

I had an email exchange with one of you regarding tomorrow's problems, so I thought I'd reproduce it here in case it helped someone else.

In my notes i have the position function for the inclined plane problems as x(t)=.5(a_x)(t^2)-(h/sin(theta)). My question is why is the last term there? It is the same as the hypotenuse of the plane, so it could be the x initial if I rotate the x-axis to be parallel to the plane, but wouldn't it make more sense to make the top of the plane the origin giving an x initial of 0? Or is the origin placed at the end of the ramp in order to more easily compute the trajectory after the block leaves the ramp/table? Sorry, I'm just a little confused about it
all.

If you call the length of the ramp along the tilted part L, then you would say sin(theta) = h/L, or L=h/sin(theta). So, the h/sin(theta) part is just the hypotenuse. We put it in this form because for the problem we did in class, we were given the height above ground level, not the ramp length.

You are also right that if you choose the origin to be the starting point, it is just x(t) = 0.5(a_x)(t^2) for the position along the ramp. I chose the end point of the ramp to make the trajectory easier, as you suggest, but it really doesn't make that much of a difference. You can put the origin at the starting position, and for the trajectory it isn't that much harder either way.

What you do need to keep in mind is that once the thing hits the bottom of the ramp, you should define a new x-y coordinate system to do the trajectory - it will be easier for that step with x & y vertical and horizontal as usual, the rotated coordinate system will just make things harder. At that point, you could actually just redefine the origin of the new system as well to be at the end of the ramp.

So basically, you are right. It is really easier to treat it as two separate problems, and give each their own coordinate system and origin - first do the ramp, then start over with the velocity you found and do the projectile.

PS 2 #8,9

For question number 9 on the current homework, you do not need to provide a sketch or a numeric solution. Since you are only asked to prove a mathematical relationship, neither is really applicable.

Similarly, you do not need a numerical solution for number 8.

Scroll down a few posts to see a hint I posted about number 9. Also, check out the notes for a very large hint (toward the end of the document).

Submitting homework electronically

I do prefer PDF, but will accept essentially any format. I can read all common image, office suite (e.g., powerpoint/word/openoffice/etc), or markup (e.g., html/tex) formats. It is unlikely that you will send me a format I cannot read.*

* That is not a challenge. Problem sets submitted in Visicalc or Appleworks will try my patience, but I will read them.

HW 1 due 15 Jan CLARIFICATION

student asks:

In problem four of the homework due Jan 15, when you say "find the
magnitude of and angle of a + b," do you mean the angle of a + b
relative to b, like in bullet number two?

Ok, that is not quite clear enough ...

For the last two cases in that problem, I mean the angle with respect to the x axis.

For example, if

\vec{c}=\vec{a}+\vec{b}

then the angle is

\tan^{-1}{\left(c_y/c_x\right)}