O Magnum Mysterium

This is beautiful, in part, because I don't understand the words.


When I first fell in love with astronomy, it was in this way. It was the stately beauty of Neptune, taken with precious little energy at the far reaches of the solar system. It was the eerie beauty of a sequence of Messier objects, representing a bizarre diversity, even by the standards of a child growing up in the modern world.

I fell in love before i knew what I was falling in love with.

With time, I began to tease apart those amazing structures. I learned the calculus that allowed me to express, in partial elegance, with partial clarity, the nature of stars. I learned the physics that gave me the beginnings of the understandings of the depth behind these things. I even learned a bit of computer programming, that I might better participate in peeling back the curtain of the unknown.

And somewhere along the way, I lost that wonder. I lost the ability to look, and just see what my eyes see, and not the questions, and work, and challenges behind those high resolution images.

Some can do so. For some, the odd combination of challenge and complexity, wedded with a firm belief that understanding is possible, gives a richer sense of beauty. These people enjoy a long, happy marriage to space -- not without its challenges. But they still feel enough to work at the relationship and find new, mature beauty as the relationship continues.

I was not. I found its very comprehensibility, and my limits at comprehension, too harsh. It was that, or it was other things -- opportunity cost, more terrestrial thoughts, hopes, and fears. So we ended our relationship, and unlike some, I never looked back and missed it. Those feelings were just gone.

And so I enjoy this song, and do not seek to look up its lyrics. For the unknown itself is what I find alluring, as long as it remains, unknown.

Maybe this is why, despite the best efforts of relatives and my therapist, I don't seek a greater connection to God. For when I do, I find my interest torn asunder by questions of doctrine, historical origins, temporal contamination of divine intent, Biblical literalism, and the like. I can only appreciate faith from a distance, and so there is where I remain, and where I am happiest.

Pernicious memories from grad school

Why doesn't it die?

I'm tired right now. I think I'm getting a cold. I tutored a student for about 5 hours today in microeconomics. He did a fantastic job; I'm so proud of him right now. I wouldn't be surprised if his semester GPA ends up at least a full point higher. I worked on some revisions for this pdf guide for AP Physics C - Mechanics. I tried, and failed, to complete some statistical analysis looking at whether students really did run out of time on the third free response question. (I'll reexamine Turkey's test tomorrow if I think it's worth it. Right now it looks like the score for the first problem is significantly higher than the other two, but there is no significant difference between the second and third questions.)

And all of a sudden, the image comes flooding back from grad school. It's a story I don't think I told anyone else -- I try really, really hard not to speak ill of specific people from my professional past. But it bothers me enough to write it.

I think my mother and (future) stepfather had come to Cornell to visit. By then, my mental health was deteriorating... it was probably early 2007. There was snow on the ground. I reluctantly took them to Space Sciences. At that point, I hadn't stopped showing up to work, but I think it was clear by then that things weren't working out.

Just in front of the building, I saw my advisor rushing outward. I called to him. He looked up quickly, and kept on walking.

Maybe he didn't recognize me. Maybe he was busy.  I've invented a hundred different reasonable explanations why he didn't acknowledge my presence, or the presence of my family.

But right then, and right now, it's hard to hang onto that.

I don't remember if I took them into the building after that. If I did, it was for a very quick tour. Another grad student from my year and in my research group expressed some surprise that she hadn't had a chance to meet my parents when they were in town.

Several months later, another grad student, also in my year and in my research group, had his father (and possibly other relatives) come into town for a visit. My advisor talked with them, and they sat in the conference room for a lunch talk.

And then, as now, I couldn't help feeling resentment. I tried to think of a hundred reasons why this happened this way with them and why it worked out differently with my family. My mental health was bad, so I couldn't play host. The dad had a PhD, and therefore knew better how academia worked. His family just cared more about what he did, and where he worked.

Anything to avoid thinking that it was directly related to the fact that he was just a better grad student than I was.

Because that's the weird thing about depression. You start seeing things that could be innocuous, but add up to a giant conspiracy against your happiness. I brought in research money into the group with the NSF; neither of the other students had that, at the time. But they got the computer workstations, while I spent time alone in an empty computer lab. They got offices on the same floor as my advisor, while I remained in an office three floors up. They got to go to AAS, and so did I-- but only admission was covered, not my airfare.

Was this their fault? Or his? Not really. I could've asked about getting a second floor office, but I didn't. (Besides, I treasured my officemates -- they helped me keep what sanity and human-ness I had left.) I could've asked for more things. I could've transferred groups if it really was a problem. But at that point I had embraced the giant bullseye I imagined on my back.

He tried, I think, to be a good advisor. Maybe it was cultural/class -- his father had been a prominent diplomat, and on his wall hung proudly an invitation to a dinner with the Prime Minister. I think he tried to reach out to me, and I pushed him away, telling him that he already had one baby to take care of; he didn't need another.

So why do I still feel this pain? Why does it go away for a long period, and resurface? Why, after all the work I've done to rebuild my mind, my ego, to give myself perspective and distance, to generate new priorities?

Why do I still feel such pain? Not hatred, nor desire for another chance. But just pain?

Sleep will probably make me feel better. But I guess some have expressed questions as far as my time there. This is another peek. Were I a stronger or better man, I would've buried this along with the other bodies in my psychological backyard.

But it returns, unsummoned. And I have no way to deal with this, right now, other than to write, and hope, by writing, I can wake up unburdened by it.

Applying general technique to solving actual problems

In the previous post, I presented a general technique for solving free-response questions, summarized here:

FRQ Guidelines Summary

I. Read the question carefully.
II. Draw a picture.
III. List important quantities and the variables you use to represent them.
IV. Diagnose the type of problem.
V. Solve the problem.
     A. Visualize and/or write down a chain of steps you need to take in order to get the answer.
     B. When writing, be neat and tidy.
     C. Take the time to show the key steps in your derivation.
     D. If you get lost and confused, stop.
     E. Use conversion factors to convert between units.
     F. Box your final answer and, if necessary, provide directions that indicate the flow of your work.
VI. Check your answer.
     A. Does it have the right units?
     B. Does it have the right sign?
     C. Does the magnitude make sense?
     D. If you take a limiting case, do you get an answer that makes sense?

(This technique applies, at some level, to multiple-choice questions as well, though because of the shorter nature of those problems, I'll discuss an abridged, more targeted technique for those in a future post.)

However, the framework is useless if you don't see it applied a few times and practice its application many, many times. With that in mind, I'll demonstrate how you use this for a specific problem. Commentary will be in red, and the actual answers will be in black.

Here's the problem. It's FRQ #3 from the 2012 AP Physics C Mechanics test (link to College Board, 2012 AP Physics C exam questions, pdf):

Looks long, and kind of is. But that's ok.


I. Read the question carefully.

If you read it carefully, you notice that you're dealing with three different types of motion:

Sliding only $\rightarrow$ translational motion, but no rotation $\rightarrow$ easy
Sliding and rotating $\rightarrow$ translational and rotational motion, without neat relationship between them $\rightarrow$ hard
Rotating with no sliding $\rightarrow$ translational and rotational motion, but with direct relationship between them, i.e. $a=r\alpha$, $v=r\omega$, etc. $\rightarrow$ medium

This means you will need to think in three different regimes. The first and third have neat and tidy formulas. The second will require a bit more work, probably involving integration. Looking ahead, it looks like some integration is involved, as well as some manipulation of equations and using constraints that may or may not be obvious.


II. Draw a picture.

Fortunately, a picture is provided. It's pretty much complete. Note that friction applies in sections II and III; otherwise, there would be no rotation. This is something that might be commonly missed in a picture created by a test taker. Also note that there is no coordinate system/sign convention stated on the figure; it's a good idea to add one. To figure out which one to use, you might need to read ahead a bit and figure out whether it would make more sense to make a given direction positive or negative.

I haven't placed a coordinate system on the image provided on the test, but I did place the coordinate system on the FBD below:

Note that I've applied a sign convention for both direction of motion and rotation.

From the FBD it's also evident that the only net force on the hoop is from friction.


III. List important quantities and the variables you use to represent them.

It's not strictly necessary to define everything; in particular, $M$, $R$, $\mu$, and $L$ don't need to be defined. Universal symbols, like $v$ and $\omega$, can also be skipped, though it doesn't hurt to state them. You also don't have to define a new variable for each region; however, you can only do this if you make it abundantly clear, through context, which region you are discussing.

Note that if you find you have to invent a new variable, you can define it later in your work; just make sure you do so before you use it. More critically, this is a good place to set initial conditions.

Given/Initial conditions:
$v(0) = v_0$ = initial translational velocity     note that it explicitly states that this is the velocity at time t=0.
$\omega(0) = 0$ = initial angular velocity     this is an implied initial condition, implied by the fact it is sliding without rotating

New variables to use:
$I=MR^2$ = rotational inertia of hoop
$N=Mg$ = normal force
$f$ = friction force


IV. Diagnose the type of problem; V. Solve the problem; VI. Check your answer.

I'm going to do IV, V, and VI by subsection. Note that not all the steps/checks will be used in a given problem.

(a)(i) Starting from Newton's second law in either translational or rotational form, as appropriate, derive a differential equation that can be used to solve for the magnitude of the following as the ring is sliding and rotating: The linear velocity $v$ of the ring as a function of time $t$.

Diagnosis: basically the problem statement. A linear mechanics problem involving Newton's Second Law, linear form. Only net force will be from friction.

Approach: $F_{net}=ma$ $\rightarrow$ calculate net force $\rightarrow$ solve for $a$ $\rightarrow$ substitute $a$ for $\frac{dv}{dt}$

$F_{net}=ma$

$a = \frac{F_{net}}{M}$     note that I'm substituting the specific value of the mass, or $m = M$

$F_{net} = f = -\mu N = -\mu Mg$     remember that friction opposes the direction of motion; hence, minus sign

$a = \frac{-\mu Mg}{M}$

$\enclose{box}{\frac{dv}{dt} = -\mu g}$.

Units of linear acceleration ($m/s^2$): check!
Negative sign - check!


(a)(ii) Starting from Newton's second law in either translational or rotational form, as appropriate, derive a differential equation that can be used to solve for the magnitude of the following as the ring is sliding and rotating: The angular velocity $\omega$ of the ring as a function of time $t$.

Diagnosis: basically the problem statement. A rotational mechanics problem involving Newton's Second Law, rotationalform. Only net torque will be from friction.

Approach: $\tau = I\alpha \rightarrow$ calculate net force $\rightarrow$ solve for $\alpha$ $\rightarrow$ substitute $\alpha$ for $\frac{d\omega}{dt}$

$\tau_{net} = I\alpha=I\frac{d\omega}{dt}$

$\frac{d\omega}{dt} = \frac{\tau_{net}}{I}$

$\tau_{net} = Rf = R\mu Mg$     Note that $R$ is perpendicular to $f$

$\frac{d\omega}{dt} = \frac{R\mu Mg}{MR^2}$

$\frac{d\omega}{dt}= \frac{R\mu Mg}{MR^2}$

$\enclose{box}{\frac{d\omega}{dt}= \frac{\mu g}{R}}$

Units of angular acceleration ($1/s^2$): check!
Negative sign - check!


(b)(i) Derive an expression for the magnitude of the following as the ring is sliding and rotating: The linear velocity $v$ of the ring as a function of time $t$.

Diagnosis: Problem statement says "derive", which in this case means integrating the answer from (a)(i).

Approach: Start from differential equation in (a)(i) $\rightarrow$ Separate variable $\rightarrow$ integrate both sides $\rightarrow$ solve for constant of integration using initial condition.

From part (a)(i): $\frac{dv}{dt} = -\mu g$     $\rightarrow$ generally a good idea to start with the answer from the previous part before manipulating, and to explicitly state where you got this.

$\int dv = \int -\mu g dt$

$v = -\mu g t + C$

$v(0) = v_0$     $\leftarrow$ notice this was listed in part III to make sure we didn't forget it.

$v_0 = C$

$\enclose{box}{v = -\mu g t +v_0}$

Units of linear velocity ($m/s$): check!
Initially positive - check!


(b)(ii) Derive an expression for the magnitude of the following as the ring is sliding and rotating: The angular velocity $\omega$ of the ring as a function of time $t$.

Diagnosis: Problem statement says "derive", which in this case means integrating the answer from (a)(ii).

Approach: Start from differential equation in (a)(ii) $\rightarrow$ Separate variable $\rightarrow$ integrate both sides $\rightarrow$ solve for constant of integration using initial condition.

From part (a)(ii): $\frac{d\omega}{dt}= \frac{\mu g}{R}$     $\rightarrow$ generally a good idea to start with the answer from the previous part before manipulating, and to explicitly state where you got this.

$\int d\omega= \frac{\mu g}{R}dt$

$\omega= \frac{\mu g}{R}t+C$

$\omega(0) = 0$     $\leftarrow$ notice this was listed in part III to make sure we didn't forget it.

$\enclose{box}{\omega= \frac{\mu g}{R}t}$

Units of angular velocity ($1/s$): check!
Initially zero and thereafter positive - check!


(c) Derive an expression for the time it takes the ring to travel a distance $L$.

Diagnosis: Derive suggests some calculation, using either algebra or calculus. It's unclear which one would work. If I integrate the velocity expression from (b)(i), I will have something quadratic in $t$. That will likely be quite messy. Are there any other conditions that I can use at that point?

What is true at the point when it's traveled a distance $L$?

At that point, it starts rotating without sliding. That means $v=r\omega$. This is another implicit condition that I forgot to include in III.

Therefore, maybe I can try using that condition, and substituting expressions for $v$ and $\omega$ obtained in (b)(i) and (b)(ii). If this doesn't work, I can always try the other approach.

Approach: Use condition $v=R\omega$ for rotation without slipping $\rightarrow$ Substitute expressions for $v$ and $\omega$ from (b)(i) and (b)(ii) $\rightarrow$ solve for $t$.

For rolling without slipping after a distance $L$,

$v=R\omega$

From parts (b)(i) and (b)(ii),

$v = -\mu g t +v_0$

$\omega= \frac{\mu g}{R}t$

Substituting into the condition above,

$-\mu g t +v_0= R\frac{\mu g}{R}t$

$-\mu g t + v_0 = \mu g t$

$v_0 = 2\mu g t$

$\enclose{box}{\frac{v_0}{2\mu g} = t}$

Units of time (s): check!
Positive - check!


(d) Derive an expression for the magnitude of the velocity of the ring immediately after it has traveled the distance $L$.

Diagnosis: Given the solution from $t$, it looks like we just substitute results from (c) into the expression for velocity in (b)(i).

Approach: Take time $t$ from (c) $\rightarrow$ substitute into expression for $v$ from (b)(i).

From (c),

$t = \frac{v_0}{2\mu g}$

Substitute into velocity from (b)(i):

$v = -\mu g t +v_0$

$v = -\mu g \left(\frac{v_0}{2\mu g}\right)+v_0$

$v = -\frac{v_0}{2}+v_0$

$\enclose{box}{v = \frac{v_0}{2}}$

Units of velocity ($m/s$): check!
Positive - check!
Less than $v_0$ - check!


(e) Derive an expression for the distance $L$.

Diagnosis: It looks like neither $v$ nor $t$ appear to depend upon L directly. We might try integrating $v$ to get $x$, using a definite integral to get $L$ equal to some expression. We have the limits for $t$ from part (c).

Approach: Start with $v$ from (b)(i) $\rightarrow$ calculate definite integral of $v$ to get $x$ $\rightarrow$ evaluate definite integral using limits $x=0$ and $x=L$ and $t=0$ and $t = v_0/2$.

From (b)(i):

$v = -\mu g t +v_0$

$\frac{dx}{dt} = -\mu g t + v_0$

$\int_0^L dx = \int_0^{v_0/2\mu g} (-\mu g t + v_0)dt$

$x \vert_0^L = -\frac{\mu g}{2}t^2 + v_0 t \vert_0^{v_0/2\mu g}$

$L = -\frac{\mu g}{2}\left(\frac{v_0}{2\mu g}\right)^2+ v_0\left(\frac{v_0}{2\mu g}\right)$

$L = -\frac{v_0^2}{8\mu g}+\frac{v_0^2}{2\mu g}$

$\enclose{box}{L = \frac{3v_0^2}{8\mu g}}$

Units of length ($m$): check!
Positive - check!
Less than $v_0 t$, ($t=v_0/2\mu g$ from (c) ) $\rightarrow$ less than $2v_0^2/\mu g$ - check! (Because $v$ is decreasing, the total distance traveled, L, should be less than the initial velocity times the time it takes to travel L.)


*****

Some important notes and perspective:

This looks like way more than a person could do in 15 minutes. But keep in mind that a test taker would only really need to write down what's in black. Give it a shot -- copy only the parts of the solution in black. It should fit neatly on a page.

Based on scoring alone, this was the hardest problem given on the 2012 exam. The mean score was 2.71 out of 15, with a standard deviation of 3.69. Basically, 90% of students got less than half-credit. And in case you think this was because they ran out of time, the drop on Problem #3 on this exam was definitely larger in both absolute and relative terms than the drops for the last 11 years of tests.

A correct answer for (a) and (b) would get you 7 points. In other words, even if you had skipped everything else, you could've done better than around 90% of your peers on this question. Given that the top 25% of students usually end up with 5s (2012, unusually, saw the top 33% get 5s), this is definitely "good enough", if you were equally outstanding in other parts of the test.

General guide to answering AP physics free response questions

1. Read the question carefully.

This should go without saying, but I have met a lot of rushed (or lazy) students who just won't read the entire question.

If you have trouble keeping track of the information, underline the important pieces of information and circle what you're supposed to find. This should help if you get confused about what you are supposed to be doing.

2. Draw a picture and FBD.

Drawing a picture should be automatic for any free response question. Why? A picture does several things:

(a) It helps you understand exactly what is happening.

Can you transform a text description into a clear mental image of what is happening in the system? Are you forgetting some aspect of complexity? Are you confident you will remember whether something is frictionless or not once you start working?

If you can answer all of the above "yes", then congratulations. Your memory and reading comprehension are spectacular. You might not need to draw a picture. But for the rest of us (not them, US), a picture can help us really understand what's happening, or, perhaps as importantly, that we don't quite understand it and need to read the statement again more carefully.

(b) It helps trigger memory for similar problems.

The brain stores images as well as words. Your brain might not process a paragraph of text as similar to other bits of text floating in your head. But once you draw a picture, you might better form the connection. A picture of an object/skiier/car going down a hill might help you remember that you have solved these problems before, and possibly even help you recall what methods you should use.

(c)  It provides a way of storing some important information.

Even if you underline important pieces of information in the problem statement, you might find it easier to consoldiate all the really important bits in a picture. Instead of wondering, "what angle is that?", you can indicate it directly on the picture. It's a good way of keeping track of forces and their directions. In fact, I'd encourage a division of labor -- use the diagram to keep track of directions and coordinates, and a short list (see #3 below) to keep track of numerical quantities.

FBD: In addition to a picture of the system, construct a free body diagram for each object of interest. In other words, draw a free body diagram for every object that has motion you are asked to describe. For problems involving stacked blocks, you might need to construct a FBD for each block.

A FBD isn't needed for every problem, but it's a good idea for many of them, even problems that don't directly ask for it. It's basically mandatory for all problems using Newton's Second Law.

Note: Include a coordinate system/sign convention on both your system picture and the FBD.
3. List important quantities and the variables you use to represent them.

This is important for a couple reasons.

(a) Not all quantities have a universally accepted symbol. You don't have to define "g". But what about drag force? Do you use "$f$", or "$f_{drag}$"?

(b) It helps to organize information. As stated above, you can put some information on a diagram. But it will get very cluttered if you put every pertinent number on that diagram. I recommend keeping the directional information on the diagram, and the numerical information/defintion of variables in a short list below the picture.

(c) This helps you keep track of pieces of information that aren't explicitly stated. Statements like "starting from rest" indicate that $v_0 = 0$. Otherwise, it might be possible to overlook the quantitative information in what appears to be a piece of the problem statement text devoid of numbers.

(d) For some problems, it can help you narrow down what type of problem it is. Does it have a radius and an angular velocity? Then it probably involves rotational mechanics. For simple kinematics problems, it can help you pick which of the four equations to use. If your problem gives an initial and final velocity, an acceleration, and wants you to find a displacement, then you probably are best served using $v^2 = v_0^2 + 2a\Delta x$.

4. Diagnose the type of problem.

Yes, you do all the prep work of #1-3 first. That's because you do not want to jump in and start writing a bunch of random equations. Unless you know precisely what you're doing, you will confuse yourself.

Now, how do you diagnose which problem? If you've read the problem carefully, drawn a picture, and listed the pertinent information, then you probably have enough to tell the difference between a kinematics problem, a spring problem, a rotational mechanics problem, a drag force problem, a gravitation problem, etc.

Now the FRQs are multi-step, and integrate many different concepts into each problem. But a given step (and especially the first step) might use only one or two concepts.

This is the step where things can go wrong. But it won't if you have enough practice, training and experience. There really is no substitute for a solid conceptual background and lots of practice solving problems.

If it's a problem dealing with a spinning rod, and you don't know anything about rotational mechanics, then you won't be able to solve this problem. If you know a little about rotational mechanics, but don't know anything about a section critical for this problem (parallel axis theorem, or rotational kinematic equations, or torsional coefficients), then you might get a little farther, but not far enough. If you know conceptually all the topics, but don't have a lot of experience actually doing the free response questions, you might not get full credit if you derivations are a huge mess and you don't know the difference between "Derive, Explain, Find, Solve, and Justify" when used in one of these questions. (These keywords in the problem statement give you a sense of what kind of work you need to show in order to get full credit.)

5. Solve the problem.

If #1-4 were done properly, then this will actually be pretty easy.  A student with a solid background and plenty of practice can do steps #1-4 in about 2 minutes. This means that they have 12 minutes to solve the problem. That should be more than enough time.

More specifically, this means different things for different problems.

For graph interpretations, it might mean determining the functional form of the graph (sine, cosine, -sine, -cosine).

For many problems, it will start by choosing an appropriate equation. Once you've diagnosed the problem, and have enumerated the important quantities, then it should be a matter of choosing between a handful of equations, some of which are given to you on the formula sheet!

For some problems, it asks for a qualitative judgment. Here, again, pictures help, as well as some rough calculations.

Some problems may ask for an answer based on a limiting condition. For example, at terminal velocity, the object's acceleration is zero, which means, by Newton's Second Law, that the sum of the forces on the object is zero.

General guidelines for solving the problem:

(a) Sketch a chain of steps you need to take in order to get the answer. For example, let's say you have a roller coaster problem involving a loop track, asking you to find the normal force at the top of the loop. Your chain might look like this: solve for total energy of cart before loop $\rightarrow$ calculate  change in potential energy of cart at top of loop $\rightarrow$ use conservation of energy to find velocity at top of track $\rightarrow$ find centripedal force$\rightarrow$ subtract gravity from centripedal force to get normal force.

Some of you might want to skip this, if you are confident you can "see" the chain of steps without writing it down in words. Others might prefer writing it, in at least abbreviated form. I'll leave writing this as optional, but thinking about this before you start working is absolutely mandatory.

(b) When writing, be neat and tidy. For those of us with bad handwriting, it stinks to confuse a 7 for a 1 (or even a 5 for a f, 2, or 8 -- it happens). Take extra space if you need to -- you can always state clearly where your work is continued.

(c) Take the time to show the key steps in your derivation. Skip steps at your own risk. You don't need to explicitly show that you divided by a variable on both sides. You shouldn't show every step of algebra that you were required to show in a 7th grade class. That said, if you skip too many operations between two lines of work, you increase the risk of making an error.

(d) If you get lost and confused, stop. Go back to (a) and think about whether you are doing the right step, or even one that's necessary given the problem. Think it through conceptually. This will save you time and stress, and probably give you more points. (It's easy to get lost, then stuck, if you start manipulating equations without stopping to think about the process.

(e) Box your final answer and, if necessary, provide directions that indicate the flow of your work. This helps the grader find it. This is really important if your workspace is a bit chaotic, with bits floating in the margins and wrapping around. Although it's HIGHLY recommended your work be organized into at most two columns, left to right, arrows and directions are acceptable if absolutely necessary.

(f) Use conversion factors to convert between units. Although it's unlikely you'll need them (generally, there is SI unit consistency within the exam), use conversion factors to make absolute certain you are properly converting between units.

6. Check your answer.

There are quite a few checks you can make. Not all will apply in a given problem, but at least one should be valid.

(a) Does it have the right units?

If the units are wrong, you might have dropped them. Alternatively, maybe you forgot to multiply or divide by a critical factor. Units aren't just a way of bookeeping -- an error in units can help you trace precisely where you made a mistake. If I should end up with $m/s$ and end up with $m^2/s^2$, then I probably forgot to take a square root. If I should have $kg\ m/s$ and instead have $m/s$, I might've forgotten to multiply by mass in a momentum calculation.

(b) Does it have the right sign? 

The wrong sign could indicate a serious conceptual error. Maybe you didn't set up your coordinate system/sign convention properly. Or maybe you subtracted when you were supposed to add. Another common mistake is to not remember that the solution of $x^2 = k$ is $x = \pm \sqrt{k}$. You may actually need to choose either the positive or the negative root in a given situation -- and which one is correct will be set by the constraints of the problem.

For example, in a kinematics problem solved by using $x = x_0 + v_0 t +\frac{1/2}at^2$, the solution for $t$ will usually involve an expression that includes $\pm\sqrt{\rm something}$. In some cases, choosing the negative root leads to a negative time. If the constraints of the problem indicate that motion is considered for $t\geq 0$, then that means the solution with the $-\sqrt{\rm something}$ should be discarded.

(c) Does the magnitude make sense?

Do you get something falling with air resistance that has acceleration with magnitude $> g$? Then you probably have the wrong sign/direction on the drag force. Do you get a mass for a planet roughly equal to the mass of a person? Then you probably made a multiplicative error/forgot to square something.

(d) If you take a limiting case, do you get an answer that makes sense?

For example, if the tension in a pulley system goes to zero, does the acceleration of a hanging block go to $g$? (In other words, if you cut the cord, does the solution reflect that the block will experience freefall?) Does your complicated solution for the velocity of an object subject to air resistance reduce to the terminal velocity if $t \rightarrow \infty$?

*****

In the next post, I actually show you how to apply this to a problem. A good framework isn't worth anything if you don't see it applied and understand how it works.

Air resistance problems in Physics C

Note: this is also a test of MathJax, and my attempt to enable LaTeX in the Blogspot environment. (Easy directions here. My MathJax notes at the end of the post.)

At first glance, air resistance problems seem weird. They don't quite fit with the more familiar forces. It's sort of like friction in that it always acts in the opposite direction of motion. But it is velocity-dependent, which appears to add a layer of complexity that isn't there for other problems.

But it is doable, and within the range of students who have a reasonably solid calculus background.

(Note, however, that if a student is taking calculus for the first time concurrently with calculus-based physics, there might be a problem in which the calculus required to solve certain problems hasn't been taught yet. For best results, take a year of calculus before taking physics C. But, in practice, those taking it concurrently will have to teach themselves a bit of calculus.)

Here I look at a sample air resistance problem. I will use variables instead of specific numbers. I know a lot of students prefer numbers. Tough luck.

I will put commentary in red; what actually should appear in your answer appears in normal, black (or blue) text.

The Problem:

A block of mass $m$, initially at rest, starts sliding down a long, frictionless ramp that makes an angle $\theta$ with the ground. 

(a) Draw a free-body diagram, indicating all forces acting on the block.

(b) Calculate the terminal velocity of the block.

(c) Derive a differential equation describing the motion of the block.

(d) Solve for the velocity of the block as a function of time.

Get it? Got it? Good.

The Solution:

(a) Draw a free-body diagram, indicating all forces acting on the block.

Notes for part (a):

1. You should always draw a picture of the whole system, even if you aren't explicitly required to do so by the problem statement. A picture helps you figure out what's happening. A picture also serves as a convenient place to store bits of information - constants, variables, etc. Finally, failing anything else, it's tangible evidence that you tried, and may earn you some pity points.

2. Always define and label your coordinate system. Notice that, in this case, you want one that is tilted such that the x-axis is parallel to the surface of the plane. Hopefully, by this point of the course, you understand why -- motion will be one-dimensional in this coordinate system, and two-dimensional otherwise. Choose this one, and don't forget to label the coordinate system!

3. Drawing the free body diagram separate from the overall diagram makes it easier to draw without getting your diagram cluttered. But, if you prefer, you can draw this directly on the original image. I don't recommend you place the coordinate system directly on the block, even though the center will be the origin. (Again, it will clutter the image.)Your teacher might prefer that you do so; make sure you check.

If you do, it will look like this. Note that you can distinguish the x- and y-axis from the force vectors by making sure the lengths of the arrows are different, such that the arrowheads don't overlap.

4. Forces should always be drawn from the point where the force acts on the block. For normal and drag forces, this means the center of a surface. For weight, this is the center of the block. Yes, this matters; some past test solutions indicated taking off a half-point if this isn't done correctly. That may not sound like a lot, but keep in mind that a free response problem might be worth a total of about six points.

5. It's a good idea to define the forces if you abbreviate them in a diagram. Note that I used common abbreviations for weight (W) and normal force (N). Air resistance doesn't really have a standard abbreviation, so I use a “f” and subscript “air” to make it clear I’m talking about air resistance. It doesn't have to be lower-case; I personally like to use  the lower-case "f" for frictional forces and upper-case "F" for other forces.

You don't have to label the forces as vectors. You can get away with writing this:

normal force : $N = mg\cos \theta$
air resistance: $f_{air} = -bv$
weight: $W = mg$
(These aren't intended to be bold; the LaTeX just makes it appear so.)

Just make sure you're consistent. Don't label it a vector on one side but not on another.

Also, I follow the convention of using boldface to indicate a vector in a typed document. You can't really do bold when you're writing, so you would indicate a vector by drawing an arrow. Ex: $\overrightarrow{W}$. At this point, you know this, but for now I want these notes to err on the side of too much detail.

6. On the diagram, I drew some shaded, slanted lines on the plane surface. These lines are a standard way of noting that the contact surfaces are “frictionless”. Not at all mandatory, but potentially useful.

(b) Calculate the terminal velocity of the block. 

No motion or net force in y-direction, so all forces, velocities, and accelerations will be in the x-direction.

Terminal velocity $\rightarrow$ net force in x-direction is zero.

$F_{net} = 0$

$v_T$: terminal velocity

$mg\sin \theta -bv_{T} = 0$

$ -bv_{T} = -mg\sin \theta$

$\enclose{box}{\ v_{T} = \frac{mg\sin \theta}{b}}$

Notes for part (b):

1. Explicitly state that there is no motion in the y-direction. This allows you to ditch the $x$ and $y$ subscripts when talking about motion. It should be kind of obvious that the block isn't moving through the plane or taking off, but it doesn't hurt to mention it.

2. You should explicitly state that terminal velocity happens when the net force in the x-direction is zero, or, alternatively, when the net acceleration is zero. Note that the expression $F_{net} = 0$ is basically $F=ma$ in the special case of $a=0$. Just about every problem invokes Newton's Second Law in one form or another, so keep it in mind.

3. I explicitly defined $v_T$ as terminal velocity. It may seem nitpicky, and it probably is. But this basically makes it clear that you are solving for a specific velocity, and that this expression isn't for the general velocity of this object. In other words, you are emphasizing the fact, to the grader and yourself, that this is a constant, and not a general expression of the velocity of the object.

4. If you don't already, box your answers. It makes the job easier on the grader, which is good for everyone.

(c) Derive a differential equation describing the motion of the block.

$F_{net} = ma=m\frac{dv}{dt}$

$mg\sin \theta -bv = m\frac{dv}{dt}$

$\enclose{box}{\frac{dv}{dt}=g\sin \theta -\frac{b}{m}v}$

Notes for part (c):
1. It helps to explicitly make it clear that you are replacing $a$ with $\frac{dv}{dt}$, as demonstrated in the first line. You must do this to make this an ordinary differential equation -- an answer in terms of both $v$ and $a$ probably won't get full credit. Besides, you'll need to convert everything to one variable anyway for part (d), and writing it in terms of $a$ would involve expressing $v$ as an integral -- something not allowed in a differential equation.

2. You don't have to solve everything in terms of $\frac{dv}{dt}$. But it will help later. (You are thinking ahead, right?)

(d) Solve for the velocity of the block as a function of time.

Quick note: See note 1 for this section if you don't know why I want to take the time derivative of both sides.

$\frac{d}{dt}\left[\frac{dv}{dt}\right]=\frac{d}{dt}\left[g\sin \theta -\frac{b}{m}\right]v$

$\frac{d^2v}{dt^2}=-\frac{b}{m}\frac{dv}{dt}$

Substitute $a = \frac{dv}{dt}$ and $\frac{da}{dt} = \frac{d^2v}{dt^2}$

$\frac{da}{dt}=-\frac{b}{m}a$

Quick note: Skip the steps listed in blue if you're confident you can integrate this successfully without showing all the steps.

$\frac{da}{a}=-\frac{b}{m}dt$

$\int \frac{da}{a}=\int-\frac{b}{m}dt$         Separate variables and integrate both sides.

$\ln a = -\frac{b}{m}t + C$

$e^{\ln a} =e^{-\frac{b}{m}t + C}$

$a = Ce^{-\frac{b}{m}t}$            See note 2.

Initial condition from part (b): $a(0) = g\sin \theta -\frac{b}{m}(0)$            See note 3 if confused.

$a(0) = g\sin \theta $          

$Ce^{-\frac{b}{m}(0)}=g\sin \theta$

$C=g\sin \theta$            See note 4 if confused.

$a = g\sin \theta\ e^{-\frac{b}{m}t}$

Substitute $\frac{dv}{dt}= a$.

$\frac{dv}{dt} = g\sin \theta\ e^{-\frac{b}{m}t}$

$\int dv = g\sin \theta\ e^{-\frac{b}{m}t}dt$            Separate variables and integrate both sides.

$\int dv =\int g\sin \theta\ e^{-\frac{b}{m}t}dt$

$v = -\frac{mg}{b}\sin \theta\ e^{-\frac{b}{m}t}+C$            See note 2.

Initial condition: block starts at rest, so $v(0) = 0$.            See note 3 if confused.

$0= -\frac{mg}{b}\sin \theta\ e^{-\frac{b}{m}(0)}+C$

$0= -\frac{mg}{b}\sin \theta+C$            See note 4 if confused.

$C =  \frac{mg}{b}\sin \theta$

$v = -\frac{mg}{b}\sin \theta\ e^{-\frac{b}{m}t}+\frac{mg}{b}\sin \theta$

$\enclose{box}{v = \frac{mg}{b}\sin \theta\left(1-e^{-\frac{b}{m}t}\right)}$

Notes for part (d):

1. Why did I take the time derivative at the beginning? It's because, unless you've had a dedicated differential equations class and know how to solve a first-order ordinary differential equation with an additive constant, you probably need to go through these steps to make sure you end up with the correct constant coefficient. At time of writing, I'm not sure when (or if) this is covered in AP calculus AB or BC. But for those students taking AP calculus and physics C concurrently, it's possible they will have had no experience with solving these kinds of problems. (Some of my students have confirmed this.) Therefore, I'd recommend this more detailed derivation.

2. It pays to be careful. Note that the constant of integration for $a(t)$ involves a multiplicative constant, while the constant of integration for $v(t)$ is an additive constant. If you don't understand why it's a multiplicative constant for $a(t)$, go back and look at the derivation, and review the section in your calculus book that covers derivatives and integrals involving the natural logarithm and exponential functions.

3. Note that the initial conditions. At $t=0$, the initial velocity is zero, or $v(0)=0$.

The initial acceleration is not zero, however. Because the initial velocity is zero, we can substitute into the expression for $v$ from part (b).

4. $e^{(0)} = 1$

5. Note that, as $t\rightarrow \infty$,$v \rightarrow \frac{mg}{b}\sin\theta$, which is precisely what you calculated as the terminal velocity in part (b). More generally, substituting in limiting cases like $t=0$ and $t=\infty$ are really good ways of checking that you didn't screw up.

*****
Summary:

I know, I know -- the problem looks too long. The calculation for (d) is a bit longer than most of the derivations you will have to do on the free response. But it's a good problem. And if you take out all my red commentary, you'll find that the actual solution should fit within the space allotted.

Everything in red text will get more or less automatic as you practice questions. Assuming I consolidate this and other materials into an e-book at some point, a lot of the text will be consolidated into earlier chapters, with possible hyperlinks to the relevant section if someone is confused.

Do let me know if you've found this useful, or if you would make any changes. If you have a preferred way to simplify the long-ish derivation in (c), I'd love to hear it. If you think it's overkill to be picky about various aspects of notation or explanations in solutions, I'd also like to hear it.

*****
MathJax note: It appears that MathJax scales the font size of mathematical expressions dynamically according to the size of adjoining text. It presently looks too small. Workarounds that involve scaling commands in the html are evidently not recommended, at least according to some of the posts I saw online. For Blogger, it is sufficient to highlight the LaTeX code and set the font size to "Large", though that plays a bit of havoc with line spacing.

Also, it took me a while to realize that I had to edit the html to activate an additional package, "enclose.js", to box my answers. This actually took a lot more time to figure out than actually installing the base package. There are a lot of optional packages that are only activated if you edit the html to call them on the MathJax server. More details on the optional packages here: http://docs.mathjax.org/en/latest/tex.html

The 48 Laws of Power: Meditations

The 48 Laws of Power is one of my favorite books. (It made my 15 books list.) It contains a bit of psychology, historical anecdotes, and some suggestions on how to behave in certain situations.

It is not a good book -- some might call it an evil book. It is with some irony that I'm considering writing a series of posts going through them. Call it an effort in personal development, call it a waste of time. Call it misplaced yearnings for religion. I find it a fun book, even though I think I would never live in the manner that it suggests. It's too different from who I am today.

But therein lies the draw. Could I adopt this mindset? And once adopted, could I safely cast it off? I think that I have labored among a false view of the world and myself. Would this simply be another false perspective? For those who don't know, I could succinctly call it "popular Machiavelli", though that probably doesn't quite capture it. Machiavelli had an eye for how to govern a principality or a republic; this book largely suggests a nation of one, and makes few attempts to suggest a more noble or elevated ambition than self-interest.

As I have previously written, I don't think there's a grand unified theory of human psychology or behavior, and I don't think it's valuable to labor under the assumption that it does exist, and with further research, it will reveal itself to us. So no single lens is helpful.

I think the philosophy this book outlines, sometimes seriously, sometimes in a manner that suggests it's meant to sell books to greedy wanna-be captains of industry, is one I could not willingly adopt. But it is intriguing. And, perhaps, there are times -- many times -- in which this framework is better than the blithely ignorant liberal/social justice/streak of selfishness hodgepodge that is my current cognitive framework.

If for no other reason, a structured approach, a daily devotional of sorts, might prove helpful to order my thoughts, whether they be aligned or in opposition. Moreover, I need to train my mind to be a bit more strategic.

I may indeed try to make this daily. Each "law" is relatively short. If I'm ambitious, I might supplement the discussion with some other examples of transgressions, observances, and possible counterexamples.

Next Post: Law 1: Never Outshine The Master (pending)