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.

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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.

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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