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):
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:
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.
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