AP Physics C: Mechanics · Topic 6.2

Topic 6.2: Torque and Work

Unit 6: Energy and Momentum of Rotating Systems10-15% of the multiple-choice section

Work done on a rigid system by a torque is the integral of the torque with respect to angular position, the area under a graph of torque against angle. AP Physics 1 prints torque times angular displacement instead. The integral is what lets a Physics C question use a varying torque.

AP Physics: Unit 6 (topics 6.2 Torque and Work). Topic 6.2 of the current AP Physics C: Mechanics course and exam description, inside Unit 6, weighted 10 to 15% of the multiple-choice section. One learning objective, 6.2.A, and three essential-knowledge statements: 6.2.A.1 on a torque transferring energy only if exerted over an angular displacement, 6.2.A.2 with the relevant equation W as the integral of tau d theta between two angles, and 6.2.A.3 on the work as the area under a torque against angular position graph. Suggested skills 1.A, 2.A, 2.C and 3.B. No boundary statement: Unit 6 prints exactly one, under Topic 6.5. Calculus differentiator against the identically titled AP Physics 1 Topic 6.2: the integral replaces the algebra-based sheet's W = tau delta theta, so a torque that varies with angular position is fair game. Sheet detail verified against the printed Table of Information: the rotational work line is written with a centred multiplication dot and no vector arrows and no limits, while the translational line W = integral of F dot dr carries arrows on both vectors and limits a and b. Not printed on the C: Mechanics sheet: W = tau delta theta, P = tau omega, and the rotational work-energy theorem (the printed work-energy line is written for forces and parallel displacements). Printed but often missed because equation-sheet transcriptions omit it: the cross-product magnitude AB sin theta, in the Vectors table of the same appendix.

What Topic 6.2 requires

Topic 6.2 of AP Physics C: Mechanics carries one learning objective and three essential-knowledge statements.

Learning objective 6.2.A: describe the work done on a rigid system by a given torque or collection of torques.

Statement	What it says
6.2.A.1	A torque can transfer energy into or out of an object or rigid system if the torque is exerted over an angular displacement
6.2.A.2	The amount of work done on a rigid system by a torque is related to the magnitude of that torque and the angular displacement through which the rigid system rotates during the interval in which that torque is exerted. Relevant equation: $W = \int_{\theta_1}^{\theta_2} \tau\, d\theta$
6.2.A.3	Work done on a rigid system by a given torque can be found from the area under the curve of a graph of the torque as a function of angular position

The suggested skills are 1.A (create diagrams, tables, charts, or schematics to represent physical situations), 2.A (derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway), 2.C (compare physical quantities between scenarios or at different times and locations within a single scenario) and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim).

Skill 2.A carries the largest single share of the multiple-choice section at 25 to 30%. Its presence here is a fair description of what Topic 6.2 questions look like: set up an integral, evaluate it symbolically, and only then substitute numbers.

Topic 6.2 prints no boundary statement. Unit 6 prints exactly one across all six topics, under Topic 6.5.

The integral, and the question it lets an exam ask

This is the topic where the two rotation courses genuinely part company, and the difference is one symbol wide.

The AP Physics 1 equation sheet prints

W = \tau\Delta\theta

and the AP Physics C: Mechanics sheet prints

W = \int \tau \cdot d\theta

A product is only correct when the thing being multiplied holds still. $\tau\Delta\theta$ assumes the torque is the same at every angle in the sweep, or that you already know its average. The integral makes no such assumption, so a Physics C question is free to hand you a torque that depends on angular position and ask for the energy transferred. That is a category of problem the algebra-based course cannot pose at all.

The most common physical source of an angle-dependent torque is the most ordinary one: gravity on something pivoted off center. A rod hinged at one end and released from horizontal feels a gravitational torque $\tau = Mg\frac{L}{2}\cos\theta$ about the hinge, where $\theta$ is measured from the horizontal. That torque starts at its largest value and falls to zero as the rod swings through the vertical. Multiplying the starting torque by the total angle overestimates the work by more than half. Worked example 1 shows by exactly how much.

Two details of the printed line are worth reading carefully, because they are easy to over-interpret. The sheet writes the rotational work with a centered dot, $\int \tau \cdot d\theta$ , but with no vector arrows on either symbol. The translational line in the left-hand column of the same table, $W = \int_a^b \vec{F} \cdot d\vec{r}$ , does carry arrows on both, and limits. So the rotational line is a product of scalars written with a multiplication dot, not a dot product you are expected to expand into components. The other detail: the sheet omits the limits that essential knowledge 6.2.A.2 supplies. Put them in. An indefinite integral is not an amount of energy.

Three different quantities get called theta

More Topic 6.2 errors come from this than from the calculus. Three angles appear within two lines of each other, and they are not the same angle.

The angle inside the torque. This course prints torque as a cross product, $\vec{\tau} = \vec{r} \times \vec{F}$ . The Vectors table in the same appendix prints $\lvert \vec{A} \times \vec{B} \rvert = AB\sin\theta$ , so the magnitude is $rF\sin\theta$ with $\theta$ measured between the position vector and the force. This angle decides how much of the force turns the object.
The angle you integrate over. In $W = \int \tau\, d\theta$ , $\theta$ is the angular position of the rigid system, and $d\theta$ is how far it turns. This angle decides how far the torque acts.
The angle in the translational work integral. Expanding the dot product in $W = \int \vec{F} \cdot d\vec{r}$ brings in $\cos\theta$ between the force and the displacement. That is a third angle again.

Angle 1 uses sine and angle 3 uses cosine, which is the surface symptom. The underlying reason is that they measure different pairs of things. A useful habit on any Topic 6.2 problem: write $\tau(\theta)$ explicitly as a function before you integrate, and check that the $\theta$ in the function and the $\theta$ in the $d\theta$ mean the same thing. In the falling-rod example they do, because the geometry has been set up so that the same angle both locates the rod and sets the lever arm. When a problem gives you a force at a fixed angle to a rotating arm, they do not.

All three angles are in radians the moment they enter the work integral. Radians are what make $\mathrm{N \cdot m}$ times angle come out in joules. Degrees give an answer too large by a factor of $180/\pi$ , and it will look plausible.

No angular displacement, no work, and the sign of the work

Statement 6.2.A.1 sets the condition for any energy transfer at all: a torque can transfer energy into or out of an object or rigid system if the torque is exerted over an angular displacement.

So the conditional is doing real work in that sentence. Lean on a wrench against a seized bolt as hard as you like: there is a torque, the bolt does not turn, $d\theta = 0$ , and the integral is zero. No energy reaches the bolt. Your arm gets tired for physiological reasons that are outside the model.

The phrase "into or out of" is the other half. Work by a torque is a signed scalar:

The torque and the rotation in the same sense makes $\tau$ and $d\theta$ share a sign, the integrand is positive, and the system gains energy.
The torque opposing the rotation makes them opposite in sign, the integrand is negative, and the system loses energy. A brake pad is the standard case.
A torque that reverses partway through gives an integral with a positive piece and a negative piece. The angular speed peaks where the torque crosses zero, not where the torque is largest. Worked example 2 is built around that.

Declare a positive rotation sense before you start and keep it to the end. Once you have, the net work follows from the net torque:

W_{\text{net}} = \int \tau_{\text{net}}\, d\theta = \int I\alpha\, d\theta = \int I \frac{d\omega}{dt} \frac{d\theta}{dt}\, dt = \int I\omega\, d\omega = \tfrac{1}{2}I\omega_f^2 - \tfrac{1}{2}I\omega_i^2

That is the rotational work-energy theorem, and it is not printed on the equation sheet in that form. The sheet's work-energy line, $\Delta K = \sum W_i = \sum F_{\parallel, i}\, d_i$ , is written for forces and parallel displacements. Notice the qualifier the derivation quietly needs: $I$ had to come out of the integral, so it holds when rotational inertia is constant. Essential knowledge 6.3.C.2.ii attaches the same condition to the parallel result for angular momentum. For a system that changes shape while it turns, this shortcut is not available and the energy accounting has to be done term by term.

The area under a torque against angular position graph (6.2.A.3)

Statement 6.2.A.3 is 6.2.A.2 drawn instead of written: work done on a rigid system by a given torque can be found from the area under the curve of a graph of the torque as a function of angular position.

What to actually do with such a graph:

Area, not slope. The slope of a torque against angle graph has units of $\mathrm{N \cdot m/rad}$ and no name in this course. Do not reach for it.
Signed area. Area below the axis is negative work. Sum the pieces with their signs; do not add magnitudes.
Break it into shapes. Exam graphs are built from triangles, rectangles and trapezoids precisely so that the area is exact without calculus. Use the geometry, then state that it is the integral.
Find where the curve crosses zero. If the system started from rest, the maximum angular speed is at the crossing, because that is where the running total of the area stops growing. This is the most frequently asked feature of such a graph and it costs one line of algebra.
The vertical axis has to be torque, not force. A force against position graph gives translational work. Check the axis label before you compute anything.

The companion graph in the next topic is torque against time, whose area is the angular impulse rather than the work. Same axis label, different horizontal axis, different quantity, different units. The next section says why that matters.

Skill 1.A is listed for this topic and is a Science Practice 1 skill, so it is assessed on the free-response section rather than the multiple-choice section. Practice 1 carries 20 to 35% of the free-response score. Being able to sketch the torque against angle graph that matches a described situation, and to shade the region that answers the question, is directly what that skill asks for.

Torque times time is a different integral, and it is not work

The AP Physics C: Mechanics sheet prints these two lines within a few rows of each other in the rotational column:

W = \int \tau \cdot d\theta \qquad \qquad \Delta L = \int \tau\, dt

Same torque, different variable of integration, completely different quantity. One is energy in joules, the other is angular momentum in $\mathrm{kg \cdot m^2/s}$ . They are not convertible into each other, and neither is a shortcut for the other.

The reason both exist is that a torque acting over an angle transfers energy, while a torque acting over a time transfers angular momentum. A given torque can do a great deal of one and very little of the other. Hold a wheel steady against a torque for a minute: zero work, large angular impulse absorbed by whatever is holding it. Snap a wheel through a full turn very quickly: substantial work, small angular impulse.

Where this bites in practice is a problem that gives you $\tau(t)$ and asks for the work. You cannot integrate $\tau$ over $t$ and call the result work, and you cannot multiply the angular impulse by an angular speed either. The chain that does work is:

Integrate $\tau(t)$ over time to get $\Delta L$ , hence $\omega(t)$ if $I$ is constant.
Integrate $\omega(t)$ over time to get $\theta(t)$ if you want the angle.
Either evaluate $W = \int \tau\, \omega\, dt$ , using $d\theta = \omega\, dt$ , or skip straight to $W_{\text{net}} = \Delta K_{\text{rot}}$ .

Worked example 3 runs that chain and shows that the two plausible shortcuts miss by a factor of two and by 50% respectively. Topic 6.3 owns the time integral in full.

Rotational power is a derivative, and it is not printed

Power is Unit 3 content in this course, not Unit 6 required content, but Topic 6.2 questions ask for it often enough to be worth two paragraphs.

The sheet prints two power lines and neither is rotational:

P_{\text{avg}} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t} \qquad \qquad P_{\text{inst}} = \frac{dW}{dt}

There is no $P = \tau\omega$ anywhere on the AP Physics C: Mechanics equation sheet. Derive it in one line from what is there. Differentiate the work integral with respect to time:

P_{\text{inst}} = \frac{dW}{dt} = \frac{d}{dt}\int \tau\, d\theta = \tau \frac{d\theta}{dt} = \tau\omega

using $\omega = d\theta/dt$ , which is printed. The translational analogue on the AP Physics 1 sheet, $P_{\text{inst}} = F_{\parallel}v$ , is the same statement for forces.

The distinction between the two printed lines matters more here than in the translational case, because a wheel spun up by a constant torque has an angular speed that grows linearly, so its instantaneous power grows linearly too. Average and instantaneous power differ by a factor of two over such an interval, starting from rest. Reporting one when the question asked for the other is a straightforward loss of a point, and the subscripts on the sheet are the reminder.

What the equation sheet prints for Topic 6.2

Checked line by line against the Table of Information in the AP Physics C: Mechanics course and exam description:

Equation	On the C: Mechanics sheet
$W = \int \tau \cdot d\theta$	yes, without limits and without vector arrows
$W = \int_a^b \vec{F} \cdot d\vec{r}$	yes, with limits and with arrows on both vectors
$\vec{\tau} = \vec{r} \times \vec{F}$	yes
$\lvert \vec{A} \times \vec{B} \rvert = AB\sin\theta$	yes, in the Vectors table of the same appendix
$K_{\text{rot}} = \frac{1}{2}I\omega^2$	yes
$\omega = \frac{d\theta}{dt}$	yes
$\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}} = \frac{\tau_{\text{net}}}{I_{\text{sys}}}$	yes
$P_{\text{avg}} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}$ and $P_{\text{inst}} = \frac{dW}{dt}$	yes
$W = \tau\Delta\theta$	no, that is the AP Physics 1 sheet
$P = \tau\omega$	no
$W_{\text{net}} = \Delta K_{\text{rot}}$	no, the printed work-energy line is written for forces
$\tau = rF\sin\theta$ as a standalone line	no, but it follows from the two cross-product lines above

The fourth row is the one people get wrong in both directions. `equations.ts`-style transcriptions of the sheet routinely omit the Vectors, Calculus, Geometry and Trigonometry tables, so it is easy to conclude that the cross-product magnitude is absent. It is printed, on the third page of the Table of Information, alongside $\vec{A} \cdot \vec{B} = AB\cos\theta$ , the unit-vector form of a vector, and component addition. The calculus rules you need to evaluate a polynomial torque integral are printed on that same page.

The row that does cost marks is $W = \tau\Delta\theta$ . It is a correct result for a constant torque, and you may use it, but it is not on your sheet and it is not in your framework. If you write it down on a free-response question with a varying torque, it is wrong rather than merely unsupported.

If you are in AP Physics 1, this is not your page

AP Physics 1 has a Topic 6.2 with the same title, and its treatment stops where this one starts. Its sheet prints the product form, its questions supply constant torques, and its graph statement is the same 6.2.A.3 sentence read as a triangle-and-rectangle exercise rather than as an integral.

The [AP Physics 1 Topic 6.2 page](/ap-physics-1/unit-6-energy-and-momentum-of-rotating-systems/6-2-torque-and-work) is for students in the algebra-based course, who need torque times angular displacement and the area under a graph. This page is for students in AP Physics C: Mechanics, who need the integral, the sign handling when a torque reverses, and the derivation of rotational power from the printed instantaneous-power definition. If you are in Physics 1, none of the integration on this page is on your exam, and the product form is genuinely all your questions will need.

AP Physics C: Mechanics is a calculus-based, college-level course, equivalent to a first course in an introductory college sequence in calculus-based physics. Its stated prerequisite is that students have taken or are concurrently taking calculus.

Related pages: Topic 6.1 is where the energy this work transfers ends up, and Topic 6.3 owns the other integral of the same torque. The torque guide owns the step-by-step routine for finding a torque in the first place, the work-energy theorem guide owns the translational version, and the torque calculator checks arithmetic on a single force. The Unit 6 hub lists all six topics.

A hinged rod falling: a torque that changes as it turns

A uniform rod of mass $M = 0.75$ kg and length $L = 0.90$ m is hinged at one end and held horizontal, then released from rest. Take $g = 9.8\ \mathrm{m/s^2}$ . Find (a) the gravitational torque about the hinge as a function of the angle $\theta$ below the horizontal, (b) the work gravity does on the rod as it swings to vertical, (c) the rod's angular speed at that moment, and (d) what $W = \tau\Delta\theta$ would have given.

Set the convention: $\theta$ is measured from the horizontal, increasing as the rod falls, and positive torque turns the rod in that same sense. The rod's weight acts at its center of mass, a distance $L/2$ from the hinge.
(a) The angle between the position vector to the center of mass and the downward weight is $90^\circ - \theta$ , so $\lvert \vec{r} \times \vec{F} \rvert = \frac{L}{2}(Mg)\sin(90^\circ - \theta) = Mg\frac{L}{2}\cos\theta$ .
Numerically the maximum, at $\theta = 0$ , is $Mg\frac{L}{2} = (0.75)(9.8)(0.45) = 3.3075\ \mathrm{N \cdot m}$ . It falls to zero at $\theta = 90^\circ$ .
(b) Apply 6.2.A.2 with limits in radians: $W = \int_0^{\pi/2} Mg\frac{L}{2}\cos\theta\, d\theta = Mg\frac{L}{2}\left[\sin\theta\right]_0^{\pi/2} = Mg\frac{L}{2}$ .
$W = 3.3075 \approx 3.31\ \mathrm{J}$ . Cross-check it against the printed $\Delta U_g = mg\Delta y$ : the center of mass drops $L/2 = 0.45$ m, so the potential energy falls by $(0.75)(9.8)(0.45) = 3.3075$ J. The two agree exactly, which they must, because gravity is the only thing doing work.
(c) The rod starts at rest and rotates about the hinge, so $W_{\text{net}} = \frac{1}{2}I\omega^2$ with $I = \frac{1}{3}ML^2 = \frac{1}{3}(0.75)(0.81) = 0.2025\ \mathrm{kg \cdot m^2}$ .
$\omega = \sqrt{\dfrac{2(3.3075)}{0.2025}} = \sqrt{32.667} = 5.7155 \approx 5.72\ \mathrm{rad/s}$ .
Symbolic check: $\frac{1}{2}\left(\frac{1}{3}ML^2\right)\omega^2 = Mg\frac{L}{2}$ gives $\omega = \sqrt{3g/L}$ , with no mass in it. $\sqrt{3(9.8)/0.90} = 5.7155\ \mathrm{rad/s}$ , matching. Skill 2.A wants that symbolic form before the number.
(d) Using the product form with the largest torque: $(3.3075)(\pi/2) = 5.195\ \mathrm{J}$ , too big by 57%. Using it with the correct average torque works, but the correct average is $W/\Delta\theta = 3.3075/1.5708 = 2.106\ \mathrm{N \cdot m}$ , which is $\frac{2}{\pi}$ of the maximum and is not a number you could have guessed. The integral hands it to you; the product form asks you to already know it.

(a) $\tau(\theta) = Mg\frac{L}{2}\cos\theta$ , maximum $3.31\ \mathrm{N \cdot m}$ at the horizontal and zero at the vertical. (b) $W = Mg\frac{L}{2} = 3.31$ J, equal to the drop in the center of mass's gravitational potential energy. (c) $\omega = \sqrt{3g/L} = 5.72$ rad/s. (d) Torque times angle with the maximum torque gives 5.20 J, 57% too large.

Reading signed work off a torque against angle graph

A wheel with rotational inertia $I = 1.5\ \mathrm{kg \cdot m^2}$ starts from rest. The torque on it varies with angular position as follows: it rises linearly from 0 to $6.0\ \mathrm{N \cdot m}$ over the first 3.0 rad, holds at $6.0\ \mathrm{N \cdot m}$ from 3.0 to 5.0 rad, then falls linearly to $-2.0\ \mathrm{N \cdot m}$ at 8.0 rad. Find (a) the total work done through 8.0 rad, (b) the angular speed there, (c) the angle at which the angular speed is greatest and its value.

Convention: positive torque acts in the direction the wheel turns, so positive area is energy in.
(a) Statement 6.2.A.3 says the work is the area under the curve. Break the region into pieces.
0 to 3.0 rad, a triangle: $\frac{1}{2}(3.0)(6.0) = 9.0\ \mathrm{J}$ .
3.0 to 5.0 rad, a rectangle: $(2.0)(6.0) = 12.0\ \mathrm{J}$ .
5.0 to 8.0 rad, a line from $+6.0$ to $-2.0$ , so its slope is $-8.0/3.0 = -2.667\ \mathrm{N \cdot m/rad}$ and it crosses zero at $\theta = 5.0 + 6.0/2.667 = 7.25$ rad.
Positive triangle from 5.0 to 7.25 rad: $\frac{1}{2}(2.25)(6.0) = 6.75\ \mathrm{J}$ . Negative triangle from 7.25 to 8.0 rad: $\frac{1}{2}(0.75)(2.0) = 0.75\ \mathrm{J}$ , counted as $-0.75$ J because it lies below the axis.
$W_{\text{total}} = 9.0 + 12.0 + 6.75 - 0.75 = 27.0\ \mathrm{J}$ .
(b) Starting from rest, $\frac{1}{2}I\omega^2 = 27.0$ J, so $\omega = \sqrt{2(27.0)/1.5} = \sqrt{36} = 6.0\ \mathrm{rad/s}$ .
(c) The running total of the area is still growing wherever the torque is positive and starts shrinking once it goes negative, so the angular speed peaks at the zero crossing, $\theta = 7.25$ rad, not at 3.0 rad where the torque first reaches its maximum.
The work up to that point is $9.0 + 12.0 + 6.75 = 27.75\ \mathrm{J}$ , so $\omega_{\max} = \sqrt{2(27.75)/1.5} = \sqrt{37} = 6.083 \approx 6.08\ \mathrm{rad/s}$ .
Sanity check the size of the difference: the wheel loses only 0.75 J over the last stretch, which is 2.7% of its energy, and speed goes as the square root of energy, so a 1.4% drop in $\omega$ is exactly what the numbers show.

(a) $W = 27.0$ J, the signed area, with the last $0.75$ J subtracted. (b) $\omega = 6.0$ rad/s at 8.0 rad. (c) The peak is at $\theta = 7.25$ rad, where the torque crosses zero, with $\omega_{\max} = 6.08$ rad/s.

A torque given as a function of time, and the two shortcuts that fail

A wheel with rotational inertia $I = 0.80\ \mathrm{kg \cdot m^2}$ is at rest at $t = 0$ . A torque $\tau(t) = (0.60\ \mathrm{N \cdot m/s^2})t^2$ acts on it for 3.0 s, with no other torque. Find (a) the angular momentum and angular speed at $t = 3.0$ s, (b) the total angle turned, (c) the work done on the wheel, evaluated two independent ways, and (d) what the two tempting shortcuts would have given.

Convention: the torque and the resulting rotation are both positive, and $I$ is constant so the rotational form of Newton's second law applies as written in 6.3.C.2.ii.
(a) The time integral is angular impulse, not work: $\Delta L = \int_0^{3.0} 0.60 t^2\, dt = 0.60\left[\frac{t^3}{3}\right]_0^{3.0} = 0.60(9.0) = 5.4\ \mathrm{kg \cdot m^2/s}$ .
With $I$ constant and $\omega_0 = 0$ , $\omega(3.0) = \Delta L / I = 5.4/0.80 = 6.75\ \mathrm{rad/s}$ .
(b) Generally $\omega(t) = \frac{1}{I}\int_0^t 0.60 t'^2\, dt' = \frac{0.20 t^3}{0.80} = 0.25 t^3$ , which gives 6.75 rad/s at $t = 3.0$ s as a check. Then $\theta(t) = \int_0^t 0.25 t'^3\, dt' = 0.0625 t^4$ , so $\theta(3.0) = 0.0625(81) = 5.0625\ \mathrm{rad}$ .
(c) Route one, the work integral with $d\theta = \omega\, dt$ : $W = \int_0^{3.0} \tau(t)\,\omega(t)\, dt = \int_0^{3.0} (0.60t^2)(0.25t^3)\, dt = 0.15\left[\frac{t^6}{6}\right]_0^{3.0} = 0.15(121.5) = 18.225\ \mathrm{J}$ .
Route two, the rotational work-energy theorem, valid here because $I$ is constant: $W = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.80)(6.75)^2 = 0.40(45.5625) = 18.225\ \mathrm{J}$ . The two agree to every digit, which is the check worth doing.
$W \approx 18.2\ \mathrm{J}$ .
(d) Shortcut one, multiplying the angular impulse by the final angular speed: $(5.4)(6.75) = 36.45\ \mathrm{J}$ , exactly twice the right answer. It fails for the same reason $Ft$ times $v$ is not work: the wheel spent most of the interval slower than 6.75 rad/s.
Shortcut two, the final torque times the total angle: $\tau(3.0) = 0.60(9.0) = 5.4\ \mathrm{N \cdot m}$ , and $(5.4)(5.0625) = 27.3\ \mathrm{J}$ , 50% too large, because the torque was smaller than its final value for the whole interval.
The units are the tell. Angular impulse is $\mathrm{kg \cdot m^2/s}$ and angular speed is $\mathrm{rad/s}$ , so their product carries $\mathrm{s^{-2}}$ and only looks like joules because the radian is dimensionless. Checking units would not have caught shortcut one. Only knowing which integral answers which question does.

(a) $\Delta L = 5.4\ \mathrm{kg \cdot m^2/s}$ and $\omega = 6.75$ rad/s. (b) $\theta = 5.06$ rad. (c) $W = 18.2$ J, from the work integral and from the change in rotational kinetic energy, agreeing exactly. (d) Angular impulse times final angular speed gives 36.45 J, twice too large; final torque times total angle gives 27.3 J, 50% too large.

Frequently asked questions

What is the formula for work done by a torque in AP Physics C?

Essential knowledge 6.2.A.2 of AP Physics C: Mechanics gives it as the integral of the torque with respect to angular position, evaluated between the starting and finishing angles. The equation sheet prints it as W equals the integral of tau times d theta, without limits, so supply the limits yourself. The angle must be in radians. For the special case of a constant torque the integral collapses to torque times angular displacement, which is what the AP Physics 1 sheet prints, but the calculus-based course does not restrict itself to that case.

Why is work done by a torque an integral in AP Physics C but a product in AP Physics 1?

Because a product assumes the torque never changes during the rotation. Multiplying torque by angular displacement is only correct for a constant torque, or if you already know the average torque over the sweep. The integral makes no such assumption, so an AP Physics C question can give you a torque that depends on angular position, such as gravity acting on a rod pivoted at one end, and ask how much energy it transfers. The algebra-based course has no route to that number, which is why its sheet prints the product form instead.

Can a torque do zero work?

Yes, in two different ways. Essential knowledge 6.2.A.1 says a torque transfers energy into or out of a rigid system only if the torque is exerted over an angular displacement, so a torque applied to something that does not rotate does no work at all. Pushing on a seized bolt is the standard case. Separately, a torque acting over a rotation can do zero net work if it is positive over part of the sweep and negative over the rest, so that the signed areas cancel. The angular speed then returns to its starting value even though the torque was never zero.

How do you find work from a torque versus angular position graph?

Take the area between the curve and the horizontal axis, counting area below the axis as negative. Essential knowledge 6.2.A.3 states this directly. Exam graphs are built from triangles, rectangles and trapezoids so the area comes out exactly. Two habits pay off: find where the curve crosses zero, because that is where the angular speed of a system starting from rest is greatest, and check the horizontal axis label, since the area under torque against time is the angular impulse instead, measured in kilogram metres squared per second rather than joules.

Is rotational power on the AP Physics C Mechanics equation sheet?

No. The sheet prints average power as work divided by time interval, or energy change divided by time interval, and instantaneous power as the derivative of work with respect to time. Torque times angular speed does not appear. Derive it in one line: differentiate the work integral with respect to time and use the printed definition of angular velocity as the derivative of angular position, which turns the derivative of the integral of tau d theta into tau times omega. For a constant torque starting from rest, the instantaneous power at the end of an interval is twice the average power over it.

What is the difference between the integral of torque over angle and the integral of torque over time?

They answer different questions and they are both printed on the AP Physics C: Mechanics sheet. Integrating a torque over angular position gives the work done, in joules, and it changes the rotational kinetic energy. Integrating the same torque over time gives the angular impulse, in kilogram metres squared per second, and it changes the angular momentum, which is essential knowledge 6.3.C.2.i. Neither converts into the other. Holding a wheel still against a torque delivers a large angular impulse and zero work, and there is no valid shortcut that turns one into the other.

Why is torque measured in newton metres and not in joules?

Because torque is not energy. A newton metre of torque and a newton metre of work have the same base units only because the radian, which appears in the work but not in the torque, is dimensionless. The physical distinction is that torque is a cross product of a position vector with a force, so it is at right angles to both, while work is a product of a torque with the angle turned and has no direction at all. Keeping the units named differently, newton metres for torque and joules for work, is the convention that stops the two being interchanged.