Topic 6.2: Torque and Work

What Topic 6.2 requires

Topic 6.2 has one learning objective, 6.2.A: describe the work done on a rigid system by a given torque or collection of torques. Three essential knowledge statements sit under it, none of them with sub-statements.

• 6.2.A.1 says 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 says 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. The relevant equation the CED prints alongside it is $W = \tau \Delta\theta$.
• 6.2.A.3 says work done on a rigid system by a given torque can be found from the area under the curve of a graph of torque as a function of angular position.

The CED prints no boundary statement for this topic. Boundary statements are the device the CED uses to mark the content boundaries of the AP Physics courses, and Topic 6.2 carries none.

The suggested skills are 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; and 3.A, create experimental procedures that are appropriate for a given scientific question. The Topic 6.2 page and the Unit 6 Unit at a Glance table print the same five.

That list tells you what the questions look like. Skill 1.B and essential knowledge 6.2.A.3 are the same idea seen twice, so expect a torque versus angle graph. Skill 2.A means an algebraic derivation with no numbers in it. Skills 2.C and 2.D mean scaling questions: what happens to the work if the torque doubles and the angle halves.

Unit 6 carries 5 to 8 percent of the multiple-choice section, tied with Unit 7 for the smallest share of the eight units, and the CED suggests roughly 8 to 14 class periods for the whole unit.

The equation, term by term

$W$ is work, in joules. There is no separate rotational unit of energy: the joule a motor puts into a flywheel is the same joule a hand puts into a sliding crate.

$\tau$ is the torque about the axis, in newton meters. It is the constant torque over the interval in question. If the torque changes as the system turns, this equation no longer applies as written and you fall back on the area rule in 6.2.A.3.

$\Delta\theta$ is the angular displacement in radians. Not degrees, not revolutions.

The units are worth a second look, because they explain the radian requirement. A newton meter multiplied by a radian is a newton meter, which is a joule, only because the radian is a ratio of arc length to radius and therefore dimensionless. Feed the equation degrees and the arithmetic silently produces a number that is too large by a factor of $180/\pi$, about 57.3, with no unit clash to warn you.

It is also worth naming the family resemblance to Topic 3.2. The translational sheet equation is $W = F_{\parallel} d = Fd\cos\theta$: a force multiplied by the displacement along it. The rotational version is a torque multiplied by the angular displacement it turns through. Same structure, rotational symbols.

Two different angles both called theta

Here is a unit trap rather than a physics one, and it is the kind that lets you do every step of the physics correctly and still hand in a wrong number. Two equations you will use in the same problem both contain a symbol that looks like an angle, and they want different units.

The $\theta$ inside $\tau = rF\sin\theta$ is an angle between two directions. Nothing is rotating through it; it is a geometric fact about how the force is aimed relative to the lever arm. Angles of that kind are quoted in degrees on this site and at every calculator input here, and your calculator wants degree mode when you take the sine. That convention is described on the torque calculator, and the geometry itself belongs to Topic 5.3.

The $\Delta\theta$ in $W = \tau \Delta\theta$ is a different animal. It is an angular displacement, which essential knowledge 5.1.A.1 defines as the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis. It is a distance traveled, expressed as an angle, and radians are what make it behave like a distance. One full turn is $2\pi$ radians, about 6.28.

So a single problem can legitimately read: take the sine of 60 degrees to get the torque, then convert a 90 degree turn to 1.571 radians to get the work. That is not an inconsistency. The two angles are measuring different things. Worked example two below does exactly this.

Energy in, energy out, or neither (6.2.A.1)

Essential knowledge 6.2.A.1 puts the condition first: a torque transfers energy into or out of a system if the torque is exerted over an angular displacement. No rotation means no energy transfer, however hard you push.

Leaning on a wrench that is clamped on a seized bolt is the clean example. The torque is large, you are getting tired, and the work done on the bolt is exactly zero because $\Delta\theta = 0$. It is the rotational twin of pushing on a wall, which does no work in Topic 3.2 for the same reason.

For the sign, you have to declare a positive sense of rotation first. Essential knowledge 5.1.A.1.ii says one direction of angular displacement about an axis of rotation, clockwise or counterclockwise, is typically indicated as mathematically positive, with the other direction becoming mathematically negative. On this page, counterclockwise is positive. Once that is fixed:

• A torque in the same sense as the rotation does positive work. Energy goes into the system, and if it is the only torque acting, $\frac{1}{2} I \omega^2$ rises.
• A torque opposing the rotation does negative work. Energy leaves the system and the rotation slows. Friction at a bearing and a brake pad both live here.
• A torque acting while nothing turns does zero work.

Two cautions on the sign. First, both $\tau$ and $\Delta\theta$ carry the sign, and the product is what matters: a negative torque acting through a negative angular displacement does positive work, which is what happens when a motor drives a wheel clockwise. Second, work is a scalar even though torque is not, so once you have the sign, add the works from different torques as ordinary signed numbers.

Net torque and the rotational work-energy theorem

For a rigid system turning about a fixed axis, every torque acts through the same $\Delta\theta$, because the whole body turns together. So the total work is the net torque multiplied by that angular displacement:

And that net work is what changes the rotational kinetic energy:

That second line is not printed on the AP Physics 1 equation sheet in that form. What the sheet prints is $\Delta K = \sum W_i = \sum F_{\parallel, i} d_i$ in the translational group and $K = \frac{1}{2} I \omega^2$ in the rotational group. The rotational statement is something you assemble, and skill 2.A for this topic is exactly the ability to assemble it.

Here is the assembly, using only printed equations. Start with the rotational second law $\alpha_{sys} = \tau_{net}/I_{sys}$, so $\tau_{net} = I\alpha$. Take the rotational kinematics equation $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$ and rearrange it to $\alpha \Delta\theta = \frac{1}{2}(\omega^2 - \omega_0^2)$. Multiply:

Printed equations only, one line of algebra, and you have the rotational work-energy theorem. It is also a free consistency check on any numerical answer: solve a spin-up problem by energy and again by rotational kinematics, and the two must agree to every digit. Worked example one does that.

One condition to keep in view: this holds for a rigid system turning about a fixed axis. If the object also travels, as a rolling wheel does, the translational kinetic energy term is in play too, which is what Topic 6.5 is about.

Work as the area under a torque versus angle graph (6.2.A.3)

Essential knowledge 6.2.A.3 says work done by a given torque can be found from the area under the curve of a graph of torque as a function of angular position. This is the version of the topic that survives a varying torque, and skill 1.B, creating quantitative graphs with appropriate scales and units, is listed right beside it.

Label the axes before anything else. Torque in newton meters goes up, angular position in radians goes across. Then one grid square that is 1 N m tall and 1 rad wide is worth exactly 1 J, and you can count area instead of calculating it.

The method after that is arithmetic. Cut the region at every kink so what is left is rectangles and triangles, work out each piece with units attached, and subtract any area that lies below the horizontal axis, because that is a torque opposing the rotation doing negative work. Worked example three runs a graph with all three pieces.

Check the horizontal axis label every single time. A graph of torque against time looks identical, and its area is not work. The area under a torque versus time graph is angular impulse, $\Delta L = \tau \Delta t$, which changes angular momentum rather than kinetic energy. Both equations are printed on the sheet, both are torque multiplied by something, and they are not interchangeable:

The same distinction exists in straight-line motion, where the area under a force versus position graph is work and the area under a force versus time graph is impulse. If you already have that one straight, this is the same rule wearing rotational symbols.

Rotational power

Power is the rate at which energy is transferred, and nothing about that definition is specific to straight lines. Essential knowledge 3.5.A.1 defines power as the rate at which energy changes with respect to time, either by transfer into or out of a system or by conversion from one type to another within a system, and 3.5.A.3 gives average power as the total work done divided by the time during which that work was done. Both cover rotational work without amendment.

So the printed equation $P_{avg} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}$ is already all you need. Put $W = \tau \Delta\theta$ into it for a constant torque, and notice that the angular displacement divided by the time interval is the average angular speed:

Shrink the interval and the same statement becomes the instantaneous power, $P = \tau\omega$.

Be clear about the status of that result. **$P = \tau\omega$ is not printed on the AP Physics 1 equation sheet.** The sheet's two power equations are $P_{avg} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}$ and $P_{inst} = F_{\parallel} v = Fv\cos\theta$, and the second one is the translational case, a force parallel to a velocity, which is what essential knowledge 3.5.A.4 describes. You can derive $\tau\omega$ in one line whenever you want it, but do not go looking for it on the sheet during the exam.

The units check out either way: a newton meter multiplied by a radian per second is a joule per second, a watt.

One useful consequence for constant-torque problems. If $\tau$ is constant and the system starts from rest, $\omega$ grows linearly with time, so the instantaneous power grows linearly too, and the average power over the spin-up is exactly half the final power. More on the underlying definitions at Topic 3.5.

What the equation sheet gives you, and what it does not

Everything on this page that is printed on the AP Physics 1 equation sheet, checked against the sheet rather than recalled:

• $W = \tau \Delta\theta$, in the rotational group.
• $\tau = r_{\perp} F = rF\sin\theta$, which is how you get $\tau$ in the first place.
• $K = \frac{1}{2} I \omega^2$, the energy the work changes.
• $\alpha_{sys} = \frac{\sum \tau}{I_{sys}} = \frac{\tau_{net}}{I_{sys}}$ and $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$, the two ingredients of the derivation above.
• $\Delta L = \tau \Delta t$, the neighbor that is easy to grab by mistake.
• $P_{avg} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}$ and $P_{inst} = F_{\parallel} v = Fv\cos\theta$, both in the translational group.

And what is not printed, which matters just as much:

• No rotational power equation. $P = \tau\omega$ has to be derived.
• No rotational work-energy theorem written as $W_{net} = \frac{1}{2} I \omega^2 - \frac{1}{2} I \omega_0^2$. The sheet's $\Delta K = \sum W_i$ line is stated with translational symbols.
• No shape specific rotational inertias. If a problem needs $I$ for a disk, a rod, or a sphere, the value comes from the problem statement, not from the sheet.

None of those absences means the idea is off the exam. The sheet is a table of equations; the CED is the list of what you are responsible for.

Traps, and where Topic 6.2 leads

Five to watch for, roughly in the order they tend to bite:

1. Leaving $\Delta\theta$ in degrees or in revolutions. Every answer comes out too large by 57.3 or too small by 6.28, and no unit in the working looks wrong.
2. Quoting a torque in joules. A newton meter and a joule have the same base units, but a torque is never an energy. Keep torques in N m and works in J.
3. Forgetting the condition in 6.2.A.1. A torque with no rotation does no work at all.
4. Reading the area off a torque versus time graph and calling it work. That area is angular impulse.
5. Using the rotational inertia about the wrong axis in the energy step. $I$ is defined about an axis, as Topic 5.4 sets out.

From here, Topic 6.1 is the energy this work changes, and Topic 6.3 is the other thing torque can change: apply it over a time interval instead of an angle and you get angular impulse. Topic 6.5 puts both energy terms into the same problem.

If the torque itself is the shaky part rather than the work, go back to how to calculate torque first, then the work-energy theorem for the straight-line statement this topic mirrors.