AP Physics C: Mechanics · Topic 5.6

Topic 5.6: Newton's Second Law in Rotational Form

Q: What is Newton's second law in rotational form?

It says a system's angular acceleration equals the net torque exerted on it divided by its rotational inertia. Essential knowledge 5.6.A.2 states it in words as well: the rate at which the angular velocity of a rigid system changes is directly proportional to the net torque exerted on the rigid system and is in the same direction, and the angular acceleration is inversely proportional to the rotational inertia. Both AP Physics equation sheets print it solved for the angular acceleration rather than as torque equals rotational inertia times angular acceleration, and both print the middle step showing the numerator as a sum of torques. The rotational inertia in the denominator must be the whole system's about the same axis you used for the torques. Essential knowledge 5.6.A.1 gives the qualitative version: angular velocity changes when the net torque is not equal to zero.

Q: Is Topic 5.6 different in AP Physics C and AP Physics 1?

The required content is not. All three essential-knowledge statements, 5.6.A.1, 5.6.A.2 and 5.6.A.3, are printed word for word in both course and exam descriptions with the same equation, and neither course prints a boundary statement for this topic. The suggested skills differ more here than for any other Unit 5 topic. AP Physics C: Mechanics lists five: create quantitative graphs including plotting data, derive a symbolic expression, predict new values using functional dependence, create experimental procedures, and justify a claim using evidence. AP Physics 1 lists four: create diagrams and schematics, derive a symbolic expression, compare quantities between scenarios, and justify a claim using evidence. What also differs is the input: in the calculus-based course the net torque can be a function of time, which turns the law into something you integrate, and the rotational inertia can be one you derived from an integral rather than one supplied on the exam.

Q: Why are the two string tensions different when a pulley has mass?

Because an unbalanced torque is what changes the pulley's angular velocity. Essential knowledge 5.6.A.1 says angular velocity changes when the net torque is not equal to zero, and 5.6.A.2 gives the angular acceleration as the net torque divided by the rotational inertia. If the string pulled equally hard on both sides of a pulley of equal radius, the two torques about the axle would cancel, the net torque would be zero, and the pulley could not speed up. Since it does speed up, the tensions must be unequal, with the larger one on the side that drives the rotation. This is sample multiple-choice question 10 in the AP Physics C: Mechanics course and exam description, aligned to skill 3.C and essential knowledge 5.6.A.1, and its keyed answer is that the tension on one side is smaller because an unbalanced clockwise torque is needed to accelerate the wheel clockwise. The single-tension shortcut is safe only for a massless pulley.

Q: How do I handle a torque that changes with time?

Integrate rather than substitute. Divide the net torque by the rotational inertia to get the angular acceleration as a function of time, then use the definition of angular acceleration from essential knowledge 5.1.A.3 backwards: the change in angular velocity is the integral of the angular acceleration over time. The constant-angular-acceleration equations cannot be used at all, because essential knowledge 5.1.A.4.i introduces them with the explicit condition of constant angular acceleration, and substituting one instantaneous value into them does not approximate the right answer. One useful consequence: the angular velocity is largest at the moment the net torque crosses zero, not at the moment the torque is largest, because the maximum of angular velocity is where its derivative vanishes. The AP Physics C: Mechanics equation sheet does not print the angular integral forms even though it prints both linear ones, so write them yourself.

Q: Which axis should I use for Newton's second law in rotational form?

Any axis will work, but the torques and the rotational inertia must both be taken about the same one, and that consistency is what most wrong answers get wrong. For a body rolling down a ramp, two choices are standard and the CED's own scoring guidelines credit both: the center of mass, where friction is the only force with a torque, or the contact point with the ramp, where friction and the normal force both have zero lever arm and gravity supplies the whole torque. The center-of-mass choice uses the body's ordinary rotational inertia; the contact-point choice needs the parallel axis theorem to shift that value. A useful default is the axis about which the body physically turns, since forces applied there contribute no torque. Whichever you pick, name it in the first line and use it for every term.

Q: Does the angular acceleration point in the same direction as the net torque?

Yes, and essential knowledge 5.6.A.2 says so directly: the rate at which the angular velocity changes is directly proportional to the net torque and is in the same direction. There is a scope subtlety worth knowing in AP Physics C: Mechanics. Topic 5.3 in that course prints no boundary statement and gives torque a genuine vector direction found with a right-hand rule, while the boundary statements under Topics 5.1 and 5.2 say the directions of angular displacement, angular velocity and angular acceleration as vectors will not be assessed on the exam, and limit descriptions of those directions to clockwise and counterclockwise with respect to a given axis. So you may be asked for a torque's direction as a vector, and the resulting angular acceleration is described as clockwise or counterclockwise. In a planar problem the two descriptions agree: a torque out of the page produces a counterclockwise speeding-up.

Q: Why does a hoop roll down a ramp more slowly than a solid sphere?

Because more of its mass sits far from the rotation axis, so a larger share of the gravitational potential energy released goes into rotation rather than translation. Working it from Newton's second law in rotational form gives the acceleration as the component of gravity along the ramp divided by one plus the ratio of the rotational inertia to mass times radius squared. That ratio is the only thing distinguishing shapes of equal mass and radius, and it is larger for a hoop than for a sphere. The AP Physics C: Mechanics course description uses exactly this setup in its sample free-response Question 4, and the example response argues it conceptually: because an object's rotational inertia is related to the average distance of the object's mass from its rotational axis, the hoop will move the slowest and the solid sphere the fastest. Mass and radius cancel out of the time entirely, so only the shape matters.

Unit 5: Torque and Rotational Dynamics10-15% of the multiple-choice section

A system's angular acceleration equals the net torque on it divided by its rotational inertia, and it points the same way as the net torque. In AP Physics C: Mechanics that torque can be a function of time, which turns the law into something you integrate rather than evaluate once.

AP Physics: Unit 5 (topics 5.6 Newton's Second Law in Rotational Form). AP Physics C: Mechanics Unit 5, Topic 5.6. One learning objective, 5.6.A: describe the conditions under which a system's angular velocity changes. Three essential-knowledge statements: 5.6.A.1 (angular velocity changes when the net torque exerted on the object or system is not equal to zero); 5.6.A.2 (the rate at which the angular velocity of a rigid system changes is directly proportional to the net torque exerted on the rigid system and is in the same direction, and the angular acceleration of the rigid system is inversely proportional to the rotational inertia of the rigid system, with relevant equation alpha_sys = sum tau / I_sys = tau_net / I_sys); and 5.6.A.3 (to fully describe a rotating rigid system, linear and rotational analyses may need to be performed independently). NO BOUNDARY STATEMENT; Topic 5.6 is one of two in Unit 5 with none, along with Topic 5.3. All three statements and the absence of a boundary statement are matched exactly in the AP Physics 1 CED, so the required content of this topic does not differ between the two courses. The suggested skills differ more than for any other Unit 5 topic: this course lists five, 1.B, 2.A, 2.D, 3.A and 3.C, while AP Physics 1 lists four, 1.A, 2.A, 2.C and 3.C, sharing 2.A and 3.C. Both equation sheets print the law solved for angular acceleration, in the same three-part form as 5.6.A.2, and neither prints it as tau_net = I alpha. The C: Mechanics sheet prints delta-L = integral tau dt where the AP Physics 1 sheet prints delta-L = tau delta-t, and the C: Mechanics rotational column prints no angular integral forms for omega or theta even though its translational column prints both linear ones. CED sample multiple-choice question 10 aligns to skill 3.C, LO 5.6.A and EK 5.6.A.1: two blocks connected by a string over a circular wheel of mass m rotating freely about its center, keyed answer D, T1 < T2 because an unbalanced clockwise torque is needed to accelerate the wheel clockwise. CED sample free-response Question 4 aligns to 5.6.A among other objectives; its scoring guidelines award a point for a multistep derivation using Newton's second law in rotational form to determine a translational acceleration and a point for a net torque and rotational inertia consistent with the chosen axis, whether the contact point or the center of mass, with a scoring note that a derivation correctly applying conservation of energy can earn those same points. Both Unit 5 sample instructional activities assigned to Topic 5.6 are experiments, matching skill 3.A: activity 4 determines a yo-yo's rotational inertia from its measured downward acceleration, mass and axle radius, and activity 5 asks students to plan a drop of two toilet paper rolls, one free and one unrolling, that land simultaneously.

What Topic 5.6 requires

Topic 5.6 carries one learning objective, 5.6.A: describe the conditions under which a system's angular velocity changes. It lists five suggested skills, as does Topic 5.4; the other four Unit 5 topics list four each. They 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.D (predict new values or factors of change of physical quantities using functional dependence between variables), 3.A (create experimental procedures that are appropriate for a given scientific question) and 3.C (justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws).

5.6.A.1 Angular velocity changes when the net torque exerted on the object or system is not equal to zero.

5.6.A.2 The rate at which the angular velocity of a rigid system changes is directly proportional to the net torque exerted on the rigid system and is in the same direction. The angular acceleration of the rigid system is inversely proportional to the rotational inertia of the rigid system. Its relevant equation is

\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}} = \frac{\tau_{\text{net}}}{I_{\text{sys}}}

5.6.A.3 To fully describe a rotating rigid system, linear and rotational analyses may need to be performed independently.

Topic 5.6 prints no boundary statement. It is one of two Unit 5 topics with none, along with Topic 5.3. AP Physics 1's Topic 5.6 prints none either.

Statement 5.6.A.1 is the converse of Topic 5.5's first law: a nonzero net torque is what changes an angular velocity. Statement 5.6.A.2 quantifies how much. Statement 5.6.A.3 is a procedural warning that costs more points than either of the others, and section six below is about it.

The equation as the sheets actually print it

Neither AP Physics equation sheet prints $\tau_{\text{net}} = I\alpha$ . Both print it solved for the angular acceleration, in the same three-part form as statement 5.6.A.2:

\alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}} = \frac{\tau_{\text{net}}}{I_{\text{sys}}}

That arrangement is deliberate and it mirrors the translational line on both sheets, which is also solved for the acceleration: $\vec{a}_{\text{sys}} = \frac{\sum\vec{F}}{m_{\text{sys}}} = \frac{\vec{F}_{\text{net}}}{m_{\text{sys}}}$ . Read left to right, both say the same thing: the net influence divided by the resistance gives the response. The subscript "sys" on all three symbols is a reminder that the rotational inertia in the denominator must be the whole system's about the axis you chose, not one part's.

The middle expression is worth keeping. Writing $\sum\tau$ rather than $\tau_{\text{net}}$ reminds you that the numerator is a signed sum you have to assemble, torque by torque, from Topic 5.3.

Here is what changes between the two courses on this line: nothing. Both sheets print it identically, and both course descriptions print statement 5.6.A.2 identically. The difference lives entirely in what you are allowed to put into each slot, which the next two sections cover.

One forward pointer, because it shows where this law goes. The C: Mechanics sheet prints $\Delta L = \int \tau\, dt$ in its rotational column, while the AP Physics 1 sheet prints $\Delta L = \tau\Delta t$ . That is the same difference between an integral and a product that separates the two courses everywhere else, and it belongs to Unit 6 rather than here. It is also the cleanest way to handle a time-varying torque, which worked example 1 does the long way, from 5.6.A.2 and the definition of angular acceleration.

Identical required content, and a different pair of exams behind it

All three essential-knowledge statements above, 5.6.A.1, 5.6.A.2 and 5.6.A.3, are printed word for word in the AP Physics 1 course and exam description under the same numbers, with the same equation. Neither course prints a boundary statement. On the statement of the law, the two frameworks are the same document, as they are at Topics 5.2 and 5.5.

Where they differ is in the suggested skills, and of the three Unit 5 topics whose required content matches, this is where the two lists diverge most: Topic 5.2's are identical, Topic 5.5's share two of four, and Topic 5.6's share two out of five and four.

Skill	AP Physics C: Mechanics 5.6	AP Physics 1 5.6
1.A create diagrams and schematics	not listed	listed
1.B create quantitative graphs, including plotting data	listed	not listed
2.A derive a symbolic expression	listed	listed
2.C compare quantities between scenarios	not listed	listed
2.D predict new values using functional dependence	listed	not listed
3.A create experimental procedures	listed	not listed
3.C justify a claim using evidence	listed	listed

Five skills against four, sharing two. The three that this course adds all point the same way. Skill 3.A, creating experimental procedures, together with 1.B, plotting data on a quantitative graph, says that Topic 5.6 is where you are expected to measure a rotational inertia rather than be handed one. Both of the CED's Unit 5 sample instructional activities for Topic 5.6 are experiments, and worked example 2 is one of them. Skill 2.D says you are expected to predict how the answer scales when a variable changes, rather than only to recompute it.

And then there is what goes into the equation. The numerator is a sum of torques, which Topic 5.3 defines in this course as cross products with directions, and which can be functions of time. The denominator is a rotational inertia, which Topic 5.4 lets you derive with an integral where AP Physics 1's boundary statement says extended-body values "will be provided within the exam". Same equation, different inputs.

Direction, and the one place Unit 5's boundary statements pull against each other

Statement 5.6.A.2 says the rate of change of angular velocity is directly proportional to the net torque "and is in the same direction". That clause is printed in both courses, and it means something different in each.

In AP Physics C: Mechanics, torque has a genuine direction. Topic 5.3 prints no boundary statement and gives you $\vec{\tau} = \vec{r} \times \vec{F}$ , a magnitude of $rF\sin\theta$ , a result perpendicular to both inputs, and a right-hand rule.

But Topics 5.1 and 5.2 both print this: "AP Physics C: Mechanics expects students to be able to mathematically manipulate the magnitudes of angular displacement, angular velocity, and angular acceleration using vector conventions. However, the directions of said vectors will not be assessed on the exam." And: "Descriptions of the directions of rotational kinematics quantities for a point or rigid body are limited to clockwise and counterclockwise with respect to a given axis of rotation."

So the CED hands you a right-hand rule for the cause and then tells you the effect will be described as clockwise or counterclockwise. That is not a contradiction, and it is worth reading exactly rather than smoothing over:

You may be asked for the direction of a torque as a vector, using the right-hand rule.
You will not be asked for the direction of the resulting angular acceleration as a vector along the axis.
In a planar problem the two questions collapse anyway. A torque out of the page produces an angular acceleration out of the page, which is a counterclockwise speeding-up as drawn, and that is the answer either framing wants.

In AP Physics 1 the same clause in 5.6.A.2 can only carry the weaker reading, because that course's Topic 5.3 boundary statement puts the direction of torque outside the course entirely. So a sentence printed identically in two documents means a little more in one of them.

The practical habit is the same in both courses: declare a positive sense of rotation at the top of the solution, keep every torque signed relative to it, and let the sign of $\alpha$ come out of the arithmetic.

When the torque changes with time

This is the situation the algebra-based course cannot pose, and it follows from combining 5.6.A.2 with the definition of angular acceleration from Topic 5.1.

If the net torque is a function of time, then so is the angular acceleration:

\alpha(t) = \frac{\tau_{\text{net}}(t)}{I}

and since $\alpha = d\omega/dt$ , the angular velocity is no longer something you evaluate but something you integrate:

\omega(t) = \omega_0 + \frac{1}{I}\int_0^t \tau_{\text{net}}(t')\, dt'

Three consequences to have ready.

The constant-angular-acceleration equations do not apply. Statement 5.1.A.4.i introduces them with the condition "for constant angular acceleration", and a time-varying torque violates it. Substituting an instantaneous value of $\alpha$ into them does not approximate the right answer.
The angular velocity peaks where the torque crosses zero, not where it peaks. The maximum of $\omega$ sits where $d\omega/dt = \alpha = 0$ , which by 5.6.A.2 is where the net torque vanishes. Worked example 1 has a torque that peaks at 6.0 s and an angular velocity that keeps rising until 12.0 s.
You write the integral yourself. The rotational column of the C: Mechanics sheet prints $\omega = d\theta/dt$ and $\alpha = d\omega/dt$ but no angular integral forms, even though the translational column prints both $\Delta x = \int v_x(t)\, dt$ and $\Delta v_x = \int a_x(t)\, dt$ . Statement 5.1.A.4 licenses the transfer by saying the angular quantities demonstrate the same mathematical relationships as the linear ones.

A torque can also vary with angle rather than with time, as it does for a rod swinging down under gravity, where the lever arm of the weight shrinks as the rod approaches vertical. That case is usually handled with energy rather than by integrating the second law, and the CED's own scoring guidelines allow the energy route, as the last section notes.

Linear and rotational analyses, performed independently

Statement 5.6.A.3 is one sentence: to fully describe a rotating rigid system, linear and rotational analyses may need to be performed independently. It is the sentence that turns most Topic 5.6 problems into simultaneous equations, and skipping it is the most expensive habit in this topic.

The standard shape of a coupled problem is three relations:

Newton's second law for each object that translates, one equation per object, from $\vec{a}_{\text{sys}} = \vec{F}_{\text{net}}/m_{\text{sys}}$ .
Newton's second law in rotational form for each body that rotates, from 5.6.A.2, with the torques taken about that body's axis.
A constraint linking them, which is almost always $a = r\alpha$ from statement 5.2.A.2 when a string does not slip or a body rolls without slipping.

Count unknowns and equations before solving. A yo-yo has $a$ , $\alpha$ and the string tension $T$ , three unknowns, and the three relations above close it exactly. That is worked example 2.

One thing the constraint is not: an assumption you get for free. "The string does not slip" and "rolls without slipping" are the phrases that license $a = r\alpha$ , and a problem that omits them is telling you something.

The most tested consequence of 5.6.A.3 is that a pulley with mass makes the tensions on its two sides unequal. If the string pulled equally hard on both sides, the two torques about the pulley's axis would cancel, the net torque would be zero, and by 5.6.A.1 the pulley's angular velocity could not change. Since it does change, the tensions must differ. The CED tests exactly this, as the assessment section below describes.

Traps

The rotational inertia must be about the same axis you took the torques about. Both are chosen together. Using a center-of-mass value with torques taken about a contact point, or the reverse, is the most common way to get a plausible wrong answer here.

Do not carry a single tension across a massive pulley. The one-tension shortcut is valid only when the pulley's rotational inertia is negligible.

The constant-angular-acceleration equations require a constant net torque and a constant rotational inertia. A time-varying torque breaks the first; a system whose mass distribution changes breaks the second.

Rotational inertia is not a fixed property of an object the way mass is. The same rod has $\frac{1}{3}ML^2$ about its end and $\frac{1}{12}ML^2$ about its center, a factor of four apart, so the same torque produces four times the angular acceleration about one axis as about the other.

Write $\sum\tau$ before you write $\tau_{\text{net}}$ . The sheet prints both for a reason: the numerator is assembled from individual torques, each with its own sign relative to your declared positive sense.

A force applied at the axis contributes nothing. Its position vector is zero. On a pulley mounted at its center, the axle force never appears in the torque equation, which is why choosing that axis is the right move.

Radians, always. The constraint $a = r\alpha$ and everything derived from it assume angular quantities in radians.

Do not skip the constraint equation. With three unknowns and two equations, the missing one is nearly always $a = r\alpha$ .

How Topic 5.6 is assessed

Topic 5.6 turns up in the CED's own sample materials four times: one of the two Unit 5 sample multiple-choice questions, the sample free-response set's Question 4, and both of the two Unit 5 sample instructional activities assigned to it. No other Unit 5 topic appears in as many; Topic 5.3 is next with three.

Sample multiple-choice question 10 is aligned to skill 3.C, learning objective 5.6.A and essential knowledge 5.6.A.1. Two blocks of mass $m_1$ and $m_2$ are connected by a thin string passing over a circular wheel of mass $m$ , attached to a vertical stand and rotating freely about its center. The string does not slip and exerts forces $T_1$ and $T_2$ on the blocks. Released from rest, the system undergoes an angular acceleration and the wheel rotates clockwise. The question asks which statement about the magnitudes of $T_1$ and $T_2$ is correct, and the keyed answer is D: $T_1 < T_2$ because an unbalanced clockwise torque is needed to accelerate the wheel clockwise. Three of the four options assert the tensions are equal or get the inequality backwards.

Sample free-response Question 4, the Qualitative/Quantitative Translation question worth 8 points, aligns to objectives 1.3.A, 3.4.B, 5.4.A, 5.6.A, 5.4.B and 6.1.A with skills 3.B, 3.C, 2.A and 2.D. A hollow sphere, a uniform solid sphere and a hoop of the same mass and radius roll without slipping down a ramp, and students rank the times, derive $t = \sqrt{2L(I + MR^2)/(MgR^2\sin\theta)}$ , and then predict what a liquid-filled shell does when the liquid inside does not rotate with it. Worked example 3 works that derivation.

Two things in its scoring guidelines are worth knowing. One point is awarded for a multistep derivation that uses Newton's second law in rotational form to determine a translational acceleration, and one for a net torque and rotational inertia consistent with the chosen axis, whether that axis is the contact point with the ramp or the sphere's center of mass. A scoring note then adds that a derivation that correctly applies conservation of energy can earn those same points. Two routes, both credited. The example response for the ranking says that because an object's rotational inertia is related to the average distance of the object's mass from its rotational axis, the hoop will move the slowest and the solid sphere the fastest.

Both Topic 5.6 sample instructional activities are experiments, matching skill 3.A. Activity 4 is a Desktop Experiment Task: let a yo-yo fall and unroll, use a meterstick and stopwatch to determine its downward acceleration, then measure its mass and the radius of its axle and use rotational dynamics to determine its rotational inertia. Worked example 2 is that experiment. Activity 5 is a Create a Plan task: drop two rolls of toilet paper, one falling freely and one unrolling as it falls, and determine the difference in heights so that both, released simultaneously from rest, land at the same time.

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

All three essential-knowledge statements in this topic are printed identically in your framework, and neither course prints a boundary statement here, so the law itself is common ground. If you are in AP Physics 1, read the [AP Physics 1 Topic 5.6 page](/ap-physics-1/unit-5-torque-and-rotational-dynamics/5-6-newtons-second-law-in-rotational-form) instead. It is written to your exam, whose suggested skills for this topic are 1.A, 2.A, 2.C and 3.C. This page is for AP Physics C: Mechanics students, whose skill list adds 1.B, 2.D and 3.A.

What this page adds is what a calculus-based course can put on either side of the equation: a net torque that is a function of time, so the law becomes something you integrate; a rotational inertia you derived from $I = \int r^2\, dm$ rather than were given; and torques that are cross products with directions you can defend. Worked example 1 has no AP Physics 1 equivalent.

AP Physics C: Mechanics is a calculus-based introductory college-level course equivalent to the first course in a college calculus-based physics sequence, and its only stated prerequisite is that students have taken or are concurrently taking calculus. The course also requires that 25 percent of instructional time be spent in hands-on laboratory work, which is why skill 3.A appears here.

From here, Unit 6 takes this law two ways: multiply by angular displacement and integrate for rotational work, or multiply by time and integrate for angular impulse and angular momentum. Within Unit 5, Topic 5.5 is the zero-net-torque case of this page. The how to calculate torque guide owns the procedure, the torque calculator checks arithmetic, and the Unit 5 hub lists all six topics.

A flywheel driven by a torque that changes with time

A flywheel with rotational inertia $I = 0.85\ \mathrm{kg \cdot m^2}$ starts from rest and is driven by a net torque $\tau_{\text{net}}(t) = (6.0\ \mathrm{N \cdot m/s})t - (0.50\ \mathrm{N \cdot m/s^2})t^2$ , with $t$ in seconds. Find (a) when the torque is largest and when it returns to zero, (b) an expression for the angular velocity, (c) the angular velocity at the two times from part (a), and (d) when the flywheel is spinning fastest.

Declare the convention: the positive sense of rotation is the one in which the driving torque acts, so every quantity below is positive while the drive is on.
(a) Factor the torque: $\tau_{\text{net}} = t(6.0 - 0.50t)$ , which is zero at $t = 0$ and at $t = 12.0$ s. Between those it is positive. Its maximum is at the midpoint of the two roots, $t = 6.0$ s, where $\tau_{\text{net}} = (6.0)(6.0) - (0.50)(36) = 36 - 18 = 18\ \mathrm{N \cdot m}$ .
(b) Apply 5.6.A.2 at every instant: $\alpha(t) = \tau_{\text{net}}(t)/I$ . Because the torque depends on $t$ , so does $\alpha$ , and the constant-angular-acceleration equations of 5.1.A.4.i do not apply anywhere in this problem.
Since $\alpha = d\omega/dt$ from 5.1.A.3, integrate: $\omega(t) = \dfrac{1}{I}\displaystyle\int_0^t \tau_{\text{net}}(t')\, dt' = \dfrac{1}{I}\left[3.0t^2 - \dfrac{t^3}{6}\right]$ , starting from rest.
(c) At $t = 6.0$ s: the bracket is $3.0(36) - 216/6 = 108 - 36 = 72\ \mathrm{N \cdot m \cdot s}$ , so $\omega = 72/0.85 = 84.7\ \mathrm{rad/s}$ .
At $t = 12.0$ s: the bracket is $3.0(144) - 1728/6 = 432 - 288 = 144\ \mathrm{N \cdot m \cdot s}$ , so $\omega = 144/0.85 = 169\ \mathrm{rad/s}$ .
(d) The flywheel spins fastest where $d\omega/dt = \alpha = 0$ , and by 5.6.A.2 that is where the net torque is zero. That happens at $t = 12.0$ s, not at $t = 6.0$ s where the torque peaks.
This is the point of the whole example. At $t = 6.0$ s the torque is at its largest and the flywheel is still speeding up as fast as it ever will; from 6.0 s to 12.0 s the torque falls but stays positive, so the flywheel keeps gaining angular velocity, more slowly. The angular velocity is still rising at the moment the torque is falling.
Check the size of the second half's contribution. The angular velocity doubles exactly between 6.0 s and 12.0 s, from 84.7 to 169 rad/s, so the shrinking half of the torque pulse delivers exactly as much as the growing half. That is a property of the symmetric parabola, and it is a good check that the integral was set up right.
Cross-check the integral by differentiating back: $\frac{d}{dt}\left[3.0t^2 - t^3/6\right] = 6.0t - 0.50t^2$ , which is the given torque. Then $\alpha =$ that over $I$ , as 5.6.A.2 requires.

(a) Largest at $t = 6.0$ s with $18\ \mathrm{N \cdot m}$ ; back to zero at $t = 12.0$ s. (b) $\omega(t) = \frac{1}{I}(3.0t^2 - t^3/6)$ . (c) $84.7\ \mathrm{rad/s}$ at 6.0 s and $169\ \mathrm{rad/s}$ at 12.0 s. (d) Fastest at $t = 12.0$ s, where the torque vanishes, not where it peaks.

The CED's yo-yo experiment, worked as a measurement of rotational inertia

This is sample instructional activity 4 from the AP Physics C: Mechanics Unit 5 guide. A yo-yo of mass $m = 0.115$ kg falls and unrolls from a string wrapped around its axle of radius $r = 4.5$ mm. A meterstick and stopwatch give its downward acceleration as $a = 0.42\ \mathrm{m/s^2}$ . Derive an expression for the yo-yo's rotational inertia in terms of the measured quantities, evaluate it, and check that the result is physically reasonable. Use $g = 9.8\ \mathrm{m/s^2}$ .

Statement 5.6.A.3 says linear and rotational analyses may need to be performed independently, so set up both. Take downward as positive for the translation and the corresponding unwinding sense as positive for the rotation.
Translation. Two forces act: gravity $mg$ down and the string tension $T$ up. So $mg - T = ma$ , giving $T = m(g - a)$ .
Rotation about the yo-yo's own center. Gravity acts at the center of mass and so exerts no torque about it; the tension acts at the axle surface, a perpendicular distance $r$ away, so $\tau_{\text{net}} = Tr$ . By 5.6.A.2, $\alpha = Tr/I$ .
The constraint. The string does not slip, so the string's linear acceleration equals the axle surface's tangential acceleration, and 5.2.A.2 gives $a = r\alpha$ , that is $\alpha = a/r$ . Three unknowns, $T$ , $\alpha$ and $I$ , and three relations.
Combine. $I = \dfrac{Tr}{\alpha} = \dfrac{m(g-a)r}{a/r} = mr^2\left(\dfrac{g}{a} - 1\right)$ . That is the expression the experiment is designed to fill in, and skill 2.A is what it exercises.
Evaluate the bracket first: $g/a = 9.8/0.42 = 23.333$ , so $g/a - 1 = 22.333$ .
Then $mr^2 = (0.115)(0.0045)^2 = (0.115)(2.025 \times 10^{-5}) = 2.329 \times 10^{-6}\ \mathrm{kg \cdot m^2}$ .
$I = (2.329 \times 10^{-6})(22.333) = 5.20 \times 10^{-5}\ \mathrm{kg \cdot m^2}$ .
Check it for plausibility, which is what skill 3.C asks for. If the yo-yo's body were a uniform disk of radius $R$ , then $I = \frac{1}{2}mR^2$ , so $R = \sqrt{2I/m} = \sqrt{2(5.20 \times 10^{-5})/0.115} = \sqrt{9.045 \times 10^{-4}} = 0.030$ m. A 3.0 cm radius is a reasonable yo-yo, so the measurement is consistent with the object.
Read the functional dependence, which is skill 2.D. The bracket $g/a - 1$ is large precisely because $a$ is small: a yo-yo that falls slowly has a large rotational inertia relative to $mr^2$ . In the limit $a \to g$ the bracket goes to zero and the object is in free fall with no string torque at all; in the limit $a \to 0$ the bracket diverges, which is the infinitely reluctant yo-yo.
Note the sensitivity, which is the real experimental lesson. $I$ depends on $r^2$ , and the axle radius is a few millimetres, so a 5% error in measuring $r$ becomes a 10% error in $I$ . Measuring the axle carefully matters more than measuring the mass.

$I = mr^2\left(\dfrac{g}{a} - 1\right) = 5.20 \times 10^{-5}\ \mathrm{kg \cdot m^2}$ . The value corresponds to a uniform disk of radius about 3.0 cm, which is consistent with a real yo-yo. Because $I$ scales as the axle radius squared, that radius is the measurement to take most carefully.

The CED's rolling-shapes derivation, from Newton's second law in rotational form

This is the CED's sample free-response Question 4. A body of mass $M$ and radius $R$ , with rotational inertia $I$ about its center, is released from rest and rolls without slipping down a ramp of length $L$ inclined at angle $\theta$ . Derive the time to reach the bottom, show it matches the CED's stated result $t = \sqrt{2L(I + MR^2)/(MgR^2\sin\theta)}$ , and use it to rank a hoop, a hollow sphere and a uniform solid sphere of equal mass and radius for $L = 2.0$ m and $\theta = 25^\circ$ .

Choose the axis at the body's center of mass and set up both analyses, per 5.6.A.3. Take down the ramp as positive.
Translation. Three forces act: the component of gravity down the ramp, $Mg\sin\theta$ ; the normal force, perpendicular to the surface; and static friction $f$ up the ramp. So $Mg\sin\theta - f = Ma$ .
Rotation about the center of mass. Gravity acts at the center and the normal force's line of action passes through it, so neither exerts a torque about that axis. Friction is the only torque, with lever arm $R$ . By 5.6.A.2, $\alpha = fR/I$ , so $f = I\alpha/R$ .
Constraint. Rolling without slipping gives $a = R\alpha$ , so $\alpha = a/R$ and $f = Ia/R^2$ .
Substitute into the translational equation: $Mg\sin\theta = Ma + \dfrac{Ia}{R^2} = a\left(M + \dfrac{I}{R^2}\right)$ , so $a = \dfrac{Mg\sin\theta}{M + I/R^2} = \dfrac{MgR^2\sin\theta}{MR^2 + I}$ .
The acceleration is constant, so $L = \frac{1}{2}at^2$ applies, giving $t = \sqrt{2L/a} = \sqrt{\dfrac{2L(MR^2 + I)}{MgR^2\sin\theta}}$ . That is the CED's stated expression, term for term.
The scoring guidelines award a point for a multistep derivation that uses Newton's second law in rotational form to determine a translational acceleration, and a point for a net torque and rotational inertia consistent with the axis chosen, which may be the contact point or the center of mass. A scoring note adds that a derivation correctly applying conservation of energy earns the same points, so the energy route is equally credited.
Simplify for the ranking by writing $I = \beta MR^2$ . Then $a = \dfrac{g\sin\theta}{1 + \beta}$ and $t = \sqrt{\dfrac{2L(1+\beta)}{g\sin\theta}}$ . Only $\beta$ distinguishes the three shapes, and $t$ rises with $\beta$ .
Numbers, with $\sin 25^\circ = 0.42262$ so $g\sin\theta = 4.1417\ \mathrm{m/s^2}$ . The coefficients $\beta = 2/5$ for a uniform solid sphere, $2/3$ for a hollow sphere and $1$ for a hoop are standard results, not printed in either CED and not on either equation sheet; a solid sphere is not on the Topic 5.4 boundary statement's list of expected derivations, which is why the CED's own question keeps $I$ symbolic.
Solid sphere: $a = 4.1417/1.4 = 2.958\ \mathrm{m/s^2}$ and $t = \sqrt{4.0/2.958} = 1.16$ s. Hollow sphere: $a = 4.1417/(5/3) = 2.485\ \mathrm{m/s^2}$ and $t = \sqrt{4.0/2.485} = 1.27$ s. Hoop: $a = 4.1417/2 = 2.071\ \mathrm{m/s^2}$ and $t = \sqrt{4.0/2.071} = 1.39$ s.
So the hoop takes the longest and the solid sphere the shortest, which is what the CED's example response says: because an object's rotational inertia is related to the average distance of its mass from the rotational axis, the hoop moves the slowest and the solid sphere the fastest. Note that mass and radius cancel out of $t$ entirely; only the shape matters.
Part C of the CED's question asks about a thin shell filled with liquid that does not rotate with it. Almost none of the mass rotates, so $I$ is nearly zero, $\beta \to 0$ , and $t \to \sqrt{2L/(g\sin\theta)} = \sqrt{4.0/4.1417} = 0.98$ s. It beats the solid sphere, because gravitational potential energy that is not spent on rotation is available for translation.

$t = \sqrt{2L(I + MR^2)/(MgR^2\sin\theta)}$ , equivalently $\sqrt{2L(1+\beta)/(g\sin\theta)}$ with $I = \beta MR^2$ . For $L = 2.0$ m at $25^\circ$ : solid sphere 1.16 s, hollow sphere 1.27 s, hoop 1.39 s, and a liquid-filled shell whose contents do not rotate 0.98 s. Mass and radius cancel; only the shape matters.

Frequently asked questions

What is Newton's second law in rotational form?

It says a system's angular acceleration equals the net torque exerted on it divided by its rotational inertia. Essential knowledge 5.6.A.2 states it in words as well: the rate at which the angular velocity of a rigid system changes is directly proportional to the net torque exerted on the rigid system and is in the same direction, and the angular acceleration is inversely proportional to the rotational inertia. Both AP Physics equation sheets print it solved for the angular acceleration rather than as torque equals rotational inertia times angular acceleration, and both print the middle step showing the numerator as a sum of torques. The rotational inertia in the denominator must be the whole system's about the same axis you used for the torques. Essential knowledge 5.6.A.1 gives the qualitative version: angular velocity changes when the net torque is not equal to zero.

Is Topic 5.6 different in AP Physics C and AP Physics 1?

The required content is not. All three essential-knowledge statements, 5.6.A.1, 5.6.A.2 and 5.6.A.3, are printed word for word in both course and exam descriptions with the same equation, and neither course prints a boundary statement for this topic. The suggested skills differ more here than for any other Unit 5 topic. AP Physics C: Mechanics lists five: create quantitative graphs including plotting data, derive a symbolic expression, predict new values using functional dependence, create experimental procedures, and justify a claim using evidence. AP Physics 1 lists four: create diagrams and schematics, derive a symbolic expression, compare quantities between scenarios, and justify a claim using evidence. What also differs is the input: in the calculus-based course the net torque can be a function of time, which turns the law into something you integrate, and the rotational inertia can be one you derived from an integral rather than one supplied on the exam.

Why are the two string tensions different when a pulley has mass?

Because an unbalanced torque is what changes the pulley's angular velocity. Essential knowledge 5.6.A.1 says angular velocity changes when the net torque is not equal to zero, and 5.6.A.2 gives the angular acceleration as the net torque divided by the rotational inertia. If the string pulled equally hard on both sides of a pulley of equal radius, the two torques about the axle would cancel, the net torque would be zero, and the pulley could not speed up. Since it does speed up, the tensions must be unequal, with the larger one on the side that drives the rotation. This is sample multiple-choice question 10 in the AP Physics C: Mechanics course and exam description, aligned to skill 3.C and essential knowledge 5.6.A.1, and its keyed answer is that the tension on one side is smaller because an unbalanced clockwise torque is needed to accelerate the wheel clockwise. The single-tension shortcut is safe only for a massless pulley.

How do I handle a torque that changes with time?

Integrate rather than substitute. Divide the net torque by the rotational inertia to get the angular acceleration as a function of time, then use the definition of angular acceleration from essential knowledge 5.1.A.3 backwards: the change in angular velocity is the integral of the angular acceleration over time. The constant-angular-acceleration equations cannot be used at all, because essential knowledge 5.1.A.4.i introduces them with the explicit condition of constant angular acceleration, and substituting one instantaneous value into them does not approximate the right answer. One useful consequence: the angular velocity is largest at the moment the net torque crosses zero, not at the moment the torque is largest, because the maximum of angular velocity is where its derivative vanishes. The AP Physics C: Mechanics equation sheet does not print the angular integral forms even though it prints both linear ones, so write them yourself.

Which axis should I use for Newton's second law in rotational form?

Any axis will work, but the torques and the rotational inertia must both be taken about the same one, and that consistency is what most wrong answers get wrong. For a body rolling down a ramp, two choices are standard and the CED's own scoring guidelines credit both: the center of mass, where friction is the only force with a torque, or the contact point with the ramp, where friction and the normal force both have zero lever arm and gravity supplies the whole torque. The center-of-mass choice uses the body's ordinary rotational inertia; the contact-point choice needs the parallel axis theorem to shift that value. A useful default is the axis about which the body physically turns, since forces applied there contribute no torque. Whichever you pick, name it in the first line and use it for every term.

Does the angular acceleration point in the same direction as the net torque?

Yes, and essential knowledge 5.6.A.2 says so directly: the rate at which the angular velocity changes is directly proportional to the net torque and is in the same direction. There is a scope subtlety worth knowing in AP Physics C: Mechanics. Topic 5.3 in that course prints no boundary statement and gives torque a genuine vector direction found with a right-hand rule, while the boundary statements under Topics 5.1 and 5.2 say the directions of angular displacement, angular velocity and angular acceleration as vectors will not be assessed on the exam, and limit descriptions of those directions to clockwise and counterclockwise with respect to a given axis. So you may be asked for a torque's direction as a vector, and the resulting angular acceleration is described as clockwise or counterclockwise. In a planar problem the two descriptions agree: a torque out of the page produces a counterclockwise speeding-up.

Why does a hoop roll down a ramp more slowly than a solid sphere?

Because more of its mass sits far from the rotation axis, so a larger share of the gravitational potential energy released goes into rotation rather than translation. Working it from Newton's second law in rotational form gives the acceleration as the component of gravity along the ramp divided by one plus the ratio of the rotational inertia to mass times radius squared. That ratio is the only thing distinguishing shapes of equal mass and radius, and it is larger for a hoop than for a sphere. The AP Physics C: Mechanics course description uses exactly this setup in its sample free-response Question 4, and the example response argues it conceptually: because an object's rotational inertia is related to the average distance of the object's mass from its rotational axis, the hoop will move the slowest and the solid sphere the fastest. Mass and radius cancel out of the time entirely, so only the shape matters.