AP Physics C: Mechanics · Topic 5.4

Topic 5.4: Rotational Inertia

Q: What is the difference between rotational inertia in AP Physics C and AP Physics 1?

One essential-knowledge statement. AP Physics C: Mechanics prints 5.4.A.4, which says that for a solid that can be considered as a collection of differential masses, the solid's rotational inertia can be calculated as the integral of the squared perpendicular distance from each mass element to the axis, against the mass element. AP Physics 1's Topic 5.4 has no statement with that number. The other five statements, covering what rotational inertia measures, the single-object form, the sum over a collection, the fact that the center-of-mass axis minimizes the value, and the parallel axis theorem, are printed word for word in both. The boundary statements then diverge sharply: AP Physics 1 caps the calculation at systems of five or fewer objects in a two-dimensional configuration and says extended-body values will be provided on the exam, while AP Physics C: Mechanics names a specific list of bodies whose derivations you are expected to perform.

Q: Which rotational inertia derivations does AP Physics C Mechanics expect?

The Topic 5.4 boundary statement names them exactly. It says AP Physics C: Mechanics only expects students to use calculus in the derivations of the rotational inertia of thin rods of uniform or nonuniform density about an arbitrary axis perpendicular to the rod, as well as derivations of the rotational inertia of a thin cylindrical shell, disk, or rigid bodies that can be considered to be made up of coaxial rings or shells about an axis that passes through their centers, giving annular rings as the example. A second paragraph adds that students should have a qualitative understanding of the factors that affect rotational inertia, for example how rotational inertia is greater when mass is farther from the axis of rotation, which is why a hoop has more rotational inertia than a solid puck of the same mass and radius. A solid sphere is not on the list, and neither is a rod about an axis along its own length.

Q: Is the parallel axis theorem only on AP Physics C?

No. The parallel axis theorem is essential knowledge 5.4.B.2 in both AP Physics C: Mechanics and AP Physics 1, printed with the same wording and the same equation relating the rotational inertia about any axis to the center-of-mass value plus the mass times the square of the distance between the two parallel axes. It is also printed on both equation sheets. What differs is what you can put into it: in the calculus-based course the center-of-mass value may be one you derived yourself with the integral form, while in the algebra-based course extended-body values are supplied on the exam. Two conditions apply in both courses and are often dropped: the two axes must be parallel, and one of them must pass through the center of mass. Hopping between two arbitrary parallel axes takes two applications, routed through the center of mass.

Q: Is moment of inertia the same as rotational inertia?

They are two names for the same quantity, but only one of them appears in the AP frameworks. The phrase moment of inertia appears zero times in the AP Physics C: Mechanics course and exam description and zero times in the AP Physics 1 one. The phrase rotational inertia appears 39 times in the C: Mechanics document and 28 times in the AP Physics 1 document, and it is the term used in the learning objectives, the essential-knowledge statements, the boundary statements and the equation sheet's symbol definitions, where the letter I is defined as rotational inertia. Many textbooks and most of the internet say moment of inertia. If you are searching for AP-aligned material or reading an exam question, expect rotational inertia, and use that term in a free-response answer.

Q: How do I set up the integral for rotational inertia?

Three steps. First, fix the axis and write the perpendicular distance from a general mass element to that axis in terms of one variable; essential knowledge 5.4.A.4 specifies that this distance is measured perpendicular to the axis. For a rod along the x axis rotating about a perpendicular axis at the origin, that distance is simply x. Second, write the mass element in terms of the same variable: for a rod it is the linear mass density times the length element, and the equation sheet prints the linear mass density as the derivative of mass with respect to length. Third, and most important, choose the element so that all of it lies at a single distance from the axis. That is why a disk is integrated as a stack of concentric rings, each of area equal to its circumference times its width, rather than as strips. The boundary statement's phrase about bodies made of coaxial rings or shells is naming precisely this technique.

Q: Why does a hoop have more rotational inertia than a disk of the same mass?

Because rotational inertia weights every piece of mass by the square of its distance from the axis, and a hoop puts all of its mass at the largest possible distance while a disk spreads mass all the way in to the center. Essential knowledge 5.4.A.1 says rotational inertia is related to both the mass of the system and the distribution of that mass relative to the axis of rotation, and the Topic 5.4 boundary statement gives this exact comparison as its example of the qualitative understanding expected, phrased as a hoop having more rotational inertia than a solid puck of the same mass and radius. Integrating both gives the hoop the full mass times radius squared and the disk one half of that, so the ratio is exactly 2 for any mass and any radius. The mass that matters most is the mass farthest out.

Q: Is the rotational inertia integral on the AP Physics C Mechanics equation sheet?

Yes. The rotational column of the mechanics table prints the integral of the squared perpendicular distance against the mass element, directly below the sum form written as the total rotational inertia equalling the sum of the individual rotational inertias equalling the sum of each mass times its distance squared, and directly above the parallel axis theorem. The AP Physics 1 sheet prints the sum form and the parallel axis theorem but not the integral. Two supporting equations for continuous bodies are on the C: Mechanics sheet only, both in its translational column: the linear mass density as the derivative of mass with respect to length, and the center of mass as the integral of position against mass element divided by total mass. One required equation is on neither sheet: the single-object form giving rotational inertia as mass times the square of the perpendicular distance, which is essential knowledge 5.4.A.2 in both courses.

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

AP Physics C: Mechanics adds one essential-knowledge statement AP Physics 1 does not have: for a solid treated as a collection of differential masses, rotational inertia is an integral of the squared distance to the axis. The sum over point masses and the parallel axis theorem are in both courses.

AP Physics: Unit 5 (topics 5.4 Rotational Inertia). AP Physics C: Mechanics Unit 5, Topic 5.4. Two learning objectives. 5.4.A, describe the rotational inertia of a rigid system relative to a given axis of rotation, supported by 5.4.A.1 (rotational inertia measures a rigid system's resistance to changes in rotation and is related to the mass of the system and the distribution of that mass relative to the axis of rotation), 5.4.A.2 (I = m r^2 for an object rotating a perpendicular distance r from an axis), 5.4.A.3 (I_tot = sum of I_i = sum of m_i r_i^2) and 5.4.A.4 (for a solid that can be considered as a collection of differential masses dm, I = integral of r^2 dm, where r is the perpendicular distance from dm to the axis of rotation). 5.4.B, describe the rotational inertia of a rigid system rotating about an axis that does not pass through the system's center of mass, supported by 5.4.B.1 (a rigid system's rotational inertia in a given plane is at a minimum when the rotational axis passes through the system's center of mass) and 5.4.B.2 (the parallel axis theorem, I' = I_cm + M d^2). Two-paragraph boundary statement: 'AP Physics C: Mechanics only expects students to use calculus in the derivations of the rotational inertia of thin rods of uniform or nonuniform density about an arbitrary axis perpendicular to the rod, as well as derivations of the rotational inertia of a thin cylindrical shell, disk, or rigid bodies that can be considered to be made up of coaxial rings or shells about an axis that passes through their centers (e.g., annular rings).' and 'Students should have a qualitative understanding of the factors that affect rotational inertia; for example, how rotational inertia is greater when mass is farther from the axis of rotation, which is why a hoop has more rotational inertia than a solid puck of the same mass and radius.' AP Physics 1's Topic 5.4 prints 5.4.A.1, 5.4.A.2, 5.4.A.3, 5.4.B.1 and 5.4.B.2 identically and has no 5.4.A.4; its boundary statement caps calculation at systems of five or fewer objects in a two-dimensional configuration, says extended-body rotational inertias will be provided within the exam, and repeats the qualitative paragraph with 'solid disk' in place of 'solid puck'. Suggested skills 1.B, 2.A, 2.C, 3.A, 3.B; AP Physics 1's Topic 5.4 suggests 1.B, 2.B, 2.C, 3.A, 3.B, differing only in 2.A against 2.B. Printed on the C: Mechanics sheet: the sum form as I_tot = sum I_i = sum m_i r_i^2, the integral form, the parallel axis theorem, plus lambda = d/d-ell of m(ell) and r_cm = integral r dm over integral dm in the translational column. The AP Physics 1 sheet prints the sum form as I = sum m_i r_i^2 and the parallel axis theorem, and not the integral form, the linear mass density or the center-of-mass integral. I = m r^2 is printed on neither sheet. Neither sheet prints a table of rotational inertias for common shapes. The phrase 'moment of inertia' appears zero times in both mechanics CEDs; 'rotational inertia' appears 39 times in the C: Mechanics CED and 28 times in the AP Physics 1 CED. CED sample free-response Question 4 aligns to 5.4.A and 5.4.B among other objectives and keeps I symbolic throughout.

What Topic 5.4 requires

Topic 5.4 carries two learning objectives, one of only two topics in Unit 5 that do. Its 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/or locations within a single scenario), 3.A (create experimental procedures that are appropriate for a given scientific question) and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim).

5.4.A: Describe the rotational inertia of a rigid system relative to a given axis of rotation.

5.4.A.1 Rotational inertia measures a rigid system's resistance to changes in rotation and is related to the mass of the system and the distribution of that mass relative to the axis of rotation.
5.4.A.2 The rotational inertia of an object rotating a perpendicular distance $r$ from an axis is described by $I = mr^2$ .
5.4.A.3 The total rotational inertia of a collection of objects about an axis is the sum of the rotational inertias of each object about that axis:

I_{\text{tot}} = \sum I_i = \sum m_i r_i^2

5.4.A.4 For a solid that can be considered as a collection of differential masses, $dm$ , the solid's rotational inertia can be calculated using

I = \int r^2\, dm

where $r$ is the perpendicular distance from $dm$ to the axis of rotation.

5.4.B: Describe the rotational inertia of a rigid system rotating about an axis that does not pass through the system's center of mass.

5.4.B.1 A rigid system's rotational inertia in a given plane is at a minimum when the rotational axis passes through the system's center of mass.
5.4.B.2 The parallel axis theorem uses the following equation to relate the rotational inertia of a rigid system about any axis that is parallel to an axis through its center of mass:

I' = I_{\text{cm}} + Md^2

One note on vocabulary before anything else, because it affects what you should be searching for. The phrase "moment of inertia" appears zero times in the AP Physics C: Mechanics course and exam description and zero times in the AP Physics 1 one. The phrase "rotational inertia" appears 39 times in the C: Mechanics document and 28 times in the AP Physics 1 document. They mean the same quantity, and your textbook may well say moment of inertia, but every learning objective, essential-knowledge statement, boundary statement and equation-sheet symbol definition in both courses uses the framework's term.

The one statement AP Physics 1 does not have

AP Physics 1 has a Topic 5.4 with the same title, the same two learning objectives, and five of this course's six essential-knowledge statements printed word for word: 5.4.A.1, 5.4.A.2, 5.4.A.3, 5.4.B.1 and 5.4.B.2 are all identical in the two documents, parallel axis theorem included.

AP Physics C: Mechanics adds 5.4.A.4, and AP Physics 1's Topic 5.4 has no statement with that number. That single addition is the whole content difference, and it changes what a problem is allowed to look like.

I_{\text{tot}} = \sum m_i r_i^2 \qquad \text{against} \qquad I = \int r^2\, dm

These are not two notations for one operation. The sum can only describe mass that sits at a countable list of distances from the axis: five beads on a frame, four blocks at the corners of a square. The integral describes mass spread continuously along a rod or through a disk, where there is no next mass element to add. Every extended body is the second case, and the sum simply cannot reach it.

What AP Physics 1 does instead is stated in its own boundary statement, quoted in the next section: it caps the sum at five objects and says the rotational inertias of extended rigid systems will be provided within the exam. So both courses ask you to use $I$ for a hoop, a disk and a rod. Only one of them asks you to produce those values.

That difference shows up in the suggested skills too, and by exactly one entry. Both courses list 1.B, 2.C, 3.A and 3.B for Topic 5.4. The fifth is 2.A, derive a symbolic expression, in AP Physics C: Mechanics, and 2.B, calculate or estimate an unknown quantity, in AP Physics 1. Derive against calculate, one skill code apart, in the one topic where the frameworks differ by one statement.

The boundary statement, and exactly which derivations you owe

This is the boundary statement that decides how much integration you actually do. Both paragraphs, whole:

"AP Physics C: Mechanics only expects students to use calculus in the derivations of the rotational inertia of thin rods of uniform or nonuniform density about an arbitrary axis perpendicular to the rod, as well as derivations of the rotational inertia of a thin cylindrical shell, disk, or rigid bodies that can be considered to be made up of coaxial rings or shells about an axis that passes through their centers (e.g., annular rings)."

"Students should have a qualitative understanding of the factors that affect rotational inertia; for example, how rotational inertia is greater when mass is farther from the axis of rotation, which is why a hoop has more rotational inertia than a solid puck of the same mass and radius."

The first paragraph is a finite list. Read it as one:

Thin rods, of uniform or nonuniform density, about an arbitrary axis perpendicular to the rod. Not just the end and not just the center: any perpendicular axis, and the density may vary along the length.
A thin cylindrical shell.
A disk.
Rigid bodies that can be considered to be made up of coaxial rings or shells about an axis through their centers, with annular rings given as the example.

A solid sphere is not on that list. Neither is a rectangular plate, nor a rod about an axis along its own length. Worked examples 1 and 2 below work two of the four listed cases.

The second paragraph is not decoration and it is assessed. It says a hoop has more rotational inertia than a solid puck of the same mass and radius, because its mass is farther from the axis. Worked example 2 shows the factor is exactly 2.

Now the AP Physics 1 boundary statement for the same topic, also whole, because the contrast is the point:

"AP Physics 1 only expects students to calculate the rotational inertia for systems of five or fewer objects arranged in a two-dimensional configuration."

"Students do not need to know the rotational inertia of extended rigid systems, as these will be provided within the exam. Students should have a qualitative understanding of the factors that affect rotational inertia; for example, how rotational inertia is greater when mass is farther from the axis of rotation, which is why a hoop has more rotational inertia than a solid disk of the same mass and radius."

The qualitative paragraph is nearly identical in the two courses. The one wording change is that AP Physics 1 says "solid disk" where AP Physics C: Mechanics says "solid puck". Everything before it is different: five point objects against a named list of integrals.

Three forms, and how to tell which one a body calls for

The CED hands you three expressions and expects you to pick. The test is what the mass distribution looks like.

Form	Statement	Use it when
$I = mr^2$	5.4.A.2	one object, all of it at one perpendicular distance $r$ from the axis
$I_{\text{tot}} = \sum I_i = \sum m_i r_i^2$	5.4.A.3	several objects, each at its own distance
$I = \int r^2\, dm$	5.4.A.4	mass spread continuously, so there is no list of distances

Setting up the integral is three decisions, and the third is where most of the errors live.

Choose the axis and write $r$ as a function of position. Statement 5.4.A.4 says $r$ is the perpendicular distance from $dm$ to the axis of rotation. For a rod on the $x$ axis rotating about a perpendicular axis at the origin, $r = x$ . For a disk rotating about its central axis, every point of a ring of radius $r$ is at that same distance, so $r$ is the ring's radius.
Write $dm$ in terms of that same variable. For a rod, $dm = \lambda\, dx$ , and the equation sheet prints $\lambda = \frac{d}{d\ell}m(\ell)$ for exactly this purpose. For a disk built from rings, $dm = \sigma(2\pi r\, dr)$ , since a ring of radius $r$ and width $dr$ has area $2\pi r\, dr$ .
Choose the element so that every part of it sits at one $r$ . This is the step that makes a disk easy and a sphere hard. Slice the disk into rings and each ring is entirely at one distance, so $r^2$ comes out of the element cleanly. Slice it into strips instead and different parts of one strip sit at different distances, and the integral becomes two-dimensional. The boundary statement's phrase "rigid bodies that can be considered to be made up of coaxial rings or shells" is naming this trick.

A useful sanity check on any result: the answer must be a pure number times the mass times a length squared. Write that length as $R_{\max}$ , the greatest distance from the axis to any part of the body, and the pure number cannot exceed 1, because no mass element is farther out than that. A hoop about its central axis gives exactly 1, the extreme case where every mass element sits at $R_{\max}$ . A disk gives $\frac{1}{2}$ , and a rod about a perpendicular axis through its end gives $\frac{1}{3}$ of $ML^2$ with $R_{\max} = L$ . The same rod about its center gives $\frac{1}{12}ML^2$ , which is $\frac{1}{3}$ of $M R_{\max}^2$ because $R_{\max}$ is only $L/2$ there. Compare like with like before you use the check.

The parallel axis theorem is in both courses, and it runs backwards

Statement 5.4.B.2 gives the parallel axis theorem as $I' = I_{\text{cm}} + Md^2$ , where $d$ is the distance between the two parallel axes. It is printed in the AP Physics 1 course and exam description under the same number, and it is printed on the AP Physics 1 equation sheet. It is not something the calculus-based course adds. If a study resource tells you the parallel axis theorem is Physics C material, check the AP Physics 1 sheet: it is there, between $I = \sum m_ir_i^2$ and $\alpha_{\text{sys}} = \sum\tau/I_{\text{sys}}$ .

What differs is what you can feed it. In this course, $I_{\text{cm}}$ can be something you derived with 5.4.A.4 a moment earlier.

Statement 5.4.B.1 says a rigid system's rotational inertia in a given plane is at a minimum when the rotational axis passes through the system's center of mass. Notice that 5.4.B.1 and 5.4.B.2 are the same fact stated twice, once in words and once in symbols: $Md^2$ is never negative, so moving the axis off the center of mass can only increase $I$ . Use 5.4.B.1 as a check on every 5.4.B.2 calculation. If your answer for an off-center axis comes out smaller than the center-of-mass value, something is wrong.

Three ways the theorem earns its place:

Forwards, the usual direction: you know $I_{\text{cm}}$ and want the value about a parallel axis a distance $d$ away. Add $Md^2$ .
Backwards: you integrated about a convenient axis, such as a rod's end, and want the center-of-mass value. Subtract $Md^2$ rather than integrating again. Worked example 1 does this and then checks it against a direct integration.
In a composite body: each part contributes its own value about the common axis, which usually means each part's own $I_{\text{cm}}$ plus its own $m d^2$ . Worked example 3 does this.

Two restrictions the statement carries and people drop. The two axes must be parallel, and one of them must pass through the center of mass. You cannot use it to hop between two arbitrary axes in one step; go through the center of mass in two.

What each sheet prints

Equation	Statement	C: Mechanics sheet	AP Physics 1 sheet
$I = mr^2$	5.4.A.2	no	no
the sum form	5.4.A.3	yes, as $I_{\text{tot}} = \sum I_i = \sum m_i r_i^2$	yes, as $I = \sum m_i r_i^2$
$I = \int r^2\, dm$	5.4.A.4	yes	no, and there is no such statement
$I' = I_{\text{cm}} + Md^2$	5.4.B.2	yes	yes
$\lambda = \frac{d}{d\ell}m(\ell)$	Unit 1	yes	no
$\vec{r}_{\text{cm}} = \frac{\int \vec{r}\, dm}{\int dm}$	Unit 2	yes	no

Four observations, each checked line by line rather than recalled.

The single-object form $I = mr^2$ is not printed on either sheet, even though it is required content in both courses. It is the one-term case of the sum, which is presumably why.

The two sheets write the sum form differently. The C: Mechanics sheet writes the full three-part chain, $I_{\text{tot}} = \sum I_i = \sum m_i r_i^2$ , matching statement 5.4.A.3 exactly. The AP Physics 1 sheet writes only $I = \sum m_i r_i^2$ . The physics is the same; the C version makes the middle step, adding each part's own rotational inertia, explicit.

No table of rotational inertias for common shapes is printed on either sheet. AP Physics 1's boundary statement says those values will be provided within the exam. This course's boundary statement instead names the shapes whose derivations you are expected to be able to perform.

The two supporting tools for a continuous body are on the C sheet only. The linear mass density relation and the center-of-mass integral both sit in the translational column of the C: Mechanics mechanics table, and neither appears on the AP Physics 1 sheet. They are what make a nonuniform rod tractable. The C sheet's CALCULUS box supplies the power rule and the integral of $x^n$ on top of that.

Traps

Rotational inertia belongs to a body and an axis together. There is no such thing as the rotational inertia of a rod. There is the rotational inertia of that rod about a perpendicular axis through its end, which is $\frac{1}{3}ML^2$ , and about a perpendicular axis through its center, which is $\frac{1}{12}ML^2$ , a factor of four apart. Name the axis first, every time.

$r$ is the perpendicular distance to the axis, not the distance to the center of mass and not a coordinate. Statement 5.4.A.4 says so in its own last clause. For a rod rotating about an axis running along its own length, every $r$ is zero and so is $I$ , which is why that case is not on the boundary statement's list: there is nothing to integrate.

Do not square the sum. $\sum m_i r_i^2$ squares each distance and then adds. It is not $(\sum m_i r_i)^2$ and it is not $M r_{\text{cm}}^2$ .

Choose an element that lies at one distance. A disk sliced into rings integrates in one line; a disk sliced into strips does not.

Check the parallel axis theorem's direction. Adding $Md^2$ moves you away from the center of mass; subtracting it moves you toward it. Statement 5.4.B.1 says the center-of-mass value is the minimum, so if your off-center answer is the smaller of the two, you have the sign backwards.

Both axes must be parallel and one must pass through the center of mass. Two arbitrary parallel axes need two applications, via the center of mass.

A uniform density is an assumption, not a default. The boundary statement puts nonuniform rods explicitly inside the course, so $\lambda$ may be a function of position, and then the mass, the center of mass and the rotational inertia all need their own integrals.

Say rotational inertia. The framework never says moment of inertia, in either course, and the term you use should match the term the question uses.

How Topic 5.4 is assessed

Skill 3.A, creating experimental procedures appropriate for a given scientific question, is one of the five suggested skills for Topic 5.4 in both courses, and it is a reminder that this is a measurable quantity. AP Physics C: Mechanics requires that 25 percent of instructional time be spent in hands-on laboratory work. Skill 1.B, creating quantitative graphs with appropriate scales and units including plotting data, is also suggested here, which is what a lab determination of $I$ typically ends in: a linearized graph whose slope carries the rotational inertia.

The CED's sample free-response Question 4, the Qualitative/Quantitative Translation question worth 8 points, is aligned to learning 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. It places a hollow sphere, a uniform solid sphere and a hoop of the same mass $M$ and the same radius $R$ at the top of a ramp of length $L$ at angle $\theta$ , releases them from rest to roll without slipping, and asks students to rank the times taken to reach the bottom and to justify the ranking with conceptual reasoning beyond algebraic solutions.

The scoring guidelines' example response is worth reading for how Topic 5.4 is expected to be argued rather than computed: it says the more rotational inertia the object has, the more gravitational potential energy is converted into rotational kinetic energy, leaving less translational kinetic energy, and 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. The whole argument runs on statement 5.4.A.1, and no numerical coefficient appears in it.

That matters for how you prepare. Coefficients like $\frac{2}{5}MR^2$ for a solid sphere are standard textbook results, but they are not printed in either CED, they are not on either equation sheet, and a solid sphere is not on this course's list of expected derivations. The CED's own question keeps $I$ symbolic from beginning to end and derives $t = \sqrt{2L(I + MR^2)/(MgR^2\sin\theta)}$ . The Topic 5.6 page works that derivation.

The AP Classroom Progress Check for Unit 5 runs about 18 multiple-choice questions and 4 free-response questions, one of each type.

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

Five of the six essential-knowledge statements in this topic are identical in your framework, so most of the physics above is common ground. What is not is the integral. Your boundary statement says AP Physics 1 only expects you to calculate rotational inertia for systems of five or fewer objects arranged in a two-dimensional configuration, and that you do not need to know the rotational inertia of extended rigid systems because they will be provided within the exam. Nothing in worked examples 1, 2 or 3 below is required of you.

If you are in AP Physics 1, read the [AP Physics 1 Topic 5.4 page](/ap-physics-1/unit-5-torque-and-rotational-dynamics/5-4-rotational-inertia) instead. It is written to your framework, it works the sum over point masses, and it covers the parallel axis theorem, which is in your course too. This page is for AP Physics C: Mechanics students.

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. Topic 5.4 is where that prerequisite does the most visible work in Unit 5.

From here, Topic 5.5 sets the sum of torques to zero and Topic 5.6 divides the net torque by the $I$ you just derived. The mass versus rotational inertia comparison covers why the two are not analogues in every respect, and the Unit 5 hub lists all six topics.

A uniform rod about three different perpendicular axes

A thin uniform rod has mass $M = 0.72$ kg and length $L = 0.90$ m. Find its rotational inertia about a perpendicular axis (a) through one end, by direct integration, (b) through its center, using the parallel axis theorem backwards, and (c) through a point one quarter of the length from the center, checked by direct integration.

The Topic 5.4 boundary statement puts this problem inside the course by name: thin rods of uniform or nonuniform density about an arbitrary axis perpendicular to the rod.
Set up the element. The rod is uniform, so its linear mass density is constant at $\lambda = M/L = 0.72/0.90 = 0.80\ \mathrm{kg/m}$ , and $dm = \lambda\, dx$ . Placing the axis at the origin and the rod along $+x$ , every element sits a perpendicular distance $r = x$ from the axis.
(a) Apply 5.4.A.4: $I_{\text{end}} = \int_0^L x^2 \lambda\, dx = \lambda\frac{L^3}{3} = \frac{M}{L}\cdot\frac{L^3}{3} = \frac{ML^2}{3}$ .
$I_{\text{end}} = \frac{(0.72)(0.90)^2}{3} = \frac{(0.72)(0.81)}{3} = \frac{0.5832}{3} = 0.1944\ \mathrm{kg \cdot m^2}$ .
(b) Rather than integrating again, run 5.4.B.2 backwards. The end is a distance $d = L/2 = 0.45$ m from the center of mass, so $I_{\text{cm}} = I_{\text{end}} - Md^2 = 0.1944 - (0.72)(0.45)^2 = 0.1944 - 0.1458 = 0.0486\ \mathrm{kg \cdot m^2}$ .
Check it against the standard symbolic result: $\frac{ML^2}{12} = \frac{(0.72)(0.81)}{12} = \frac{0.5832}{12} = 0.0486\ \mathrm{kg \cdot m^2}$ . The two agree exactly. Note the ratio, $0.1944/0.0486 = 4$ : the same rod, two axes, a factor of four.
Also check 5.4.B.1, which says the center-of-mass value is the minimum in that plane. $0.0486 < 0.1944$ , as required.
(c) Forwards this time, with $d = L/4 = 0.225$ m: $I' = I_{\text{cm}} + Md^2 = 0.0486 + (0.72)(0.225)^2 = 0.0486 + (0.72)(0.050625) = 0.0486 + 0.03645 = 0.08505\ \mathrm{kg \cdot m^2}$ .
Verify by integrating directly. With that axis at the origin the rod runs from $x = -L/4 = -0.225$ m to $x = 3L/4 = 0.675$ m, so $I = \lambda\int_{-0.225}^{0.675} x^2\, dx = (0.80)\frac{(0.675)^3 - (-0.225)^3}{3} = (0.80)\frac{0.307547 + 0.011391}{3} = (0.80)(0.106313) = 0.08505\ \mathrm{kg \cdot m^2}$ .
The theorem and the integral agree to every digit, and the value sits between the center value 0.0486 and the end value 0.1944, which is where an axis between the center and the end has to put it.

(a) $I_{\text{end}} = ML^2/3 = 0.194\ \mathrm{kg \cdot m^2}$ . (b) $I_{\text{cm}} = ML^2/12 = 0.0486\ \mathrm{kg \cdot m^2}$ , four times smaller, obtained by subtracting $Md^2$ rather than integrating again. (c) $0.0851\ \mathrm{kg \cdot m^2}$ at a quarter length from the center, confirmed by direct integration.

A disk from coaxial rings, and the hoop the boundary statement compares it with

A uniform solid disk has mass $M = 1.60$ kg and radius $R = 0.18$ m. Find (a) its rotational inertia about its central axis by integration, (b) the value for a thin hoop of the same mass and radius, and (c) the value for an annular ring of the same mass with inner radius $0.090$ m and outer radius $0.18$ m.

The boundary statement names all three of these: a disk, a thin cylindrical shell, and rigid bodies made up of coaxial rings, with annular rings as its own example.
(a) Choose the element so that every part of it is at one distance from the axis. A ring of radius $r$ and width $dr$ qualifies, since every point on it sits exactly $r$ from the central axis. Its area is its circumference times its width, $2\pi r\, dr$ .
The surface mass density is uniform, $\sigma = M/(\pi R^2)$ , so $dm = \sigma(2\pi r\, dr)$ .
Apply 5.4.A.4: $I = \int_0^R r^2 \sigma(2\pi r)\, dr = 2\pi\sigma\int_0^R r^3\, dr = 2\pi\sigma\frac{R^4}{4}$ . Substituting $\sigma$ gives $I = 2\pi\frac{M}{\pi R^2}\cdot\frac{R^4}{4} = \frac{1}{2}MR^2$ .
$I_{\text{disk}} = \frac{1}{2}(1.60)(0.18)^2 = \frac{1}{2}(1.60)(0.0324) = 0.0259\ \mathrm{kg \cdot m^2}$ .
(b) A thin hoop needs no integral at all, because every element is already at $r = R$ . Pull $R^2$ out of 5.4.A.4 and what is left is $\int dm = M$ , so $I_{\text{hoop}} = MR^2 = (1.60)(0.0324) = 0.0518\ \mathrm{kg \cdot m^2}$ .
Compare them: $I_{\text{hoop}}/I_{\text{disk}} = 2$ exactly, for any mass and any radius. That is the boundary statement's qualitative claim made quantitative, and note the CED words it with a solid puck where the AP Physics 1 version says a solid disk.
(c) Same integral, different limits. $I = 2\pi\sigma\int_a^b r^3\, dr = \frac{\pi\sigma}{2}(b^4 - a^4)$ with $\sigma = M/[\pi(b^2 - a^2)]$ , so $I = \frac{M(b^4 - a^4)}{2(b^2 - a^2)} = \frac{1}{2}M(a^2 + b^2)$ , using the difference of two squares.
$I_{\text{annulus}} = \frac{1}{2}(1.60)[(0.090)^2 + (0.18)^2] = (0.80)(0.0081 + 0.0324) = (0.80)(0.0405) = 0.0324\ \mathrm{kg \cdot m^2}$ .
Sanity check all three against 5.4.A.1, which says rotational inertia depends on how the mass is distributed relative to the axis. The disk has mass right down to the center and gives the smallest value, $0.0259$ ; the annulus has its inner half hollowed out and gives $0.0324$ ; the hoop has everything at the rim and gives the largest, $0.0518$ . Also check the limits of the annulus formula: $a \to 0$ recovers $\frac{1}{2}MR^2$ and $a \to b$ recovers $MR^2$ .

(a) $I_{\text{disk}} = \frac{1}{2}MR^2 = 0.0259\ \mathrm{kg \cdot m^2}$ . (b) $I_{\text{hoop}} = MR^2 = 0.0518\ \mathrm{kg \cdot m^2}$ , exactly twice as large for any mass and radius. (c) $I_{\text{annulus}} = \frac{1}{2}M(a^2 + b^2) = 0.0324\ \mathrm{kg \cdot m^2}$ , between the two.

A composite body needing all three statements at once

A uniform rod of mass $0.40$ kg and length $0.75$ m is pivoted about a perpendicular axis at one end. A uniform disk of mass $0.25$ kg and radius $0.060$ m is fixed to the far end of the rod, with the disk's center on the rod's axis line and the disk in the plane of rotation, so it turns with the rod. Find the system's total rotational inertia about the pivot, and compare it with treating the disk as a point mass.

Statement 5.4.A.3 is the frame for the whole problem: the total rotational inertia of a collection of objects about an axis is the sum of the rotational inertias of each object about that axis. Every part must be computed about the same axis, the pivot, before anything is added.
Rod, about the pivot. This is worked example 1's case: $I_{\text{rod}} = \frac{1}{3}M_rL^2 = \frac{1}{3}(0.40)(0.75)^2 = \frac{1}{3}(0.40)(0.5625) = \frac{0.225}{3} = 0.0750\ \mathrm{kg \cdot m^2}$ .
Disk, about its own center. From worked example 2, $I_{\text{disk,cm}} = \frac{1}{2}m_dr^2 = \frac{1}{2}(0.25)(0.060)^2 = \frac{1}{2}(0.25)(0.0036) = 4.50 \times 10^{-4}\ \mathrm{kg \cdot m^2}$ .
Disk, about the pivot. Its center sits a distance $d = L = 0.75$ m from the pivot, and the two axes are parallel, so 5.4.B.2 applies: $I_{\text{disk,pivot}} = I_{\text{disk,cm}} + m_dd^2 = 4.50 \times 10^{-4} + (0.25)(0.75)^2$ .
$(0.25)(0.5625) = 0.140625$ , so $I_{\text{disk,pivot}} = 0.000450 + 0.140625 = 0.141075\ \mathrm{kg \cdot m^2}$ .
Add, per 5.4.A.3: $I_{\text{tot}} = 0.0750 + 0.141075 = 0.216075 \approx 0.216\ \mathrm{kg \cdot m^2}$ .
Now the comparison. Treating the disk as a point mass at the rod's end means dropping its own $I_{\text{cm}}$ and using 5.4.A.2 alone: $I_{\text{tot}} \approx 0.0750 + 0.140625 = 0.215625\ \mathrm{kg \cdot m^2}$ .
The point-mass shortcut understates the true value by $4.50 \times 10^{-4}\ \mathrm{kg \cdot m^2}$ , which is 0.21%. It is a good approximation here for one reason only: the disk's radius, $0.060$ m, is small next to its distance from the pivot, $0.75$ m, and the two terms go as $r^2$ against $d^2$ .
Check the direction of the error against 5.4.B.1, which says the center-of-mass axis gives the minimum. Ignoring the disk's own spread of mass can only lose rotational inertia, never add it, so the shortcut must come out low. It does.
Where the shortcut would fail: swap the numbers so the disk's radius is comparable with its distance from the pivot, and the neglected term stops being negligible. If the disk's radius were $0.30$ m instead, its own term would be $0.01125\ \mathrm{kg \cdot m^2}$ against $0.140625$ , about 8%.

$I_{\text{tot}} = 0.216\ \mathrm{kg \cdot m^2}$ , from the rod's $0.0750$ plus the disk's $0.141$ about the pivot. Treating the disk as a point mass gives $0.2156\ \mathrm{kg \cdot m^2}$ , low by 0.21%, because the disk's radius is small compared with its distance from the axis.

Frequently asked questions

What is the difference between rotational inertia in AP Physics C and AP Physics 1?

One essential-knowledge statement. AP Physics C: Mechanics prints 5.4.A.4, which says that for a solid that can be considered as a collection of differential masses, the solid's rotational inertia can be calculated as the integral of the squared perpendicular distance from each mass element to the axis, against the mass element. AP Physics 1's Topic 5.4 has no statement with that number. The other five statements, covering what rotational inertia measures, the single-object form, the sum over a collection, the fact that the center-of-mass axis minimizes the value, and the parallel axis theorem, are printed word for word in both. The boundary statements then diverge sharply: AP Physics 1 caps the calculation at systems of five or fewer objects in a two-dimensional configuration and says extended-body values will be provided on the exam, while AP Physics C: Mechanics names a specific list of bodies whose derivations you are expected to perform.

Which rotational inertia derivations does AP Physics C Mechanics expect?

The Topic 5.4 boundary statement names them exactly. It says AP Physics C: Mechanics only expects students to use calculus in the derivations of the rotational inertia of thin rods of uniform or nonuniform density about an arbitrary axis perpendicular to the rod, as well as derivations of the rotational inertia of a thin cylindrical shell, disk, or rigid bodies that can be considered to be made up of coaxial rings or shells about an axis that passes through their centers, giving annular rings as the example. A second paragraph adds that students should have a qualitative understanding of the factors that affect rotational inertia, for example how rotational inertia is greater when mass is farther from the axis of rotation, which is why a hoop has more rotational inertia than a solid puck of the same mass and radius. A solid sphere is not on the list, and neither is a rod about an axis along its own length.

Is the parallel axis theorem only on AP Physics C?

No. The parallel axis theorem is essential knowledge 5.4.B.2 in both AP Physics C: Mechanics and AP Physics 1, printed with the same wording and the same equation relating the rotational inertia about any axis to the center-of-mass value plus the mass times the square of the distance between the two parallel axes. It is also printed on both equation sheets. What differs is what you can put into it: in the calculus-based course the center-of-mass value may be one you derived yourself with the integral form, while in the algebra-based course extended-body values are supplied on the exam. Two conditions apply in both courses and are often dropped: the two axes must be parallel, and one of them must pass through the center of mass. Hopping between two arbitrary parallel axes takes two applications, routed through the center of mass.

Is moment of inertia the same as rotational inertia?

They are two names for the same quantity, but only one of them appears in the AP frameworks. The phrase moment of inertia appears zero times in the AP Physics C: Mechanics course and exam description and zero times in the AP Physics 1 one. The phrase rotational inertia appears 39 times in the C: Mechanics document and 28 times in the AP Physics 1 document, and it is the term used in the learning objectives, the essential-knowledge statements, the boundary statements and the equation sheet's symbol definitions, where the letter I is defined as rotational inertia. Many textbooks and most of the internet say moment of inertia. If you are searching for AP-aligned material or reading an exam question, expect rotational inertia, and use that term in a free-response answer.

How do I set up the integral for rotational inertia?

Three steps. First, fix the axis and write the perpendicular distance from a general mass element to that axis in terms of one variable; essential knowledge 5.4.A.4 specifies that this distance is measured perpendicular to the axis. For a rod along the x axis rotating about a perpendicular axis at the origin, that distance is simply x. Second, write the mass element in terms of the same variable: for a rod it is the linear mass density times the length element, and the equation sheet prints the linear mass density as the derivative of mass with respect to length. Third, and most important, choose the element so that all of it lies at a single distance from the axis. That is why a disk is integrated as a stack of concentric rings, each of area equal to its circumference times its width, rather than as strips. The boundary statement's phrase about bodies made of coaxial rings or shells is naming precisely this technique.

Why does a hoop have more rotational inertia than a disk of the same mass?

Because rotational inertia weights every piece of mass by the square of its distance from the axis, and a hoop puts all of its mass at the largest possible distance while a disk spreads mass all the way in to the center. Essential knowledge 5.4.A.1 says rotational inertia is related to both the mass of the system and the distribution of that mass relative to the axis of rotation, and the Topic 5.4 boundary statement gives this exact comparison as its example of the qualitative understanding expected, phrased as a hoop having more rotational inertia than a solid puck of the same mass and radius. Integrating both gives the hoop the full mass times radius squared and the disk one half of that, so the ratio is exactly 2 for any mass and any radius. The mass that matters most is the mass farthest out.

Is the rotational inertia integral on the AP Physics C Mechanics equation sheet?

Yes. The rotational column of the mechanics table prints the integral of the squared perpendicular distance against the mass element, directly below the sum form written as the total rotational inertia equalling the sum of the individual rotational inertias equalling the sum of each mass times its distance squared, and directly above the parallel axis theorem. The AP Physics 1 sheet prints the sum form and the parallel axis theorem but not the integral. Two supporting equations for continuous bodies are on the C: Mechanics sheet only, both in its translational column: the linear mass density as the derivative of mass with respect to length, and the center of mass as the integral of position against mass element divided by total mass. One required equation is on neither sheet: the single-object form giving rotational inertia as mass times the square of the perpendicular distance, which is essential knowledge 5.4.A.2 in both courses.