AP Physics C: Mechanics · Topic 6.1

Topic 6.1: Rotational Kinetic Energy

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

Rotational kinetic energy is one half the rotational inertia times the angular speed squared. The AP Physics C equation matches AP Physics 1's. What changes is where the rotational inertia comes from: here you integrate r squared over the mass distribution instead of adding point masses.

AP Physics: Unit 6 (topics 6.1 Rotational Kinetic Energy). Topic 6.1 of the current AP Physics C: Mechanics course and exam description, inside Unit 6, which is weighted 10 to 15% of the multiple-choice section at about 13 to 19 class periods. One learning objective, 6.1.A, and five essential-knowledge statements: 6.1.A.1 with the equation K_rot = (1/2)I omega squared, 6.1.A.1.i on that energy being the summed translational kinetic energy of the parts, 6.1.A.1.ii on the total being the center-of-mass rotational term plus the translational term, 6.1.A.2 on spinning with the center of mass at rest, and 6.1.A.3 on the quantity being a scalar. Suggested skills 1.C, 2.C, 3.B and 3.C. No boundary statement: Unit 6 prints exactly one, under Topic 6.5. Calculus differentiator against the identically titled AP Physics 1 Topic 6.1: the equation is the same on both sheets, but this course's sheet also prints I as the integral of r squared dm and lambda as the derivative of m with respect to length, so a non-uniform mass distribution is fair game here and out of reach in the algebra-based course. Not printed on the C: Mechanics sheet: the total kinetic energy of a rolling body as a single sum (that is essential knowledge 6.5.A.1), K_rot in terms of angular momentum squared, and any table of rotational inertias for named shapes. Sample free-response Question 4, a Qualitative/Quantitative Translation question worth 8 points, aligns to 6.1.A among five other objectives.

What Topic 6.1 requires

Topic 6.1 of AP Physics C: Mechanics carries one learning objective and five essential-knowledge statements. That is the whole of the required content, and it fits in a paragraph.

Learning objective 6.1.A: describe the rotational kinetic energy of a rigid system in terms of the rotational inertia and angular velocity of that rigid system.

Statement	What it says
6.1.A.1	The rotational kinetic energy of an object or rigid system is related to the rotational inertia and angular velocity of the rigid system, and is given by $K_{\text{rot}} = \frac{1}{2}I\omega^2$
6.1.A.1.i	The rotational inertia of an object about a fixed axis can be used to show that the rotational kinetic energy of that object is equivalent to its translational kinetic energy, which is its total kinetic energy
6.1.A.1.ii	The total kinetic energy of a rigid system is the sum of its rotational kinetic energy due to its rotation about its center of mass and the translational kinetic energy due to the linear motion of its center of mass
6.1.A.2	A rigid system can have rotational kinetic energy while its center of mass is at rest, due to the individual points within the rigid system having linear speed and therefore kinetic energy
6.1.A.3	Rotational kinetic energy is a scalar quantity

The suggested skills for this topic are 1.C (create qualitative sketches of graphs), 2.C (compare physical quantities between two or more scenarios or at different times and locations within a single scenario), 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim) and 3.C (justify or support a claim using evidence). Neither 2.A nor 2.B is listed. Do not read that as a promise: the framework's exam-weighting page states that required course content can be assessed with any skill, and the sample free-response question that aligns to 6.1.A is built around a derivation.

Topic 6.1 prints no boundary statement. Unit 6 prints exactly one across all six topics, and it sits under Topic 6.5.

The equation is the same as AP Physics 1's. What feeds it is not.

Say it plainly, because the honest answer is more useful than a manufactured difference: $K_{\text{rot}} = \frac{1}{2}I\omega^2$ is the same equation in both courses. The AP Physics C: Mechanics sheet prints it with the subscript, $K_{\text{rot}}$ ; the AP Physics 1 sheet prints it as $K = \frac{1}{2}I\omega^2$ . The physics behind those two lines is identical.

The calculus enters one step earlier, in the $I$ . AP Physics 1's sheet prints two routes to a rotational inertia, the sum over point masses

I = \sum m_i r_i^2

and the parallel-axis theorem $I' = I_{\text{cm}} + Md^2$ , which shifts a value you already have to a new axis. Neither reaches a continuous body whose density varies. The AP Physics C: Mechanics sheet prints both of those and adds the one that does:

I = \int r^2\, dm

along with the tool for turning a rod into a $dm$ , $\lambda = \dfrac{d}{d\ell}m(\ell)$ .

That single extra line is the whole difference in what a Topic 6.1 question can ask. A calculus-based question can hand you a rod whose linear mass density varies along its length, or a disk whose density varies with radius, and ask for the rotational kinetic energy. The algebra-based course has no route to that number. Everything downstream, the one half, the $\omega^2$ , the scalar nature, is unchanged.

Second, smaller difference: $\omega = \dfrac{d\theta}{dt}$ is printed here and is not on the AP Physics 1 sheet, so an angular position given as a function of time is a legitimate starting point. Differentiate to get $\omega(t)$ , then square it.

Where one half I omega squared comes from, and what 6.1.A.1.i is saying

Statement 6.1.A.1.i is worded in a way that stops people. Here it is whole:

"The rotational inertia of an object about a fixed axis can be used to show that the rotational kinetic energy of that object is equivalent to its translational kinetic energy, which is its total kinetic energy."

The derivation it is pointing at is short. Chop the rigid body into mass elements $dm$ . Every element sits a distance $r$ from the fixed axis and, because the body is rigid, every element shares the same $\omega$ , so each moves with linear speed $v = r\omega$ . Add up the ordinary translational kinetic energy of the pieces:

K = \int \tfrac{1}{2} v^2\, dm = \int \tfrac{1}{2} (r\omega)^2\, dm = \tfrac{1}{2}\omega^2 \int r^2\, dm = \tfrac{1}{2}I\omega^2

$\omega$ comes out of the integral because it is the same for every element of a rigid body. What is left inside is the definition of rotational inertia. So the rotational kinetic energy is not a second, separate kind of energy sitting alongside the translational kinetic energy of the parts. It is that energy, re-bookkept in terms of one number per body instead of one number per particle. That is what "equivalent to its translational kinetic energy, which is its total kinetic energy" means for a body on a fixed axis.

Two consequences worth carrying:

$I$ depends on the axis, so $K_{\text{rot}}$ depends on the axis. The same object at the same $\omega$ has different rotational kinetic energy about different axes, because the mass elements are at different distances and so move at different speeds. Worked example 3 below puts a number on that.
The step $v = r\omega$ requires $\omega$ in radians per second. The radian is dimensionless, which is why $\mathrm{kg \cdot m^2}$ times $\mathrm{rad^2/s^2}$ comes out in joules and not in something with a stray unit attached.

A rigid system can spin with its center of mass at rest (6.1.A.2)

Statement 6.1.A.2 exists because the intuition that kinetic energy means going somewhere is wrong for rotation. Its reason clause is the interesting part: a rigid system can have rotational kinetic energy while its center of mass is at rest "due to the individual points within the rigid system having linear speed and, therefore, kinetic energy."

A flywheel bolted to a bench has $v_{\text{cm}} = 0$ and $\vec{p} = 0$ . Its linear momentum is exactly zero. Its kinetic energy is not, because the mass elements are moving even though their velocities cancel in the vector sum that defines $\vec{v}_{\text{cm}}$ . Momentum cancels because it is a vector; kinetic energy cannot cancel because $v^2$ is never negative.

The framework's vocabulary appendix draws the same picture for a wheel on a fixed axle: "At any given instant in time, Point A is traveling with a greater translational speed and in a different direction than Point B." It adds that if the rotation matters to the analysis, the wheel cannot be modeled as an object and has to be modeled as a rigid system, which that appendix defines as a system that does not change shape but whose different points may move in different directions and with different speeds.

That definition is the licence for Topic 6.1. You may treat the whole wheel as one thing with one $I$ and one $\omega$ precisely because the shape does not change, which is what let $\omega$ come out of the integral above.

Total kinetic energy is a sum, and both terms are about the center of mass

Statement 6.1.A.1.ii is the one that shows up in almost every rolling and ramp problem:

K_{\text{total}} = \tfrac{1}{2}M v_{\text{cm}}^2 + \tfrac{1}{2}I_{\text{cm}}\omega^2

Read the subscripts. The translational term uses the speed of the center of mass and the total mass. The rotational term uses the rotational inertia about the center of mass, not about a contact point, not about a pivot. Mixing an $I$ about one axis with a $v_{\text{cm}}$ about another double counts.

Topic 6.5 states the same idea as $K_{\text{tot}} = K_{\text{trans}} + K_{\text{rot}}$ in essential knowledge 6.5.A.1, and that equation is not printed on the equation sheet. The sheet gives you $K = \frac{1}{2}mv^2$ and $K_{\text{rot}} = \frac{1}{2}I\omega^2$ on separate lines; adding them is a step you take.

There is a second, equivalent bookkeeping that the sheet does support, and it is worth knowing because it makes some problems shorter. If an object rotates about a fixed axis that is not through its center of mass, you can either use the split above or use a single rotation about that axis with the parallel-axis theorem, $I' = I_{\text{cm}} + Md^2$ , which is printed. The two give the same number. Worked example 3 checks that they do, digit for digit.

Rotational kinetic energy is a scalar (6.1.A.3)

Statement 6.1.A.3 is one sentence: rotational kinetic energy is a scalar quantity. It earns its place in the framework because Unit 6 is otherwise full of vectors. Angular momentum is defined here as a cross product, $\vec{L} = \vec{r} \times \vec{p}$ , angular impulse has the direction of the torque that delivered it, and torque itself is printed on this course's sheet as $\vec{\tau} = \vec{r} \times \vec{F}$ . Against that background, energy stands out for having no direction at all.

Practical consequences:

Two wheels spinning opposite ways have angular momenta that cancel and kinetic energies that add. This is the single most useful sanity check in the unit.
A negative $\omega$ gives a positive $K_{\text{rot}}$ , because $\omega$ is squared. If a sign convention makes your kinetic energy come out negative, the arithmetic is wrong, not the convention.
You never resolve rotational kinetic energy into components, and there is no such thing as the rotational kinetic energy "in the x direction".

Be careful with the neighbouring quantity. Work done by a torque is also a scalar, and it does carry a sign, because it is the transfer, not the store. A torque opposing the rotation does negative work and takes energy out. That is Topic 6.2, and it does not contradict 6.1.A.3.

What the equation sheet prints for Topic 6.1

The lines that matter for this topic, checked against the Table of Information in the AP Physics C: Mechanics course and exam description rather than recalled:

Equation	On the C: Mechanics sheet
$K_{\text{rot}} = \frac{1}{2}I\omega^2$	yes
$K = \frac{1}{2}mv^2$	yes
$I = \int r^2\, dm$	yes
$I_{\text{tot}} = \sum I_i = \sum m_i r_i^2$	yes
$I' = I_{\text{cm}} + Md^2$	yes
$\lambda = \frac{d}{d\ell}m(\ell)$	yes
$\vec{r}_{\text{cm}} = \frac{\int \vec{r}\, dm}{\int dm}$	yes
$\omega = \frac{d\theta}{dt}$	yes
$v = r\omega$	yes
$K_{\text{tot}} = K_{\text{trans}} + K_{\text{rot}}$	no, it is essential knowledge 6.5.A.1 only
$K_{\text{rot}} = \frac{L^2}{2I}$	no
rotational inertias of named shapes	no

The last row is the one that changes how you prepare. No table of rotational inertias for common shapes appears anywhere in the Table of Information, which is three pages: constants and conversion factors with the prefixes, unit symbols, trigonometric values and exam conventions; the mechanics equations; and the geometry, trigonometry, vectors, calculus and identities tables. There is no $\frac{1}{2}MR^2$ printed for a disk and no $\frac{2}{5}MR^2$ printed for a solid sphere.

When an exam question needs one, it tends to supply it in the stem. Sample multiple-choice question 14 in the framework does exactly that: it states that the rotational inertia of a uniform disk of mass $m$ and radius $r$ is $\frac{1}{2}mr^2$ , then asks a conservation-of-angular-momentum question. The other route the exam expects is the one Topic 6.1 is built on: derive the rotational inertia from $I = \int r^2\, dm$ . Worked example 2 does that for a solid cylinder in four lines.

How Topic 6.1 is tested, and where it leads

Unit 6 is weighted 10 to 15% of the multiple-choice section of the AP Physics C: Mechanics exam, at about 13 to 19 class periods. On that section, skill 2.C is weighted 10 to 15%, 3.B is 15 to 25% and 3.C is 5 to 10%. Science Practice 1 is not assessed there at all, so the 1.C listed for this topic points at the free-response section, where Practice 1 carries 20 to 35%.

The sample free-response set's Question 4 is a Qualitative/Quantitative Translation question worth 8 points, and 6.1.A is one of the six learning objectives it aligns to. It races a hollow sphere, a uniform solid sphere and a hoop of equal mass and radius down a ramp. Its published example response for part A argues in exactly the language of this topic, that more rotational inertia means more of the gravitational potential energy is converted into rotational kinetic energy and "less translational kinetic energy" is left. Its scoring guidelines then credit two routes through part B, a Newton's-second-law-in-rotational-form derivation and, per an explicit scoring note, a derivation that correctly applies conservation of energy.

The unit's Preparing for the AP Exam note is blunt about how that gets graded. It says that when writing justifications for claims, simply referencing an equation, law, or physical principle is not sufficient, and gives as its example that stating one disk is rolling faster than another because of "conservation of energy" is not a complete enough answer to earn credit on the free-response section. Students must clearly and concisely explain the steps in their reasoning that lead from the equation, law, or principle to the justification of their claim. On a Topic 6.1 question that means naming the split, naming which term grows, and naming why.

The unit's first sample instructional activity is also a 6.1 activity: race a ring, a disk and a low-friction cart loaded to the same mass down identical inclines, watch the cart win, and explain it first with forces and then with energy.

From here, Topic 6.2 supplies the mechanism that changes this energy, and Topic 6.5 is where the two-term split does most of its work.

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

AP Physics 1 has a Topic 6.1 with the same title, and on this particular topic the two courses are closer than anywhere else in Unit 6. The equation is the same, the scalar statement is the same, and the spinning-in-place idea is the same.

The honest split: the [AP Physics 1 Topic 6.1 page](/ap-physics-1/unit-6-energy-and-momentum-of-rotating-systems/6-1-rotational-kinetic-energy) is for students in the algebra-based course, who will only ever meet $I$ as a sum over point masses or as a number handed to them. This page is for students in AP Physics C: Mechanics, who are expected to produce $I$ from $\int r^2\, dm$ and to answer symbolically. Neither page is a reading-level variant of the other, and if you are in Physics 1 the integral here is not on your exam.

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

Other pages that help here: the conservation of energy guide owns the step-by-step energy-accounting routine, the rotational kinematics guide owns the angular equations of motion, and mass compared with rotational inertia is the two-minute version of why the axis matters. The kinetic energy calculator checks translational arithmetic. The Unit 6 hub lists all six topics and every equation in the unit.

A rod whose density is not uniform

A thin rod of length $L = 1.20$ m has linear mass density $\lambda(x) = \lambda_0\left(1 + \frac{x}{L}\right)$ with $\lambda_0 = 0.50$ kg/m, where $x$ is measured from the light end. It rotates about a fixed axis through the light end, perpendicular to the rod, at $\omega = 8.0$ rad/s. Find (a) the rod's mass, (b) its rotational inertia about that axis, (c) its rotational kinetic energy, and (d) how that compares with a uniform rod of the same mass and length on the same axis.

Set the convention: $x$ runs from 0 at the axis to $L$ at the heavy end, and the mass element is $dm = \lambda(x)\, dx$ . The sheet's $\lambda = \frac{d}{d\ell}m(\ell)$ is that relation read backwards.
(a) $M = \int_0^L \lambda_0\left(1 + \frac{x}{L}\right) dx = \lambda_0\left[x + \frac{x^2}{2L}\right]_0^L = \lambda_0\left(L + \frac{L}{2}\right) = \frac{3}{2}\lambda_0 L$ .
$M = 1.5(0.50)(1.20) = 0.90$ kg.
(b) $I = \int r^2\, dm = \int_0^L x^2 \lambda_0\left(1 + \frac{x}{L}\right) dx = \lambda_0\left[\frac{x^3}{3} + \frac{x^4}{4L}\right]_0^L = \lambda_0 L^3\left(\frac{1}{3} + \frac{1}{4}\right) = \frac{7}{12}\lambda_0 L^3$ .
With $L^3 = 1.728\ \mathrm{m^3}$ : $I = \frac{7}{12}(0.50)(1.728) = 0.504\ \mathrm{kg \cdot m^2}$ .
(c) $K_{\text{rot}} = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.504)(8.0)^2 = \frac{1}{2}(0.504)(64) = 16.128 \approx 16\ \mathrm{J}$ .
(d) A uniform rod of the same 0.90 kg and 1.20 m about its end has $I_{\text{unif}} = \frac{1}{3}ML^2 = \frac{1}{3}(0.90)(1.44) = 0.432\ \mathrm{kg \cdot m^2}$ , giving $K = \frac{1}{2}(0.432)(64) = 13.824 \approx 14\ \mathrm{J}$ .
The ratio is $0.504/0.432 = 7/6 \approx 1.17$ . Same mass, same length, same axis, same angular speed, 17% more kinetic energy, because this rod's mass sits farther from the axis on average.
Sanity check the direction of that result against the center of mass. $x_{\text{cm}} = \frac{1}{M}\int_0^L x\,\lambda(x)\,dx = \frac{\lambda_0}{M}\left(\frac{L^2}{2} + \frac{L^2}{3}\right) = \frac{5L}{9} = 0.667$ m, which is beyond the midpoint at 0.600 m. Mass shifted away from the axis, so $I$ and $K_{\text{rot}}$ both had to rise.
Units: $\mathrm{kg \cdot m^2}$ times $\mathrm{rad^2/s^2}$ is $\mathrm{kg \cdot m^2/s^2}$ , which is joules, because the radian is dimensionless.

(a) $M = 0.90$ kg. (b) $I = \frac{7}{12}\lambda_0 L^3 = 0.504\ \mathrm{kg \cdot m^2}$ . (c) $K_{\text{rot}} = 16$ J. (d) A uniform rod of the same mass and length stores 14 J at the same angular speed, a factor of $7/6$ less, because its mass sits closer to the axis on average.

Deriving a cylinder's rotational inertia, then splitting its energy

A uniform solid cylinder of mass $M = 2.4$ kg and radius $R = 0.15$ m rolls without slipping with $v_{\text{cm}} = 3.0$ m/s. (a) Derive $I_{\text{cm}}$ from $I = \int r^2\, dm$ rather than quoting it. (b) Find the translational and rotational kinetic energies and the total. (c) State the fraction of the total that is rotational, and show that the fraction does not depend on $M$ , $R$ or $v$ .

(a) Slice the cylinder into thin coaxial shells of radius $r$ and thickness $dr$ . Every point in one shell is the same distance $r$ from the axis, so no approximation is involved. Writing the mass per unit face area as $\sigma = \dfrac{M}{\pi R^2}$ , a shell of circumference $2\pi r$ has $dm = \sigma\, 2\pi r\, dr$ .
$I = \int_0^R r^2 \left(\frac{M}{\pi R^2}\right) 2\pi r\, dr = \frac{2M}{R^2}\int_0^R r^3\, dr = \frac{2M}{R^2}\cdot\frac{R^4}{4} = \frac{1}{2}MR^2$ .
$I_{\text{cm}} = \frac{1}{2}(2.4)(0.15)^2 = \frac{1}{2}(2.4)(0.0225) = 0.027\ \mathrm{kg \cdot m^2}$ .
(b) The rolling condition of essential knowledge 6.5.B.1 gives $\omega = v_{\text{cm}}/R = 3.0/0.15 = 20\ \mathrm{rad/s}$ .
$K_{\text{trans}} = \frac{1}{2}Mv_{\text{cm}}^2 = \frac{1}{2}(2.4)(9.0) = 10.8\ \mathrm{J}$ .
$K_{\text{rot}} = \frac{1}{2}I_{\text{cm}}\omega^2 = \frac{1}{2}(0.027)(400) = 5.4\ \mathrm{J}$ .
$K_{\text{total}} = 10.8 + 5.4 = 16.2\ \mathrm{J}$ , which is statement 6.1.A.1.ii applied directly.
(c) The rotational share is $5.4/16.2 = 1/3$ . Symbolically, substitute $\omega = v_{\text{cm}}/R$ into the rotational term: $K_{\text{rot}} = \frac{1}{2}I_{\text{cm}}\frac{v_{\text{cm}}^2}{R^2}$ , so $\dfrac{K_{\text{rot}}}{K_{\text{trans}}} = \dfrac{I_{\text{cm}}}{MR^2}$ .
For a solid cylinder $I_{\text{cm}}/MR^2 = \frac{1}{2}$ exactly, so the rotational share of the total is $\dfrac{1/2}{1 + 1/2} = \dfrac{1}{3}$ for every solid cylinder, whatever its mass, radius or speed. That dimensionless ratio is what decides every ramp race in this unit, and it is the quantity the sample free-response Question 4 is really about.

(a) $I_{\text{cm}} = \frac{1}{2}MR^2 = 0.027\ \mathrm{kg \cdot m^2}$ , derived by integrating over coaxial shells. (b) $K_{\text{trans}} = 10.8$ J, $K_{\text{rot}} = 5.4$ J, total 16.2 J. (c) One third, and the fraction is fixed by $I_{\text{cm}}/MR^2 = \frac{1}{2}$ alone, independent of mass, radius and speed.

One disk, one angular speed, two axes

A uniform disk of mass $M = 1.5$ kg and radius $R = 0.20$ m spins at $\omega = 12$ rad/s. Find its kinetic energy (a) about a fixed axis through its center, perpendicular to the disk, and (b) about a parallel fixed axis through a point on its rim. (c) Recompute part (b) using the center-of-mass split of 6.1.A.1.ii and check the two agree.

Take $I_{\text{cm}} = \frac{1}{2}MR^2$ for a uniform disk about its central axis, derived in worked example 2 for the cylinder, which is the same integral.
(a) $I_{\text{cm}} = \frac{1}{2}(1.5)(0.20)^2 = \frac{1}{2}(1.5)(0.040) = 0.030\ \mathrm{kg \cdot m^2}$ .
$K = \frac{1}{2}(0.030)(12)^2 = \frac{1}{2}(0.030)(144) = 2.16\ \mathrm{J}$ .
(b) For the rim axis, use the printed parallel-axis theorem with $d = R$ : $I' = I_{\text{cm}} + Md^2 = 0.030 + (1.5)(0.040) = 0.030 + 0.060 = 0.090\ \mathrm{kg \cdot m^2}$ .
$K = \frac{1}{2}(0.090)(144) = 6.48\ \mathrm{J}$ , three times the answer to part (a) at the identical angular speed.
(c) Now do it the other way. About the rim axis the center of mass is not at rest: it travels a circle of radius $R$ , so $v_{\text{cm}} = \omega R = (12)(0.20) = 2.4\ \mathrm{m/s}$ .
$K_{\text{trans}} = \frac{1}{2}Mv_{\text{cm}}^2 = \frac{1}{2}(1.5)(2.4)^2 = \frac{1}{2}(1.5)(5.76) = 4.32\ \mathrm{J}$ .
$K_{\text{rot}} = \frac{1}{2}I_{\text{cm}}\omega^2 = 2.16\ \mathrm{J}$ , the same rotation about the center of mass as in part (a), because a rigid body turning about the rim axis at $12$ rad/s is also turning about its own center at $12$ rad/s.
$K_{\text{total}} = 4.32 + 2.16 = 6.48\ \mathrm{J}$ , matching part (b) exactly.
The two methods are the same bookkeeping. What you must not do is add them: using $I'$ about the rim and a translational term is double counting, because $Md^2$ already is the translational term. That is the defect the subscripts in 6.1.A.1.ii are there to prevent.

(a) 2.16 J. (b) 6.48 J, exactly three times as much, because $I' = 3I_{\text{cm}}$ for a disk about its rim. (c) The split gives $4.32 + 2.16 = 6.48$ J, identical. Use one method or the other, never both at once.

Frequently asked questions

What is rotational kinetic energy in AP Physics C Mechanics?

It is the kinetic energy an object has because it is spinning, and essential knowledge 6.1.A.1 of AP Physics C: Mechanics gives it as one half the rotational inertia times the square of the angular velocity. The equation is printed on the course's equation sheet as K with a rot subscript equals one half I omega squared. Angular velocity must be in radians per second. The quantity is measured in joules, exactly like translational kinetic energy, and statement 6.1.A.3 records that it is a scalar, so it has no direction and is never negative.

How is rotational kinetic energy different in AP Physics C than in AP Physics 1?

The equation itself is not different. Both courses print one half the rotational inertia times angular speed squared, and both mean the same thing by it. What differs is how you are expected to obtain the rotational inertia. The AP Physics 1 sheet offers only the sum of each point mass times its distance squared, so that course can only handle point masses or a value handed to you in the question. The AP Physics C: Mechanics sheet also prints the integral of r squared with respect to mass, plus the linear mass density as a derivative, so a Physics C question can give you a rod or disk whose density varies with position and ask for its kinetic energy.

How do you find rotational kinetic energy when the mass is not uniformly distributed?

Integrate first, then square. Write the mass element in terms of the given density: for a rod, the mass element is the linear mass density at that position times a small length; for a disk, it is the areal density times the area of a thin ring. Substitute into the integral of r squared with respect to mass, where r is the distance from the rotation axis, and evaluate over the whole body to get the rotational inertia. Only then put that number into one half I omega squared. Both the integral for rotational inertia and the linear mass density definition are printed on the AP Physics C: Mechanics equation sheet.

Can an object have kinetic energy if its center of mass is not moving?

Yes, and essential knowledge 6.1.A.2 states it directly: a rigid system can have rotational kinetic energy while its center of mass is at rest, because the individual points within the system have linear speed and therefore kinetic energy. A flywheel on a fixed axle has exactly zero linear momentum, since momentum is a vector and the mass elements' velocities cancel in pairs, but its kinetic energy is not zero, because speed is squared before it is added and squares never cancel.

Are rotational inertias of common shapes given on the AP Physics C equation sheet?

No. The AP Physics C: Mechanics Table of Information runs to three pages, covering constants, prefixes, unit symbols, trigonometric values and exam conventions, then the mechanics equations, then geometry, trigonometry, vectors, calculus and identities. No table of rotational inertias for named shapes appears on any of them. What is printed is the machinery for producing one: the integral of r squared with respect to mass, the sum over point masses, and the parallel-axis theorem. Exam questions that need a specific shape's value tend to state it in the stem, as the framework's own sample multiple-choice question 14 does for a uniform disk.

How do you find the total kinetic energy of a rolling object?

Add two terms, per essential knowledge 6.1.A.1.ii: one half the total mass times the speed of the center of mass squared, plus one half the rotational inertia about the center of mass times the angular speed squared. Both subscripts matter. The rotational inertia has to be the one about the center of mass, not about the contact point, or you count the translational motion twice. If the object rolls without slipping you can then substitute the angular speed as the center-of-mass speed divided by the radius, which turns the total into the translational term multiplied by one plus the ratio of the rotational inertia to mass times radius squared.

Why does the same object have different rotational kinetic energy about different axes?

Because rotational inertia is defined relative to an axis, and kinetic energy inherits that. Move the axis and the mass elements sit at different distances from it, so at the same angular speed they move at different linear speeds and carry different amounts of energy. A uniform disk about an axis through its rim has three times the rotational inertia it has about its center, by the parallel-axis theorem, so at the same angular speed it has three times the kinetic energy. Both answers are correct about their own axes, which is why a rotational energy quoted without a stated axis is incomplete.