AP Physics C: Mechanics Unit 6: Energy and Momentum of Rotating Systems Notes

Introduction

Calculus connects torque to rotational work and angular impulse, while conservation laws explain rolling, changing mass distributions, and orbital motion.

Rotation uses familiar mechanics principles with one added discipline: every torque, inertia, and angular momentum must be defined about an axis or origin. Integrating torque over angle transfers energy; integrating torque over time changes angular momentum. Those two areas answer different physical questions.

Rolling motion combines translation and rotation through a no-slip constraint. Orbital motion combines inverse-square gravity with energy and angular momentum. In both settings, the constraint determines which compact relationships are valid and when the full vector or integral law is needed.

Mass distribution and angular speed

rotational kinetic energy

Torque integrated over angle

rotational work

Torque integrated over time

angular impulse

Negligible external angular impulse

angular-momentum conservation

No-slip or central-force constraint

rolling and orbital relations

Current exam guidance: Aligned to the current 2026–27 AP Physics C: Mechanics framework, with calculus-based modeling, derivations, experimental reasoning, and explicit assumptions.

6.1

Rotational Kinetic Energy

Why does rotational kinetic energy depend on the squared distance of each mass element from the axis?

Short answer

Because a rigid body’s mass element moves at $v=r\omega$ , integrating $\tfrac12v^2\,dm$ gives $K_{rot}=\tfrac12I\omega^2$ with $I=\int r^2\,dm$ .

Know firstRigid-body kinematics · Rotational inertia integrals · Center-of-mass motion

Energy is distributed throughout the body

Choose an axis and divide a rigid body into infinitesimal mass elements. An element a perpendicular distance $r$ from the axis has speed $v=r\omega$ . Its kinetic energy is $dK=\tfrac12(r\omega)^2\,dm$ . Since every element shares the same angular speed, integration gives $K_{rot}=\tfrac12\omega^2\int r^2\,dm=\tfrac12I\omega^2$ . The $r^2$ weighting means a small amount of distant mass can contribute strongly to both inertia and energy.

For general planar motion, separate translation of the center of mass from rotation about it: $K=\tfrac12Mv_{cm}^2+\tfrac12I_{cm}\omega^2$ . This decomposition avoids double counting. Using an inertia about an instantaneous contact point and then adding center-of-mass translation would count the same motion twice.

The comparison depends on what remains fixed. At fixed $\omega$ , larger $I$ means larger energy. At fixed angular momentum, $K=L^2/(2I)$ , so larger $I$ means smaller energy. Rotational kinetic energy is a nonnegative scalar: reversing the sign of $\omega$ changes the rotation direction but not $K_{rot}$ . State the constraint before drawing a proportionality conclusion.

Physics model

Rotation stores kinetic energy across the whole mass distribution

Change signed angular velocity and see that reversing rotation leaves the quadratic rotational energy unchanged.

Angular velocity5 rad/s

Rotational inertia: 2 kg·m²
Angular velocity: 5 rad/s
Direction: counterclockwise
Rotational energy: 25 J

rigid bodysigned angular velocityrotational energy

K_{rot}=\tfrac12I\omega^2

Read this figure without the graphic

Change signed angular velocity and see that reversing rotation leaves the quadratic rotational energy unchanged.

Current quantity	Value
Rotational inertia	2 kg·m²
Angular velocity	5 rad/s
Direction	counterclockwise
Rotational energy	25 J
Model relation	$K_{rot}=\tfrac12I\omega^2$

Large motor-generator flywheel assembly at Lawrence Berkeley Laboratory in 1961. — A flywheel stores energy in distributed rotating mass
The wheel stores rotational kinetic energy throughout its mass distribution. Reversing the rotation changes angular velocity’s sign but not the stored kinetic energy.
Use this visual to see
Connect rotational kinetic energy to both inertia and angular speed.
What it cannot show
The photograph does not give rotational inertia, angular speed, stored energy, power rating, or operating direction.

Integrate the particle energies

System: A rigid body rotating about a stated fixed axis
Frame: Inertial frame in which the axis is stationary
Assumptions: Rigid body; all mass elements share angular speed $\omega$
Givens: Mass distribution $dm$ and perpendicular distance $r$

Write $dK=\tfrac12v^2\,dm$ .
Use rigid-body kinematics $v=r\omega$ .
Factor out $\omega^2$ and identify $I=\int r^2\,dm$ .

Result

K_{rot}=\int \frac12(r\omega)^2\,dm=\frac12I\omega^2

Units:

I

in kg m

^2

\omega

in rad/s, and

K

in J

Boundary cases

$r=0$ mass contributes to total mass but not to $I$ about that axis.
The center-of-mass decomposition requires $I_{cm}$ , not an arbitrary parallel-axis inertia.

Each radial mass bin contributes one-half times its mass times squared radius times squared angular speed.

Build a complete kinetic-energy ledger

Choose the inertial frame and axis.
Decide whether the center of mass translates.
Evaluate $I$ about the center of mass for the rotational term.
Add translational and rotational terms once each.
Check the stated fixed quantity before making a factor comparison.

Keep these three ideas

Rotational energy is the integral of every mass element’s translational energy.
$K_{rot}$ is scalar and unchanged when the direction of rotation reverses.
Planar rigid-body motion can require both center-of-mass translation and rotation.
Fixed- $\omega$ and fixed- $L$ comparisons produce opposite inertia trends.

Energy and Momentum of Rotating Systems