AP Physics 1 · Unit 6 · Topic 6.1

Rotational Kinetic Energy

A rotating rigid system has kinetic energy because every mass element away from the axis has linear speed. Rotational kinetic energy collects all of those particle energies into one expression using the system's rotational inertia and angular velocity.

Titration response · Rotational Kinetic EnergyThe equivalence region produces the steepest change; its location depends on stoichiometry and acid-base strength.

Learning Goals

After this lesson, you should be able to:

derive \(K_{\mathrm{rot}}=\frac12I\omega^2\) from particle kinetic energy;
calculate rotational kinetic energy about a stated axis;
explain how angular speed and mass distribution affect stored kinetic energy;
compare rigid systems without unnecessary numerical calculation;
distinguish pure fixed-axis rotation from center-of-mass translation plus rotation;
interpret \(K\)-versus-\(\omega\) and \(K\)-versus-\(\omega^2\) graphs;
apply energy accounting to a rotating rigid system; and
justify why rotational kinetic energy is a nonnegative scalar.

1. The Rotational Kinetic-Energy Model

For a rigid system rotating about a fixed axis,

\[\boxed{K_{\mathrm{rot}}=\frac12I\omega^2}\]

\(I\) is the rotational inertia about that same axis and \(\omega\) is the angular velocity. The SI unit is the joule:

\[(\mathrm{kg\,m^2})(\mathrm{rad/s})^2=\mathrm{kg\,m^2/s^2}=\mathrm J\]

Radians are dimensionless in this unit check. The equation describes energy associated with motion; rotational inertia itself is not energy.

2. One Angular Speed, Many Linear Speeds

**Rotational Kinetic Energy**Topic-matched fine pixel-art model · use the lesson's equations and labeled diagrams for exact scientific meaning.

3. Derivation from the Particles in a Rigid System

Divide a rigid system into particles \(m_i\) at perpendicular distances \(r_i\) from the axis. Each particle has tangential speed \(v_i=\omega r_i\), so the total kinetic energy is

\[K=\sum_i\frac12m_iv_i^2=\sum_i\frac12m_i(\omega r_i)^2\]

All particles share the same angular speed, allowing \(\omega^2\) to be factored out:

\[K=\frac12\left(\sum_i m_ir_i^2\right)\omega^2\]

Because \(I=\sum_i m_ir_i^2\),

\[K_{\mathrm{rot}}=\frac12I\omega^2\]

Thus rotational kinetic energy is not a separate mysterious kind of energy; it is the sum of the ordinary linear kinetic energies of the system's moving parts.

4. Rotational Kinetic Energy Is a Scalar

Clockwise and counterclockwise angular velocities may receive opposite signs, but \(\omega\) is squared:

\[K_{\mathrm{rot}}(+\omega)=K_{\mathrm{rot}}(-\omega)\]

Energy therefore has no clockwise or counterclockwise direction. It is nonnegative for an ordinary rigid system because \(I\ge0\) and \(\omega^2\ge0\). Reversing rotation without changing angular-speed magnitude does not change rotational kinetic energy, although an interaction may be required to accomplish the reversal.

5. Functional Dependence and Ratio Reasoning

Change	\(K_{\mathrm{rot}}\) factor	Reason
\(I\to2I\), fixed \(\|\omega\|\)	\(2\)	Energy is directly proportional to \(I\)
\(\|\omega\|\to2\|\omega\|\), fixed \(I\)	\(4\)	Energy depends on \(\omega^2\)
\(I\to I/2\), \(\|\omega\|\to2\|\omega\|\)	\(2\)	\((1/2)(2^2)=2\)
\(\omega\to-\omega\), fixed \(I\)	\(1\)	Direction changes, speed does not

For two states of the same or different systems,

\[\frac{K_2}{K_1}=\frac{I_2}{I_1}\left(\frac{\omega_2}{\omega_1}\right)^2\]

6. Worked Example: Uniform Disk

A \(6.0\,\mathrm{kg}\) uniform disk of radius \(0.40\,\mathrm m\) rotates at \(12\,\mathrm{rad/s}\) about its center. Use the provided formula \(I=\frac12MR^2\).

\[I=\frac12(6.0)(0.40)^2=0.48\,\mathrm{kg\,m^2}\]

\[K_{\mathrm{rot}}=\frac12(0.48)(12)^2=34.6\,\mathrm J\]

Answer: The disk has \(34.6\,\mathrm J\) of rotational kinetic energy. The axis matters because it determines the appropriate \(I\).

7. Worked Example: A Point-Mass System

Two \(0.50\,\mathrm{kg}\) masses are attached to a light rigid bar, each \(0.30\,\mathrm m\) from the central axis. The system rotates at \(8.0\,\mathrm{rad/s}\).

\[I=2mr^2=2(0.50)(0.30)^2=0.090\,\mathrm{kg\,m^2}\]

\[K_{\mathrm{rot}}=\frac12(0.090)(8.0)^2=2.88\,\mathrm J\]

Directly adding the two particle energies gives the same result because each particle has \(v=\omega r=(8.0)(0.30)=2.4\,\mathrm{m/s}\):

\[2\left[\frac12(0.50)(2.4)^2\right]=2.88\,\mathrm J\]

8. Mass Distribution Matters

A uniform disk and thin hoop have the same mass \(M\), radius \(R\), and angular-speed magnitude. With the provided formulas \(I_{\mathrm{disk}}=\frac12MR^2\) and \(I_{\mathrm{hoop}}=MR^2\),

\[\frac{K_{\mathrm{hoop}}}{K_{\mathrm{disk}}}=\frac{I_{\mathrm{hoop}}}{I_{\mathrm{disk}}}=2\]

The hoop has twice the rotational kinetic energy at the same \(\omega\) because more mass is far from the axis. This comparison holds only when the quantities stated as equal really are held fixed; a different constraint, such as equal energy, gives a different angular-speed comparison.

9. Equal Energy Does Not Mean Equal Angular Speed

If two rotating systems have the same rotational kinetic energy,

\[\frac12I_A\omega_A^2=\frac12I_B\omega_B^2\]

\[\frac{|\omega_A|}{|\omega_B|}=\sqrt{\frac{I_B}{I_A}}\]

If \(I_A=4I_B\), system A needs only half the angular-speed magnitude to have the same rotational kinetic energy. The square root is essential.

10. Converting Revolutions per Minute

The energy equation requires angular speed in radians per second. A flywheel rotating at \(600\,\mathrm{rev/min}\) has

\[\omega=600\frac{\mathrm{rev}}{\mathrm{min}}\left(\frac{2\pi\,\mathrm{rad}}{1\,\mathrm{rev}}\right)\left(\frac{1\,\mathrm{min}}{60\,\mathrm s}\right)=20\pi\,\mathrm{rad/s}\]

If \(I=0.20\,\mathrm{kg\,m^2}\), then

\[K_{\mathrm{rot}}=\frac12(0.20)(20\pi)^2\approx395\,\mathrm J\]

Substituting \(600\) directly as though it were radians per second would overestimate the energy dramatically.

11. The Center of Mass Can Be at Rest

A wheel spinning on a fixed axle has \(v_{\mathrm{cm}}=0\), yet points away from the axis move and possess kinetic energy. Its kinetic energy is

\[K_{\mathrm{total}}=K_{\mathrm{rot}}=\frac12I\omega^2\]

“The object is not going anywhere” refers only to its center of mass. It does not imply that all its particles are stationary.

12. Translation Plus Rotation About the Center of Mass

For a rigid system whose center of mass translates while the system rotates about its center of mass, the kinetic energies add:

\[\boxed{K_{\mathrm{total}}=\frac12Mv_{\mathrm{cm}}^2+\frac12I_{\mathrm{cm}}\omega^2}\]

An \(8.0\,\mathrm{kg}\) wheel has \(v_{\mathrm{cm}}=3.0\,\mathrm{m/s}\), \(I_{\mathrm{cm}}=0.36\,\mathrm{kg\,m^2}\), and \(\omega=10\,\mathrm{rad/s}\):

\[K_{\mathrm{trans}}=\frac12(8.0)(3.0)^2=36\,\mathrm J\]

\[K_{\mathrm{rot}}=\frac12(0.36)(10)^2=18\,\mathrm J\]

\[K_{\mathrm{total}}=54\,\mathrm J\]

No rolling condition was assumed here; \(v_{\mathrm{cm}}\) and \(\omega\) were given independently. Topic 6.5 later adds the no-slip relationship when appropriate.

13. Graphs Reveal the Model

At fixed \(I\), a graph of \(K_{\mathrm{rot}}\) versus \(\omega\) is an upward-opening parabola symmetric about \(\omega=0\). A graph of \(K_{\mathrm{rot}}\) versus \(\omega^2\) is linear:

\[K_{\mathrm{rot}}=\left(\frac I2\right)\omega^2\]

Its slope is \(I/2\), so \(I=2(\text{slope})\). If two systems produce lines with slopes \(s_A\) and \(s_B\), then \(I_A/I_B=s_A/s_B\). A nonzero vertical intercept suggests unremoved background energy or a measurement offset rather than the ideal model.

14. Energy Accounting for a Rotating System

Choose the system boundary before writing an energy equation. If a rigid body begins from rest and loses \(24\,\mathrm J\) of potential energy while \(4.0\,\mathrm J\) is transferred to thermal energy, then \(20\,\mathrm J\) remains for rotational kinetic energy:

\[\frac12I\omega_f^2=20\,\mathrm J\]

For \(I=0.40\,\mathrm{kg\,m^2}\),

\[\omega_f=\sqrt{\frac{2(20)}{0.40}}=10\,\mathrm{rad/s}\]

Do not silently set lost potential energy equal to kinetic energy if energy also leaves the selected mechanical system.

15. Investigation: Test the \(K_{\mathrm{rot}}\) Model

Use a low-friction rotating platform, two movable equal masses, ruler, balance, and video or rotary-motion sensor.

Measure each movable mass and its distance from the axis; include the platform's known or separately determined rotational inertia.
Record angular speed for several trials and calculate \(\omega^2\).
Calculate rotational kinetic energy by summing \(\frac12m_iv_i^2\) for tracked mass points, using their measured tangential speeds.
Plot the independently summed particle energy against \(\omega^2\).
Compare twice the best-fit slope with \(I=\sum m_ir_i^2\) plus the platform contribution.
Move equal masses farther outward and repeat.
Identify uncertainty from axis location, radius measurement, sensor timing, platform inertia, wobble, and friction.

The two calculation routes should agree within uncertainty because the rotational expression is a compact version of the particle-energy sum.

16. Common Traps and Key Takeaways

Use rotational inertia about the stated axis.
Square the angular speed; doubling \(|\omega|\) quadruples energy.
Do not assign a direction to energy.
Clockwise and counterclockwise motion can have equal kinetic energy.
A stationary center of mass does not imply zero rotational energy.
Add translational and center-of-mass rotational terms when both motions occur.
Do not count particle kinetic energy and \(\frac12I\omega^2\) twice; they represent the same rotational energy.
Convert rpm to rad/s before substitution.
State which quantities are held fixed when comparing mass distributions.

Checkpoint

A \(4.0\,\mathrm{kg}\), \(0.30\,\mathrm m\)-radius rigid system moves with center-of-mass speed \(5.0\,\mathrm{m/s}\) and rotates at \(8.0\,\mathrm{rad/s}\). Case A is a provided uniform disk with \(I_{\mathrm{cm}}=\frac12MR^2\); case B is a provided thin hoop with \(I_{\mathrm{cm}}=MR^2\). For each case, find translational, rotational, and total kinetic energy. Compare the cases at the given equal angular speed, then determine the hoop's angular speed if it instead has the same rotational kinetic energy as the disk. Explain why reversing either rotation leaves every energy value unchanged, and sketch both systems on a \(K_{\mathrm{rot}}\)-versus-\(\omega^2\) graph.

Concept scope cross-checked with the College Board AP Physics 1 Course and Exam Description and the open educational references OpenStax University Physics: Moment of Inertia and Rotational Kinetic Energy and OpenStax College Physics for AP Courses: Rotational Kinetic Energy. Explanations, derivation, diagram, numerical examples, tables, graph analysis, checkpoint, and investigation are independently written for Insomnia Studyroom.

Official curriculum reference: College Board AP Physics 1 course page. The explanation and worked example are independently written for this study site.