Unit 6 · Topic 6.1

6.1 Rotational Kinetic Energy

A spinning object has rotational kinetic energy $K = \tfrac{1}{2}I\omega^2$ , even if its center of mass stays still. An object that moves and spins at the same time has translational kinetic energy plus rotational kinetic energy about its center of mass.

Key terms

rotational kinetic energy
rotational inertia
angular velocity
translational kinetic energy
total kinetic energy

What changed in the 2024 update

Before the Fall 2024 course update, this topic was part of the Rotation unit (old Unit 5). It's now in Unit 6, Energy and Momentum of Rotating Systems, so older videos may file it under rotation.

Where the formula comes from

A spinning wheel's center doesn't move, but every other bit of the wheel does. Each tiny piece of mass m at distance r from the axis moves at speed v = rω, so it has kinetic energy ½mv² = ½mr²ω².

Add up all the pieces. They share the same ω, so $K_{\text{rot}} = \sum \tfrac{1}{2}m_i r_i^2\omega^2 = \tfrac{1}{2}\left(\sum m_i r_i^2\right)\omega^2 = \tfrac{1}{2}I\omega^2$

Rotational kinetic energy is just the ordinary kinetic energy of all the moving pieces, written in a form that's easier to use. It's a scalar, measured in joules, and it's never negative.

Moving and spinning at once

A rolling ball or a thrown, spinning rod has two kinds of kinetic energy. The total is $K_{\text{total}} = \tfrac{1}{2}Mv_{\text{cm}}^2 + \tfrac{1}{2}I_{\text{cm}}\omega^2$ The first term is for the center of mass moving. The second is for spinning about the center of mass.

There's a shortcut for an object turning about a fixed axle or pivot. Use $K = \tfrac{1}{2}I_{\text{pivot}}\omega^2$ with I about the pivot, and that's the whole kinetic energy. Don't add $\tfrac{1}{2}Mv_{\text{cm}}^2$ on top: the parallel axis term inside $I_{\text{pivot}}$ already counts it.

How K depends on I and ω

K grows with the square of ω. Double the spin rate and the rotational kinetic energy is four times as large.

At the same ω, the object with more rotational inertia stores more energy. A hoop spinning at the same rate as a solid disk of equal mass and radius has twice the kinetic energy, because MR² is twice ½MR². This is why flywheels, which store energy, put most of their mass near the rim.

Change	Effect on rotational K
double ω	×4
double I (same ω)	×2
double the radius of a disk (same M and ω)	×4

Energy with rotation

Energy conservation from Unit 3 still works. You just add rotational kinetic energy to your list of energies. When a pivoted rod swings down, gravitational potential energy turns into rotational kinetic energy. Measure the rod's height change at its center of mass.

In energy bar charts, show translational and rotational kinetic energy as separate bars when both are present.

Systems with pulleys work the same way. When a block hangs from a string wrapped around a pulley that has mass, the falling block's lost potential energy splits between the block's kinetic energy and the pulley's rotational kinetic energy. With a = Rα and v = Rω linking them, you can find the block's speed after it falls a height h without ever finding the tension.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Energy in a flywheel
A flywheel is a uniform solid disk of mass 50 kg and radius 0.40 m. It spins at 300 rad/s. How much kinetic energy does it store?
Show the solutionHide the solution
1. Step 1: $I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(50)(0.40)^2 = 4.0$ kg·m².
2. Step 2: $K = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}(4.0)(300)^2 = 1.8\times10^{5}$ J.
Answer: 1.8 × 10⁵ J (180 kJ)
Example 2Calculator allowed
Kinetic energy of a rolling ball
A solid sphere (I = ⅖MR²) of mass 0.50 kg rolls without slipping at 2.0 m/s. Find its translational, rotational and total kinetic energy.
Show the solutionHide the solution
1. Step 1: Translational: $\tfrac{1}{2}Mv^2 = \tfrac{1}{2}(0.50)(2.0)^2 = 1.0$ J.
2. Step 2: Rolling without slipping means $\omega = \dfrac{v}{R}$ (topic 6.5). Rotational: $\tfrac{1}{2}\left(\tfrac{2}{5}MR^2\right)\left(\dfrac{v}{R}\right)^2 = \tfrac{1}{5}Mv^2 = 0.40$ J. The radius cancels.
3. Step 3: Total: 1.0 + 0.40 = 1.4 J. So 2/7 of the ball's kinetic energy is rotational.
Answer: 1.0 J translational, 0.40 J rotational, 1.4 J total
Example 3Calculator allowed
Swinging rod (classic trap)
A uniform rod of length 1.2 m is pivoted at one end, held horizontal and released from rest. How fast is it turning when it hangs straight down? How fast is its free end moving?
Show the solutionHide the solution
1. Step 1: The center of mass falls L/2 = 0.60 m. Energy: $Mg\dfrac{L}{2} = \tfrac{1}{2}I_{\text{pivot}}\omega^2$ , with $I_{\text{pivot}} = \tfrac{1}{3}ML^2$ .
2. Step 2: Solve: $\omega = \sqrt{\dfrac{3g}{L}} = \sqrt{\dfrac{3(9.8)}{1.2}} \approx 4.95$ rad/s. The free end moves at $L\omega \approx 5.9$ m/s.
3. Step 3: The trap is writing $\tfrac{1}{2}Mv_{\text{cm}}^2 + \tfrac{1}{2}I_{\text{pivot}}\omega^2$ . That counts the center-of-mass motion twice. Use either $\tfrac{1}{2}I_{\text{pivot}}\omega^2$ alone, or $\tfrac{1}{2}Mv_{\text{cm}}^2 + \tfrac{1}{2}I_{\text{cm}}\omega^2$ . Both give $\tfrac{1}{6}ML^2\omega^2$ .
Answer: ω ≈ 4.95 rad/s; the free end moves at about 5.9 m/s

Common mistakes

Saying an object has no kinetic energy because its center of mass is at rest. A spinning wheel on a fixed axle has rotational kinetic energy.
Adding ½Mv²_cm to ½I_pivot ω² for an object on a fixed pivot. That double-counts.
Forgetting the rotational term for a rolling object, which makes your predicted speed too high.
Using the height change of the end of a swinging rod instead of its center of mass.

On the exam

Ranking questions about objects spinning or rolling often come down to how the mass is spread out. Compare the fraction of I to MR², not the masses.
When you justify a claim with energy, name the energy forms and say which grows and which shrinks. Just writing "conservation of energy" doesn't earn credit.

Connected topics

Videos

Topic 6.1 - Rotational Kinetic Energy
Lessons With LondotWatch on YouTube (opens in a new tab)
Rotational kinetic energy of rigid systems | AP Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Rotational Kinetic Energy Made Easy (with Examples) | AP Physics 1 - Unit 6 Lesson 1
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Moment of Inertia Introduction and Rotational Kinetic Energy Derivation
Flipping PhysicsWatch on YouTube (opens in a new tab)
37.1 Kinetic Energy of Translation and Rotation
MIT OpenCourseWareWatch on YouTube (opens in a new tab)
Rotational Kinetic Energy and Moment of Inertia Examples & Physics Problems
The Organic Chemistry TutorWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 6.1 Rotational Kinetic Energy. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A uniform solid disk of mass 2.0 kg and radius 0.30 m spins at 10 rad/s about its central axis. What is its kinetic energy?

Question 2 of 4Calculator allowed

Uniform disks A and B have the same mass. B has twice A's radius and spins at half A's angular speed. What is the ratio of B's rotational kinetic energy to A's?

Question 3 of 4Calculator allowed

A uniform solid disk rolls without slipping along a level floor. What fraction of its total kinetic energy is rotational?

Question 4 of 4Calculator allowed

A uniform rod of mass 0.60 kg and length 1.0 m rotates at 5.0 rad/s about a fixed axis through one end, perpendicular to the rod. What is its kinetic energy?

0 of 4 answered