Physics I: Mechanics › Rotation › full formula sheet

Krot = ½Iω²

Say it: “the rotational kinetic energy equals one-half the moment of inertia times the angular velocity squared”

Rotational kinetic energy

½mv², rebuilt for spin — the energy hiding in every flywheel, planet, and rolling ball.

Notation on this page: Krot is the spin kinetic energy in joules; I the moment of inertia (kg·m²) about the spin axis; ω the angular velocity in rad/s.

Before this lesson: Moment of inertia, Parallel-axis theorem

Where it comes from

A spinning body is a crowd of particles, each with its own ½mivi². The problem: every particle has a different vi (rim bits fly, hub bits crawl). The fix: trade v for ω, the one speed they all share:

Before reading on: particle A at radius R, particle B at R/2, same mass, same ω. What’s the ratio of their kinetic energies? (Hint: v = rω.)
KA/KB
=
(½m(Rω)²) / (½m(Rω/2)²) = 4
Same ω, double the radius → 4× the energy. Energy, like I, obeys the r² law.

So spin energy concentrates at the rim — the same r² story as the moment of inertia. Summing every particle’s ½mi(riω)² and factoring out the shared ω² is exactly the derivation below, and the Σ miri² that appears is I.

Derivation

Add up ½mivi² over every particle. Write each vi = riω, factor out the ω they all share, and the moment of inertia appears on its own.

K
=
Σ ½ mi vi²
Step 1 — start honest. Kinetic energy is ½mv² per particle, summed. No shortcuts yet.
=
Σ ½ mi (riω)²
Step 2 — trade v for ω. Each particle circles at vi = riω. Now the speeds differ only through ri.
=
½ ω² Σ mi ri²
Step 3 — factor the shared ω. Rigid rotation: one ω for the whole body, so ω² factors out of the sum.
Krot
=
½ I ω²
Step 4 — name the sum. Σ miri² = I. The ½ survived from ½mv² — rotational KE keeps its half, just like linear KE. ∎

Why ω and not v? Because ω is the body’s collective variable — every particle shares it, so it factors out. No single v exists for a spinning body (rim vs hub), which is exactly why ½mv² can’t be applied to the whole body at once.

How to use it

The procedure, every time:

  1. Get I for the shape about the spin axis (shape lessons + parallel-axis theorem).
  2. Get ω in rad/s. rpm → multiply by 2π/60 (≈ 0.1047). Degrees per second → multiply by π/180.
  3. Compute ½Iω². Units: kg·m² × 1/s² = joules ✓.
  4. Energy methods: Krot slots into conservation of energy alongside ½mv² and mgh — see the rolling lesson for the full split.

The rpm trap

120 rpm is not ω = 120. ω = 120 × 2π/60 = 4π ≈ 12.57 rad/s. Plugging rpm straight into ½Iω² inflates K by (60/2π)² ≈ 91× — the single most expensive slip in rotation energy problems.

Common mistake: K = Iω² (dropping the ½). The half survived the derivation — it’s the same half as in ½mv². “Half I omega squared”: say the “half” every time.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: a flywheel disk

  1. I. Solid disk M = 20 kg, R = 0.5 m ⇒ I = ½ × 20 × 0.25 = 2.5 kg·m².
  2. ω. 30 rad/s (already in rad/s).
  3. K. Krot = ½ × 2.5 × 900 = 1125 J.
  4. Read it. Over a kilojoule of spin — enough to lift ~115 kg by a meter. Flywheels are serious batteries.
Common mistake: squaring I instead of ω, or computing Iω² = 2250 J (dropped half). The square is on the angular velocity; the half is non-negotiable.
Your turn — Disk M = 10 kg, R = 0.4 m, ω = 25 rad/s. Krot = ?

Answer: 250 J. I = 0.5 × 10 × 0.16 = 0.8; K = 0.5 × 0.8 × 625 = 250 J.

Example 2 — rpm conversion: a spinning hoop

  1. I. Hoop M = 2 kg, R = 0.3 m ⇒ I = 2 × 0.09 = 0.18 kg·m².
  2. Convert. 120 rpm → ω = 120 × 2π/60 = 4π ≈ 12.57 rad/s.
  3. K. Krot = ½ × 0.18 × (12.57)² = 0.09 × 157.9 = 14.2 J.
  4. The trap, dodged. Plugging 120 straight in would claim ~1300 J — 91× too big. Convert first, always.
Common mistake: ω = rpm/60 = 2 (“revolutions per second”) — still not rad/s. A revolution is 2π radians; revolutions-per-second ≠ radians-per-second.
Your turn — Hoop M = 1 kg, R = 0.5 m at 60 rpm. Krot = ?

Answer: ≈ 4.9 J. I = 0.25; ω = 60 × 2π/60 = 2π ≈ 6.283 rad/s; K = 0.5 × 0.25 × 39.48 = 4.93 ≈ 4.9 J.

Before reading on: a 3-kg, 2-m rod spun about its center at 4 rad/s. I = 1.0. Is K closer to 4 J, 8 J, or 16 J? Estimate, then compute.

Example 3 — a spinning rod

  1. I. Rod M = 3 kg, L = 2 m, center axis ⇒ I = (1/12) × 3 × 4 = 1.0 kg·m².
  2. K. ω = 4 rad/s ⇒ Krot = ½ × 1.0 × 16 = 8 J.
  3. Compare. Spun about its end (I = 4.0), the same ω would hold 32 J — 4× the energy. Axis choice is energy choice.
Common mistake: using the end I (4.0) for a center-spun rod — or vice versa. The axis in the problem statement picks the I; there are no do-overs.
Your turn — Rod M = 2 kg, L = 3 m, center axis, ω = 6 rad/s. Krot = ?

Answer: 27 J. I = (1/12) × 2 × 9 = 1.5; K = 0.5 × 1.5 × 36 = 27 J.

Example 4 — hoop vs disk: energy storage showdown

  1. Setup. M = 4 kg, R = 0.5 m, ω = 10 rad/s for both.
  2. Hoop. I = 4 × 0.25 = 1.0 ⇒ K = ½ × 1.0 × 100 = 50 J.
  3. Disk. I = 0.5 ⇒ K = 25 J.
  4. The lesson. Same mass, same size, same spin — the hoop stores twice the energy. Flywheel engineers aren’t being decorative: rim mass is energy mass.
Common mistake: “same M, R, ω → same energy.” K scales with I, and I is shape-dependent. The shape is the energy rating.
Your turn — M = 6 kg, R = 0.2 m, ω = 20 rad/s. Hoop K and disk K?

Answer: 48 J and 24 J. Hoop: I = 6 × 0.04 = 0.24, K = 0.5 × 0.24 × 400 = 48; disk: half, 24 J.

Memorization tips

  • Say it aloud: “half I omega squared” — lead with the “half” so you never drop it.
  • Map onto ½mv²: I ↔ m, ω ↔ v. If you know linear KE, you know rotational KE — swap the cast.
  • rpm → rad/s first: ×2π/60 (≈ ÷9.55). Make it step zero of every energy computation.
  • K scales with I: “which stores more spin energy?” is always “which has the bigger I?” — the hoop wins, the sphere loses.
  • Units check: (kg·m²)(1/s²) = kg·m²/s² = J ✓. If it isn’t joules, ω wasn’t in rad/s.

Final challenge

Five mixed questions — basics, judgment calls, and the traps, all in one. Score 5/5 and rotational kinetic energy is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is rotational kinetic energy?

Krot = ½·I·ω²: the kinetic energy stored in a body’s spin. It is the rotational analog of ½·m·v², with I replacing m and ω replacing v.

Why does it use ω instead of v?

Different bits of a spinning body move at different speeds (v = rω), so there is no single v. But every bit shares one ω — factoring it out of the summed ½mivi² leaves ½·I·ω².

Must ω be in rad/s?

Yes. The derivation uses v = rω with ω in radians per second. Convert rpm first: ω = rpm · 2π/60. Using rpm directly gives nonsense units and numbers.

Can a body have translational and rotational KE at once?

Yes — a rolling wheel has K = ½mvcm² + ½Iω²: the center of mass translates while the body spins about it. The rolling-without-slipping lesson covers this.

Which shape stores the most spin energy for fixed M, R, ω?

The hoop: K = ½Iω² scales with I, and the hoop’s I = MR² is the largest. That’s why flywheels are hoops — maximum energy per kilogram.

More from the codex

Live

Physics I formula sheet

All the mechanics formulas — kinematics, forces, energy — printable and quiz-ready.

Open sheet → Live

Formula Sheet Builder

Mix and match any sections into your own printable sheet.

Open tool → Live

Prompt Simulator

Practice prompt engineering with deterministic scoring.

Open tool →