Physics I: Mechanics › Rotation › full formula sheet

I = MR²

Say it: “the moment of inertia of a hoop equals its mass times its radius squared”

Hoop / ring

All the mass at the rim — the laziest spinner of its size, and the simplest I formula there is.

Notation on this page: M is the hoop’s total mass, R its radius. This formula is for a thin ring spinning about its central symmetry axis (the axle through the middle, perpendicular to the ring’s plane).

Before this lesson: Moment of inertia

Where it comes from

The general rule is I = Σ miri². A thin hoop is the one shape where the sum is trivial — because there is nothing to vary:

Before reading on: a hoop and a solid disk, same mass M, same radius R. Every bit of the hoop is at distance R; the disk’s mass is spread from 0 to R. Which has the bigger I — and roughly by what factor?
hoop
every mi at ri = R
The defining property of a thin ring: one radius for all the mass.
disk
mass spread over 0 ≤ r ≤ R
Inner mass pays small r² — it barely contributes. The disk must come out smaller.

The hoop wins the laziness contest: with everything parked at the maximum distance, no shape of the same M and R resists spinning more. That’s exactly why flywheels are built as rims — and why a bicycle wheel is mostly rim plus light spokes.

Derivation

Apply I = Σ miri² to a thin ring. Since every particle sits at the same distance R from the central axis, the ri² is common to all terms — factor it out and the sum collapses to the total mass.

I
=
Σ mi ri²
Step 1 — start general. The definition: sum over every bit of mass.
=
Σ mi R²
Step 2 — use thinness. For a thin ring about its central axis, ri = R for every particle. Replace each ri with R.
=
R² Σ mi
Step 3 — factor. R² is the same in every term, so it factors out of the sum — the one-line miracle that only works because the ring is thin.
=
MR²
Step 4 — total mass. Σ mi = M, the whole hoop’s mass. ∎

Why doesn’t this work for a disk? Step 2 fails: a disk’s particles sit at many different radii, so ri² can’t factor out. The sum has to be done properly (an integral), and the inner mass drags the answer down to ½MR² — the next lesson.

How to use it

The procedure, every time:

  1. Confirm it’s a thin ring about its central axis: bicycle wheel, hula hoop, flywheel rim, wedding band. Mass concentrated at one radius.
  2. Read off M and R. R is the ring’s radius — the distance from center to the band.
  3. Compute I = MR². No fractions, no constants — the simplest shape formula on the sheet.
  4. Plug into Στ = Iα or K = ½Iω² as the problem demands.

When NOT to use it

Spinning about a diameter (like a coin flipping)? Different formula — not on this sheet, don’t improvise. Thick ring (a donut with real width)? The mass isn’t all at one R. Solid disk? That’s ½MR² — half the laziness.

Common mistake: using MR² for anything round. A solid cylinder, a sphere, a disk — each has its own fraction. MR² is the maximum; reaching for it by default overestimates every other shape.

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 bicycle wheel

  1. Model. Wheel as a thin hoop: M = 1.5 kg, R = 0.35 m, central axis.
  2. Compute. I = MR² = 1.5 × 0.35² = 1.5 × 0.1225 = 0.184 kg·m² (3 s.f.).
  3. Sanity check. Units kg·m² ✓; a light wheel gives a small I ✓.
Common mistake: squaring M instead of R, or forgetting to square at all (1.5 × 0.35 = 0.525 with wrong units). The square belongs on the distance.
Your turn — Hoop M = 2 kg, R = 0.4 m. I = ?

Answer: 0.32 kg·m². I = 2 × 0.16 = 0.32 kg·m².

Example 2 — spin it up: α from a torque

  1. I. Hoop M = 4 kg, R = 0.5 m ⇒ I = 4 × 0.25 = 1.0 kg·m².
  2. Torque. Tangential push F = 6 N at the rim: τ = 6 × 0.5 = 3 N·m.
  3. α. α = τ/I = 3/1.0 = 3 rad/s².
Common mistake: dividing torque by mass (3/4) instead of by I. Στ = Iα wants the moment of inertia — mass alone is never the divisor in rotation.
Your turn — Hoop M = 3 kg, R = 0.4 m; tangential F = 6 N at rim. α = ?

Answer: 5 rad/s². I = 3 × 0.16 = 0.48; τ = 6 × 0.4 = 2.4; α = 2.4/0.48 = 5 rad/s².

Before reading on: a hoop and a disk, same M, R, and spin rate ω. Which carries more rotational kinetic energy? Ratio?

Example 3 — rotational kinetic energy of the hoop

  1. I. Hoop M = 2 kg, R = 0.3 m ⇒ I = 2 × 0.09 = 0.18 kg·m².
  2. Spin. ω = 10 rad/s.
  3. K. Krot = ½Iω² = 0.5 × 0.18 × 100 = 9 J.
  4. Compare. A disk with the same M, R, ω would carry only 4.5 J — the hoop stores twice the spin energy. (This is why flywheels are hoops.)
Common mistake: using ω in rpm without converting. 10 rad/s ≠ 10 rpm. Always convert: ω(rad/s) = rpm × 2π/60.
Your turn — Hoop M = 1 kg, R = 0.5 m, ω = 8 rad/s. Krot = ?

Answer: 8 J. I = 1 × 0.25 = 0.25; K = 0.5 × 0.25 × 64 = 8 J.

Example 4 — hoop vs disk, head to head

  1. Setup. M = 6 kg, R = 0.2 m for both.
  2. Hoop. I = 6 × 0.04 = 0.24 kg·m².
  3. Disk. I = ½ × 6 × 0.04 = 0.12 kg·m².
  4. The lesson. Same mass, same size — the hoop resists twice as hard. Under the same torque it spins up half as fast; at the same ω it carries twice the energy.
Common mistake: memorizing “round things are ½MR².” Only solid disks are. Hoops are the full MR² — the 1 vs ½ distinction is worth marks.
Your turn — M = 10 kg, R = 0.3 m. Hoop I and disk I?

Answer: hoop 0.9, disk 0.45 kg·m². Hoop: 10 × 0.09 = 0.9; disk: half that, 0.45.

Memorization tips

  • Say it aloud: “hoop: M R squared — no fraction.” The absence of a fraction is the thing to remember.
  • The ladder: hoop (1) > disk (½) > sphere (࡫) — for fixed M, R, more rim-mass means bigger I. The hoop sits at the top.
  • One-radius picture: close your eyes and see every atom of the ring at distance R. If you can picture that, the derivation is one line and you’ll never blank on it.
  • Thin is load-bearing: the formula dies the moment mass spreads over multiple radii. Thick ring, disk, sphere — different formulas.
  • Axis check: central axle, perpendicular to the ring. A flipping coin (diameter axis) is a different problem entirely.

Final challenge

Five mixed questions — basics, judgment calls, and the traps, all in one. Score 5/5 and the hoop 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 the moment of inertia of a hoop?

I = M·R² about its central axis: every bit of a thin hoop sits at distance R from the axis, so the sum of m·r² is just R² times the total mass.

Why is the hoop’s I larger than a disk’s?

A hoop keeps all its mass at the maximum distance R, while a disk spreads mass inward where r² contributions are smaller. For the same M and R, the hoop’s I = MR² is twice the disk’s ½MR².

Which axis does I = MR² apply to?

The symmetry axis through the center, perpendicular to the ring’s plane — the axle a wheel spins on. About any other axis (e.g. a diameter) the formula is different.

Does the thickness of the ring matter?

The formula assumes a thin ring: all mass at essentially one radius R. A thick ring needs a slightly different formula; for bicycle wheels and hula hoops the thin-ring formula is the right one.

What are real examples of hoops in rotation problems?

Bicycle wheels (rim-dominated), hula hoops, flywheels, and gymnastic rings. Any time the mass is concentrated at the rim, I = MR² is the model.

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 →