Physics I: Mechanics › Rotation › full formula sheet

I = ½MR²

Say it: “the moment of inertia of a solid disk equals one-half M R squared”

Disk / solid cylinder

Half the hoop’s laziness — because half the mass lives where r² barely counts.

Notation on this page: M is the total mass, R the disk’s radius. The axis is the central symmetry axis (the axle). A solid cylinder about its own long axis uses the same formula — length doesn’t matter.

Before this lesson: Moment of inertia, Hoop / ring

Where it comes from

A disk is a hoop that filled in its middle. The rim bits still pay full R², but every inner bit pays less — and there’s a lot of inner material. The ½ is the exact bookkeeping of that discount:

Before reading on: the average of r² over the disk — is it closer to R², R²/2, or R²/4? The answer is the fraction in the formula. Guess first.
hoop
all mass at R ⇒ ⟨r²⟩ = R²
No discount: I = MR².
disk
mass spread 0–R ⇒ ⟨r²⟩ = R²/2
The r²-average over the area is exactly half — so I = M⟨r²⟩ = ½MR².

Intuition for the half: area grows with r (there’s more material in outer rings), but r² weights the center at nearly zero. The two effects balance at exactly one half — which the integral below proves.

Derivation

Slice the disk into thin concentric rings. Each ring at radius r is itself a mini-hoop with mass dm = σ·2πr dr, contributing dI = r² dm. Integrate from center to rim.

dI
=
r² dm = r² (σ·2πr dr) = 2πσ r³ dr
Step 1 — one ring. A thin ring at radius r is a hoop: dI = (its mass)×r². Its mass is density × area = σ·(2πr dr).
I
=
∫₀R 2πσ r³ dr = 2πσ [r⁴/4]₀R
Step 2 — add the rings. Integrate r³ from 0 to R. The antiderivative of r³ is r⁴/4 — this is where the ½ is born (2π/4 = π/2).
=
(πσR²) R²/2
Step 3 — regroup. 2πσ·R⁴/4 = (πσR²)·R²/2. The bracket πσR² should look familiar…
=
M R²/2
Step 4 — total mass. πσR² = (density)×(area) = M. So I = ½MR². ∎

Why doesn’t a cylinder’s length appear? Stack disks along the axis: each disk has I = ½mdiskR² about the shared axis, and I adds — Σ½miR² = ½R²Σmi = ½MR². Length changes M (more disks), never the ½.

How to use it

The procedure, every time:

  1. Confirm solid + symmetry axis: CD, coin spinning flat, pulley wheel, solid roller — uniform disk about its central axle.
  2. Read off M and R. Radius, not diameter.
  3. Compute I = ½MR². The ½ is the whole difference from the hoop — don’t drop it.
  4. Use it in Στ = Iα, K = ½Iω², or L = Iω.

Hoop or disk? Decide in one glance

Mass at the rim only (wheel, ring, band) → hoop, MR². Mass filled through the middle (coin, CD, solid roller) → disk, ½MR². When a problem says “disk” it means solid; “hoop” or “ring” means hollow.

Common mistake: using the diameter as R. A 12-cm CD has R = 0.06 m — plugging in 0.12 m quadruples I. Halve diameters on sight.

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 5-kg disk

  1. Identify. Solid disk, M = 5 kg, R = 0.2 m, central axis.
  2. Compute. I = ½ × 5 × 0.2² = 0.5 × 5 × 0.04 = 0.1 kg·m².
  3. Compare. A hoop of the same M, R would be 0.2 — the disk is exactly half as lazy.
Common mistake: writing I = MR² = 0.2 by “round thing” habit. The ½ is the disk’s identity — say “disk, half” as you write it.
Your turn — Solid disk M = 8 kg, R = 0.25 m. I = ?

Answer: 0.25 kg·m². I = 0.5 × 8 × 0.0625 = 0.25 kg·m².

Example 2 — tiny numbers: a CD

  1. Identify. CD as a solid disk: M = 0.015 kg (15 g), R = 0.06 m.
  2. Compute. I = 0.5 × 0.015 × 0.06² = 0.5 × 0.015 × 0.0036 = 2.7 × 10⁻⁵ kg·m².
  3. Read it. Microscopic I — which is why a CD spins up to thousands of rpm with a whisper of torque.
Common mistake: mixing grams and meters (15 × 6² = 540 with garbage units). Convert to kg and m first; tiny objects punish unit laziness.
Your turn — Disk M = 0.02 kg, R = 0.05 m. I = ?

Answer: 2.5 × 10⁻⁵ kg·m². I = 0.5 × 0.02 × 0.0025 = 0.000025 = 2.5 × 10⁻⁵ kg·m².

Before reading on: a 20 N tangential push at the rim of a 10-kg, 0.5-m disk. Will α be closer to 2, 8, or 20 rad/s²? Estimate, then compute.

Example 3 — angular acceleration of a pulley disk

  1. I. M = 10 kg, R = 0.5 m ⇒ I = 0.5 × 10 × 0.25 = 1.25 kg·m².
  2. Torque. F = 20 N tangential at rim: τ = 20 × 0.5 = 10 N·m.
  3. α. α = 10/1.25 = 8 rad/s².
Common mistake: computing τ with the diameter (20 × 1.0 = 20) while computing I with the radius — inconsistent R’s. Pick radius once, use it everywhere.
Your turn — Disk M = 6 kg, R = 0.4 m; tangential F = 12 N at rim. α = ?

Answer: 10 rad/s². I = 0.5 × 6 × 0.16 = 0.48; τ = 12 × 0.4 = 4.8; α = 4.8/0.48 = 10 rad/s².

Example 4 — spin energy of the disk

  1. I. M = 4 kg, R = 0.3 m ⇒ I = 0.5 × 4 × 0.09 = 0.18 kg·m².
  2. Spin. ω = 20 rad/s.
  3. K. Krot = ½ × 0.18 × 400 = 36 J.
  4. Compare. The hoop version would hold 72 J — the disk’s inner mass is “dead weight” for spin energy.
Common mistake: K = Iω² (dropping the ½) — 72 J, double the truth. Rotational KE keeps its half, just like ½mv².
Your turn — Disk M = 3 kg, R = 0.2 m, ω = 10 rad/s. Krot = ?

Answer: 3 J. I = 0.5 × 3 × 0.04 = 0.06; K = 0.5 × 0.06 × 100 = 3 J.

Memorization tips

  • Say it aloud: “disk: half M R squared.” Stress the “half” — it’s the only thing separating it from the hoop.
  • The ½’s origin story: average of r² over the disk = R²/2. If you remember why it’s half, you’ll never write the hoop’s 1 by accident.
  • Cylinder = stack of disks: length never appears. If a problem gives you the cylinder’s length and you’re computing I about its axis, that number is a decoy.
  • Radius, not diameter: halve diameters on sight, before any arithmetic.
  • Ladder check: hoop (1) > disk (½) > sphere (࡫). If your disk answer ever exceeds MR², it’s wrong.

Final challenge

Five mixed questions — basics, judgment calls, and the traps, all in one. Score 5/5 and the disk 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 solid disk?

I = ½·M·R² about its central symmetry axis: half the hoop’s value, because the disk’s mass is spread from the center out to R instead of all sitting at the rim.

Where does the ½ come from?

From integrating r² over the disk’s area: I = ∫r² dm from 0 to R gives ½MR². Equivalently, the average value of r² over a uniform disk is R²/2.

Does I = ½MR² work for a solid cylinder too?

Yes — for rotation about the cylinder’s symmetry axis. A solid cylinder is a stack of disks sharing one axis, and each disk contributes ½m·r², so the total is ½MR² regardless of length.

Which axis does the formula apply to?

The symmetry axis through the center, perpendicular to the disk’s face (the axle). Spinning about a diameter uses a different formula.

Disk vs hoop race: which wins rolling downhill?

The disk: its smaller I (half the hoop’s) means less energy locked in rotation, so more goes into translation — it reaches the bottom faster. Full race math is in the Rolling without slipping lesson.

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