Physics I: Mechanics › Rotation › full formula sheet

Στ = Iα

Say it: “the net torque equals the moment of inertia times the angular acceleration”

Newton’s 2nd law for rotation

F = ma, rebuilt for spinning things — net twist in, angular acceleration out.

Notation on this page: Στ is the net (signed sum) torque in N·m; I is the moment of inertia in kg·m² (the body’s rotational mass); α is the angular acceleration in rad/s².

Before this lesson: Torque

Where it comes from

You already own the linear version: F = ma. Net force in, acceleration out; mass resists. Rotation needs the same machine with rotated parts:

Before reading on: a merry-go-round — same shove, but now loaded with kids. Does it spin up faster or slower? Which quantity plays the role of mass here?
linear
ΣF = m a
Net force → acceleration. Mass m resists.
rotation
Στ = I α
Net torque → angular acceleration. I resists — the moment of inertia is literally “rotational mass.”

The merry-go-round answers the prediction: loaded with kids it spins up slower under the same shove — its I grew, so the same Στ buys less α. And just like F = ma, the Σ is doing real work: one torque shoving clockwise while friction drags counterclockwise means only their difference accelerates anything.

Derivation

Start with one particle of mass m circling at radius r, pushed tangentially. Apply F = ma to it, then multiply the whole equation by r — the r turns force into torque and mass into moment of inertia. Then add up every particle in the body.

Ft
=
m at
Step 1 — Newton on one particle. Only the tangential force component changes the speed around the circle (the radial part just bends the path). at is the tangential acceleration.
r Ft
=
r m at
Step 2 — multiply by r. Multiplying an equation by r changes nothing. But r Ft is now a torque (τ = rFt for a tangential push), and at = rα ties the arc to the angle.
τ
=
(m r²) α
Step 3 — rename. r m (rα) = (mr²)α. The combination mr² is that particle’s moment of inertia — its personal resistance to being spun.
Στ
=
(Σ miri²) α = I α
Step 4 — sum the body. A rigid body is many particles sharing one α (it rotates as a unit), so α factors out of the sum. Σ miri² is the body’s total moment of inertia I. ∎

Why does one α serve the whole body? “Rigid” means the distances between particles never change — every particle sweeps the same angle in the same time, so every particle shares the same ω and the same α. That shared α is what lets it factor out of the sum in Step 4.

How to use it

The procedure, every time:

  1. Name the body and its axis. I belongs to a specific body about a specific axis — a rod about its center and about its end have different I’s.
  2. List every torque with its sign. Applied pushes, friction, gravity, tension — each gets r, F, θ, and a +/− for direction. Then Στ.
  3. Get I. Point masses: Σ miri². Shaped bodies: the shape formulas (hoop, disk, rod, sphere — each has its own lesson).
  4. Divide: α = Στ / I. The sign of α follows the sign of Στ — angular acceleration points the way the net torque twists.
  5. Sanity-check units: N·m / (kg·m²) = 1/s² = rad/s² ✓
α = Στ / Ibigger twist → faster spin-up; bigger I → lazier responseSay it: “the angular acceleration equals the net torque over the moment of inertia”
Common mistake: plugging in a single torque instead of the net torque — e.g. computing α from the applied push while ignoring the opposing friction torque. Σ is not decoration: opposing twists subtract before anything accelerates.

Worked examples

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

Example 1 — spin up a merry-go-round

  1. Body and I. Treat it as a solid disk: M = 100 kg, R = 1.5 m ⇒ I = ½MR² = 0.5 × 100 × 2.25 = 112.5 kg·m².
  2. Torque. You push tangentially at the rim with F = 200 N: τ = rF sin 90° = 1.5 × 200 = 300 N·m. (Only one torque — it is the net.)
  3. Divide. α = 300 / 112.5 = 2.67 rad/s².
  4. Read it. Every second, the spin rate grows by 2.67 rad/s — about 25 rpm per second. A heavy wheel is lazy: the same 200 N on a bicycle wheel would give a far bigger α.
Common mistake: using the diameter (3 m) as r in both I and τ. The formulas want the radius — R = 1.5 m. Diameter-instead-of-radius is the most expensive slip in rotation problems: it quadruples I.
Your turn — Disk M = 80 kg, R = 2 m; tangential push F = 240 N at the rim. α = ?

Answer: 3 rad/s². I = ½ × 80 × 4 = 160 kg·m²; τ = 240 × 2 = 480 N·m; α = 480/160 = 3 rad/s².

Example 2 — braking a grinding wheel (hoop)

  1. Body and I. Grinding wheel as a hoop: M = 3 kg, R = 0.4 m ⇒ I = MR² = 3 × 0.16 = 0.48 kg·m².
  2. Torque. Friction drags tangentially at the rim with F = 6 N, opposing the spin: τ = −6 × 0.4 = −2.4 N·m.
  3. Divide. α = −2.4 / 0.48 = −5 rad/s² — the wheel loses 5 rad/s of spin every second.
  4. Read the sign. Negative α means angular deceleration here (opposing the current ω), not “spinning backwards.”
Common mistake: reporting “α = 5 rad/s²” and dropping the minus. The sign carries the physics — it says the wheel is slowing, not speeding up. Keep it.
Your turn — Hoop M = 2 kg, R = 0.5 m; tangential friction F = 4 N at the rim. α = ?

Answer: −4 rad/s². I = 2 × 0.25 = 0.5 kg·m²; τ = −4 × 0.5 = −2 N·m; α = −2/0.5 = −4 rad/s².

Before reading on: a motor applies +12 N·m but bearing friction drags −4 N·m. Is the angular acceleration driven by 12, by 16, or by 8? Commit, then check.

Example 3 — net torque really means net

  1. Body and I. Solid disk: M = 10 kg, R = 0.5 m ⇒ I = ½ × 10 × 0.25 = 1.25 kg·m².
  2. Sum the torques. Motor: +12 N·m. Friction: −4 N·m. Στ = 12 − 4 = +8 N·m (not 12, not 16).
  3. Divide. α = 8 / 1.25 = 6.4 rad/s².
  4. Compare. Ignoring friction would claim 12/1.25 = 9.6 rad/s² — 50% too high. The Σ earned its keep.
Common mistake: adding torque magnitudes (12 + 4 = 16) when they oppose. Signs are directions — opposing twists subtract, always.
Your turn — I = 2 kg·m²; torques +9 N·m and −3 N·m. α = ?

Answer: 3 rad/s². Στ = 9 − 3 = 6 N·m; α = 6/2 = 3 rad/s².

Example 4 — hoop vs disk: same push, different laziness

  1. Setup. Two wheels, M = 5 kg, R = 0.3 m each; same torque τ = 10 N·m. One is a hoop (I = MR²), one a solid disk (I = ½MR²).
  2. Hoop. I = 5 × 0.09 = 0.45 ⇒ α = 10/0.45 ≈ 22.2 rad/s².
  3. Disk. I = 0.225 ⇒ α = 10/0.225 ≈ 44.4 rad/s².
  4. The lesson. Same mass, same size, same twist — the disk spins up twice as fast, because its mass sits closer to the axle (smaller I). I is not just “how much” mass, but where it is.
Common mistake: assuming equal mass means equal I. It never does across shapes — always check which I formula matches the body and axis.
Your turn — τ = 8 N·m on a hoop and a disk, each M = 4 kg, R = 0.5 m. Both α’s?

Answer: hoop 8 rad/s², disk 16 rad/s². Ihoop = 4 × 0.25 = 1.0 → α = 8; Idisk = 0.5 → α = 16. Same 2:1 ratio.

Memorization tips

  • Say it aloud: “the net torque equals the moment of inertia times the angular acceleration.”
  • Map it onto F = ma: τ ↔ F, I ↔ m, α ↔ a. If you can recite the linear version, you already know the rotational one — just swap the cast.
  • I is “rotational mass”: big I = lazy spinner. Whenever a problem asks “which spins up faster,” compare I’s, not masses.
  • The Σ is the exam: most Στ = Iα problems are secretly torque-sign problems. List every torque with its sign first.
  • Unit check in 3 seconds: (N·m)/(kg·m²) = 1/s² = rad/s². If your I has wrong units, α will too.
  • α follows Στ’s sign: positive net torque → positive (counterclockwise) angular acceleration. Direction is information — don’t drop it.

Final challenge

Five mixed questions — basics, judgment calls, and the traps, all in one. Score 5/5 and the rotational second law 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 Newton’s second law for rotation?

The net torque on a rigid body equals its moment of inertia times its angular acceleration: Στ = Iα. It is the rotational version of F = ma.

How does Στ = Iα relate to F = ma?

Term by term: net torque replaces net force, moment of inertia replaces mass, and angular acceleration replaces linear acceleration. Same structure, rotational quantities.

Why is it net torque in the formula?

Opposing torques fight each other, just like opposing forces. Only the signed sum, Στ, determines the angular acceleration — a single torque in isolation misleads.

What are the units of angular acceleration?

Radians per second squared (rad/s²). Check: torque (N·m) divided by moment of inertia (kg·m²) gives 1/s², i.e. rad/s².

Does Στ = Iα work for a single particle?

Yes: for one particle I = m·r², so Στ = m·r²·α. The full rigid-body law is just this summed over every particle in the body.

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 →