Physics I: Mechanics › Rotation › full formula sheet

I = Σ miri²

Say it: “the moment of inertia is the sum of each mass times its distance from the axis, squared”

Moment of inertia

Rotational mass — and why a kilogram at the rim fights you harder than a kilogram at the hub.

Notation on this page: I is the moment of inertia (kg·m²); mi is the mass of particle i; ri is its perpendicular distance from the rotation axis (not from some other point).

Before this lesson: Newton’s 2nd (rotation)

Where it comes from

In Στ = Iα, the I arrived as Σ miri² — a sum over the body’s particles. But why squared? One r comes from torque’s lever arm, the other from the arc rule at = rα: distance gets counted twice, so it matters quadratically.

Before reading on: two 1-kg masses on a light rod — one at 1 m from the pivot, one at 2 m. Which contributes more to I, and by what factor? Guess the ratio before you compute it.
mass at 1 m
1 × 1² = 1 kg·m²
The near mass contributes its face value.
mass at 2 m
1 × 2² = 4 kg·m²
Four times the resistance — from twice the distance. The r² punishes far-flung mass.
total I
=
1 + 4 = 5 kg·m²
Moments of inertia add: each particle pays miri² into the common pot.

This is the deep reason flywheels are rims, not disks, and why figure skaters pull their arms in: moving mass inward collapses its r² contribution, so the same body gets dramatically easier to spin.

Derivation

We already derived τ = (mr²)α for one particle in the previous lesson. The moment of inertia is simply what’s left when you name that combination — then extend it to a whole body and to continuous matter.

τi
=
(mi ri²) α
Step 1 — one particle. From the last lesson: torque on particle i equals (miri²) times the shared α. Define Ii = miri².
Στi
=
(Σ miri²) α
Step 2 — sum the body. Rigidity gives every particle the same α, so it factors out. The bracket is the body’s total I.
I
=
Σ miri²
Step 3 — name it. This sum is the moment of inertia: “rotational mass” as a single number. ∎
I
=
∫ r² dm
Step 4 — go continuous. For a solid body, the particles blur into a continuum: the sum becomes an integral over mass elements dm at distance r. This integral is what produces the hoop, disk, rod, and sphere formulas in the next lessons.

Why is r measured from the axis, not from the pivot point or the center of mass? Because ri entered as the radius of particle i’s circle of motion — and circles are centered on the rotation axis. Measure from anywhere else and you’re computing a different (wrong) I.

How to use it

The procedure, every time:

  1. Fix the axis. Say it out loud: “I about this axis.” Everything downstream depends on this choice.
  2. Point masses: for each mass, find its perpendicular distance ri from the axis, compute miri², and add.
  3. Shaped bodies: don’t integrate — use the shape formulas (each has its own lesson): hoop MR², disk ½MR², rod-about-center ML²/12, rod-about-end ML²/3, solid sphere 2MR²/5.
  4. Composite bodies: I adds. A dumbbell is two point masses; a wheel with spokes is hoop + rods — sum the pieces about the same axis.
  5. Check units: kg·m². If you see kg·m or kg/m², something went wrong upstream.
Itotal = I1 + I2 + …moments of inertia add — but only about the same axisSay it: “the total moment of inertia is the sum of the parts, all about one axis”
Common mistake: measuring r from the wrong place — e.g. from one end of the rod when the axis is at its center. r is always distance from the rotation axis. Redraw the axis first; measure second.

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: two point masses

  1. List the particles. m1 = 2 kg at r1 = 0.3 m; m2 = 3 kg at r2 = 0.5 m from the axis.
  2. Each pays miri². I1 = 2 × 0.09 = 0.18; I2 = 3 × 0.25 = 0.75 (kg·m²).
  3. Add. I = 0.18 + 0.75 = 0.93 kg·m².
  4. Notice. The 3-kg mass dominates — not just because it’s heavier, but because it’s farther out: its r² is nearly triple.
Common mistake: computing miri (forgetting the square) — here that would give 0.6 + 1.5 = 2.1 with wrong units (kg·m). The square is the whole point; check your units to catch it.
Your turn — 1 kg at 0.4 m and 4 kg at 0.2 m. I = ?

Answer: 0.32 kg·m². I = 1 × 0.16 + 4 × 0.04 = 0.16 + 0.16 = 0.32. The lighter-but-farther mass ties the heavier-but-nearer one — r² at work.

Example 2 — a dumbbell about its center

  1. Model it. Two 5-kg spheres at the ends of a 1.2 m massless rod; axis through the center, perpendicular to the rod.
  2. Distances. Each mass sits r = 0.6 m from the axis.
  3. Add. I = 2 × (5 × 0.6²) = 2 × 5 × 0.36 = 3.6 kg·m².
  4. Read it. All 10 kg lives at 0.6 m — nearly four times the I of Example 1’s 5 kg total, because placement beats amount.
Common mistake: using the full 1.2 m as r for each mass. r is pivot-to-mass: half the rod, 0.6 m. Using 1.2 m would quadruple the answer.
Your turn — Two 2-kg masses at the ends of a 2 m massless rod, axis through center. I = ?

Answer: 4 kg·m². r = 1 m each: I = 2 × 2 × 1² = 4 kg·m².

Before reading on: three masses 1, 2, 3 kg at 0.1, 0.2, 0.3 m. Does the heaviest mass dominate I? Estimate the split before computing.

Example 3 — three masses, one sum

  1. Term by term. I = 1×0.1² + 2×0.2² + 3×0.3².
  2. Compute. = 1×0.01 + 2×0.04 + 3×0.09 = 0.01 + 0.08 + 0.27 = 0.36 kg·m².
  3. Read the split. The 3-kg mass supplies 0.27 of 0.36 — 75% of the total. Heaviest and farthest is a double win for I.
Common mistake: adding the masses first (1+2+3 = 6 kg) and then multiplying by something. I is a sum of products miri² — each mass keeps its own r.
Your turn — A single 2-kg mass at 0.5 m from the axis. I = ?

Answer: 0.5 kg·m². I = 2 × 0.25 = 0.5 kg·m². One term, one product — the formula at its simplest.

Example 4 — same mass, different placement

  1. Case A. All 4 kg at r = 0.5 m: I = 4 × 0.25 = 1.0 kg·m².
  2. Case B. 2 kg at 0.5 m, 2 kg at 0.25 m: I = 2×0.25 + 2×0.0625 = 0.5 + 0.125 = 0.625 kg·m².
  3. Compare. Identical total mass — but Case B is 37.5% easier to spin. Where the mass sits matters more than how much there is.
  4. The moral. This is why the shape lessons exist: a hoop (all mass at R) always out-resists a disk (mass spread inward) of the same M and R.
Common mistake: thinking “4 kg is 4 kg” and assigning both cases the same I. Mass without position is not a moment of inertia.
Your turn — 3 kg at 0.6 m vs 3 kg at 0.3 m. Both I’s?

Answer: 1.08 vs 0.27 kg·m². 3 × 0.36 = 1.08; 3 × 0.09 = 0.27. Halving the distance quarters I — the r² law in one line.

Memorization tips

  • Say it aloud: “I is the sum of m r squared.” Short, exact, unforgettable.
  • The r² mantra: “double the distance, quadruple the resistance.” One sentence that predicts every placement question.
  • I adds: composite bodies are sums of parts — but every part must be about the same axis.
  • Axis first: before any number, write “I about ___ axis.” Half of all I errors are really axis errors.
  • Units are the tripwire: kg·m². If your answer isn’t, you forgot a square or measured r wrong.
  • Shape ladder: hoop MR² > disk ½MR² > sphere 2MR²/5 — for fixed M and R, more central mass means smaller I. The ladder is the r² law.

Final challenge

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

The moment of inertia I is a body’s resistance to angular acceleration — “rotational mass.” For point masses it is the sum of m·r² over every particle, where r is each particle’s distance from the rotation axis.

Why is r squared in the moment of inertia?

One factor of r comes from torque’s lever arm and the other from the arc-length relation at = rα. Physically it means distance from the axis matters quadratically: doubling r quadruples that mass’s contribution.

Does the moment of inertia depend on the axis?

Yes — completely. I is defined relative to a specific rotation axis: a rod spun about its center has I = ML²/12 but about its end has I = ML²/3, four times larger.

What are the units of moment of inertia?

Kilogram-meters squared (kg·m²): mass times distance squared.

How do I compute I for a continuous object like a disk?

Replace the sum with an integral, I = ∫r² dm, or use the standard shape formulas: hoop MR², disk ½MR², rod about center ML²/12, rod about end ML²/3, solid sphere 2MR²/5.

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