Physics I: Mechanics › Rotation › full formula sheet

I = Icm + Md²

Say it: “the moment of inertia about any parallel axis equals the center-of-mass moment plus M d squared”

Parallel-axis theorem

Move the axle, pay the toll — every shift away from the center of mass costs exactly Md².

Notation on this page: Icm is the moment of inertia about a parallel axis through the center of mass; d is the distance between the two parallel axes; M is the body’s total mass.

Before this lesson: Moment of inertia, Rod about center

Where it comes from

Every shape formula so far was about a center axis. But doors hinge at edges, pendulums pivot at tops, wheels roll about contact points. Re-integrating for every shifted axis would be miserable — the theorem says you never have to:

Before reading on: a rod about its center is ML²/12. Slide the axis to the end (d = L/2). Predict the new I from Icm + Md² — then check it against the ML²/3 you already know.
rod, end axis
I = ML²/12 + M(L/2)² = ML²/12 + ML²/4
= (1 + 3)ML²/12 = ML²/3 ✓ — matches the direct integral exactly.

One line, no integral — and it proves the 4× center-to-end factor you memorized earlier (ML²/4 is 3× ML²/12, plus the original). The theorem turns every “weird axis” problem into arithmetic.

Derivation

Measure each particle’s position xi from the CM, and put the new axis at distance d from the CM (parallel). Particle i’s distance from the new axis is (xi − d) — square it, sum over particles, and watch the middle term die.

I
=
Σ mi (xi − d)²
Step 1 — shift coordinates. Distance from the new axis is (xi − d), with xi measured from the CM. Square it: the definition of I about the new axis.
=
Σ mixi² − 2d Σ mixi + d² Σ mi
Step 2 — expand. (xi − d)² = xi² − 2dxi + d². Three sums appear.
=
Icm − 2d (0) + Md²
Step 3 — the cross term dies. Σ mixi = 0 by the definition of the center of mass (first moment about the CM vanishes). Σ mixi² = Icm; Σ mi = M.
I
=
Icm + Md²
Step 4 — done. Only the CM-based start makes the middle term vanish — which is exactly why the theorem must start at the CM. ∎

Why can’t I shift from a non-CM axis? Step 3’s cancellation used Σ mixi = 0, true only with x measured from the CM. Start anywhere else and the −2d Σ mixi term survives — the one-line formula breaks. Route through the CM, always.

How to use it

The procedure, every time:

  1. Start at the CM. Look up (or compute) Icm for the shape about its center axis — the shape lessons give you these.
  2. Measure d: the distance between the CM axis and your target axis. Axes must be parallel.
  3. Add Md². I = Icm + Md². The toll is always positive — shifted axes are always lazier.
  4. Never shift twice directly: going from axis A to axis B (neither at CM)? Go A → CM → B: IB = IA − MdA² + MdB².
Imin = Icmthe center-of-mass axis is the laziest-axis minimum — every shift addsSay it: “the C M axis gives the smallest moment of inertia; shifting away only adds”
Common mistake: shifting between two non-CM axes with M×(their separation)². Wrong — the formula only spans from the CM. Detour through Icm first.

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: rod about its end (again, fast)

  1. Icm. Rod M = 2 kg, L = 3 m: Icm = ML²/12 = (1/12) × 2 × 9 = 1.5 kg·m².
  2. d. End axis is L/2 = 1.5 m from the center: d = 1.5 m.
  3. Toll. Md² = 2 × 2.25 = 4.5 kg·m².
  4. Total. I = 1.5 + 4.5 = 6.0 kg·m² = ML²/3 ✓ — the integral’s answer, with no integral.
Common mistake: d = L (the full length) instead of L/2 (center-to-end). d is axis-to-axis: CM axis to end axis is half the rod.
Your turn — Rod M = 4 kg, L = 2 m, axis at one end. I = ?

Answer: 5.33 kg·m². Icm = (1/12) × 4 × 4 = 1.333; d = 1 m; Md² = 4; I = 5.333 kg·m² (= ML²/3 ✓).

Example 2 — disk about a rim point

  1. Icm. Disk M = 3 kg, R = 0.4 m: Icm = ½MR² = 0.5 × 3 × 0.16 = 0.24 kg·m².
  2. d. Rim axis is one radius from the center: d = 0.4 m.
  3. Total. I = 0.24 + 3 × 0.16 = 0.24 + 0.48 = 0.72 kg·m² = (3/2)MR².
  4. Read it. Triple the center value — a disk pivoted at its edge is three times as lazy. (Physical pendulums live here.)
Common mistake: adding MR² (the hoop’s formula) as the toll instead of Md² with the disk’s own d. The toll is always M×d² — shape-independent; the shape lives in Icm.
Your turn — Disk M = 2 kg, R = 0.5 m, axis at a rim point (parallel). I = ?

Answer: 0.75 kg·m². Icm = 0.5 × 2 × 0.25 = 0.25; Md² = 2 × 0.25 = 0.5; I = 0.75 kg·m².

Before reading on: a hoop (Icm = MR²) pivoted at a point on its rim, d = R. Guess the total before computing — then notice what fraction the toll is.

Example 3 — hoop about a rim point

  1. Icm. Hoop M = 1 kg, R = 0.6 m: Icm = 1 × 0.36 = 0.36 kg·m².
  2. Toll. Md² = 1 × 0.36 = 0.36 — equal to Icm itself!
  3. Total. I = 0.36 + 0.36 = 0.72 kg·m² = 2MR².
  4. Notice. For the hoop the toll doubles I — the shift costs as much as the shape itself. Rim-heavy shapes pay the biggest tolls.
Common mistake: forgetting the theorem applies to the hoop too (“it’s already all at R”). The axis still moved — every shape pays Md².
Your turn — Hoop M = 2.5 kg, R = 0.4 m, axis at rim point. I = ?

Answer: 0.8 kg·m². Icm = 2.5 × 0.16 = 0.4; Md² = 0.4; I = 0.8 = 2MR² kg·m².

Example 4 — the trap: shifting between two non-CM axes

  1. Setup. Rod M = 1.2 kg, L = 2 m; axis at L/4 from one end (0.5 m from that end).
  2. Route through the CM. Icm = (1/12) × 1.2 × 4 = 0.4 kg·m².
  3. d. Axis is L/2 − L/4 = L/4 = 0.5 m from the CM axis.
  4. Total. I = 0.4 + 1.2 × 0.25 = 0.4 + 0.3 = 0.7 kg·m².
  5. The trap, sprung. Shifting “directly” from the end formula (ML²/3) with d = L/4 would give 1.6 + 0.3 = 1.9 — wrong. The theorem only spans from the CM.
Common mistake: I = Iend + M×(L/4)². The Md² toll is measured from the CM axis, never from another shifted axis. When in doubt, go home to the CM first.
Your turn — Rod M = 3 kg, L = 2 m; axis at L/3 from one end. I = ?

Answer: 1.333 kg·m². Icm = (1/12) × 3 × 4 = 1.0; d = L/2 − L/3 = L/6 = 1/3 m; Md² = 3 × 1/9 = 0.333; I = 1.333 kg·m².

Memorization tips

  • Say it aloud: “I equals I-C-M plus M d squared.” Three beats, one formula.
  • CM is home base: every shift starts at the center-of-mass axis. Tattoo “route through the CM” on the inside of your eyelids.
  • d is axis-to-axis: not pivot-to-end, not pivot-to-CM-of-something-else. Draw both axes, measure between them.
  • The toll is always positive: Md² ≥ 0 — shifted axes are always lazier. If your shifted I comes out smaller, the sign went wrong somewhere.
  • Parallel is load-bearing: tilted axes need the inertia tensor (beyond Physics I). If the axes aren’t parallel, the theorem doesn’t apply.
  • The rod-end check: ML²/12 + M(L/2)² = ML²/3. If you can do this line, you own the theorem.

Final challenge

Five mixed questions — basics, judgment calls, and the traps, all in one. Score 5/5 and the parallel-axis theorem 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 parallel-axis theorem?

I = Icm + M·d²: the moment of inertia about any axis parallel to a center-of-mass axis equals the CM moment of inertia plus the total mass times the squared distance between the axes.

Why must the axis pass through the center of mass first?

The derivation’s cross term (−2d·Σmixi) vanishes only because coordinates are measured from the CM, where Σmixi = 0 by definition. Starting from any other axis leaves a leftover term and the simple formula fails.

Can I shift between two non-CM axes directly?

No — always route through the CM: compute Icm first, then shift to the target axis. Shifting directly between two arbitrary axes with M·d² (using their separation) gives the wrong answer.

Why does shifting the axis always increase I?

Because M·d² ≥ 0 always. The CM axis is the minimum-inertia axis for a given direction — every parallel shift adds positive inertia. The CM is the “easiest” axis to spin about.

Do the axes have to be parallel?

Yes — “parallel-axis” is load-bearing. The theorem relates two axes pointing the same direction; for tilted axes you need the full inertia tensor, which is beyond Physics I.

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