Rod About Center Moment of Inertia (I = ML²/12) Explained — Derivation, Examples & Quizzes

Q: What is the moment of inertia of a rod about its center?

I = (1/12)*M*L^2 for a thin uniform rod rotating about an axis through its center, perpendicular to the rod.

Q: Where does the 1/12 come from?

From integrating x^2 along the rod from -L/2 to L/2: the integral of x^2 gives x^3/3, and evaluating at symmetric limits +-L/2 produces L^3/12, times mass per unit length M/L.

Q: Does the rod's thickness matter?

The formula assumes a THIN rod: all mass at essentially one line. A thick bar needs a small correction for its cross-section; for meter sticks, batons, and seesaws the thin-rod formula is the right one.

Where it comes from

Spin a rod about its middle and every bit of mass circles at some distance x from the center — but x maxes out at L/2, and most of the rod is closer than that. The 1/12 is the r²-average over the rod’s length:

Before reading on: compare with the rod about its end (I = ML²/3). Same rod — which I is bigger, and by what factor? What does that say about where the mass sits relative to each axis?

about center

mass spans −L/2 … +L/2

Farthest bit pays (L/2)² = L²/4. Average of x²: L²/12.

about end

mass spans 0 … L

Farthest bit pays L² — four times more. Average of x²: L²/3.

The end axis sees mass up to twice as far out, and distance is squared — so its I is exactly 4× the center value. Same rod, same mass, four times the laziness: axis placement is everything.

Derivation

Lay the rod along the x-axis from −L/2 to +L/2. A slice of width dx at position x has mass dm = λ dx (with λ = M/L) and sits at distance |x| from the axis, so dI = x² dm. Integrate.

dI

=

x² dm = x² λ dx

Step 1 — one slice. Each slice is a point mass at distance |x|: dI = x² dm. Uniform rod ⇒ dm = λ dx.

I

=

∫_−L/2^+L/2 λ x² dx = λ [x³/3]_−L/2^+L/2

Step 2 — integrate. Symmetric limits about the center. Antiderivative of x² is x³/3.

=

λ [ (L/2)³/3 − (−L/2)³/3 ] = λ [ L³/24 + L³/24 ]

Step 3 — evaluate. (L/2)³ = L³/8; the two ends contribute equally (the minus cubes to a minus, and minus-a-minus adds). Total: λ L³/12.

=

(M/L) (L³/12) = ML²/12

Step 4 — total mass. λ = M/L. The 1/12 survives. ∎

Where did the symmetry help? The limits −L/2 to +L/2 made both ends contribute identically — that’s the physical statement “the axis is at the middle.” Shift the axis (next lesson: the end) and the limits become 0 to L, doubling the farthest distance and quadrupling I.

How to use it

The procedure, every time:

Confirm center axis: baton, seesaw at its middle, propeller hub, meter stick at the 50-cm mark — axis through the midpoint, perpendicular to the rod.
Read off M and full length L. Not half-length — the formula wants the whole rod.
Compute I = ML²/12.
Use it in Στ = Iα, K = ½Iω², or L = Iω.

Center or end? One question decides

Where’s the pivot? Middle → ML²/12. End (door hinge, bat handle, pendulum pivot) → ML²/3. If the problem says “pivoted at one end,” the 1/12 is the wrong formula — that’s the next lesson.

Common mistake: plugging in the half-length (L/2) as “the radius.” The formula already accounts for the geometry — it wants the full L. Using L/2 gives ML²/48, a quarter of the truth.

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 meter stick

Identify. Thin uniform rod, M = 0.15 kg, L = 1 m, axis at the 50-cm mark.
Compute. I = (1/12) × 0.15 × 1² = 0.0125 kg·m².
Sanity check. Small mass, short rod → small I ✓.

Common mistake: I = ML²/3 = 0.05 — the end formula. A meter stick at its center mark is the center case. Read the pivot location first, formula second.

Your turn — Meter stick M = 0.3 kg, L = 1 m, axis through center. I = ?

Answer: 0.025 kg·m². I = (1/12) × 0.3 × 1 = 0.025 kg·m².

Example 2 — a twirler’s baton

Identify. Baton as a thin rod: M = 0.5 kg, L = 0.8 m, spun about its center.
Compute. I = (1/12) × 0.5 × 0.8² = (1/12) × 0.5 × 0.64 = (1/12) × 0.32 = 0.0267 kg·m² (3 s.f.).
Read it. Tiny I — which is why a baton flicks around with a finger’s torque.

Common mistake: squaring M instead of L, or computing ML/12 (no square). The square is on the length — always.

Your turn — Baton M = 0.4 kg, L = 1.2 m, center axis. I = ?

Answer: 0.048 kg·m². I = (1/12) × 0.4 × 1.44 = 0.576/12 = 0.048 kg·m².

Before reading on: a 10 N push at the end of a 1.5-m, 2-kg rod (center pivot). Torque = 10 × 0.75. I = 0.375. Will α land near 2, 20, or 200 rad/s²?

Example 3 — push at the end, pivot at the center

I. M = 2 kg, L = 1.5 m ⇒ I = (1/12) × 2 × 2.25 = 0.375 kg·m².
Torque. F = 10 N perpendicular at one end; r = L/2 = 0.75 m ⇒ τ = 10 × 0.75 = 7.5 N·m.
α. α = 7.5/0.375 = 20 rad/s².
Notice. r = L/2 here, not L — the push lands half a rod from a center pivot. Torque’s r and I’s L play different roles.

Common mistake: using r = L = 1.5 m in the torque (giving 15 N·m). The pivot is at the center — the lever arm to the end is L/2.

Your turn — Rod M = 3 kg, L = 2 m, center pivot; F = 8 N perpendicular at one end. α = ?

Answer: 8 rad/s². I = (1/12) × 3 × 4 = 1.0; τ = 8 × 1 = 8; α = 8/1 = 8 rad/s².

Example 4 — scaling: what doubling does

Base. M = 1 kg, L = 1 m: I = 1/12 ≈ 0.0833 kg·m².
Double the length. L = 2 m: I = (1/12) × 1 × 4 = 1/3 ≈ 0.333 kg·m² — 4× bigger (L²!).
Double the mass instead. M = 2 kg, L = 1 m: I = 2/12 = 1/6 ≈ 0.167 kg·m² — only 2× (linear in M).
The lesson. Length is squared, mass isn’t: stretching a rod beats fattening it, four-to-two.

Common mistake: thinking “double the rod, double the I.” I scales as L² — doubling length quadruples it. Scaling questions are r²-law questions in disguise.

Your turn — Rod I = 0.5 kg·m². Triple its length (same M). New I?

Answer: 4.5 kg·m². I ∝ L²: 3² = 9 × 0.5 = 4.5 kg·m².

Memorization tips

Say it aloud: “rod center: one-twelfth M L squared.” The 12 is the odd one — that oddness makes it memorable.
The 4× rule: center = ML²/12, end = ML²/3 = 4 × center. Remember one, derive the other.
Full L, not half: the formula wants the whole rod. Whisper “full length” as you plug in.
L² scaling: double length → 4× I; triple → 9×. Mass only scales linearly.
Picture the integral: slices from −L/2 to +L/2, each paying x². If you can sketch that, the 1/12 reconstructs itself.

Final challenge

Five mixed questions — basics, judgment calls, and the traps, all in one. Score 5/5 and the center-spun rod is yours.

How to learn a formula here

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.
Work the examples with the answers covered, then uncover one step at a time and compare.
Finish with the final challenge — five mixed questions including the classic traps.
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 rod about its center?

I = (1/12)·M·L² for a thin uniform rod rotating about an axis through its center, perpendicular to the rod.

Where does the 1/12 come from?

From integrating x² along the rod from −L/2 to L/2: the integral of x² gives x³/3, and evaluating at the symmetric limits ±L/2 produces L³/12, times mass per unit length M/L.

Why is it smaller than the rod about its end?

About the center, mass extends only L/2 each way, so the farthest bits pay (L/2)². About the end, mass reaches all the way to L. The end value (1/3)ML² is exactly 4 times the center value.

Does the rod’s thickness matter?

The formula assumes a thin rod: all mass at essentially one line. A thick bar needs a small correction for its cross-section; for meter sticks, batons, and seesaws the thin-rod formula is the right one.

What are real examples?

A baton twirler’s baton, a seesaw pivoted at its middle, a propeller, a meter stick spun about its center mark.

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