How do I avoid mixing up ML^2/12 and ML^2/3?

Ask where the pivot is: middle -> 1/12, end -> 1/3. The end is always the bigger number (4x) — if your 'end' answer comes out smaller than the center one, you swapped them.

Rod About End Moment of Inertia (I = ML²/3) Explained — Derivation, Examples & Quizzes

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

I = (1/3)*M*L^2 for a thin uniform rod rotating about an axis through one end, perpendicular to the rod.

Q: Is there a shortcut from the center formula?

Yes — the parallel-axis theorem: I_end = I_cm + M*(L/2)^2 = ML^2/12 + ML^2/4 = ML^2/3. That theorem has its own lesson; this page derives the result directly by integration.

Where it comes from

Move the pivot from the middle to the end and the rod’s far half suddenly lives twice as far from the axis as anything did before. Distance is squared in I — so the price of that move is a factor of four:

Before reading on: the center value is ML²/12. The end axis doubles the maximum distance (L/2 → L). What factor should the end I be? Write your guess, then verify.

center axis

x: −L/2 … L/2 ⇒ I = ML²/12

Max distance L/2.

end axis

x: 0 … L ⇒ I = ML²/3 = 4 × ML²/12

Max distance doubled → r² quadrupled. (1/3) = 4 × (1/12) ✓

This is the same rod with the same mass — only the axis moved, and I quadrupled. If that feels dramatic, good: it’s the r² law shouting. (There’s also a shortcut — the parallel-axis theorem — coming in its own lesson; here we earn the result directly.)

Derivation

Lay the rod along the x-axis from 0 (the pivot end) to L. A slice dx at position x has dm = λ dx and sits at distance x from the axis, so dI = x² dm. Same integral as the center case — different limits.

dI

=

x² dm = x² λ dx

Step 1 — one slice. Identical to the center derivation: each slice pays x² dm.

I

=

∫₀^L λ x² dx = λ [x³/3]₀^L

Step 2 — integrate. The limits now run 0 to L: the pivot sits at one end, so mass starts at distance 0 and reaches all the way to L.

=

λ L³/3

Step 3 — evaluate. [x³/3]₀^L = L³/3. Compare the center case: its symmetric limits gave L³/12 — one quarter of this.

=

(M/L) (L³/3) = ML²/3

Step 4 — total mass. λ = M/L. ∎

Spot the difference from the center derivation: only the limits changed (−L/2…L/2 became 0…L) — the integrand is identical. The entire factor of 4 lives in the limits: that’s how completely the axis choice owns the answer.

How to use it

The procedure, every time:

Confirm end pivot: door hinge, bat handle, pendulum top pivot, diving board’s fixed end — the axis goes through one tip of the rod.
Read off M and full length L.
Compute I = ML²/3.
Use it in Στ = Iα, K = ½Iω², or L = Iω. Gravity’s torque about an end pivot is Mg·(L/2) — the weight acts at the center of mass.

The gravity-torque combo

End-pivoted rods under gravity are a classic: the rod’s weight Mg pulls at its center of mass (L/2 from the pivot), so τ = Mg(L/2) sin φ (with φ measured from the vertical). Pair it with I = ML²/3 and α follows — see Example 3.

Common mistake: using ML²/12 for a hinged door because “it’s the rod formula I remember.” There are two rod formulas; the pivot picks. End pivot → the bigger one, ML²/3.

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 about its end

Identify. Thin rod, M = 0.15 kg, L = 1 m, axis at the 0-cm end.
Compute. I = (1/3) × 0.15 × 1² = 0.05 kg·m².
Check against center. The center value was 0.0125 — this is exactly 4× ✓. The factor-4 check never lies.

Common mistake: answering 0.0125 (the center value) on an end-pivot problem. Run the 4× check: end must be bigger. If it isn’t, you grabbed the wrong formula.

Your turn — Rod M = 0.6 kg, L = 0.5 m, axis at one end. I = ?

Answer: 0.05 kg·m². I = (1/3) × 0.6 × 0.25 = 0.15/3 = 0.05 kg·m².

Example 2 — a door on its hinges

Model. Door as a thin rod (width = length): M = 20 kg, L = 0.9 m, hinge axis at one edge.
Compute. I = (1/3) × 20 × 0.9² = (1/3) × 20 × 0.81 = (1/3) × 16.2 = 5.4 kg·m².
Read it. A 20-kg door resists like 5.4 kg·m² — which is why a modest push still swings it: torque’s 0.9-m lever arm is generous.

Common mistake: modeling the door’s height as L. L is the dimension perpendicular to the hinge axis — the door’s width (0.9 m), not its height.

Your turn — Door M = 15 kg, width 1.0 m, hinges at edge. I = ?

Answer: 5.0 kg·m². I = (1/3) × 15 × 1.0 = 5.0 kg·m².

Before reading on: a horizontal rod pivoted at one end, released. Gravity pulls at the middle (L/2 out). Is the initial α bigger or smaller than g/L? Guess, then compute.

Example 3 — gravity torque on a horizontal rod

Setup. Uniform rod M = 1 kg, L = 2 m, pivoted at one end, held horizontal, released.
Torque. Weight Mg acts at the CM, L/2 = 1 m from the pivot, perpendicular: τ = 1 × 9.8 × 1 = 9.8 N·m.
I. End pivot: I = (1/3) × 1 × 4 = 1.333 kg·m².
α. α = 9.8/1.333 = 7.35 rad/s².
Check the prediction. g/L = 4.9 — the true α (7.35) is bigger. (In fact α = 3g/2L for any rod: the M and L cancel!)

Common mistake: putting gravity’s lever arm at L (the end) instead of L/2 (the CM). Weight acts at the center of mass — always L/2 for a uniform rod.

Your turn — Rod M = 2 kg, L = 1 m, end pivot, horizontal, released. Initial α = ?

Answer: 14.7 rad/s². τ = 2 × 9.8 × 0.5 = 9.8 N·m; I = (1/3) × 2 × 1 = 0.667; α = 9.8/0.667 = 14.7 rad/s². (Check: 3g/2L = 3 × 9.8/2 = 14.7 ✓)

Example 4 — center vs end, same rod, side by side

Setup. M = 3 kg, L = 1.2 m.
Center. I = (1/12) × 3 × 1.44 = 4.32/12 = 0.36 kg·m².
End. I = (1/3) × 3 × 1.44 = 4.32/3 = 1.44 kg·m².
Ratio. 1.44/0.36 = 4 ✓. Same rod, same mass — the axis alone quadruples the resistance.

Common mistake: on a test, writing both formulas but assigning them backwards. Mnemonic: end = enormous — the end value is always the bigger one.

Your turn — Rod M = 2 kg, L = 3 m. I about center and about end?

Answer: 1.5 and 6.0 kg·m². Center: (1/12) × 2 × 9 = 1.5; end: (1/3) × 2 × 9 = 6.0. Ratio 4 ✓.

Memorization tips

Say it aloud: “rod end: one-third M L squared.” Third — bigger fraction, bigger laziness.
End = enormous: the end value is always 4× the center value. If your end answer is smaller, you swapped them.
The 4× check: ML²/3 = 4 × ML²/12. Compute one, get the other free — and catch swaps instantly.
Gravity acts at L/2: for end-pivot problems under gravity, the torque lever arm is the CM distance, L/2 — not L.
α = 3g/2L: a horizontal uniform rod released from rest always starts with this angular acceleration — M cancels. A lovely check on your arithmetic.

Final challenge

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

I = (1/3)·M·L² for a thin uniform rod rotating about an axis through one end, perpendicular to the rod.

Why is it 4 times the center value?

About the end, mass reaches twice as far from the axis (up to L instead of L/2). Since distance is squared in I, doubling the reach quadruples the moment of inertia: (1/3) = 4·(1/12).

Is there a shortcut from the center formula?

Yes — the parallel-axis theorem: I_end = I_cm + M·(L/2)² = ML²/12 + ML²/4 = ML²/3. That theorem has its own lesson; this page derives the result directly by integration.

What are real examples of end-pivoted rods?

A door on its hinges, a baseball bat swung about the handle end, a pendulum rod pivoted at the top, a diving board fixed at one end.

How do I avoid mixing up ML²/12 and ML²/3?

Ask where the pivot is: middle → 1/12, end → 1/3. The end is always the bigger number (4×) — if your “end” answer comes out smaller than the center one, you swapped them.

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