📏 Moment of Inertia of a Rod Calculator

Calculate I = (1/12)mL² for a thin rod rotating about its center.

I (about center)-
I (about end)-

How to Use This Calculator

This calculator finds the moment of inertia of a thin uniform rod. Enter the mass and length, then choose whether the rotation axis passes through the center or one end.

1

Enter the mass of the rod in kilograms.

2

Enter the length in meters, end to end.

3

Select the axis location: center or end. A door hinge is an end axis; a baton twirler's spin is closer to a center axis.

4

Click Calculate. The moment of inertia appears in kg m^2.

Moment of Inertia of a Thin Rod

About center: I = (1/12) x M x L^2 About one end: I = (1/3) x M x L^2

A rod is the simplest shape where the axis location dramatically changes the answer. Spun about its center, mass is evenly spread on both sides and much of it sits near the axis, giving the small (1/12)ML^2. Pinned at one end, the far half of the rod sweeps a huge circle, and every bit of that distant mass is weighted by distance squared, giving (1/3)ML^2: four times larger.

The factor of four is not a coincidence. The parallel axis theorem says shifting the axis a distance d from the center of mass adds Md^2. Here d = L/2, so the added term is ML^2/4, which is three times the center value ML^2/12. Add them and you get 4/12 = 1/3. One theorem, one addition, and the two formulas agree.

Worked Example

GivenUniform rod M = 2 kg, L = 1.5 m
About the centerI = (1/12) x 2 x 1.5^2 = (1/12) x 2 x 2.25 = 0.375 kg m^2
About one endI = (1/3) x 2 x 2.25 = 1.5 kg m^2
Ratio check1.5 / 0.375 = 4, exactly as the parallel axis theorem predicts

Where This Comes Up

Rod moments of inertia govern anything long that rotates: baseball bats, tennis rackets, golf clubs, doors, robotic arms, and pendulums. Bat swing weight is essentially the moment of inertia about the handle end, which is why choking up (moving hands toward the barrel) makes a bat feel lighter: it shortens the effective length in the L^2 term. Clock pendulums are tuned by sliding the bob, changing both L and the mass distribution.

The deadly mistake is using the center formula for an end-pivot problem, which understates the answer by a factor of four. Always sketch the pivot point first. And remember these formulas assume a thin, uniform rod. A baseball bat is tapered and end-loaded, so treat the formula as an estimate and expect the real bat to have a larger moment of inertia than a uniform rod of the same mass and length.

Frequently Asked Questions

Why is the end-axis value four times the center-axis value?

The parallel axis theorem: moving the axis from the center to the end adds M(L/2)^2 = ML^2/4 to the center value ML^2/12. The total ML^2/3 is exactly four times ML^2/12.

What is the parallel axis theorem?

It shifts a known moment of inertia to a parallel axis: I = I_cm + Md^2, where d is the distance between the axes. It only works when you start from the center-of-mass axis.

Which axis is easier to spin the rod about?

The center axis, with the smaller moment of inertia. Less mass sits far from a central axis, so less torque is needed for the same angular acceleration.

What are the units?

Kilogram square meters (kg m^2).

Does the rod's thickness matter?

Barely. The thin-rod formulas ignore thickness. For a thick bar, a small extra term appears, but it is usually negligible.