---
title: "Moment of Inertia by Integration and the Parallel-Axis Theorem"
description: "Moment of inertia is an integral of distance squared over the mass distribution, so it depends on the axis as well as the body. The parallel-axis theorem moves a known result to any parallel axis with"
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source: https://lightmysky.com/learn/science/moment-of-inertia-by-integration-and-the-parallel-axis-theorem-mt_VFRk1WIzAK.md
retrieved: 2026-09-12
---

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# Moment of Inertia by Integration and the Parallel-Axis Theorem

Moment of inertia is an integral of distance squared over the mass distribution, so it depends on the axis as well as the body. The parallel-axis theorem moves a known result to any parallel axis without redoing the integral.

Subject: Science · Area: Forces & Motion · Ages 20 to 21
Page: https://lightmysky.com/learn/science/moment-of-inertia-by-integration-and-the-parallel-axis-theorem-mt_VFRk1WIzAK

## Ready when they can

- Sets up and evaluates the integral for a rod, a disc or a hoop about a stated axis
- Applies the parallel-axis theorem and says why the moment of inertia is smallest about the centre of mass
- Explains why the same body has different moments of inertia about different axes

## Lesson: Why shape sets spin resistance

Moment of inertia is built piece by piece: chop the body into tiny bits, multiply each bit by its squared distance from the axis, and integrate. Far flung mass counts far more, because distance enters squared. That is why the same body has different moments about different axes: move the same mass outward and I grows, so spin changes meet more resistance.

**Example.** Symmetry picks the result. A rod gives one twelfth M L squared about its centre, a hoop gives M R squared with all mass at the rim, and a disc lands in between. So for equal mass and radius, the hoop beats the disc. Just as linear kinetic energy is half m v squared, rotational kinetic energy is half I omega squared: with I equal to 3 and omega 2, the energy is 6.

The parallel-axis theorem moves a known result to any parallel axis without redoing the integral: add M times the squared shift distance to the centre of mass value. Shifting the axis always raises I, which is why I is smallest about the centre of mass. For a rod of mass 2 and length 6, the centre value is 6, and shifting 3 to the end adds 2 times 3 squared, giving 24.

**Recap.** Square the distance, integrate over the mass, shift axes with M d squared, and spin energy follows half I omega squared.

## Practice

8 questions on this page, each with its working shown.

## Needs first

- [Rotational Kinematics and the Angular Velocity Vector](https://lightmysky.com/learn/science/rotational-kinematics-and-the-angular-velocity-vector-mt_2UqK-zViqf)
- [Triple Integrals and Coordinates for Solids](https://lightmysky.com/learn/mathematics/triple-integrals-and-coordinates-for-solids-mt_RFeK8LD_jT)

## Opens up

- [Rotational Dynamics and Rolling Without Slipping](https://lightmysky.com/learn/science/rotational-dynamics-and-rolling-without-slipping-mt_1mV6ukwDxK)
