Moment of inertia

Push a light empty office chair and it spins around with almost no effort. Ask a friend to sit on it, arms stretched wide, and give the same push — the chair barely turns. Nothing about gravity or friction changed; what changed is how reluctant the system is to have its spin altered. That reluctance has a name, a formula, and — unlike mass, which is one fixed number for a body — it depends on where the axis is and how the mass is arranged around it. It is called the moment of inertia. This is the rotational twin of mass in Newton's second law. Where $F = ma$ tells you how hard it is to change a straight-line velocity, $\tau = I\alpha$ tells you how hard it is to change an angular velocity — and $I$ , the moment of inertia, plays the role mass played before. A skater pulling her arms in, a flywheel storing energy in a car engine, a door that swings easily near its handle but barely moves if you push near the hinge — all of these are moment-of-inertia stories in disguise. This article builds the definition $I = \sum m_i r_i^2$ from a single rotating particle upward, introduces the radius of gyration as a way to compress a body's whole mass distribution into one length, and lays out the standard values for rod, ring, disc and solid sphere that CBSE expects you to use, not derive. Along the way you will meet the parallel and perpendicular axis theorems — two shortcuts that let you slide a known $I$ from one axis to another without repeating any integral — and see them applied to board-style and competitive-style numbers built on the same worked examples you may already know from Class 11.

Key takeaways

Moment of inertia is rotational inertia. For a particle, $I = mr^2$ ; for a system, $I = \sum_i m_i r_i^2$ , where $r_i$ is the perpendicular distance from the axis.
It is axis-dependent, not a fixed body property. The same rigid body has a different $I$ for every different axis you choose.
Radius of gyration compresses the distribution. $k = \sqrt{I/M}$ is the single distance at which the whole mass, if it were a point, would reproduce the same $I$ .
Standard shapes are quoted, not derived, at this level. Rod ( $\tfrac{1}{12}ML^2$ through the centre, $\tfrac13 ML^2$ through one end), ring ( $MR^2$ ), disc ( $\tfrac12 MR^2$ ), solid sphere ( $\tfrac25 MR^2$ ) about their usual axes.
Two theorems move $I$ between axes. Parallel axis theorem $I = I_{cm} + Md^2$ works for any rigid body; perpendicular axis theorem $I_z = I_x + I_y$ works only for flat, planar bodies.
Moment of inertia feeds directly into rotational dynamics. It appears in $\tau = I\alpha$ , angular momentum $L = I\omega$ , and rotational kinetic energy $KE_{rot} = \tfrac12 I\omega^2$ — the rotational analogues of $F=ma$ , $p=mv$ , and $KE=\tfrac12 mv^2$ .
Key idea: moment of inertia $I=\sum m_ir_i^2$ measures a rigid body's reluctance to change its angular velocity about a chosen axis, depends on how mass is distributed relative to that axis (captured compactly by the radius of gyration $k=\sqrt{I/M}$ ), and the standard shape formulas can be shifted between axes using the parallel axis theorem $I=I_{cm}+Md^2$ and, for flat bodies only, the perpendicular axis theorem $I_z=I_x+I_y$ .

The core insight

Moment of inertia $I=\sum m_ir_i^2$ is the rotational counterpart of mass, it is always tied to a specific axis rather than being one fixed number for a body, and the parallel and perpendicular axis theorems let you slide the standard quoted values for rod, ring, disc and solid sphere onto any other convenient axis without repeating an integral.

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 Physics unit System of Particles and Rotational Motion, immediately after centre of mass and torque, and just before the rotational equations of motion. The prescribed 2025-26 scope asks you to know the definition $I=\sum m_ir_i^2$ , the meaning of radius of gyration, and the standard $I$ -values for a handful of simple bodies as given facts — NCERT explicitly removes the calculus-based derivations of these values (they were derived by integration in older syllabi) from the current Class 11 course, so nowhere in this article will an integral appear. The parallel and perpendicular axis theorems may be stated and applied freely; their proofs are likewise kept out of scope here, exactly as the board expects.

The physical picture

Start with the simplest possible rotating object: a single point mass $m$ swinging on a massless rod, at perpendicular distance $r$ from a fixed axis. Its moment of inertia about that axis is defined as

I = mr^2

where: $m$ is the particle's mass and $r$ is the perpendicular distance from the mass to the axis — not the distance from any arbitrary point, but specifically the shortest distance to the rotation axis. Notice immediately that $r$ can change even though $m$ does not: move the axis and $I$ changes, even for the very same particle. Real bodies are collections of many particles rigidly held together, so the definition simply adds up every particle's contribution:

I = \sum_{i} m_i r_i^2

where: $m_i$ is the mass of the $i$ -th particle in the body and $r_i$ is its perpendicular distance from the chosen axis, and the sum runs over every particle of the rigid body. This is the rotational parallel of how total mass $M=\sum m_i$ enters $F=Ma$ for translation — except that here every particle's contribution is weighted by the square of its distance from the axis, so mass far from the axis matters far more than the same mass sitting close to it. This is exactly why a figure skater spins faster with her arms pulled in: pulling mass closer to the spin axis shrinks $I$ sharply (it goes as $r^2$ ), and — as angular momentum $L=I\omega$ stays roughly constant with no external torque — a smaller $I$ forces $\omega$ to grow. Because $I$ depends on how mass is spread out, it helps to have one number that summarises the spread for a whole body. That number is the radius of gyration $k$ , defined through

I = Mk^2 \quad\Rightarrow\quad k = \sqrt{\frac{I}{M}}

where: $M$ is the total mass of the body and $I$ is its moment of inertia about the axis in question. Physically, $k$ is the distance from the axis at which you could concentrate the entire mass $M$ as a single point and get exactly the same moment of inertia. A body with mass spread far from the axis (like a ring) has a large $k$ ; one with mass concentrated near the axis (like a solid sphere of the same $M$ and outer radius) has a smaller $k$ .

Figure

Two bodies of equal mass and equal outer radius spinning about the same central axis, a ring with all its mass at the rim shown with a wide dashed circle marking its radius of gyration nearly equal to R, and a solid sphere with mass spread through the volume shown with a narrower dashed circle marking a smaller radius of gyration, illustrating that moment of inertia depends on mass distribution not just total mass

Standard bodies whose $I$ you must be able to recall and use at this level, always about the axis stated:

Thin uniform rod, axis perpendicular to the rod through its centre: $I = \dfrac{1}{12}ML^2$
Thin uniform rod, axis perpendicular to the rod through one end: $I = \dfrac{1}{3}ML^2$
Ring, axis through the centre, perpendicular to the plane: $I = MR^2$
Uniform disc, axis through the centre, perpendicular to the plane: $I = \dfrac12 MR^2$
Solid sphere, axis through a diameter: $I = \dfrac25 MR^2$

Two theorems let you carry these standard values across to axes that were not directly listed, without ever setting up an integral:

Parallel axis theorem (any rigid body): $I = I_{cm} + Md^2$ , where $I_{cm}$ is the moment of inertia about an axis through the centre of mass, and $d$ is the perpendicular distance from that centre-of-mass axis to the new, parallel axis.

Perpendicular axis theorem (planar, flat bodies only): $I_z = I_x + I_y$ , where $x$ and $y$ are two mutually perpendicular axes lying in the plane of the flat body, and $z$ is the axis perpendicular to the plane, all three meeting at one common point.

The physical picture, continued: why axis choice changes everything

It is worth pausing on one point that trips up nearly every beginner: moment of inertia is not a property of the body alone, the way mass is. Mass is the same number whether the body is spinning, sliding, or sitting still, and whichever axis you imagine. Moment of inertia, by contrast, is only defined once you have picked a specific axis — change the axis and, in general, $I$ changes too, even though the body and its mass have not changed at all. A rod spun about its centre has a smaller $I$ than the same rod spun about one end, purely because more of its mass now sits farther from the axis. This single idea — $I$ is an axis-dependent quantity — is the thread that runs through every worked example below.

Derivation ABuilding the moment of inertia of a two-particle system and shifting it with the parallel axis theorem

The goal of this derivation is to show, using only the definition $I=\sum m_ir_i^2$ and simple algebra (no calculus, since CBSE explicitly keeps integral derivations of standard-body formulas out of the Class 11 syllabus), exactly how the parallel axis theorem $I = I_{cm} + Md^2$ arises for the simplest possible system: two point masses on a light rigid rod. Seeing the theorem proved for a two-particle system, even though the syllabus only requires you to state and apply it for continuous bodies, makes the shift-of-axis idea concrete rather than a memorised formula, and every step below uses only the sum-of- $mr^2$ definition already introduced.

Step 1. Set up two particles of mass $m_1$ and $m_2$ on a straight light rod, at positions $x_1$ and $x_2$ measured from some convenient origin O on the rod, with the rotation axis perpendicular to the rod. By definition, the moment of inertia about O is $I_O = m_1x_1^2 + m_2x_2^2$ .

Step 2. Locate the centre of mass. Its position is $x_{cm} = \dfrac{m_1x_1 + m_2x_2}{m_1+m_2}$ , and it is convenient to write each particle's position as its distance from the centre of mass plus the centre-of-mass offset: $x_1 = x_1' + x_{cm}$ and $x_2 = x_2' + x_{cm}$ , where $x_1'$ and $x_2'$ are measured from the centre of mass itself.

Step 3. Substitute into $I_O$ and expand each square. $I_O = m_1(x_1'+x_{cm})^2 + m_2(x_2'+x_{cm})^2 = \left(m_1x_1'^2+m_2x_2'^2\right) + 2x_{cm}\left(m_1x_1'+m_2x_2'\right) + (m_1+m_2)x_{cm}^2$ . The first bracket is exactly $I_{cm}$ , the moment of inertia about the centre-of-mass axis. The middle term vanishes because $m_1x_1'+m_2x_2'=0$ by the very definition of the centre of mass — the weighted position measured from the centre of mass always sums to zero.

Step 4. Read off what remains. With $M=m_1+m_2$ and $d=x_{cm}$ the distance from the centre-of-mass axis to O, the expression collapses to $I_O = I_{cm} + Md^2$ — precisely the parallel axis theorem. The middle term dropping to zero is the whole reason the theorem takes this clean additive form: it is a general fact for any rigid body, not just two particles, because the same cancellation happens term by term inside any sum (or integral) once positions are measured from the centre of mass. Physically verified: for $m_1=m_2=1\ \text{kg}$ at $x_1'=-0.3\ \text{m}$ , $x_2'=+0.3\ \text{m}$ (so $I_{cm}=0.18\ \text{kg}\,\cdot\,\text{m}^2$ ) shifted to an axis $d=0.5\ \text{m}$ away, direct computation from $I_O=\sum m_ix_i^2$ and the formula $I_{cm}+Md^2=0.18+2(0.25)=0.68\ \text{kg}\,\cdot\,\text{m}^2$ agree exactly.

Worked examples

Board level - rod about an off-centre axis

A uniform rod of mass $M = 1.2\ \text{kg}$ and length $L = 1.5\ \text{m}$ rotates about an axis perpendicular to the rod, passing through a point $0.4\ \text{m}$ from its centre. Find its moment of inertia about this axis.

Step 1. Find the moment of inertia about the centre first, since that is the standard quoted value: $I_{cm} = \dfrac{1}{12}ML^2 = \dfrac{1}{12}(1.2)(1.5)^2 = \dfrac{1}{12}(1.2)(2.25) = 0.225\ \text{kg}\,\cdot\,\text{m}^2$ .

Step 2. Apply the parallel axis theorem, since the given axis is parallel to the centre axis and offset by $d = 0.4\ \text{m}$ : $I = I_{cm} + Md^2 = 0.225 + (1.2)(0.4)^2 = 0.225 + (1.2)(0.16) = 0.225 + 0.192$ .

Step 3. Add the two pieces: $I = \mathbf{0.417\ kg\cdot m^2}$ . This is the rod's moment of inertia about the stated off-centre axis.

Competitive level - disc, radius of gyration, and a tangent axis

A uniform circular disc of mass $M = 2\ \text{kg}$ and radius $R = 0.5\ \text{m}$ has moment of inertia about any diameter equal to $\tfrac14 MR^2$ (a standard NCERT-quoted value, obtainable from the perpendicular axis theorem applied to the disc's own-plane axis $I_z=\tfrac12MR^2$ split equally between two perpendicular diameters). Find (a) the radius of gyration about that diameter, and (b) the moment of inertia about a tangent line lying in the plane of the disc, parallel to that diameter.

Step 1. Compute the diameter moment of inertia: $I_{diameter} = \dfrac14 MR^2 = \dfrac14(2)(0.5)^2 = \dfrac14(2)(0.25) = 0.125\ \text{kg}\,\cdot\,\text{m}^2$ .

Step 2. Get the radius of gyration from $k=\sqrt{I/M}$ : $k = \sqrt{\dfrac{0.125}{2}} = \sqrt{0.0625} = \mathbf{0.25\ m}$ .

Step 3. For part (b), recognise the diameter as the centre-of-mass axis for the tangent-line problem, so the parallel axis theorem applies with offset $d=R=0.5\ \text{m}$ (the perpendicular distance from the diameter to the tangent, both lying in the disc's plane): $I_{tangent} = I_{diameter} + MR^2 = 0.125 + (2)(0.5)^2 = 0.125 + 0.5$ .

Step 4. Add: $I_{tangent} = \mathbf{0.625\ kg\cdot m^2}$ . Timing note: in a competitive exam, spot immediately that "tangent parallel to a diameter" is a parallel-axis-theorem cue with $d=R$ — do not attempt to re-derive $I_{diameter}$ from scratch; quoting the standard $\tfrac14MR^2$ result and moving straight to Step 3 should take under 60 seconds.

Common traps

Writing "the moment of inertia of a body" as though it were one fixed number, the way mass is — it always depends on the chosen axis, and a full answer must name the axis.
Confusing the perpendicular axis theorem (only valid for flat, planar bodies, relating three mutually perpendicular axes meeting at one point) with the parallel axis theorem (valid for any rigid body, relating $I_{cm}$ to a parallel axis).
Forgetting that in the parallel axis theorem, $d$ must be measured from the centre-of-mass axis specifically — you cannot chain two parallel-axis shifts back to back without routing through the centre of mass in between.
Assuming a bigger or heavier body automatically has a bigger moment of inertia — two bodies of equal mass and equal outer radius can have very different $I$ depending on how that mass is distributed (compare the ring $MR^2$ with the solid sphere $\tfrac25MR^2$ of the same $M,R$ ).
Mixing up $r$ in $I=\sum m_ir_i^2$ with the straight-line distance from an arbitrary reference point — it must specifically be the perpendicular distance from the particle to the rotation axis.

You should be able to…

Write $I = \sum m_ir_i^2$ from a single particle's $I=mr^2$ , and explain why $r_i$ must be the perpendicular distance to the axis, not any arbitrary distance.
Recall, without deriving, the standard $I$ formulas for a rod (about centre and about one end), a ring, a uniform disc, and a solid sphere, each about its stated axis.
State and correctly apply the parallel axis theorem $I=I_{cm}+Md^2$ to shift a known moment of inertia to a parallel axis offset from the centre of mass.
State and correctly apply the perpendicular axis theorem $I_z=I_x+I_y$ , and recognise that it only applies to flat, planar bodies.
Compute and physically interpret the radius of gyration $k=\sqrt{I/M}$ for a given body and axis.
Explain in words why moment of inertia depends on axis choice and on mass distribution, not merely on total mass.

Ask the tutor: unsure which axis a problem is really asking about, or want to see the parallel and perpendicular axis theorems worked through for a different shape like a hollow cylinder or a rectangular plate — ask the AI tutor to walk through more moment-of-inertia examples.

Why can't we just derive these rod, ring, disc, sphere formulas ourselves the way we derive other results?

You certainly can, and if you go on to engineering or physics you will, using integral calculus to add up every infinitesimal mass element's $m\,r^2$ contribution across the whole body. But CBSE's current Class 11 syllabus explicitly keeps that integration out of the prescribed course — you are expected to know and correctly apply the five standard results, not reproduce their derivation in an exam. Treat them the way you treat $g=9.8\ \text{m/s}^2$ : a given constant you use, not something you re-derive under time pressure.

If I know I about one axis, can I always get I about any other axis using the parallel axis theorem?

Only if one of the two axes you are relating is the centre-of-mass axis, and the two axes are parallel to each other. The formula $I=I_{cm}+Md^2$ has $I_{cm}$ baked into it as the anchor point — you cannot jump directly between two arbitrary parallel axes, neither of which passes through the centre of mass, without first calculating (or already knowing) $I_{cm}$ and then applying the theorem twice, once to each axis.

Why does pulling my arms in while spinning on a chair make me spin faster, if the syllabus does not cover angular momentum conservation in detail here?

That effect is genuinely an angular momentum story ( $L=I\omega$ stays constant with no external torque, so if $I$ drops as you pull your arms toward the spin axis, $\omega$ must rise to compensate) — and it lives properly in the next topic on angular momentum conservation. The moment-of-inertia part of the story, which does belong here, is simply that pulling mass closer to the axis reduces $I$ because $I$ depends on $r^2$ ; the reason a shrinking $I$ then speeds up the spin is the conservation law covered separately.

Is the perpendicular axis theorem tested with three-dimensional bodies like a sphere?

No — the perpendicular axis theorem $I_z=I_x+I_y$ strictly requires a flat, planar (two-dimensional, or negligibly thin) body, because it relies on every particle's coordinate along the thickness direction being essentially zero. Applying it to a genuinely three-dimensional body like a solid sphere or a thick cylinder gives a wrong answer; for those, only the parallel axis theorem (which has no such planar restriction) is available at this level.