The moment of inertia of a disk is one of the most important concepts in rotational mechanics, and students in physics and engineering repeatedly encounter it in exams, problem sets, and real‑world design work. In this detailed 2026‑style article, we’ll answer the core questions people ask:

• What is the moment of inertia of a disk formula?
• How do you derive the moment of inertia of a disk about different axes?
• How does the formula change for a disk with a central hole (annular disc)?
• What are the key formulas and numerical‑style results you can plug into your homework or exams?

1. What is the moment of inertia of a disk?

Before jumping into formulas, it helps to understand what “moment of inertia of a disk” actually means.

1.1 Basic idea of moment of inertia

The moment of inertia (I) is the rotational analogue of mass in linear motion.

• For linear motion, inertia is just mass (m).
• For rotational motion, inertia is I, and it depends not only on mass but also on how the mass is distributed relative to the axis of rotation.

So, two objects with the same mass can have different moments of inertia if one is spread out farther from the axis and the other is closer in.

1.2 Moment of inertia of a disk: Why it matters

A uniform disk (or disc) is a common ideal shape in physics problems:

• Think of a pulley, gear, flywheel, CD, or circular platform.
• In many textbook and exam questions, the “disk” is assumed to be thin, flat, and of uniform density.

For such a disk, we are typically interested in the moment of inertia about three main axes:

• About the central axis perpendicular to the plane (spinning like a CD).
• About an axis along a diameter in the plane (like a rolling wheel leaning sideways).
• About a parallel axis through the rim (using the parallel‑axis theorem).

Each of these has its own formula, and all of them start from the same underlying idea of calculus and integration.

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Moment of Inertia of Common Shapes (Disk, Ring, Rod)

Object	Axis of rotation	Moment of inertia (I)
Solid disk (uniform)	Perpendicular to plane, through centre	$I=\frac{1}{2}M{R}^2$I=21​MR2
Solid disk (uniform)	Along a diameter (in‑plane)	$I=\frac{1}{4}M{R}^2$I=41​MR2
Thin ring / hoop	Perpendicular to plane, through centre	$I=M{R}^2$I=MR2
Thin rod	Through centre, perpendicular to length	$I=\frac{1}{12}M{L}^2$I=121​ML2
Thin rod	Through one end, perpendicular to length	$I=\frac{1}{3}M{L}^2$I=31​ML2

This table is a quick reference for the standard formulas you’ll see in 2026‑style physics and engineering exams. For this article, we’ll focus on the disk rows (top two), especially the perpendicular‑through‑centre case, which is the most common “moment of inertia of a disk formula” everyone memorises.

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2. Moment of inertia of a disk formula

Let’s define the standard formula that you’ll write in your 2026 answer boxes.

2.1 Standard case: Disk about central perpendicular axis

Consider a uniform thin solid disk of:

• Mass $M$M,
• Radius $R$R,
• Rotating about an axis through its centre and perpendicular to its plane.

Then, the moment of inertia is:

$$\boxed{{\displaystyle I=\frac{1}{2}M{R}^2}}$$I=21​MR2​

This is the most famous “moment of inertia of a disk formula” and appears in virtually every rotational‑mechanics syllabus.

2.2 About a diameter (in‑plane)

For the same disk, if the axis lies in the plane of the disk and passes through its centre along a diameter, the moment of inertia is:

$$I=\frac{1}{4}M{R}^2$$I=41​MR2

This is less common in day‑to‑day use but is very important conceptually because it shows how axis‑orientation changes the moment of inertia.

2.3 Annular disc (disk with a central hole)

Now suppose the disk has a central hole: an annular disc with:

• Outer radius $R_2$R2​,
• Inner radius $R_1$R1​,
• Total mass $M$M.

Then, the moment of inertia about the central perpendicular axis is:

$$I=\frac{1}{2}M(R_2^2+R_1^2)$$I=21​M(R22​+R12​)

or, if you start from two solid disks:

$$I=I_{\text{big disk}}-I_{\text{hole}}=\frac{1}{2}{M}_{\text{big}}{R}_2^2-\frac{1}{2}{M}_{\text{hole}}{R}_1^2$$I=Ibig disk​−Ihole​=21​Mbig​R22​−21​Mhole​R12​

with $M_{\text{hole}}$Mhole​ proportional to the area of the hole.

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Moment of Inertia of Disk Formulas (2026‑Ready)

Scenario	Axis of rotation	Moment of inertia formula
Uniform solid disk	Perpendicular to plane, through centre	$I=\frac{1}{2}M{R}^2$I=21​MR2
Uniform solid disk	Along a diameter (in‑plane)	$I=\frac{1}{4}M{R}^2$I=41​MR2
Uniform solid disk	About a rim point (parallel‑axis)	$I=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$I=21​MR2+MR2=23​MR2
Annular disc (hole of radius $R_1$R1​, outer $R_2$R2​)	Central perpendicular axis	$I=\frac{1}{2}M(R_2^2+R_1^2)$I=21​M(R22​+R12​)

These are the main formulas you’ll need in 2026‑style problems. The derivation of the first one (solid disk, central axis) is what you’ll usually be asked to write in long‑answer or “derive the expression for the moment of inertia of a disk” questions.

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3. Moment of inertia of disk derivation: Central perpendicular axis

Now let’s go through the step‑by‑step derivation of the formula

$$I=\frac{1}{2}M{R}^2$$I=21​MR2

for a uniform thin solid disk rotating about an axis through its centre and perpendicular to its plane. This is the classic “moment of inertia of disk derivation” you’ll see in 2026‑style textbooks and exam instructions.

3.1 Conceptual roadmap

To compute the moment of inertia of a continuous body, we:

1. Divide the body into tiny mass elements $dm$dm.
2. For each element, compute its contribution to the moment of inertia: $dI=r^{2}dm$dI=r2dm, where $r$r is the distance from the axis.
3. Integrate over the whole body: $I=\int r^{2}dm$I=∫r2dm.

For a disk, the trick is to slice it into thin concentric rings and integrate over radius rather than area.

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3.2 Step‑by‑step derivation

We’ll work in 2026‑friendly notation you can copy directly into notes.

Given:

• Uniform thin disk:
  • Total mass: $M$M
  • Radius: $R$R
  • Axis: through centre, perpendicular to the plane.

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Step 1: Define surface mass density

Because the disk is uniform and flat, mass is spread over area.

• Surface mass density:

$$\sigma =\frac{\text{mass}}{\text{area}}=\frac{M}{\pi R^2}$$σ=areamass​=πR2M​

We’ll later relate this to tiny area elements $dA$dA.

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Step 2: Divide the disk into thin concentric rings

Imagine slicing the disk into thin concentric rings of:

• Radius $r$r (from 0 to $R$R),
• Thickness $dr$dr,

Each ring is so thin that every point on it is approximately the same distance rr from the central axis.

• Area of one ring:$$dA=\text{circumference}\times \text{thickness}=(2\pi r)dr$$dA=circumference×thickness=(2πr)dr
• Mass of that ring:

$$dm=\sigma dA=\sigma \cdot 2\pi rdr$$dm=σdA=σ⋅2πrdr

Substitute $\sigma ={\displaystyle \frac{M}{\pi R^2}}$σ=πR2M​:

$$dm=\frac{M}{\pi R^2}\cdot 2\pi rdr=\frac{2M}{R^2}rdr$$dm=πR2M​⋅2πrdr=R22M​rdr

This is the “mass element” for our integral.

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Step 3: Moment of inertia of a thin ring

For a thin ring of mass dmdm rotating about its symmetry axis, the whole ring is at distance $r$r from the axis. So:

$$dI=r^{2}dm$$dI=r2dm

Substitute $dm={\displaystyle \frac{2M}{R^2}}rdr$dm=R22M​rdr:

$$dI=r^2\cdot \frac{2M}{R^2}rdr=\frac{2M}{R^2}{r}^{3}dr$$dI=r2⋅R22M​rdr=R22M​r3dr

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Step 4: Integrate from centre to edge

The total moment of inertia is the sum (integral) of all these thin‑ring contributions:

$$I=\int dI={\int }_0^R\frac{2M}{R^2}{r}^{3}dr$$I=∫dI=∫0R​R22M​r3dr

Factor out constants:

$$I=\frac{2M}{R^2}{\int }_0^R{r}^{3}dr=\frac{2M}{R^2}\cdot {\left[\frac{r^4}{4}\right]}_0^R=\frac{2M}{R^2}\cdot \frac{R^4}{4}=\frac{2M}{R^2}\cdot \frac{R^4}{4}=\frac{2M{R}^2}{4}=\frac{1}{2}M{R}^2$$I=R22M​∫0R​r3dr=R22M​⋅[4r4​]0R​=R22M​⋅4R4​=R22M​⋅4R4​=42MR2​=21​MR2

Final result:

$$\boxed{{\displaystyle I=\frac{1}{2}M{R}^2}}$$I=21​MR2​

This is the moment of inertia of a uniform solid disk about its central perpendicular axis. The derivation is identical in 2026‑style syllabi and is the same one taught in CBSE, JEE‑style, and university‑mechanics courses.

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3.3 Why concentric rings?

You might wonder: why slice into rings instead of squares or small boxes?

• Because the axis is the centre, every ring is rotationally symmetric and every point on the ring is at the same distance rr from the axis.
• This makes the integral tractable with a single variable rr instead of a 2D areal integral.

If you tried to do it as a double integral in Cartesian coordinates, it would be much messier and is usually mentioned only for conceptual completeness.

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Moment‑of‑Inertia Derivation “Compass” — Solid Disk

Use this table as a memory‑aid compass when you practice writing the derivation in 2026‑style exam answers.

Step (in your written answer)	What to write	Key formula / idea
1. State the idea	“Moment of inertia of a continuous body is $I=\int r^{2}dm$I=∫r2dm.”	Definition of rotational inertia
2. Slice the disk	“Divide the disk into thin concentric rings of radius $r$r and thickness $dr$dr.”	Geometry choice
3. Area of ring	$dA=2\pi rdr$dA=2πrdr	Ring circumference × thickness
4. Surface density	$\sigma ={\displaystyle \frac{M}{\pi R^2}}$σ=πR2M​	Uniform mass distribution
5. Mass of ring	$dm=\sigma dA={\displaystyle \frac{2M}{R^2}}rdr$dm=σdA=R22M​rdr	Combine 3 & 4
6. Ring’s moment of inertia	$dI=r^{2}dm={\displaystyle \frac{2M}{R^2}}{r}^{3}dr$dI=r2dm=R22M​r3dr	Apply $dI=r^{2}dm$dI=r2dm
7. Integrate	$I={\int }_0^{R}dI={\int }_0^R{\displaystyle \frac{2M}{R^2}}{r}^{3}dr$I=∫0R​dI=∫0R​R22M​r3dr	Set up the integral
8. Evaluate	$I={\displaystyle \frac{2M}{R^2}}\cdot {\displaystyle \frac{R^4}{4}}={\displaystyle \frac{1}{2}}M{R}^2$I=R22M​⋅4R4​=21​MR2	Final result

If you can walk through these 8 logical steps, you’ve fully shown the “moment of inertia of a disk derivation” in a form that’s very suitable for 8–10‑mark exam questions.

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4. Non‑uniform and annular disks in 2026‑style problems

In some 2026‑style problems, the disk is not uniform, or has a hole in the centre, so let’s sketch how the same idea generalises.

4.1 Non‑uniform thin disk

Suppose the surface mass density is not constant but a function of radius: $\sigma (r)$σ(r). Then:

$$I=\int r^{2}dm={\int }_0^R{r}^2\cdot (\sigma (r)2\pi rdr)=2\pi {\int }_0^R\sigma (r)r^{3}dr$$I=∫r2dm=∫0R​r2⋅(σ(r)2πrdr)=2π∫0R​σ(r)r3dr

Once you know $\sigma (r)$σ(r) (e.g., linear, quadratic, etc.), you plug it in and integrate numerically or analytically. This is a common advanced‑style exam problem where you are told, “density varies with radius as $\sigma =kr$σ=kr” and asked to compute the moment of inertia.

4.2 Annular disc (disk with hole)

For an annular disc with:

• Inner radius $R_1$R1​,
• Outer radius $R_2$R2​,
• Mass $M$M,

you can think of it as a large solid disk of radius R2R2​ minus a small solid disk of radius R1R1​.

• Moment of inertia of big disk about centre:

$$I_{\text{big}}=\frac{1}{2}{M}_{\text{big}}{R}_2^2$$Ibig​=21​Mbig​R22​

• Mass of the hole:

$$M_{\text{hole}}=M\cdot \frac{\pi R_1^2}{\pi R_2^2}=M\cdot \frac{R_1^2}{R_2^2}$$Mhole​=M⋅πR22​πR12​​=M⋅R22​R12​​

• Moment of inertia of the hole:

$$I_{\text{hole}}=\frac{1}{2}{M}_{\text{hole}}{R}_1^2$$Ihole​=21​Mhole​R12​

So the total moment of inertia of the annular disc is:

$$I=I_{\text{big}}-I_{\text{hole}}=\frac{1}{2}{M}_{\text{big}}{R}_2^2-\frac{1}{2}{M}_{\text{hole}}{R}_1^2$$I=Ibig​−Ihole​=21​Mbig​R22​−21​Mhole​R12​

Substitute $M_{\text{big}}=M$Mbig​=M and $M_{\text{hole}}=M\cdot {\displaystyle \frac{R_1^2}{R_2^2}}$Mhole​=M⋅R22​R12​​:

I = \frac{1}{2}M R_2^2 – \frac{1}{2} \left(M

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