Moment of Inertia

Q: Why does moment of inertia depend on the axis?

Because I equals the sum of m times r squared, using the perpendicular distance r of each mass element from the chosen axis. Change the axis and every r changes, so I changes too. Mass is a scalar property of the body alone; moment of inertia is a property of the body and the axis together.

Q: When can I use the perpendicular axes theorem?

Only for planar two-dimensional bodies such as a thin disc, ring, rectangular plate, or triangular plate. It does not apply to spheres, cylinders, or any 3D body. All three axes must intersect at one common point in the plane of the lamina.

Q: What is the radius of gyration used for?

It is a compact way to represent a body's moment of inertia as if all its mass were concentrated at a single distance K from the axis. It shows up in problems comparing rotational inertia across different-shaped bodies and appears naturally in the composite moment of inertia expression I equals M K squared.

Q: How do I handle a body with a hole or missing piece?

Treat the missing piece as having negative mass. Compute the moment of inertia of the complete original body and of the removed piece, both about the same axis, then subtract. If the removed piece is not centred on that axis, use the parallel axes theorem to shift its moment of inertia to the required axis first.

PhysicsSystem Of Particles And Rotational MotionFor NEET aspirants

The moment of inertia ( $I$ ) of a rigid body about an axis is the rotational analogue of mass. For a single particle of mass $m$ at perpendicular distance $r$ from the axis, $I = m r^{2}$ . For a rigid body, $I = \sum m_{i} r_{i}^{2}$ (discrete) or $I = \int r^{2} d m$ (continuous). Moment of inertia depends on the mass distribution and the choice of axis, and it decides how hard it is to change a body's rotational state. In JEE and NEET rotational problems, the correct standard result plus the parallel and perpendicular axes theorems solve most $I$ questions in seconds.

Key Formulas - Quick Reference

Single particle: $I = m r^{2}$
System of particles: $I = \sum_{i} m_{i} r_{i}^{2}$
Continuous body: $I = \int r^{2} d m$
Radius of gyration: $I = M K^{2} \Rightarrow K = I / M$
Parallel axes: $I = I_{cm} + M d^{2}$
Perpendicular axes (planar body): $I_{z} = I_{x} + I_{y}$
Thin rod (perpendicular bisector): $I = \frac{M L ^{2}}{12}$
Ring (axis $⊥$ plane, through centre): $I = M R^{2}$
Disc (axis $⊥$ plane, through centre): $I = \frac{1}{2} M R^{2}$
Solid sphere (diameter): $I = \frac{2}{5} M R^{2}$
Hollow sphere (diameter): $I = \frac{2}{3} M R^{2}$
Solid cylinder (own axis): $I = \frac{1}{2} M R^{2}$

1. What Moment of Inertia Means

In linear motion, mass tells you how much a body resists a change in linear velocity. In rotational motion, the corresponding property is moment of inertia. A body with large $I$ is hard to spin up from rest, and once spinning is hard to stop.

Unlike mass, $I$ is not a fixed property of the body alone. It depends on:

The mass of the body
The distribution of that mass relative to the axis
The choice of axis

The same body can have many different moments of inertia, one for each axis. So every $I$ statement must specify the axis.

Definition. For a single particle of mass

m

at perpendicular distance

r

from an axis,

I = m r^{2}

. For a rigid body, sum (or integrate)

m r^{2}

over every mass element.

SI unit: $kg m^{2}$ . Dimensional formula: $[M L^{2}]$ .

2. Moment of Inertia of a System of Particles

For $n$ discrete point masses $m_{1}, m_{2}, \dots, m_{n}$ at perpendicular distances $r_{1}, r_{2}, \dots, r_{n}$ from a chosen axis,

$I = m_{1} r_{1}^{2} + m_{2} r_{2}^{2} + \dots + m_{n} r_{n}^{2} = i = 1 \sum n m_{i} r_{i}^{2}$

Two things to note. First, a particle on the axis contributes zero because $r = 0$ . Second, the distance in $m r^{2}$ is the perpendicular distance to the axis, not the distance to any specific point on the axis.

Solved Example 1

Four point masses each of mass

m

are placed at the corners of a square of side

a

. Find the moment of inertia of the system about an axis perpendicular to the plane of the square and passing through its centre.

Figure: Each of the four masses sits at diagonal distance

a / 2

from the axis through the centre

O

Solution:

The perpendicular distance of each corner from the centre of the square is $\frac{a}{2}$ .

Contribution of one mass: $m (\frac{a}{2})^{2} = \frac{m a ^{2}}{2}$ .

All four masses are at the same distance from the axis, so

$I = 4 \times \frac{m a ^{2}}{2} = 2 m a^{2}$

3. Moment of Inertia of a Continuous Rigid Body

For a continuous mass distribution the sum becomes an integral. Take a small element of mass $d m$ at perpendicular distance $r$ from the axis. Then

$I = \int r^{2} d m$

The integral runs over the entire body. In practice you express $d m$ in terms of a linear, surface, or volume mass density and pick an element whose distance from the axis is easy to write.

3.1 Thin Uniform Rod About a Perpendicular Bisector

Figure 1: Element

d x

at distance

x

from the perpendicular bisector of a uniform rod.

Mass per unit length is $λ = M / L$ , so $d m = λ d x = \frac{M}{L} d x$ . The element sits at perpendicular distance $x$ from the axis, so

$I = \int_{- L /2}^{L /2} x^{2} \cdot \frac{M}{L} d x = \frac{M}{L} [\frac{x ^{3}}{3}]_{- L /2}^{L /2} = \frac{M L ^{2}}{12}$

3.2 Uniform Circular Ring About Its Own Axis

Every element of a thin ring lies at the same perpendicular distance $R$ from the axis through the centre and perpendicular to the ring's plane, so

$I = \int R^{2} d m = R^{2} \int d m = M R^{2}$

3.3 Uniform Disc About Its Own Axis

Figure: The disc as a stack of concentric rings. The ring at radius

x

has width

d x

and area

2 π x d x

; every mass element on it lies at the same perpendicular distance

x

from the axis.

Treat the disc as a stack of concentric thin rings of radius $x$ and width $d x$ . Area of the ring is $2 π x d x$ ; surface density is $σ = M / (π R^{2})$ ; so $d m = σ \cdot 2 π x d x = \frac{2 M x d x}{R ^{2}}$ . Each ring contributes $x^{2} d m$ :

$I = \int_{0}^{R} x^{2} \cdot \frac{2 M x}{R ^{2}} d x = \frac{2 M}{R ^{2}} \cdot \frac{R ^{4}}{4} = \frac{M R ^{2}}{2}$

3.4 Standard Results Table

Body	Axis	Moment of inertia $I$
Thin rod, length $L$	Through centre, $⊥$ to rod	$\frac{M L ^{2}}{12}$
Thin rod, length $L$	Through one end, $⊥$ to rod	$\frac{M L ^{2}}{3}$
Rectangular plate, sides $a, b$	Through centre, $⊥$ to plate	$\frac{M ( a ^{2} + b ^{2} )}{12}$
Circular ring, radius $R$	Through centre, $⊥$ to plane	$M R^{2}$
Circular ring, radius $R$	Any diameter	$\frac{M R ^{2}}{2}$
Circular disc, radius $R$	Through centre, $⊥$ to plane	$\frac{M R ^{2}}{2}$
Circular disc, radius $R$	Any diameter	$\frac{M R ^{2}}{4}$
Hollow cylinder, radius $R$	Own (symmetry) axis	$M R^{2}$
Solid cylinder, radius $R$	Own (symmetry) axis	$\frac{M R ^{2}}{2}$
Hollow sphere (thin shell), radius $R$	Any diameter	$\frac{2 M R ^{2}}{3}$
Solid sphere, radius $R$	Any diameter	$\frac{2 M R ^{2}}{5}$

Memory tip. For hollow bodies the mass sits farther from the axis than for solid bodies of the same

M

and

R

, so hollow always has larger

I

. Compare: hollow sphere

\frac{2}{3} M R^{2} >

solid sphere

\frac{2}{5} M R^{2}

; hollow cylinder

M R^{2} >

solid cylinder

\frac{1}{2} M R^{2}

4. Radius of Gyration

The radius of gyration $K$ is the distance from the axis at which the entire mass of the body, if concentrated, would give the same moment of inertia as the actual distribution. Formally,

$I = M K^{2} ⟹ K = \frac{I}{M}$

For $n$ equal particles at distances $r_{1}, r_{2}, \dots, r_{n}$ from the axis,

$K = \frac{r _{1}^{2} + r _{2}^{2} + \dots + r _{n}^{2}}{n}$

which is the root-mean-square of the perpendicular distances. Two quick examples: for a solid sphere about a diameter, $K = R 2/5$ ; for a disc about its own axis, $K = R / 2$ .

5. Theorem of Parallel Axes

Figure 2: Parallel axis at perpendicular distance

d

from a parallel axis through the centre of mass.

If $I_{cm}$ is the moment of inertia about an axis through the centre of mass, then the moment of inertia about any parallel axis at perpendicular distance $d$ from it is

$I = I_{cm} + M d^{2}$

Two important points. The theorem is valid for any rigid body (2D or 3D). The reference axis must pass through the centre of mass; using any other axis will give the wrong answer.

Solved Example 2

The moment of inertia of a ring about one of its diameters is

I

. Find the moment of inertia about a tangent parallel to that diameter.

Figure: The tangent lies parallel to the diameter at perpendicular distance

R

from the centre.

Solution:

Diameter passes through the centre of mass, so $I_{cm} = I = \frac{M R ^{2}}{2}$ . The tangent is parallel to the diameter and lies at perpendicular distance $R$ from the centre. By the parallel axes theorem,

$I^{'} = I_{cm} + M R^{2} = \frac{M R ^{2}}{2} + M R^{2} = \frac{3 M R ^{2}}{2} = 3 I$

6. Theorem of Perpendicular Axes

Figure 3: Three mutually perpendicular axes through a common point

O

on a planar body. The

z

-axis is perpendicular to the plate.

For a planar body (a thin lamina), the moment of inertia about an axis perpendicular to the plane of the body equals the sum of the moments of inertia about any two mutually perpendicular axes lying in the plane and intersecting the perpendicular axis at a common point:

$I_{z} = I_{x} + I_{y}$

Two conditions matter. The body must be planar (a two-dimensional lamina); the theorem fails for 3D bodies. All three axes must pass through a common point (which need not be the centre of mass).

Solved Example 3

Find the moment of inertia of a uniform disc of mass

M

and radius

R

about a diameter.

Solution:

Take the disc in the $x y$ -plane with the $z$ -axis through its centre and perpendicular to it. Then $I_{z} = \frac{1}{2} M R^{2}$ . By symmetry, both diameters $x$ and $y$ through the centre give the same value: $I_{x} = I_{y}$ .

Perpendicular axes theorem: $I_{z} = I_{x} + I_{y} = 2 I_{x}$ , so

$I_{x} = \frac{I _{z}}{2} = \frac{M R ^{2}}{4}$

7. Moment of Inertia of Compound Bodies

When several rigid pieces are joined together, the moment of inertia of the whole about a given axis is simply the sum of the moments of inertia of each piece about the same axis:

$I_{total} = I_{1} + I_{2} + I_{3} + \dots$

The key is that every term must be about the same axis. Use the parallel axes theorem on each piece if its natural formula is about a different axis.

Solved Example 4

Two identical thin rods, each of mass

m

and length

ℓ

, are joined at right angles to form an

L

-shape. Find the moment of inertia of the

L

about an axis through the joining point, perpendicular to the plane of the

L

Figure: Axis at the joining corner, perpendicular to the plane of the

L

. Each rod rotates about a perpendicular axis at one of its ends.

Solution:

Each rod rotates about an axis through one of its ends, perpendicular to its length. For a single rod, that value is $\frac{m ℓ ^{2}}{3}$ .

Both rods contribute equally, so

$I = \frac{m ℓ ^{2}}{3} + \frac{m ℓ ^{2}}{3} = \frac{2 m ℓ ^{2}}{3}$

8. Cavity Problems

When a piece is removed from a rigid body (a hole drilled in a disc, a smaller sphere carved out of a bigger one), the moment of inertia of the remaining body is found by treating the removed piece as having negative mass:

$I_{remaining} = I_{full body} - I_{removed piece}$

Both $I$ values must be about the same axis. If the removed piece is not centred on the axis, use the parallel axes theorem to shift it.

Solved Example 5

A uniform disc of radius

2 R

has a smaller disc of radius

R

cut out from it, such that the rim of the smaller disc passes through the centre of the original disc. If the surface mass density is

σ

, find the moment of inertia of the remaining body about an axis through the centre

O

of the original disc and perpendicular to its plane.

Figure: The removed disc has radius

R

and its centre

C

sits at distance

R

from

O

, so its rim passes through

O

Solution:

Let mass of the full disc be $M_{1} = σ \cdot π (2 R)^{2} = 4 π σ R^{2}$ , and mass of the removed piece $M_{2} = σ \cdot π R^{2} = π σ R^{2}$ .

Moment of inertia of the full disc about $O$ : $I_{1} = \frac{1}{2} M_{1} (2 R)^{2} = 8 π σ R^{4}$ .

The removed disc's centre is at distance $R$ from $O$ . Its moment of inertia about its own centre is $\frac{1}{2} M_{2} R^{2}$ ; by the parallel axes theorem, about $O$ it is

$I_{2} = \frac{1}{2} M_{2} R^{2} + M_{2} R^{2} = \frac{3}{2} M_{2} R^{2} = \frac{3}{2} π σ R^{4}$

Therefore

$I_{remaining} = I_{1} - I_{2} = 8 π σ R^{4} - \frac{3}{2} π σ R^{4} = \frac{13}{2} π σ R^{4}$

8.5 Two More Applications

The next three examples show how the parallel-axes theorem, perpendicular-axes theorem, cavity subtraction, and compound-body addition combine in typical JEE and NEET questions.

Solved Example 6

A uniform rod of mass

M

and length

L

has moment of inertia

\frac{M L ^{2}}{12}

about the perpendicular axis through its centre. Use the parallel axes theorem to find its moment of inertia about a perpendicular axis through one of its ends.

Figure: The end-axis is parallel to the CM axis at perpendicular distance

L /2

Solution:

Both axes are perpendicular to the rod and parallel to each other, separated by distance $d = L /2$ . Applying the parallel axes theorem,

$I_{end} = I_{cm} + M d^{2} = \frac{M L ^{2}}{12} + M (\frac{L}{2})^{2} = \frac{M L ^{2}}{12} + \frac{M L ^{2}}{4} = \frac{M L ^{2}}{3}$

This matches the standard result for a rod about a perpendicular axis through one end.

Solved Example 7

Find the moment of inertia of a uniform disc of mass

M

and radius

R

about a tangent line that lies in the plane of the disc.

Figure: The tangent lies in the plane of the disc, parallel to a diameter, at perpendicular distance

R

from the centre.

Solution:

Step 1 — disc about a diameter. From Solved Example 3 (perpendicular axes theorem), the moment of inertia of the disc about any diameter is

$I_{diameter} = \frac{M R ^{2}}{4}$

Step 2 — shift to the tangent. The tangent lies in the plane of the disc, parallel to a diameter, at perpendicular distance $R$ from the centre of mass. By the parallel axes theorem,

$I_{tangent} = I_{diameter} + M R^{2} = \frac{M R ^{2}}{4} + M R^{2} = \frac{5 M R ^{2}}{4}$

Note the distinction: a tangent in the plane of the disc gives $\frac{5 M R ^{2}}{4}$ , whereas a tangent perpendicular to the plane (parallel to the disc's own axis) gives $\frac{M R ^{2}}{2} + M R^{2} = \frac{3 M R ^{2}}{2}$ .

Solved Example 8

A uniform disc of mass

M

and radius

R

has a circular hole of radius

R /2

cut from it, with the hole's centre

C

lying at distance

R /2

from the disc's centre

O

. Two point particles, each of mass

M /4

, are then fixed to the rim of the disc at diametrically opposite points on the diameter perpendicular to

O C

. Find the moment of inertia of the composite body about an axis through

O

perpendicular to the plane of the disc.

Figure: Hole centred at

C

on the horizontal diameter; two particles on the perpendicular diameter, each at distance

R

from

O

Solution:

Build the composite piece by piece. Every moment of inertia below is taken about the axis through $O$ perpendicular to the plane.

Step 1 — full disc (before cutting):

$I_{full} = \frac{1}{2} M R^{2}$

Step 2 — mass of the removed piece. Surface density $σ = \frac{M}{π R ^{2}}$ . The hole has area $π (R /2)^{2} = \frac{π R ^{2}}{4}$ , so its mass is

$m_{h} = σ \cdot \frac{π R ^{2}}{4} = \frac{M}{4}$

Step 3 — moment of inertia of the removed piece about $O$ . About its own centre $C$ (perpendicular to plane) it is $\frac{1}{2} m_{h} (R /2)^{2} = \frac{M R ^{2}}{32}$ . Shifting to $O$ by the parallel axes theorem (distance $R /2$ ):

$I_{hole about O} = \frac{M R ^{2}}{32} + m_{h} (\frac{R}{2})^{2} = \frac{M R ^{2}}{32} + \frac{M R ^{2}}{16} = \frac{3 M R ^{2}}{32}$

Step 4 — disc with hole (cavity subtraction):

$I_{disc with hole} = \frac{M R ^{2}}{2} - \frac{3 M R ^{2}}{32} = \frac{16 M R ^{2} - 3 M R ^{2}}{32} = \frac{13 M R ^{2}}{32}$

Step 5 — add the two particles. Each particle of mass $M /4$ lies at distance $R$ from the axis, so together

$I_{particles} = 2 \cdot \frac{M}{4} \cdot R^{2} = \frac{M R ^{2}}{2} = \frac{16 M R ^{2}}{32}$

Step 6 — total by compound-body addition:

$I_{total} = \frac{13 M R ^{2}}{32} + \frac{16 M R ^{2}}{32} = \frac{29 M R ^{2}}{32}$

This one problem blends every technique from this note: parallel axes (Step 3), cavity subtraction (Step 4), and compound-body addition (Step 5).

9. Factors on Which $I$ Depends (and Doesn't)

Moment of inertia changes when you change any of these:

The total mass of the body
The distribution of mass (shape and size)
The axis of rotation

Moment of inertia does not change if the body is:

Shifted parallel to the axis of rotation (translation along the axis)
Rotated about the axis while keeping every mass element at the same distance from the axis

Common Mistakes to Avoid

Watch out

Applying the parallel axes theorem without one axis passing through the centre of mass. The reference must be the CM axis, not any convenient axis.
Using the perpendicular axes theorem on a 3D body. It works only for planar (2D) laminae.
Using distance from a point instead of perpendicular distance from the axis in $m r^{2}$ .
Forgetting that a particle on the axis contributes zero moment of inertia.
Confusing $I = M R^{2} /2$ (disc about own axis) with $I = M R^{2} /4$ (disc about a diameter). Always name the axis.
Adding moments of inertia of two joined pieces when they are about different axes. Bring both to the same axis first.
In cavity problems, forgetting to shift the removed piece's $I$ to the same axis before subtracting.
Treating radius of gyration as if it locates a physical point. It is a fictitious distance for the "equivalent single-particle" picture.

Frequently Asked Questions

What is the physical meaning of moment of inertia?

It measures a body's resistance to angular acceleration about a specified axis. The larger $I$ , the more torque you need to produce a given angular acceleration, just as more force is needed to accelerate a heavier mass linearly.

Why does moment of inertia depend on the axis?

Because $I = \sum m_{i} r_{i}^{2}$ uses the perpendicular distance $r_{i}$ of each mass element from the chosen axis. Change the axis and every $r_{i}$ changes, so $I$ changes too. Mass is a scalar property of the body alone; moment of inertia is a property of the body and the axis together.

Is moment of inertia a scalar or a vector?

For rotation about a fixed axis, we use it as a scalar. More generally, the moment of inertia of a 3D body about an arbitrary axis is described by a tensor (the inertia tensor), but this level of detail is not part of the JEE and NEET syllabus.

When can I use the perpendicular axes theorem?

Only for planar (two-dimensional) bodies such as a thin disc, ring, rectangular plate, or triangular plate. It does not apply to spheres, cylinders, or any 3D body. All three axes must intersect at one common point in the plane of the lamina.

Does the parallel axes theorem work between any two parallel axes?

No. One of the two parallel axes must pass through the centre of mass. The formula $I = I_{cm} + M d^{2}$ gives the moment of inertia about the other axis. If neither axis passes through the CM, apply the theorem twice, once for each axis, using the CM axis as an intermediate.

Why is the moment of inertia of a hollow body larger than a solid body of the same $M$ and $R$ ?

Because in a hollow body all the mass is at the outer radius, so every mass element contributes the maximum $m r^{2}$ . In a solid body much of the mass lies at smaller $r$ , contributing less. That is why a hollow sphere ( $\frac{2}{3} M R^{2}$ ) has a larger $I$ than a solid sphere ( $\frac{2}{5} M R^{2}$ ) of the same $M$ and $R$ .

What is the radius of gyration used for?

It is a compact way to represent a body's $I$ as if all its mass were concentrated at a single distance $K$ from the axis. It shows up in problems where you compare rotational inertia across different-shaped bodies, and it appears naturally in the moment of inertia of composite bodies expressed as $I = M K^{2}$ .

How do I handle a body with a hole or missing piece?

Treat the missing piece as an object of negative mass. Compute the moment of inertia of the complete original body and of the removed piece, both about the same axis, then subtract. If the removed piece is not centred on that axis, use the parallel axes theorem to shift its $I$ to the required axis first.

Previous year questions on Moment of Inertia

25 questions from past papers, each with a step-by-step solution.

Show all 25 questions

Ready to master System Of Particles And Rotational Motion?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test

1. What Moment of Inertia Means

2. Moment of Inertia of a System of Particles

3. Moment of Inertia of a Continuous Rigid Body

3.1 Thin Uniform Rod About a Perpendicular Bisector

3.2 Uniform Circular Ring About Its Own Axis

3.3 Uniform Disc About Its Own Axis

3.4 Standard Results Table

4. Radius of Gyration

5. Theorem of Parallel Axes

6. Theorem of Perpendicular Axes

7. Moment of Inertia of Compound Bodies

8. Cavity Problems

8.5 Two More Applications

9. Factors on Which I Depends (and Doesn't)

Common Mistakes to Avoid

Frequently Asked Questions

What is the physical meaning of moment of inertia?

Why does moment of inertia depend on the axis?

Is moment of inertia a scalar or a vector?

When can I use the perpendicular axes theorem?

Does the parallel axes theorem work between any two parallel axes?

Why is the moment of inertia of a hollow body larger than a solid body of the same M and R?

What is the radius of gyration used for?

How do I handle a body with a hole or missing piece?

Previous year questions on Moment of Inertia

Ready to master System Of Particles And Rotational Motion?

9. Factors on Which $I$ Depends (and Doesn't)

Why is the moment of inertia of a hollow body larger than a solid body of the same $M$ and $R$ ?