Physics
Rotational Motion
Moment of Inertia

Rotational Motion · JEE & NEET Physics

Moment of Inertia: notes and previous year questions

How hard a body is to spin up: I = Σmr², rods, rings, discs and spheres, the radius of gyration, rotational kinetic energy and the rolling race.

63 JEE Main questions (2003–2026)
3 JEE Advanced questions (2000–2015)
16 NEET questions (2004–2024)

Moment of Inertia in short

Moment of inertia measures how hard it is to change a body's spin.
I = Σmr², with r the perpendicular distance from the axis; unit kg m².
I plays the part of mass for spinning: τ = Iα, KE = ½Iω², L = Iω.
I depends on the mass, how far out it sits (as r²), and the axis.

1What is moment of inertia?

Give two rotors the same push. The one with its mass near the axis spins up quickly; the one with the same mass far out hardly gets going. The moment of inertia $I$ measures how hard it is to change a body's spin.

I = \sum m_i r_i^2

r is the perpendicular distance of each mass from the axis. Unit: kg m².

It plays the same part for spinning that mass plays for moving in a line:

Moving in a line	Spinning
mass $m$	moment of inertia $I$
$F = ma$	$\tau = I\alpha$
$KE = \frac12 mv^2$	$KE = \frac12 I\omega^2$
$p = mv$	$L = I\omega$

$I$ depends on three things: how much mass, how far the mass is from the axis (counted as distance squared), and which axis the body spins about. So, unlike mass, one body has many moments of inertia.

2Point masses

Each small mass adds $mr^2$ . Doubling the distance makes it four times harder to spin; a mass on the axis adds nothing. Bodies on the same axis simply add their values of $I$ .

3Rods

Cut a uniform rod (mass $M$ , length $L$ ) into slices. A slice of width $dx$ at distance $x$ has mass $dm = \frac{M}{L}dx$ .

I = \int_{-L/2}^{L/2} x^2\,\frac{M}{L}\,dx = \frac{ML^2}{12}

Axis through the middle, across the rod.

I = \int_0^L x^2\,\frac{M}{L}\,dx = \frac{ML^2}{3}

Axis through one end: four times as much.

With the same push, a rod held at the middle spins up four times faster than one held at an end. That is why you grip a stick in the middle to twirl it.

4Rings and discs

In a thin ring every bit of mass is at distance $R$ from the central axis, so $I = MR^2$ .

A disc is a stack of thin rings. A ring of radius $r$ and width $dr$ has area $2\pi r\,dr$ , so its mass is $dm = \frac{M}{\pi R^2}\,2\pi r\,dr = \frac{2M}{R^2}r\,dr$ . Then

I = \int_0^R r^2\,\frac{2M}{R^2}r\,dr = \frac{MR^2}{2}

Body	Axis	I
Ring	through the centre, ⟂ to its plane	$MR^2$
Ring	a diameter	$MR^2/2$
Disc	through the centre, ⟂ to its face	$MR^2/2$
Disc	a diameter	$MR^2/4$

5Spheres, cylinders and plates

Body	Axis	I
Solid cylinder	its own axis	$MR^2/2$ (any length)
Thin hollow cylinder	its own axis	$MR^2$
Thick hollow cylinder, radii $R_1$ , $R_2$	its own axis	$M(R_1^2 + R_2^2)/2$
Solid cylinder, length $L$	across, through the centre	$M(R^2/4 + L^2/12)$
Solid sphere	a diameter	$2MR^2/5$
Hollow sphere	a diameter	$2MR^2/3$
Rectangular plate $a \times b$	⟂ to it, through the centre	$M(a^2 + b^2)/12$
Rectangular plate $a \times b$	through the centre, along side $a$	$Mb^2/12$

The pattern: the farther the mass sits from the axis, the bigger $I$ . For the same mass and radius: ring ( $1$ ) > hollow sphere ( $\frac23$ ) > disc ( $\frac12$ ) > solid sphere ( $\frac25$ ), in units of $MR^2$ . A hollow sphere has $\frac53$ of a solid sphere's $I$ .

6Radius of gyration

The radius of gyration $k$ is the radius of a thin ring of the same mass that has the same moment of inertia:

I = Mk^2, \qquad k = \sqrt{I/M}

Body and axis	I	k
Rod, through the middle	$ML^2/12$	$L/(2\sqrt3) \approx 0.29L$
Rod, through one end	$ML^2/3$	$L/\sqrt3 \approx 0.58L$
Ring, ⟂ to its plane	$MR^2$	$R$
Disc (or solid cylinder), ⟂ to its face	$MR^2/2$	$R/\sqrt2 \approx 0.71R$
Solid sphere	$2MR^2/5$	$\sqrt{2/5}\,R \approx 0.63R$
Hollow sphere	$2MR^2/3$	$\sqrt{2/3}\,R \approx 0.82R$

7Rotational kinetic energy

Every bit of a spinning body moves at $v = \omega r$ , with kinetic energy $\frac12 m(\omega r)^2$ . Adding them all gives $\frac12\left(\sum mr^2\right)\omega^2$ :

KE_{rot} = \tfrac12 I\omega^2

ω in rad/s. From rpm: ω = 2π × rpm ÷ 60.

ExampleExamples

Rod, 2 kg and 1 m, at 10 rad/s about its middle: $I = \frac16$ kg m², $KE = \frac12 \times \frac16 \times 100 \approx 8.33$ J.

Disc, 4 kg and 0.5 m, at 120 rpm: $\omega = 4\pi$ rad/s, $I = 0.5$ kg m², $KE = 4\pi^2 \approx 39.5$ J.

To store 16 J in a flywheel with $I = 0.5$ kg m²: $\frac12 \times 0.5\,\omega^2 = 16$ , so $\omega = 8$ rad/s.

A ring and a disc of the same $M$ , $R$ and $\omega$ : the ring has twice the kinetic energy. Stopping a ring takes work $\frac12 MR^2\omega^2$ .

A disc of 2 kg and 20 cm at 300 rpm: $\omega = 10\pi$ rad/s, $I = 0.04$ kg m², $KE = 2\pi^2 \approx 19.7$ J.

8The rolling race

A body rolling without slipping has $v = R\omega$ , so its kinetic energy is shared between moving and spinning:

KE = \tfrac12 Mv^2\left(1 + \frac{I}{MR^2}\right)

Rolling from rest down a height $h$ , $Mgh$ equals this, so

v = \sqrt{\frac{2gh}{1 + I/MR^2}}

The smaller $I/MR^2$ , the faster: solid sphere $\sqrt{10gh/7}$ > disc $\sqrt{4gh/3}$ > ring $\sqrt{gh}$ . Their speeds are in the ratio $\sqrt{30} : \sqrt{28} : \sqrt{21}$ . Mass and radius cancel, and a solid sphere beats a hollow one.

Summary

Key ideas

Moment of inertia measures how hard it is to change a body's spin.
I = Σmr², with r the perpendicular distance from the axis; unit kg m².
I plays the part of mass for spinning: τ = Iα, KE = ½Iω², L = Iω.
I depends on the mass, how far out it sits (as r²), and the axis.
Moments of inertia about the same axis add.
A rod about its end has four times its I about the middle.
The farther out the mass, the bigger I: ring > hollow sphere > disc > solid sphere.
A disc is MR²/2 about its central axis but MR²/4 about a diameter.
The radius of gyration k is where a ring of the same mass would give the same I.
Rolling bodies share energy between moving and spinning; a smaller I/MR² rolls down faster.

Every equation

Point masses: $I = \sum m_ir_i^2$
Continuous body: $I = \int r^2\,dm$
Rod, middle: $I = ML^2/12$
Rod, end: $I = ML^2/3$
Ring, central axis: $I = MR^2$
Ring, diameter: $I = MR^2/2$
Disc or solid cylinder, central axis: $I = MR^2/2$
Disc, diameter: $I = MR^2/4$
Thick hollow cylinder: $I = M(R_1^2 + R_2^2)/2$
Solid cylinder, across: $I = M(R^2/4 + L^2/12)$
Solid sphere: $I = 2MR^2/5$
Hollow sphere: $I = 2MR^2/3$
Rectangular plate, ⟂: $I = M(a^2 + b^2)/12$
Radius of gyration: $I = Mk^2$
Rotational kinetic energy: $KE = \tfrac12 I\omega^2$
rpm to rad/s: $\omega = 2\pi n/60$
Rolling kinetic energy: $KE = \tfrac12 Mv^2(1 + I/MR^2)$
Rolling down a height h: $v = \sqrt{2gh/(1 + I/MR^2)}$

Previous year questions with solutions

Real JEE and NEET questions on moment of inertia. Try each one before you open the solution.

Q1NEET 2023One correct option

The ratio of radius of gyration of a solid sphere of mass $M$ and radius $R$ about its own axis to the radius of gyration of the thin hollow sphere of same mass and radius about its axis is :-

A $5:3$
B $2:5$
C $\sqrt{5}:\sqrt{3}$
D $\sqrt{3}:\sqrt{5}$

Show answer and solution

Answer: Option D

From the table, $k_{\mathrm{solid}} = \sqrt{\dfrac{2}{5}}\,R$ and $k_{\mathrm{hollow}} = \sqrt{\dfrac{2}{3}}\,R$ , so $\dfrac{k_{\mathrm{solid}}}{k_{\mathrm{hollow}}} = \sqrt{\dfrac{2/5}{2/3}} = \sqrt{\dfrac{3}{5}}$ , that is $\sqrt{3} : \sqrt{5}$ . The solid sphere has the smaller $k$ , as it must: much of its mass lies deep inside.

The trap is C, the same ratio upside down, which would make the solid sphere the harder one to turn. A, $5 : 3$ , is the ratio of the moments of inertia, upside down and without the square root.

Q2JEE Main 2021Numerical answer

A uniform thin bar of mass 6 kg and length 2.4 meter is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is _______ $\times$ 10⁻¹ kg m².

Show answer and solution

Answer: 8

The bar makes six equal sides, each $0.4\ \mathrm{m}$ long with mass $1\ \mathrm{kg}$ . A regular hexagon is six equilateral triangles meeting at the centre, so the centre is one triangle's height, $\dfrac{\sqrt{3}}{2}a$ , from the midpoint of each side: $d^{2} = \dfrac{3}{4}(0.4)^{2} = 0.12\ \mathrm{m^{2}}$ . Each side: $\dfrac{1}{12}(1)(0.16) + 1(0.12) = 0.0133 + 0.12 = 0.1333\ \mathrm{kg\,m^{2}}$ . Six sides: $0.8\ \mathrm{kg\,m^{2}} = 8 \times 10^{-1}\ \mathrm{kg\,m^{2}}$ .

The trap is shifting each side by the distance from the centre to a corner, $0.4\ \mathrm{m}$ , instead of to the side's midpoint; that gives $6(0.0133 + 0.16) = 1.04\ \mathrm{kg\,m^{2}}$ .

Q3NEET 2021One correct option

From a circular ring of mass 'M' and radius 'R' an arc corresponding to a 90 ${}^{\circ }$ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the center of the ring and perpendicular to the plane of the ring is 'K' times 'MR²'. Then the value of 'K' is :

A $\frac{1}{8}$
B $\frac{3}{4}$
C $\frac{7}{8}$
D $\frac{1}{4}$

Show answer and solution

Answer: Option B

Removing a $90^{\circ}$ arc removes a quarter of the ring, and with it a quarter of the mass, leaving $\dfrac{3M}{4}$ . Every bit of what remains is still at distance $R$ from the axis, so $I = \dfrac{3M}{4}R^{2}$ and $K = \dfrac{3}{4}$ . It is the semicircle again: only the mass that is left and its distance from the axis count.

The trap is $\dfrac{1}{4}$ , the share removed rather than the share left. $\dfrac{7}{8}$ would need only an eighth of the ring to have gone, and nothing about the gap changes the distance of the rest from the centre.

Practice questions, easy to hard

Three questions from the moment of inertia practice ladder: one easy, one medium, one hard.

Q4Numerical answer

The last build gave a uniform rod about its centre, $\dfrac{1}{12}ML^{2}$ , and the same integral with limits $0$ to $L$ gives a uniform rod about one end, $\dfrac{1}{3}ML^{2}$ . The integral also handles a rod whose mass is not spread evenly. If the mass per metre $\lambda(x)$ changes along the rod, with $x$ measured from the axis, then $M = \int \lambda\,dx$ and $I = \int x^{2}\lambda\,dx$ .

A rod lies along the $x$ -axis from $x = 0$ to $x = 1\ \mathrm{m}$ , and its mass per metre is $\lambda = 6x$ (in $\mathrm{kg/m}$ , with $x$ in metres), so it gets heavier towards the far end. What is its moment of inertia, in $\mathrm{kg\,m^{2}}$ , about the axis through the end $x = 0$ perpendicular to the rod?

Show answer and solution

Answer: 1.5 kg m²

$I = \int_{0}^{1} x^{2}(6x)\,dx = 6\left[\dfrac{x^{4}}{4}\right]_{0}^{1} = 1.5\ \mathrm{kg\,m^{2}}$ . For comparison, the mass is $M = \int_{0}^{1} 6x\,dx = 3\ \mathrm{kg}$ , and a uniform $3\ \mathrm{kg}$ rod of the same length about its end would have $\dfrac{1}{3}(3)(1)^{2} = 1\ \mathrm{kg\,m^{2}}$ .

The trap is exactly that $1$ : using the uniform-rod result on a rod that is not uniform. This rod keeps more of its mass out at the far end, so its $I$ must come out larger, and it does; here $I = \dfrac{1}{2}ML^{2}$ .