JEE Mains Physics · System of Particles and Rotational Motion

Moment of Inertia and Radius of Gyration

Moment of inertia, I = Σmr², measures how hard a body is to spin about an axis; the radius of gyration k is the distance at which all the mass would give the same I, so I = Mk².

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Why this matters

Eighteen PYQs, twelve of them asking for a number, and three from 2026. Ten use the standard results directly: compare bodies of the same mass and radius, or find a radius of gyration. Eight change the body first: the same material at a new radius or thickness, a piece carved out, a rod bent into a ring, a sphere remoulded into new shapes.

Concept 1 of 2: Moments of inertia of standard bodies

Moment of inertia adds up each bit of mass times the square of its distance from the axis. Mass far from the axis counts for a lot. That is why a ring, with all its mass on the rim, has the largest I for a given mass and radius, and a solid sphere, with mass packed near the centre, has one of the smallest.

Definition

$I = \sum m_ir_i^{2}$ , with $r_i$ the perpendicular distance from the AXIS (not from a point).
Radius of gyration: $I = Mk^{2}$ , so $k = \sqrt{I/M}$ .
Same mass and radius: ring about its axis $MR^{2}$ > hollow sphere $\tfrac{2}{3}MR^{2}$ > disc or solid cylinder about the axis $\tfrac{1}{2}MR^{2}$ (equal to a ring about a diameter) > solid sphere $\tfrac{2}{5}MR^{2}$ .
A disc has three symmetry axes through its centre, but k about a diameter is R/2 and about the normal axis R/√2: k depends on the axis.
A semicircular ring about the normal axis through its centre has all its mass at R, so I = MR², the same as a full ring of that mass.
A body with a large I keeps turning: a fan switched off slows only gradually.

Body	Axis	Moment of inertia	Radius of gyration
Thin ring, radius R	Through the centre, normal to the plane	$MR^{2}$	$R$
Thin ring, radius R	A diameter	$\tfrac{1}{2}MR^{2}$	$R/\sqrt{2}$
Disc, radius R	Through the centre, normal to the plane	$\tfrac{1}{2}MR^{2}$	$R/\sqrt{2}$
Disc, radius R	A diameter	$\tfrac{1}{4}MR^{2}$	$R/2$ Half the normal-axis value, by the perpendicular-axis theorem.
Solid cylinder, radius R	Its own axis	$\tfrac{1}{2}MR^{2}$	$R/\sqrt{2}$
Thin-walled hollow cylinder, radius R	Its own axis	$MR^{2}$	$R$
Solid cylinder, radius R, length L	Through the centre, normal to its axis	$M\left(\tfrac{R^{2}}{4} + \tfrac{L^{2}}{12}\right)$	$\sqrt{\tfrac{R^{2}}{4} + \tfrac{L^{2}}{12}}$
Solid sphere, radius R	A diameter	$\tfrac{2}{5}MR^{2}$	$\sqrt{2/5}\,R$
Thin hollow sphere, radius R	A diameter	$\tfrac{2}{3}MR^{2}$	$\sqrt{2/3}\,R$
Thin rod, length L	Through the centre, normal to the rod	$\tfrac{1}{12}ML^{2}$	$L/(2\sqrt{3})$
Thin rod, length L	Through one end, normal to the rod	$\tfrac{1}{3}ML^{2}$	$L/\sqrt{3}$
Rectangular plate, sides a and b	Through the centre, normal to the plate	$\tfrac{1}{12}M(a^{2} + b^{2})$	$\sqrt{(a^{2} + b^{2})/12}$
Semicircular ring, radius R	Through the centre, normal to the plane	$MR^{2}$	$R$

M is the body's mass. Every value is about an axis through the centre of mass, except the rod about its end.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · JEE Mains 2022 — 25 June · Q114Moderate

Example 1 · System of Particles and Rotational Motion · Moment of Inertia and Radius of Gyration

Moment of Inertia (M.I.) of four bodies having same mass '

M

' and radius '

2R

' are as follows:

I_{1}=

M.I. of solid sphere about its diameter

I_{2}=

M.I. of solid cylinder about its axis

I_{3}=

M.I. of solid circular disc about its diameter

I_{4}=

M.I. of thin circular ring about its diameter If

2\left( I_{2}+I_{3} \right)+I_{4}=x

I_{1}

then the value of

x

will be

Disc about a diameter is not MR²/2

MR²/2 is a disc about its normal axis. About a diameter it is MR²/4. A ring about a diameter is MR²/2, which is easy to confuse with the disc.

The radius of gyration is squared in I

I = Mk², so k = √(I/M). A ratio of radii of gyration is the square root of a ratio of moments of inertia (for equal masses), not the ratio itself.

Read the radius off the stem

If the bodies have radius 2R, every I picks up a factor of 4. A stem that gives a diameter needs halving before it goes into MR².

Drill 9 more on moments of inertia of standard bodies

Concept 2 of 2: Moment of inertia when a body is resized or reshaped

A moment of inertia is a shape factor times mass times a length squared. When the size or the material changes, the mass changes too: mass is density times volume. So write the mass in terms of density and dimensions first, then put it into the standard formula.

Definition

Mass $= \rho \times$ volume. A disc: $M = \rho\pi R^{2}t$ , so $I = \tfrac{1}{2}MR^{2} = \tfrac{1}{2}\rho\pi R^{4}t$ .
Same material and thickness: $I \propto R^{4}$ . Same material, equal I: $R_1^{4}t_1 = R_2^{4}t_2$ .
Same MASS: $I \propto R^{2}$ , whatever the material.
A coaxial cylinder of radius R/3 and length L/2 carved from a cylinder of mass M has mass $M \times \tfrac{1}{9} \times \tfrac{1}{2} = M/18$ and $I = \tfrac{1}{2}(M/18)(R/3)^{2} = MR^{2}/324$ , so the original's I is 162 times the carved piece's.
A rod of length L bent into a ring: $2\pi R = L$ , so $R = L/2\pi$ and $I = ML^{2}/4\pi^{2}$ .
Remoulding keeps volume: a sphere of an eighth of the mass has half the radius.

Disc in terms of density, and mass from volume

I_{\text{disc}} = \tfrac{1}{2}MR^{2} = \tfrac{1}{2}\rho\pi R^{4}t \qquad M = \rho V

Worked example

A disc of radius 10 cm and thickness 4 mm has moment of inertia I about its axis. A disc of the same material has radius 20 cm and thickness 1 mm. Find its moment of inertia about its axis.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 1 · Q5Moderate

Example 2 · System of Particles and Rotational Motion · Moment of Inertia and Radius of Gyration

A solid sphere of mass M and radius R is divided into two unequal parts. The smaller part having mass M/8 is converted into a sphere of radius r and the larger part is converted into a circular disc of thickness

t

and radius

2R

. If

I_{1}

is moment of inertia of a sphere having radius r about an axis through its centre and

I_{2}

is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia

I_{2}/I_{1}=

\_\_\_\_

The new piece keeps the old mass

A piece cut or remoulded from a body has its own mass, found from its volume. Putting the original M into the new piece's formula is the usual wrong answer.

I ∝ R⁴ only for the same material and thickness

For equal masses, I goes as R². The fourth power appears only when the mass itself grows with R², as for discs cut from the same sheet.

Drill 7 more on moment of inertia when a body is resized or reshaped

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Moment of inertia when a body is resized or reshaped
Disc in terms of density, and mass from volume
$I_{\text{disc}} = \tfrac{1}{2}MR^{2} = \tfrac{1}{2}\rho\pi R^{4}t \qquad M = \rho V$

Reference tables (1)

Moments of inertia of standard bodies13 rows

Body	Axis	Moment of inertia	Radius of gyration
Thin ring, radius R	Through the centre, normal to the plane	$MR^{2}$	$R$
Thin ring, radius R	A diameter	$\tfrac{1}{2}MR^{2}$	$R/\sqrt{2}$
Disc, radius R	Through the centre, normal to the plane	$\tfrac{1}{2}MR^{2}$	$R/\sqrt{2}$
Disc, radius R	A diameter	$\tfrac{1}{4}MR^{2}$	$R/2$ Half the normal-axis value, by the perpendicular-axis theorem.
Solid cylinder, radius R	Its own axis	$\tfrac{1}{2}MR^{2}$	$R/\sqrt{2}$
Thin-walled hollow cylinder, radius R	Its own axis	$MR^{2}$	$R$
Solid cylinder, radius R, length L	Through the centre, normal to its axis	$M\left(\tfrac{R^{2}}{4} + \tfrac{L^{2}}{12}\right)$	$\sqrt{\tfrac{R^{2}}{4} + \tfrac{L^{2}}{12}}$
Solid sphere, radius R	A diameter	$\tfrac{2}{5}MR^{2}$	$\sqrt{2/5}\,R$
Thin hollow sphere, radius R	A diameter	$\tfrac{2}{3}MR^{2}$	$\sqrt{2/3}\,R$
Thin rod, length L	Through the centre, normal to the rod	$\tfrac{1}{12}ML^{2}$	$L/(2\sqrt{3})$
Thin rod, length L	Through one end, normal to the rod	$\tfrac{1}{3}ML^{2}$	$L/\sqrt{3}$
Rectangular plate, sides a and b	Through the centre, normal to the plate	$\tfrac{1}{12}M(a^{2} + b^{2})$	$\sqrt{(a^{2} + b^{2})/12}$
Semicircular ring, radius R	Through the centre, normal to the plane	$MR^{2}$	$R$

M is the body's mass. Every value is about an axis through the centre of mass, except the rod about its end.

Watch out for (5)

Disc about a diameter is not MR²/2
→ Moments of inertia of standard bodies
The radius of gyration is squared in I
→ Moments of inertia of standard bodies
Read the radius off the stem
→ Moments of inertia of standard bodies
The new piece keeps the old mass
→ Moment of inertia when a body is resized or reshaped
I ∝ R⁴ only for the same material and thickness
→ Moment of inertia when a body is resized or reshaped

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