Moment of Inertia | JEE Notes

Hold a long stick near its centre and twist it about your hand. Now hold it near an end. The mass has not changed, but turning it feels different because more of its mass lies farther from the axis. Moment of inertia measures that distribution.

About a specified axis, $I = \int r_{⊥}^{2} d m$ , where $r_{⊥}$ is the shortest distance from the mass element to the axis. Its SI unit is $k g m^{2}$ . A body has no single “moment of inertia” without an axis. A point mass on the axis contributes zero; moving the same mass to twice the distance gives four times the contribution.

The radius of gyration $k$ is defined by $I = M k^{2}$ . It is the distance at which the entire mass would have to be concentrated to produce the same inertia. It is not generally a physical radius. For a disc about its symmetry axis, $k = R / \sqrt{2}$ ; for a ring, $k = R$ . Distinguish this $k$ from unrelated constants appearing in individual problems.

Explore in 3D: Inertia explorer

A useful scaling rule follows immediately from the definition. If every distance from the axis doubles while each mass element keeps its mass, inertia becomes four times as large. If all lengths of a three-dimensional object double while density stays fixed, its mass becomes eight times larger and its inertia becomes thirty-two times larger. Always state whether a scaling comparison holds mass or density fixed.

Derivation A · A uniform slender rod

Let a rod of mass $M$ and length $L$ lie along $x$ , centred at the origin. For an axis through the centre and perpendicular to the rod, $r_{⊥} = | x |$ . Uniform linear density is $λ = M / L$ , so $d m = λ d x$ . Thus

I_{C} = \frac{M}{L} \int_{- L / 2}^{L / 2} x^{2} d x = \frac{M}{L} {[\frac{x^{3}}{3}]}_{- L / 2}^{L / 2} = \frac{M L^{2}}{12} .

With a perpendicular axis through an end, integrate from $0$ to $L$ : $I_{e n d} = (M / L) \int_{0}^{L} x^{2} d x = M L^{2} / 3$ . Slender means the rod’s transverse size is negligible relative to its length; a real thick rod has additional contributions when an axis probes its thickness.

Derivation B · A uniform disc and annulus

Divide a disc into concentric thin rings. With uniform surface density $σ = M / (π R^{2})$ , a ring of radius $r$ and width $d r$ has $d m = σ (2 π r d r)$ . Every point on that ring is approximately distance $r$ from the central normal axis:

I_{z} = \int_{0}^{R} r^{2} (2 π σ r d r) = \frac{π σ R^{4}}{2} = \frac{12}{M} R^{2} .

For an annulus with inner radius $a$ and outer radius $b$ , use $σ = M / [π (b^{2} - a^{2})]$ and integrate from $a$ to $b$ . Factor $b^{4} - a^{4} = (b^{2} - a^{2}) (b^{2} + a^{2})$ to obtain $I = \frac{12}{M} (a^{2} + b^{2})$ . It becomes a disc for $a = 0$ and approaches a thin ring’s $M R^{2}$ when both radii approach $R$ while total mass is held fixed.

Watch the animation: Why a disc is half a ring

Derivation C · A uniform solid sphere

Slice a sphere perpendicular to its $z$ -axis. A slice at coordinate $z$ is a thin disc of radius $s = \sqrt{R^{2} - z^{2}}$ . For uniform volume density $ρ$ , its mass is $d m = ρ π (R^{2} - z^{2}) d z$ . The disc result gives $d I_{z} = {\frac{12}{s}}^{2} d m$ , so

I_{z} = \frac{ρ π}{2} \int_{- R}^{R} (R^{2} - z^{2})^{2} d z = \frac{ρ π}{2} \cdot \frac{16 R^{5}}{15} = \frac{8 π ρ R^{5}}{15} .

Substitute $M = 4 π ρ R^{3} / 3$ : $I_{z} = 2 M R^{2} / 5$ . Symmetry makes this value the same about every diameter. The sphere is not a planar lamina; you must not use the perpendicular-axis theorem on it.

For comparison, a uniform thin spherical shell has $x^{2} + y^{2} + z^{2} = R^{2}$ for every mass element. Symmetry gives $\int x^{2} d m = \int y^{2} d m = \int z^{2} d m = M R^{2} / 3$ ; therefore $I_{z} = \int (x^{2} + y^{2}) d m = 2 M R^{2} / 3$ . A hollow spherical shell is not a thin circular ring.

Parallel-axis theorem: the centre-of-mass condition is essential

Choose a central $z$ -axis and a parallel axis displaced by $d$ along $x$ . Then

I_{O} = \int [(x - d)^{2} + y^{2}] d m = I_{C} - 2 d \int x d m + M d^{2} = I_{C} + M d^{2} .

The cross term vanishes because the original axis passes through the centre of mass, so $\int x d m = 0$ . The axes must be parallel, and $d$ is their perpendicular separation. You cannot start from any convenient noncentral axis and blindly add $M d^{2}$ . For two noncentral parallel axes, locate the centre first: $I_{A} - I_{B} = M (d_{A}^{2} - d_{B}^{2})$ .

Since $M d^{2} \geq 0$ , the central axis has the smallest inertia among all axes parallel to it. This gives an immediate reasonableness check. A shifted axis cannot have a smaller inertia than its parallel central counterpart.

Watch the animation: Parallel axes: the cost of moving away

Perpendicular-axis theorem: a planar statement

For a lamina lying entirely in the $x y$ -plane, choose perpendicular axes $x$ and $y$ in that plane, meeting at O, and a normal $z$ -axis through O. Then

I_{x} = \int y^{2} d m, I_{y} = \int x^{2} d m, I_{z} = \int (x^{2} + y^{2}) d m = I_{x} + I_{y} .

The common point need not be the centre of mass, but all three axes must meet. For a uniform disc, in-plane symmetry gives $I_{x} = I_{y}$ , so the inertia about a diameter is $M R^{2} / 4$ . For a three-dimensional object, $I_{x}$ and $I_{y}$ include extra $z^{2}$ contributions; the lamina identity therefore does not apply.

Uniform body	Specified axis	Moment of inertia
Thin ring, radius $R$	Central axis normal to ring	$M R^{2}$
Thin ring, radius $R$	A diameter in its plane	$M R^{2} / 2$
Thin disc, radius $R$	Central normal axis / diameter	$M R^{2} / 2$ / $M R^{2} / 4$
Solid cylinder, radius $R$ , length $ℓ$	Longitudinal symmetry axis	$M R^{2} / 2$
Solid cylinder, radius $R$ , length $ℓ$	Central axis perpendicular to its length	$M (3 R^{2} + ℓ^{2}) / 12$
Thick cylindrical shell, radii $a, b$	Longitudinal symmetry axis	$M (a^{2} + b^{2}) / 2$
Solid sphere / thin spherical shell	Any diameter	$2 M R^{2} / 5$ / $2 M R^{2} / 3$
Slender rod, length $L$	Perpendicular through centre / end	$M L^{2} / 12$ / $M L^{2} / 3$
Rectangular lamina, sides $a, b$	Central normal axis	$M (a^{2} + b^{2}) / 12$
Rectangular lamina, sides $a, b$	Central in-plane axis parallel to side $a$	$M b^{2} / 12$

In the table, $M$ always means the mass of the particular body described in that row. For composite objects, first convert all masses to a consistent density or to the stated constituent masses. A removed piece is subtracted about the same axis as the original object; subtracting two central moments about different centres is a common serious error.

Worked example 06 · A rod rotating about an oblique end axis

Problem. A uniform slender rod of mass $M$ and length $L$ makes an angle $β$ with a fixed axis passing through one end. Find its moment of inertia about that axis.

Step 1: identify the perpendicular distance. An element a distance $x$ along the rod is only $x \sin β$ from the axis. The distance along the rod is not the lever distance.

Step 2: integrate.

I = \int_{0}^{L} (x \sin β)^{2} \frac{M}{L} d x = \frac{M L^{2}}{3} \sin^{2} β .

Step 3: verify by shifting a parallel axis. The central parallel axis has $I_{C} = M L^{2} \sin^{2} β / 12$ , and its separation from the end axis is $d = (L / 2) \sin β$ . Adding $M d^{2}$ reproduces the result.

Checks. The unit is $k g m^{2}$ . At $β = 90^{\circ}$ , recover $M L^{2} / 3$ ; at $β = 0$ , an ideal zero-thickness rod lies entirely on the axis and $I = 0$ . A thick rod would not give zero in the latter limit.

Worked example 07 · A disc with an off-centre circular hole

Problem. A uniform disc originally has mass $M$ , radius $R$ , and centre O. A circular hole of radius $R / 2$ is cut with its centre H at $x = R / 2$ , tangent internally to the outer edge. Find the remaining body’s centre of mass and inertia about a normal axis through its own centre C.

Step 1: assign the removed mass. Equal surface density means the removed mass is $m_{h} = M / 4$ , so $M_{r e m} = 3 M / 4$ .

Step 2: subtract its first moment.

x_{C} = \frac{M (0) - (M / 4) (R / 2)}{3 M / 4} = - \frac{R}{6} .

The remaining centre shifts away from the hole, as expected.

Step 3: subtract inertias about O. The removed disc’s inertia about O is $\frac{12}{(} M / 4) (R / 2)^{2} + (M / 4) (R / 2)^{2} = 3 M R^{2} / 32$ . Thus

I_{O} = \frac{12}{M} R^{2} - \frac{3 M R^{2}}{32} = \frac{13 M R^{2}}{32} .

Step 4: shift back to the remaining centre.

I_{C} = I_{O} - M_{r e m} (R / 6)^{2} = (\frac{13}{32} - \frac{1}{48}) M R^{2} = \frac{37 M R^{2}}{96} .

Checks. $I_{C} < I_{O}$ , as the central axis must minimise inertia among parallel axes. The answer has inertia units and is positive. Expressed using the remaining mass, $I_{C} = (37 / 72) M_{r e m} R^{2}$ ; do not confuse that mass with the original $M$ .

Worked example 08 · Nonuniform density defeats the memorised disc formula

Problem. A thin circular plate of radius $R$ has surface density $σ (r) = c r^{2}$ , where $c > 0$ . Its total mass is $M$ . Find the inertia about its central normal axis and its radius of gyration.

Step 1: normalise the density. A thin ring has mass $d m = σ (r) 2 π r d r = 2 π c r^{3} d r$ . Therefore $M = 2 π c \int_{0}^{R} r^{3} d r = π c R^{4} / 2$ , giving $c = 2 M / (π R^{4})$ .

Step 2: calculate the weighted second moment.

I = \int_{0}^{R} r^{2} (2 π c r^{3} d r) = \frac{π c R^{6}}{3} = \frac{23}{M} R^{2} .

Step 3: extract the radius of gyration. $k = \sqrt{I / M} = R \sqrt{2 / 3}$ .

Checks. Since mass is concentrated more strongly near the rim than in a uniform disc, $I > M R^{2} / 2$ . Since all mass lies within radius $R$ , $I < M R^{2}$ . Both bounds hold. Density units require $[c] = k g m^{- 4}$ , making $d m$ a mass. The familiar $M R^{2} / 2$ formula requires uniform density.

A reliable inertia workflow

Draw the axis explicitly. Decide whether the object is uniform. Choose mass elements whose distance from the axis is simple. Express $d m$ using the correct linear, surface or volume density. Integrate or use an applicable standard result. Shift only between parallel axes, using the actual centre of mass. Add or subtract constituents about one common axis. Finally inspect dimensions and the limits of zero size, zero removed mass, or coincident axes.

Checkpoint · A uniform disc has a concentric hole. Does its inertia increase or decrease?

The comparison must specify what is held fixed. Cutting material from an existing disc decreases its absolute inertia about the same axis because positive mass contributions are removed. Comparing an annulus and a disc with the same total mass and outer radius gives the annulus a larger inertia, because its mass lies farther out. A question without the mass condition is incomplete.

Extra results worth memorising

Uniform body	Axis	$I$
Square plate, side $a$	Normal through centre	$M a^{2} / 6$
Square plate	Any in-plane line through centre, including a diagonal	$M a^{2} / 12$
Disc	Tangent in plane	$\frac{54}{M} R^{2}$
Disc	Normal through rim	$\frac{32}{M} R^{2}$
Ring	Tangent in plane	$\frac{32}{M} R^{2}$
Ring	Normal through rim	$2 M R^{2}$
Solid sphere	Tangent	$\frac{75}{M} R^{2}$
Spherical shell	Tangent	$\frac{53}{M} R^{2}$
Solid cone	Its axis	$\frac{3}{10} M R^{2}$
Half disc	Normal through the centre of the full circle	$\frac{12}{M} R^{2}$
Cube, side $a$	Through centre, normal to a face	$M a^{2} / 6$
Rod, $λ = k x$	Normal through $x = 0$	$\frac{12}{M} L^{2}$

JEE Main's favourite moves. Rod bent into a ring ( $R = L / 2 π$ ). Rod cut and rearranged (each piece has $M / 2$ , $L / 2$ ). Identical bodies at corners (add own inertia, then $m d^{2}$ ). Sphere recast into a disc (mass kept, volume kept).

Moment of inertia: derive, shift and combine

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