Moment of inertia

Push a stationary bicycle wheel at its rim and it takes real effort to get it spinning — that reluctance to start (or stop) rotating is moment of inertia, the rotational twin of mass. But unlike mass, moment of inertia is not a single number stamped on a body: a pen is easy to spin about its long axis and hard to spin end-over-end, because the same mass sits at very different distances from the two axes. This article builds the idea from a single point mass up to a rigid body, introduces the radius of gyration as a way to picture any body's rotational inertia as an equivalent point mass, and gives the standard no-derivation formulae for rings, cylinders, rods, discs and spheres that the ISC Class 11 syllabus expects you to apply directly in numerical problems.

The core insight

Moment of inertia $I=\sum m_i r_i^2$ measures a body's resistance to angular acceleration about a chosen axis, and it grows with the square of how far the mass sits from that axis.

The core idea

From one particle to a whole body. A single particle of mass $m$ moving in a circle of radius $r$ about an axis has moment of inertia

I = mr^2

where: $m$ is the particle's mass and $r$ is its perpendicular distance from the axis, not its distance from any origin. For a rigid body made of many mass elements $m_i$ sitting at distances $r_i$ from the axis, the contributions simply add:

I = \sum_i m_i r_i^2

This sum is the definition of moment of inertia. It plays the same role in rotational dynamics that mass plays in Newton's second law: $\tau = I\alpha$ is the rotational version of $F = ma$ , and $L = I\omega$ is the rotational version of linear momentum $p = mv$ . But there is a crucial difference — mass alone tells you nothing about $I$ ; you also need to know exactly how that mass is arranged relative to the axis, since the distance enters as $r^2$ . Double the distance of a mass element from the axis and its contribution to $I$ quadruples, not doubles.

The axis is not optional. Because $I=\sum m_i r_i^2$ depends entirely on the $r_i$ values, the same rigid body has a different moment of inertia for every different axis you choose to rotate it about. A thin rod spun about its centre, its end, or its long axis gives three completely different values of $I$ even though the mass never changes. Quoting a value of $I$ without naming the axis is therefore incomplete — always state the axis. SI unit of $I$ is $\text{kg m}^2$ , and its dimensional formula is $[ML^2]$ (no dependence on time, unlike force or torque).

Radius of gyration. For a body of total mass $M$ , it is often useful to write its moment of inertia in the compact form

I = MK^2 \qquad\Rightarrow\qquad K = \sqrt{\dfrac{I}{M}}

where: $K$ is the radius of gyration, an imagined distance from the axis at which, if the entire mass $M$ were concentrated as one point, it would reproduce the same $I$ as the actual, spread-out body. $K$ is not the location of any real mass — it is a bookkeeping distance that lets you compare rotational inertias of differently shaped bodies on equal footing, purely from their $I$ and $M$ .

Figure

Two discs of equal mass and radius, one solid and one a thin ring, each redrawn as an equivalent point mass on a circle at its own radius of gyration

Standard formulae (about the symmetry axis, no derivation required by the syllabus):

Body	Moment of inertia about symmetry axis
Ring (radius $R$ , mass $M$ )	$I = MR^2$
Solid cylinder / disc (radius $R$ , mass $M$ )	$I = \tfrac12 MR^2$
Hollow cylinder, thin-walled (radius $R$ , mass $M$ )	$I = MR^2$
Thin uniform rod (length $L$ , mass $M$ ), perpendicular axis through centre	$I = \tfrac{1}{12}ML^2$
Solid sphere (radius $R$ , mass $M$ ), about a diameter	$I = \tfrac25 MR^2$
Hollow sphere, thin-walled (radius $R$ , mass $M$ ), about a diameter	$I = \tfrac23 MR^2$

You are expected to know and apply these directly; deriving them by integration lies outside this syllabus.

Derivation ARadius of gyration from the defining sum, for a two-mass system

The goal is to show, using nothing beyond the defining sum $I=\sum m_i r_i^2$ and simple algebra, exactly how a real distribution of mass collapses into a single equivalent distance $K$ — the calculation that the compact formula $I = MK^2$ is quietly hiding, worked out for the simplest possible case of two point masses so every step stays concrete and checkable by hand.

Step 1. Set up the system. Take two equal point masses, each of mass $m$ , fixed at the two ends of a light (massless) rigid rod of length $2a$ , so each mass sits at distance $r=a$ from an axis through the midpoint, perpendicular to the rod. Total mass of the system is $M = 2m$ .

Step 2. Apply the defining sum directly. There are only two mass elements, so $I=\sum_i m_i r_i^2$ has just two terms: $I = m(a)^2 + m(a)^2 = 2ma^2$ .

Step 3. Rewrite in terms of the total mass $M=2m$ . Substituting $m = M/2$ gives $I = 2\left(\dfrac{M}{2}\right)a^2 = Ma^2$ .

Step 4. Read off the radius of gyration using $I=MK^2$ . Comparing $I = Ma^2$ with $I = MK^2$ gives $K^2 = a^2$ , so $K = a$ . For this particular system the equivalent point-mass distance turns out to equal the actual distance $a$ at which each mass sits — a useful check, since both masses genuinely are at that same distance from the axis, so "replacing" them with a single mass $M$ at distance $K=a$ changes nothing about $I$ . For bodies whose mass is spread over a range of distances (a disc, a rod, a sphere) the same procedure — sum $m_ir_i^2$ , divide by $M$ , take the square root — still applies, but $K$ then lands somewhere between the smallest and largest $r_i$ present, never outside that range.

Worked example

A uniform disc of mass $2\ \text{kg}$ and radius $0.5\ \text{m}$ rotates about an axis through its centre, perpendicular to its plane. Find (a) its moment of inertia and (b) its radius of gyration about this axis.

Step 1. Identify the correct standard formula. The axis is the disc's symmetry axis, so $I = \tfrac12 MR^2$ applies directly — no need to build the sum from scratch.

Step 2. Substitute the given values. $I = \tfrac12(2)(0.5)^2 = \tfrac12(2)(0.25) = 0.25\ \text{kg m}^2$ .

Step 3. Find the radius of gyration from $I = MK^2$ . $K = \sqrt{I/M} = \sqrt{0.25/2} = \sqrt{0.125} = 0.354\ \text{m}$ . So the disc behaves, rotationally, as if its whole $2\ \text{kg}$ were concentrated in a thin ring of radius $0.354\ \text{m}$ — noticeably smaller than the disc's actual edge at $0.5\ \text{m}$ , because most of a disc's mass lies closer to the centre than to the rim.

Answer: $I = 0.25\ \text{kg m}^2$ , $K = 0.354\ \text{m}$ .

Competitive stretch

A thin uniform rod of mass $1.2\ \text{kg}$ and length $1.5\ \text{m}$ rotates about a perpendicular axis through its centre. A ring of the same mass $1.2\ \text{kg}$ is found to have the same moment of inertia about its own symmetry axis. Find the radius of the ring.

Step 1. Compute the rod's moment of inertia. $I_{rod} = \tfrac{1}{12}ML^2 = \tfrac{1}{12}(1.2)(1.5)^2 = \tfrac{1}{12}(1.2)(2.25) = 0.225\ \text{kg m}^2$ .

Step 2. Set the ring's moment of inertia equal to it. For the ring, $I_{ring} = MR^2$ with the same mass $M=1.2\ \text{kg}$ , so $1.2\,R^2 = 0.225$ .

Step 3. Solve for $R$ . $R^2 = 0.225/1.2 = 0.1875$ , so $R = \sqrt{0.1875} = 0.433\ \text{m}$ .

Answer: a ring of radius $0.433\ \text{m}$ matches the rod's rotational inertia about these two axes.

Common traps

Treating moment of inertia as a fixed property of a body, like mass — it always depends on the chosen axis and must be quoted with reference to that axis.
Mixing up which axis a standard formula belongs to: $I=\tfrac12MR^2$ for a disc or solid cylinder is about its symmetry axis (through the centre, perpendicular to the flat face), not a diameter or an edge.
Confusing the solid-sphere ( $\tfrac25MR^2$ ) and hollow-sphere ( $\tfrac23MR^2$ ) formulae — the hollow sphere has more of its mass farther from the centre, so its $I$ is larger for the same $M$ and $R$ .
Assuming distance and moment of inertia are linearly related — because of the $r^2$ factor, doubling a mass element's distance from the axis quadruples its contribution to $I$ .

Ask the tutor: curious why flywheels deliberately put mass at the rim, or want more practice matching a body and axis to the right standard formula? Ask the AI tutor for extra examples.

Why does the same body have different moments of inertia for different axes, when its mass never changes?

Because $I=\sum m_i r_i^2$ depends on the distances $r_i$ of each mass element from the chosen axis, not just on how much mass exists. Rotate the axis and every $r_i$ changes, even though the total mass $M$ stays fixed. A pen spun about its long axis has its mass sitting very close to that axis (small $r_i$ , small $I$ ); the same pen spun end-over-end has its mass spread far along the length (large $r_i$ , large $I$ ). Always specify the axis when you state a moment of inertia.

Is the radius of gyration a real physical distance inside the body?

Not necessarily. $K$ is defined purely by $I=MK^2$ , so it is whatever distance makes that equation true for the body's actual $I$ and $M$ — it need not correspond to where any real mass sits. For a thin ring, every particle genuinely is at distance $R$ , so $K=R$ exactly. For a solid disc, mass is spread from the centre out to $R$ , and the calculation shows $K=R/\sqrt2\approx0.71R$ , a distance that lies inside the disc but is not the location of any labelled point. Treat $K$ as a convenient equivalent-point-mass distance, not a marked spot.

Do I need to derive the standard formulae for the ring, disc, rod and sphere for the ISC exam?

No. The ISC Class 11 syllabus for this topic explicitly asks only that you know and correctly apply the standard formulae — $I=MR^2$ for a ring or thin hollow cylinder, $I=\tfrac12MR^2$ for a disc or solid cylinder, $I=\tfrac{1}{12}ML^2$ for a rod about its centre, $I=\tfrac25MR^2$ for a solid sphere and $I=\tfrac23MR^2$ for a hollow sphere, each about its symmetry axis. Deriving these by integration belongs to a later course. What is fair game is combining these formulae, and the definitions $I=\sum m_ir_i^2$ and $K=\sqrt{I/M}$ , in numerical problems.

Moment of inertia

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads