Physics · Engineering Physics · A2 Level

Moment of Inertia

Push a door near its hinge and almost nothing happens; push the same door near its outer edge and it swings easily. The mass being moved hasn't changed — only where that mass sits relative to the axis of rotation. That single idea is the whole story behind moment of inertia, the quantity that governs how hard it is to spin anything up.

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Mass isn't the whole story once things rotate

In straight-line motion, a single number — mass, $m$ — tells you everything you need to know about an object's resistance to acceleration under a given force. Rotation is different. An object's resistance to angular acceleration depends not just on how much mass it has, but on where that mass is located relative to the axis it's spinning about. This rotational resistance to angular acceleration is called the moment of inertia, given the symbol $I$, and it plays the same role in rotational dynamics that mass plays in linear dynamics.

For a single point mass $m$ rotating at a fixed radius $r$ from the axis, the moment of inertia is defined as:

$$I = mr^2$$

Notice immediately that $r$ is squared. This is the single most consequential feature of the whole topic: doubling an object's distance from the axis doesn't double its contribution to $I$, it quadruples it. Mass far from the axis matters enormously more than the same mass close to the axis.

Real objects aren't single point masses — they're extended bodies built from many bits of mass distributed at many different radii. To find the total moment of inertia of an extended object, you sum the contribution of every individual mass element:

$$I = \Sigma mr^2$$

Each small mass $m$ in the object contributes its own $mr^2$ term, using its own distance $r$ from the axis, and the total moment of inertia is just the sum of all of these contributions.

Worked exampleSumming point masses to find total I

Set up the system: three point masses on a light rigid rod, all free to rotate about one end.

Mass $m_1 = 2.0\text{ kg}$ at $r_1 = 0.10\text{ m}$; mass $m_2 = 1.5\text{ kg}$ at $r_2 = 0.30\text{ m}$; mass $m_3 = 0.50\text{ kg}$ at $r_3 = 0.50\text{ m}$.

Find each mass's individual contribution using $I = mr^2$.

$I_1 = 2.0 \times 0.10^2 = 0.020\text{ kg m}^2$; $I_2 = 1.5 \times 0.30^2 = 0.135\text{ kg m}^2$; $I_3 = 0.50 \times 0.50^2 = 0.125\text{ kg m}^2$.

Sum the contributions: $I = \Sigma mr^2$.

$I = 0.020 + 0.135 + 0.125 = 0.280\text{ kg m}^2$.

Notice how much the outer mass dominates despite being the smallest.

The lightest mass ($0.50\text{ kg}$, furthest out) contributes almost as much as the heaviest mass ($2.0\text{ kg}$, closest in) — because its radius is five times larger, and $r$ is squared.

I = 0.28 kg m²

What determines an object's moment of inertia

Because $I$ depends on both mass and its distribution, two objects with identical total mass and identical overall size can have very different moments of inertia, depending purely on how that mass is arranged. The factors that affect moment of inertia are:

Total mass. More mass, all else being equal, means a larger moment of inertia — this part is intuitive and mirrors ordinary mass in linear dynamics.
Distribution of mass relative to the axis. This is the factor that makes rotational dynamics genuinely different from linear dynamics. Because $I = \Sigma mr^2$, mass positioned far from the rotation axis contributes disproportionately more than the same mass positioned close to the axis, on account of the squared radius term.
Choice of rotation axis. The same rigid object has a different moment of inertia depending on which axis it rotates about — an object is far easier to spin about an axis running through its centre than about a parallel axis nearer its edge.

A vivid way to see the mass-distribution effect is to compare two objects with exactly the same total mass $m$ and the same outer radius $R$, but different internal mass distributions: a solid uniform disc, and a thin hoop (or rim) of the same mass and radius. In the solid disc, mass is spread throughout the interior, with much of it sitting well inside radius $R$. In the thin hoop, essentially all the mass sits right at the maximum radius $R$. Because the hoop's mass is, on average, further from the axis than the disc's mass, the hoop has the larger moment of inertia of the two, even though the total mass and outer size are identical. This exact comparison is the physical reasoning behind why flywheels are built with heavy rims rather than solid discs — a theme picked up in full later in this topic.

Two side-by-side circles of identical outer radius R and identical total mass m. The left circle, labelled solid disc, is shaded uniformly throughout its interior to show mass spread evenly from the centre out to the edge. The right circle, labelled thin hoop, is shaded only as a thin ring at its outer edge, showing all of its mass concentrated at radius R. Beneath each circle is a label showing its moment of inertia formula, with the hoop's formula visually larger than the disc's. — Same total mass, same outer radius, very different moment of inertia: the hoop's mass all sits at maximum radius, while the disc's mass is spread throughout its interior — this is why I(hoop) = mR² is larger than I(disc) = ½mR².

You are not expected to derive the moment of inertia of standard shapes such as discs, rings, rods or spheres from first principles using calculus — for A-level purposes, these standard expressions are simply given to you when a question requires one. What you are expected to do is reason qualitatively about how changing an object's mass or its distribution changes $I$, and apply a given formula correctly to a numerical problem.

Torque: the rotational analogue of force

Just as moment of inertia is the rotational counterpart of mass, there's a rotational counterpart of force too: torque, symbol $T$ (sometimes written $\tau$). Torque measures the turning effect of a force, and is what actually causes angular acceleration. A force applied at the rotation axis itself produces no turning effect at all — to produce a torque, a force must be applied at some perpendicular distance from the axis:

$$T = Fr$$

Here $F$ is the applied force and $r$ is the perpendicular distance from the axis to the line along which the force acts. This is exactly why a longer spanner makes a stiff nut easier to turn, and why pushing a door far from its hinges takes so much less effort than pushing near them: for the same force $F$, a larger $r$ produces a larger torque.

Newton's second law for linear motion, $F = ma$, states that a resultant force produces an acceleration in proportion to it, with mass as the constant of proportionality. Rotational motion obeys a direct parallel:

$$T = I\alpha$$

A resultant torque $T$ acting on an object produces an angular acceleration $\alpha$, with the object's moment of inertia $I$ playing the role mass plays in the linear case. A larger moment of inertia means a given torque produces a smaller angular acceleration — exactly mirroring how a larger mass means a given force produces a smaller linear acceleration.

Worked exampleTorque and angular acceleration

A flywheel of moment of inertia $I = 0.45\text{ kg m}^2$ is acted on by a resultant torque of $T = 6.3\text{ N m}$.

Find the angular acceleration produced.

Rearrange $T = I\alpha$ to make $\alpha$ the subject.

$\alpha = T / I$

Substitute the given values.

$\alpha = 6.3 / 0.45 = 14\text{ rad s}^{-2}$

α = 14 rad s⁻²

The pattern to hold onto

Throughout everything that follows in this topic, moment of inertia $I$ plays exactly the role that mass $m$ plays in linear dynamics, and torque $T$ plays exactly the role that force $F$ plays in linear dynamics. Every rotational equation you meet from here on is the direct translation of a linear equation you already know, with $m \to I$ and $F \to T$.

Key takeaways

Moment of inertia $I$ is the rotational analogue of mass: it measures an object's resistance to angular acceleration.
For a point mass, $I = mr^2$; for an extended object built from many point masses, $I = \Sigma mr^2$.
Because $r$ is squared, mass far from the rotation axis contributes far more to $I$ than the same mass close to the axis.
Moment of inertia depends on total mass, on how that mass is distributed relative to the axis, and on the choice of axis itself.
A thin hoop has a larger moment of inertia than a solid disc of the same mass and radius, because the hoop's mass sits entirely at maximum radius.
Standard formulas for common shapes are given rather than derived — the skill required is applying them, and reasoning qualitatively about mass distribution.
Torque $T = Fr$ is the rotational analogue of force, produced by a force acting at a perpendicular distance from the axis.
$T = I\alpha$ is the rotational form of Newton's second law, directly paralleling $F = ma$.

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