In short

Moment of inertia (I) is the rotational analogue of mass. For a system of particles, I = \sum m_i r_i^2, where r_i is each particle's perpendicular distance from the axis of rotation. It appears in Newton's second law for rotation as \tau = I\alpha: a larger I means more torque is needed for the same angular acceleration. Unlike mass, moment of inertia depends on which axis you rotate about. The SI unit is kg·m².

Pick up a cricket bat. Hold it by the handle and swing it — the blade arcs through smoothly. Now flip the bat around, grip it by the blade, and try to swing the handle-end at the same speed. Same bat, same 1.2 kg of wood, but swinging it feels completely different. The blade-end was sluggish and resistant; the handle-end whips around almost too easily.

The mass of the bat did not change. What changed is where that mass sits relative to your hands — the point you are swinging around. When you grip the handle, the heavy blade is far from the pivot. When you grip the blade, that same mass is close to the pivot. The farther the mass is from the axis of rotation, the harder it is to change the rotation.

That resistance — the rotational equivalent of mass — is called moment of inertia.

Why mass alone is not enough

In straight-line motion, Newton's second law says F = ma: the more mass an object has, the more force you need to accelerate it. A 60 kg person is harder to push across a floor than a 5 kg carton of mangoes.

Rotation is different. When you spin something, what matters is not just how much mass there is, but how that mass is arranged around the axis. A kumhar's (potter's) wheel with all its clay packed near the centre spins up quickly under a push. The same amount of clay spread out to the rim makes the wheel much harder to spin up — and once spinning, much harder to stop.

Think about a ceiling fan. A lightweight fan with thin aluminium blades reaches full speed in a few seconds. Replace those blades with heavy iron bars of the same total mass but extending further from the hub, and the same motor takes much longer to spin the fan up. Same mass, same motor torque, but dramatically different rotational response.

You need a quantity that captures both the amount of mass and how far that mass sits from the axis. That quantity is the moment of inertia.

Building the idea from a single particle

Start with the simplest case: a single ball of mass m tied to a string of length r, whirling in a circle around a fixed pivot.

A particle on a string rotating about a fixed pivot A ball of mass m connected by a string of length r to a fixed pivot. A tangential force F produces tangential acceleration, creating torque τ = Fr about the pivot. pivot m r F (tangential) path
A ball of mass $m$ on a string of length $r$, rotating about a fixed pivot. The tangential force $F$ — perpendicular to the string — causes the ball to speed up around the circle.

Apply Newton's second law along the tangential direction (the direction of motion around the circle):

F = m \, a_t

Why: the tangential component of force drives the ball faster or slower around the circle. The string tension points inward (centripetal) and does not contribute to tangential acceleration.

The tangential acceleration a_t relates to the angular acceleration \alpha through:

a_t = r \, \alpha

Why: when the body rotates through angle \Delta\theta, a point at distance r travels an arc of length r\,\Delta\theta. Differentiating twice with respect to time gives a_t = r\alpha. Bigger radius means faster tangential motion for the same angular rate.

Substitute into Newton's second law:

F = m \cdot r \cdot \alpha

Now multiply both sides by r. The left side becomes the torque \tau = Fr (force times the perpendicular distance from the axis):

Fr = m r^2 \, \alpha
\tau = m r^2 \, \alpha \tag{1}

Why: multiplying by r converts the linear equation (F = ma_t) into its rotational counterpart. The quantity mr^2 plays the role that mass plays in Newton's second law for rotation — it is the resistance to angular acceleration.

Compare the two laws side by side:

Linear motion Rotational motion
Force F Torque \tau
Mass m mr^2
Acceleration a Angular acceleration \alpha

For this single particle, the quantity mr^2 — mass times the square of the distance from the axis — is the moment of inertia. It measures how much torque you need per unit of angular acceleration, exactly as mass measures how much force you need per unit of linear acceleration.

Extending to many particles

A real object — a cricket bat, a wheel, a ceiling fan — is not one particle. It is a collection of particles locked together in a rigid body. When the body rotates, every particle shares the same angular acceleration \alpha (that is what "rigid" means — no part deforms relative to any other).

Each particle i, at distance r_i from the axis, needs torque \tau_i = m_i r_i^2 \, \alpha to accelerate. The total torque for the entire body is the sum:

\tau_{\text{total}} = \sum_i m_i r_i^2 \, \alpha = \left(\sum_i m_i r_i^2\right) \alpha

Why: since \alpha is the same for every particle in a rigid body, it factors out of the sum. Internal forces between particles cancel in Newton's-third-law pairs, so only external torques contribute to the total. What remains inside the parentheses depends only on the masses and their distances from the axis — a property of the body and the chosen axis, not of the motion itself.

The sum inside the parentheses is the moment of inertia of the entire body.

Moment of Inertia

The moment of inertia of a system of particles about a given axis is:

I = \sum_{i} m_i \, r_i^2

where m_i is the mass of the i-th particle and r_i is its perpendicular distance from the axis of rotation.

Newton's second law for rotation:

\tau = I \, \alpha

Reading the definition. Each particle contributes m_i r_i^2 — its mass times the square of its distance from the axis. A 1 kg mass at 2 m from the axis contributes 1 \times 4 = 4 kg·m², while the same mass at 1 m contributes only 1 \times 1 = 1 kg·m². Moving a mass twice as far from the axis does not double its contribution — it quadruples it.

This is why the cricket bat felt so different when you flipped it. The blade's mass, sitting far from the pivot, contributes enormously to I when you grip the handle. Bring that same mass closer to the pivot (grip the blade) and I drops sharply.

The unit of moment of inertia is kg·m² (kilogram metre-squared). There is no special name for this unit — it is simply mass times distance squared, exactly as the definition requires.

For a rigid body made of continuous material rather than separate particles, the sum becomes an integral: I = \int r^2 \, dm. The Going Deeper section below shows how to evaluate this for a uniform rod.

The axis matters

Here is the feature of moment of inertia that most frequently catches students: the same object has a different moment of inertia about different axes. Unlike mass, which is a single number for a body, moment of inertia is defined with respect to a specific axis.

Same dumbbell, two different rotation axes give different I Left: two 2 kg balls on a 1 m rod, axis through the centre, each ball 0.5 m away, I = 1.0 kg·m². Right: same dumbbell, axis through one end, near ball at 0 m, far ball at 1 m, I = 2.0 kg·m². Axis through centre axis 2 kg 2 kg 0.5 m 0.5 m I = 1.0 kg·m² Axis through one end axis 2 kg 2 kg r = 0 1.0 m I = 2.0 kg·m²
The same dumbbell — two 2 kg balls on a 1 m rod — has $I = 1.0$ kg·m² about the centre but $I = 2.0$ kg·m² about one end. The mass is identical; only the distances from the axis changed.

For the centre axis: both balls are 0.5 m away, so I = 2 \times (0.5)^2 + 2 \times (0.5)^2 = 0.5 + 0.5 = 1.0 kg·m².

For the end axis: one ball sits on the axis (r = 0, contributes nothing), while the other is 1 m away. I = 2 \times 0^2 + 2 \times 1^2 = 0 + 2 = 2.0 kg·m².

The moment of inertia doubled — same mass, different axis. Whenever you state a moment of inertia, you must specify which axis it refers to.

Explore the r^2 dependence

The formula I = mr^2 tells you that moment of inertia grows as the square of the distance from the axis. Doubling the distance does not double I — it quadruples it. The interactive figure below lets you drag the distance of a mass from the axis and watch how I responds, for two different masses.

Interactive: moment of inertia versus distance from axis Two parabolic curves showing I = mr² for m = 1 kg (lower curve) and m = 2 kg (upper curve). A draggable red point along the horizontal axis lets the reader explore how I changes with distance r. distance from axis r (m) moment of inertia I (kg·m²) 0 3 6 9 12 1 2 m = 1 kg m = 2 kg drag the red point along the axis
Drag the red point to change the distance $r$ from the axis. The red curve shows $I = r^2$ for a 1 kg mass; the dark curve shows $I = 2r^2$ for a 2 kg mass. Both are parabolas — doubling $r$ quadruples $I$, while doubling mass only doubles $I$.

Try dragging from r = 1 m to r = 2 m. For the 1 kg mass, I jumps from 1 to 4 kg·m² — a fourfold increase for a twofold change in distance. Now compare the two curves at r = 1 m: I goes from 1 to 2 kg·m² when the mass doubles — only a twofold increase. Distance from the axis dominates.

Worked examples

Example 1: Asymmetric dumbbell — which end do you rotate about?

A light (massless) rod of length 1.0 m has a 3 kg ball at the left end and a 1 kg ball at the right end. Find the moment of inertia about (a) an axis through the 3 kg end, and (b) an axis through the 1 kg end. Both axes are perpendicular to the rod.

Asymmetric dumbbell rotated about two different ends Top panel: axis through the 3 kg ball, 1 kg ball is 1 m away, I = 1.0 kg·m². Bottom panel: axis through the 1 kg ball, 3 kg ball is 1 m away, I = 3.0 kg·m². (a) Axis through the heavier ball axis 3 kg 1 kg r = 0 1.0 m (b) Axis through the lighter ball axis 3 kg 1 kg 1.0 m r = 0
The same asymmetric dumbbell about two axes. In (a), the heavy ball sits on the axis and only the light ball contributes to $I$. In (b), the heavy ball is far from the axis and dominates $I$.

(a) Axis through the 3 kg ball

Step 1. Identify the distances from the axis.

The 3 kg ball is on the axis: r_1 = 0. The 1 kg ball is at the far end: r_2 = 1.0 m.

Why: the axis passes directly through the 3 kg mass, so its distance from the axis is zero.

Step 2. Apply I = \sum m_i r_i^2.

I = m_1 r_1^2 + m_2 r_2^2 = 3 \times 0^2 + 1 \times (1.0)^2 = 0 + 1.0 = 1.0 \text{ kg·m}^2

Why: the 3 kg ball contributes nothing — sitting on the axis, it has r = 0, so mr^2 = 0 regardless of how heavy it is. Only the 1 kg ball, at 1 m from the axis, contributes.

(b) Axis through the 1 kg ball

Step 3. Now the 1 kg ball is on the axis (r_2 = 0) and the 3 kg ball is 1.0 m away (r_1 = 1.0 m).

I = 3 \times (1.0)^2 + 1 \times 0^2 = 3.0 + 0 = 3.0 \text{ kg·m}^2

Why: the 3 kg ball is now 1 m from the axis and contributes 3 \times 1 = 3.0 kg·m². The 1 kg ball is on the axis and contributes nothing. The heavy mass, far from the axis, dominates.

Step 4. Compare the two results.

\frac{I_{\text{light end}}}{I_{\text{heavy end}}} = \frac{3.0}{1.0} = 3

Why: rotating about the lighter end puts the heavier mass far from the axis, tripling I. The body is three times harder to spin up about the light end than about the heavy end.

Result: I_{\text{heavy end}} = 1.0 kg·m², I_{\text{light end}} = 3.0 kg·m². Rotating about the heavy end gives three times less resistance.

What this shows: If you have a choice of pivot, putting the heavier mass on the axis minimises the moment of inertia. This is why a ceiling fan's motor — the heaviest component — sits at the axis, and the lightweight blades extend outward. It is also why you instinctively choke up on a heavy cricket bat (grip closer to the blade) when you want a quicker swing.

Example 2: Spinning up a potter's wheel

A kumhar's (potter's) wheel has a moment of inertia of 2.5 kg·m² about its central axle. The potter pushes tangentially at the rim, 0.20 m from the centre, with a steady force of 50 N. Find the angular acceleration. If the potter loads more clay onto the rim and the moment of inertia doubles to 5.0 kg·m², what happens to the angular acceleration under the same push?

Top view of a potter's wheel with a tangential force at the rim A circular wheel viewed from above, radius 0.20 m. The potter pushes tangentially with 50 N at the rim, creating a torque about the central axle. I = 2.5 kg·m². I = 2.5 kg·m² r = 0.20 m F = 50 N α = ?
Top view of the potter's wheel. The force $F$ acts tangentially at the rim, creating torque $\tau = Fr$ about the central axis.

Step 1. Compute the torque.

\tau = F \times r = 50 \times 0.20 = 10 \text{ N·m}

Why: the force is tangential — perpendicular to the radius — so the full force contributes to the torque. If the push were at an angle, only the perpendicular component would count.

Step 2. Apply \tau = I\alpha to find the angular acceleration.

\alpha = \frac{\tau}{I} = \frac{10}{2.5} = 4.0 \text{ rad/s}^2

Why: this is Newton's second law for rotation, solved for \alpha. Larger I means smaller \alpha for the same torque — exactly as a larger mass gives smaller linear acceleration for the same force.

Step 3. With the clay-loaded wheel (I = 5.0 kg·m²), the same torque gives:

\alpha = \frac{10}{5.0} = 2.0 \text{ rad/s}^2

Why: doubling I halved the angular acceleration. The heavier rim resists the same push more stubbornly.

Step 4. Verify units.

[\alpha] = \frac{[\tau]}{[I]} = \frac{\text{N·m}}{\text{kg·m}^2} = \frac{\text{kg·m/s}^2 \cdot \text{m}}{\text{kg·m}^2} = \frac{1}{\text{s}^2} = \text{rad/s}^2 \checkmark

Why: the radian is dimensionless, so rad/s² has the same dimensions as 1/s². The units work out exactly.

Result: The original wheel accelerates at 4.0 rad/s². With double the moment of inertia, the same push gives 2.0 rad/s² — exactly half.

What this shows: \tau = I\alpha is the rotational twin of F = ma. The potter knows this instinctively: a wheel with clay packed at the rim is harder to spin up (large I, small \alpha), but once spinning, it holds its speed longer — which is exactly what you want for steady, even shaping.

Common confusions

If you came here to understand what moment of inertia is, apply I = \sum m_i r_i^2 and \tau = I\alpha, and solve problems — you have everything you need. What follows is for readers who want the continuous-body formulation, the integration technique for a uniform rod, and the concept of radius of gyration.

The continuous version: I = \int r^2 \, dm

A solid object — a cricket bat, a flywheel, a ceiling fan blade — is not made of discrete particles. Its mass is spread continuously through its volume. You cannot list individual m_i and r_i values; instead, you slice the body into infinitesimally small mass elements dm, each at distance r from the axis, and integrate:

I = \int r^2 \, dm

Why: this is the continuous limit of \sum m_i r_i^2. As the number of particles becomes infinite and each particle mass becomes infinitesimal, the sum becomes an integral. The physics is the same — you are still adding up mass × distance² for every piece of the body.

To evaluate the integral, express dm in terms of a geometric variable. For a uniform body of total mass M and length L, the linear mass density is \lambda = M/L, so dm = \lambda \, dx = (M/L)\, dx.

Derivation: uniform thin rod about one end

Take a thin uniform rod of mass M and length L. Place the axis of rotation at one end, perpendicular to the rod. Let x be the distance from the end where the axis sits.

Each small element of length dx at position x has mass dm = (M/L)\,dx and sits at distance r = x from the axis.

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

Why: the mass per unit length is constant (M/L) because the rod is uniform. The factor x^2 is r^2 — each element's squared distance from the axis.

I = \frac{M}{L} \left[\frac{x^3}{3}\right]_0^L = \frac{M}{L} \cdot \frac{L^3}{3}
\boxed{I_{\text{end}} = \frac{1}{3}ML^2}

Why: the standard result \int_0^L x^2\,dx = L^3/3. Elements far from the axis contribute disproportionately (because of x^2), so the effective "average" distance is larger than L/2.

Derivation: uniform thin rod about its centre

Now place the axis at the centre of the rod. Let x be measured from the centre, running from -L/2 to +L/2.

I = \frac{M}{L} \int_{-L/2}^{L/2} x^2 \, dx = \frac{M}{L} \left[\frac{x^3}{3}\right]_{-L/2}^{L/2}
= \frac{M}{L} \left(\frac{L^3}{24} + \frac{L^3}{24}\right) = \frac{M}{L} \cdot \frac{L^3}{12}
\boxed{I_{\text{centre}} = \frac{1}{12}ML^2}

Why: the integral of x^2 from -a to +a is 2a^3/3. With a = L/2, this gives 2(L/2)^3/3 = L^3/12. The centre axis gives a smaller I because no part of the rod is farther than L/2 from the axis, whereas the end axis has parts up to L away.

Notice the ratio:

\frac{I_{\text{end}}}{I_{\text{centre}}} = \frac{ML^2/3}{ML^2/12} = 4

Shifting the axis from the centre to the end quadrupled the moment of inertia for a uniform rod. This relationship is made precise by the parallel axis theorem: I_{\text{end}} = I_{\text{centre}} + Md^2, where d = L/2 is the distance between the two axes. You can verify: \frac{1}{12}ML^2 + M(L/2)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{4}{12}ML^2 = \frac{1}{3}ML^2. The full theorem is derived in Parallel and Perpendicular Axes Theorems.

Radius of gyration

Engineers sometimes write the moment of inertia as:

I = Mk^2

where k is the radius of gyration. Rearranging:

k = \sqrt{\frac{I}{M}}

The radius of gyration answers this question: if you compressed all the mass of the body into a single point at distance k from the axis, that point mass would have the same moment of inertia as the real distributed body. It is a single number summarising the "effective distance" of the mass from the axis.

For a uniform rod about its centre: k = \sqrt{ML^2/(12M)} = L/\sqrt{12} = L/(2\sqrt{3}) \approx 0.289\,L. The mass of the rod behaves as if it were concentrated about 29% of the way from the centre to the end.

For a uniform rod about one end: k = \sqrt{ML^2/(3M)} = L/\sqrt{3} \approx 0.577\,L. The effective distance is about 58% of the way along the rod — farther from the axis, because the mass is spread over a larger range of distances.

The radius of gyration is especially useful when comparing objects of different shapes and sizes. Two objects may have very different masses and shapes, but if they have the same radius of gyration about a given axis, they respond identically to the same torque (per unit mass).

Where this leads next