In short

The moment of inertia of a body depends on how its mass is distributed relative to the rotation axis. Standard results: a uniform rod about its centre has I = \tfrac{1}{12}ML^2; a ring has I = MR^2; a disc has I = \tfrac{1}{2}MR^2; a solid sphere has I = \tfrac{2}{5}MR^2; a hollow spherical shell has I = \tfrac{2}{3}MR^2. Each follows from the single integral I = \int r^2\,dm applied to the body's geometry.

A potter in Jaipur sits at a wheel, shaping clay. The wheel is a thick stone disc — about 8 kg, roughly 25 cm across. Once spinning, it keeps going for a long time. Now imagine the potter took that same 8 kg of stone and cast it as a thin ring at the rim, like a bicycle wheel. That ring would be harder to start spinning and harder to stop — even though it has the same mass and the same outer radius.

The reason is in the integral you met in the previous article:

I = \int r^2\,dm

Every tiny piece of mass is weighted by the square of its distance from the rotation axis. In the disc, much of the clay sits near the centre, where r is small — those pieces contribute almost nothing. In the ring, all the clay sits at the maximum distance R, and every gram counts at full weight. The standard results that follow — for rods, rings, discs, cylinders, and spheres — all emerge from this single integral applied to different geometries.

Uniform rod about its centre

Take a thin, straight rod of mass M and length L, spinning about an axis through its midpoint, perpendicular to its length. Think of balancing a metre ruler on your fingertip and flicking it into a horizontal spin.

Integration setup for a uniform rod about its centre A horizontal rod of length L centred at the origin. A thin shaded element dx sits at distance x from the centre. A vertical dashed line marks the rotation axis through the midpoint. axis dm x L/2 L/2 dx
A uniform rod of length $L$, with the axis through the centre. The shaded element $dm$ sits at distance $x$ from the axis.

The rod has uniform linear mass density \lambda = M/L. A thin slice of width dx at position x from the centre has mass dm = (M/L)\,dx.

Step 1. Write the integral. Position x runs from -L/2 to +L/2, with the axis at x = 0.

I = \int_{-L/2}^{L/2} x^2\,dm = \frac{M}{L}\int_{-L/2}^{L/2} x^2\,dx

Why: every slice at distance |x| from the axis contributes x^2\,dm to the moment of inertia. Summing over the entire rod gives the integral.

Step 2. Evaluate.

I = \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}

Why: x^2 is an even function, so the left half and the right half contribute equally. The two L^3/24 terms add rather than cancel.

\boxed{I_{\text{rod, centre}} = \frac{1}{12}ML^2}

Uniform rod about one end

Now clamp the axis at one end of the rod. This is a cricket bat held by the very tip of the handle and swung — all the mass lies on one side of the axis.

The element at position x from the clamped end has mass dm = (M/L)\,dx, and x now runs from 0 to L.

I = \frac{M}{L}\int_0^L x^2\,dx = \frac{M}{L} \cdot \frac{L^3}{3}

Why: the limits shift from [-L/2, L/2] to [0, L]. The integral of x^2 over [0, L] is L^3/3, which is four times L^3/12 — because mass that was at distance L/2 from the centre is now at distance L from the end.

\boxed{I_{\text{rod, end}} = \frac{1}{3}ML^2}

Four times the centre-axis value. The physical picture: shifting the axis from the centre to one end pushes more mass farther away. The far tip, which was at distance L/2, is now at distance L — and r^2 grows as the square.

Uniform ring about the central axis

A ring (or thin hoop) of mass M and radius R, spinning about the axis through its centre, perpendicular to its plane. Picture a bangle spinning on a pencil pushed through its centre.

Every particle of the ring sits at the same distance R from the axis:

I = \int r^2\,dm = R^2\int dm = R^2 M

Why: when every mass element is at the same distance R, the factor r^2 is constant and comes out of the integral. What remains is \int dm = M.

\boxed{I_{\text{ring}} = MR^2}

This is the largest possible I for a body of mass M confined within radius R. You cannot do better than placing all the mass at the maximum distance from the axis.

Uniform disc about the central axis

A flat, uniform disc of mass M and radius R, spinning about the axis through its centre — the potter's wheel.

Integration setup for a uniform disc about its central axis A circular disc viewed from above. A thin ring at radius r with width dr is highlighted in red. The axis passes through the centre, perpendicular to the disc. axis r R dr
A uniform disc of radius $R$. The shaded ring at radius $r$ (width $dr$) is the integration element. The axis passes through the centre, pointing out of the page.

The disc has uniform surface mass density \sigma = M/(\pi R^2). Break the disc into thin concentric rings. A ring at radius r with width dr has area 2\pi r\,dr and mass:

dm = \sigma \cdot 2\pi r\,dr = \frac{M}{\pi R^2} \cdot 2\pi r\,dr = \frac{2M}{R^2}\,r\,dr

Why: a thin ring's area is its circumference (2\pi r) times its width (dr). Multiplying by the surface density converts area to mass.

Every point on this ring sits at distance r from the axis, so its contribution to I is r^2\,dm:

I = \int_0^R r^2 \cdot \frac{2M}{R^2}\,r\,dr = \frac{2M}{R^2}\int_0^R r^3\,dr = \frac{2M}{R^2} \cdot \frac{R^4}{4}

Why: the r^3 in the integrand comes from r^2 (the moment-of-inertia weighting) times r (from the ring's circumference). The outer rings are heavier and farther from the axis — but mass near the centre pulls the average down.

\boxed{I_{\text{disc}} = \frac{1}{2}MR^2}

Exactly half the ring's value. Same mass, same outer radius — but the disc spreads mass inward, where r is small and the contribution to I is weak.

Solid and hollow cylinders

A solid cylinder of mass M and radius R, spinning about its central axis, is a stack of identical discs. Each disc shares the same axis, so the total I is the disc result applied to the full mass:

\boxed{I_{\text{solid cylinder}} = \frac{1}{2}MR^2}

A thin-walled hollow cylinder (a pipe with all mass at radius R) is a stack of rings:

\boxed{I_{\text{thin hollow cylinder}} = MR^2}

For a thick-walled cylinder with inner radius R_1 and outer radius R_2, integrate the thin-ring contribution from R_1 to R_2. The algebra uses the factorisation R_2^4 - R_1^4 = (R_2^2 + R_1^2)(R_2^2 - R_1^2), and the result is:

\boxed{I_{\text{thick cylinder}} = \frac{1}{2}M(R_1^2 + R_2^2)}

Set R_1 = 0 and you recover the solid cylinder. Set R_1 \to R_2 and you recover the thin-walled case.

Hollow spherical shell about a diameter

A thin spherical shell (like a thin-walled hollow rubber ball) of mass M and radius R, spinning about any diameter. The derivation is a showcase of symmetry — no complicated integrals needed.

Pick the x-axis as the rotation axis. For any mass element dm on the shell, its perpendicular distance from the x-axis is \sqrt{y^2 + z^2}, so:

I_x = \int (y^2 + z^2)\,dm

By the same logic, I_y = \int (x^2 + z^2)\,dm and I_z = \int (x^2 + y^2)\,dm. Add all three:

I_x + I_y + I_z = \int 2(x^2 + y^2 + z^2)\,dm

Why: expand each integrand. The term x^2 appears in I_y and I_z (twice total), and similarly for y^2 and z^2. Each squared coordinate appears exactly twice in the sum, giving the factor of 2.

Every point on the shell satisfies x^2 + y^2 + z^2 = R^2:

I_x + I_y + I_z = 2R^2\int dm = 2MR^2

A sphere looks identical from every direction — it has no preferred axis. By this perfect symmetry, I_x = I_y = I_z:

3I = 2MR^2

Why: instead of wrestling with a spherical-coordinate integral, you use the sphere's symmetry. The identity x^2 + y^2 + z^2 = R^2 on the surface does all the heavy lifting, and the three-way equality of moments finishes the job.

\boxed{I_{\text{hollow sphere}} = \frac{2}{3}MR^2}

Solid sphere about a diameter

Build the solid sphere from thin concentric shells, like peeling an onion inward — or like the layers of a laddu.

A shell at radius r with thickness dr has volume 4\pi r^2\,dr and mass:

dm = \rho \cdot 4\pi r^2\,dr, \qquad \text{where } \rho = \frac{3M}{4\pi R^3}

Why: \rho is the uniform volume density — total mass divided by total volume (4\pi R^3/3). The shell's volume is its surface area (4\pi r^2) times its thickness (dr).

Each shell's moment of inertia about a diameter is \frac{2}{3}r^2\,dm — the hollow-sphere result you just derived, applied at radius r with mass dm. Sum over all shells from r = 0 to r = R:

I = \int_0^R \frac{2}{3}\,r^2\,dm = \int_0^R \frac{2}{3}\,r^2 \cdot \frac{3M}{4\pi R^3} \cdot 4\pi r^2\,dr = \frac{2M}{R^3}\int_0^R r^4\,dr
= \frac{2M}{R^3} \cdot \frac{R^5}{5}

Why: the r^4 in the integrand is r^2 from the hollow-sphere formula times r^2 from the shell's surface area. The integration sums all layers from the centre outward.

\boxed{I_{\text{solid sphere}} = \frac{2}{5}MR^2}

Less than the hollow sphere's \frac{2}{3}MR^2. The solid sphere packs mass throughout the interior — including near the centre where r is small — so its effective moment of inertia is lower.

Why distribution matters — the summary

Every formula above has the form I = kMR^2 (or kML^2 for rods), where the coefficient k encodes the mass distribution:

Shape Axis I k
Thin ring / hollow cylinder Central axis MR^2 1
Hollow spherical shell Diameter \frac{2}{3}MR^2 0.667
Uniform disc / solid cylinder Central axis \frac{1}{2}MR^2 0.5
Solid sphere Diameter \frac{2}{5}MR^2 0.4
Uniform rod (centre) Centre, ⊥ to rod \frac{1}{12}ML^2 0.083
Uniform rod (end) End, ⊥ to rod \frac{1}{3}ML^2 0.333

The pattern: the more mass concentrates at the outer edge, the higher the coefficient. The ring (k = 1) puts everything at R. The disc (k = 0.5) spreads mass inward. The solid sphere (k = 0.4) spreads mass in three dimensions, pushing even more toward the centre.

The interactive figure below makes this visible. Fix M = 1 kg and drag the radius to see how I scales for each shape.

Interactive: moment of inertia versus radius for four standard shapes Four parabolic curves showing I versus R for a ring, hollow sphere, disc, and solid sphere, all with M = 1 kg. A draggable point on the R axis lets you compare I values at any radius. radius R (m) I (kg·m²), M = 1 kg 0 0.14 0.28 0.1 0.25 0.5 ring hollow sphere disc solid sphere drag the red point along the axis
Drag the red point to change $R$. All shapes have $M = 1$ kg. The ring (solid red) always has the highest $I$; the solid sphere (dashed grey) the lowest. At $R = 0.25$ m, the ring gives 0.0625 kg·m² while the disc gives 0.0313 kg·m² — the disc needs half the torque for the same angular acceleration.

Worked examples

Example 1: The potter's wheel

A potter's wheel is a uniform stone disc of mass 8 kg and radius 0.25 m. Find its moment of inertia about the central axis. If the potter applies a steady torque of 2 N·m by pushing the rim, find the angular acceleration.

Potter's wheel: a disc with torque applied Top view of a uniform disc of mass 8 kg and radius 0.25 m. A curved arrow shows torque of 2 N·m applied at the rim. A straight arrow marks the radius. R = 0.25 m τ = 2 N·m M = 8 kg
Top view of the potter's wheel (disc) with torque applied at the rim.

Step 1. Identify the shape and axis.

The wheel is a uniform disc spinning about its central axis. The formula is I = \frac{1}{2}MR^2.

Why: a potter's wheel is a flat disc rotating about the axis through its centre, perpendicular to its face — exactly the geometry derived above.

Step 2. Substitute the values.

I = \frac{1}{2} \times 8 \times (0.25)^2 = \frac{1}{2} \times 8 \times 0.0625 = 0.25 \text{ kg·m}^2

Why: use SI units — the radius must be in metres (0.25 m, not 25 cm), and the mass in kilograms.

Step 3. Find the angular acceleration using \tau = I\alpha.

\alpha = \frac{\tau}{I} = \frac{2}{0.25} = 8 \text{ rad/s}^2

Why: this is the rotational analogue of F = ma. Torque plays the role of force, moment of inertia plays the role of mass, and angular acceleration plays the role of linear acceleration.

Step 4. Interpret the result.

Starting from rest, the wheel reaches \omega = \alpha t = 8 \times 1 = 8 rad/s (about 1.3 revolutions per second) after one second of steady pushing.

Why: to convert rad/s to revolutions per second, divide by 2\pi: 8/(2\pi) \approx 1.27 rev/s.

Result: I = 0.25 kg·m²; angular acceleration \alpha = 8 rad/s².

What this shows: A potter's wheel is easy to spin up precisely because the disc shape keeps I low — half of what a ring of the same mass and radius would give. The potter benefits from the mass being spread inward rather than concentrated at the rim.

Example 2: Swinging a bamboo staff — end versus centre

A bamboo danda (staff) used in lathi training is 1.5 m long and has a mass of 0.6 kg. Compare the moment of inertia when swung from one end (like a bat) versus when spun from the centre (like a baton).

Rod with axis at one end versus at the centre Left: a rod with a dashed axis line at the left end, I = 0.45 kg·m². Right: same rod with axis at the centre, I = 0.1125 kg·m². A label shows the end-axis value is four times larger. Axis at end axis 1.5 m I = 0.45 kg·m² Axis at centre axis 1.5 m I = 0.1125 kg·m² × 4
The same rod is four times harder to swing from one end than from the centre.

Step 1. Compute I about one end.

I_{\text{end}} = \frac{1}{3}ML^2 = \frac{1}{3} \times 0.6 \times (1.5)^2 = \frac{1}{3} \times 0.6 \times 2.25 = 0.45 \text{ kg·m}^2

Why: the axis is at one end, so use I = \frac{1}{3}ML^2. The staff's mass is entirely on one side of the pivot.

Step 2. Compute I about the centre.

I_{\text{centre}} = \frac{1}{12}ML^2 = \frac{1}{12} \times 0.6 \times 2.25 = 0.1125 \text{ kg·m}^2

Why: the axis is at the midpoint, so use I = \frac{1}{12}ML^2. Mass is distributed symmetrically on both sides.

Step 3. Compare the two.

\frac{I_{\text{end}}}{I_{\text{centre}}} = \frac{0.45}{0.1125} = 4

Why: algebraically, \frac{ML^2/3}{ML^2/12} = 4 regardless of M and L. The ratio is always exactly 4 for a uniform rod.

Step 4. Physical meaning.

If you apply the same torque in both grips, the angular acceleration when holding the end is one-quarter of what it would be at the centre. The staff "feels" four times heavier when held at one tip.

Result: I_{\text{end}} = 0.45 kg·m², I_{\text{centre}} = 0.1125 kg·m². The end-axis value is exactly 4 times larger.

What this shows: The axis matters as much as the mass. Moving the pivot from the centre to one end does not change the mass at all, but it quadruples the moment of inertia. A batsman choking up on the handle — gripping closer to the blade — reduces I and can swing faster, trading reach for bat speed.

Common confusions

If you can use the formulas in the table and set up I = \int r^2\,dm for standard shapes, you have what you need for most problems. What follows is for readers who want the radius of gyration and the brute-force sphere calculation.

Radius of gyration

For any body with moment of inertia I and total mass M, the radius of gyration k is defined by:

I = Mk^2 \qquad \Longrightarrow \qquad k = \sqrt{\frac{I}{M}}

Physically, k is the distance from the axis at which you would need to concentrate all the mass (as a single point) to get the same moment of inertia. It gives you a single number that captures how "spread out" the mass is.

Shape I k
Ring MR^2 R
Disc \frac{1}{2}MR^2 R/\sqrt{2} \approx 0.707R
Hollow sphere \frac{2}{3}MR^2 R\sqrt{2/3} \approx 0.816R
Solid sphere \frac{2}{5}MR^2 R\sqrt{2/5} \approx 0.632R
Rod (centre) \frac{1}{12}ML^2 L/(2\sqrt{3}) \approx 0.289L

The ring's radius of gyration equals its actual radius — all mass is already at R. For the solid sphere, k \approx 0.63R: the effective mass concentration sits about two-thirds of the way from centre to surface.

The solid sphere by direct integration

The shell-stacking derivation in the main text is the cleanest approach. But you can also evaluate I directly in spherical coordinates (r, \theta, \phi), where \theta is the polar angle from the rotation axis:

I = \int \rho\,(r\sin\theta)^2\,r^2\sin\theta\,dr\,d\theta\,d\phi

Why: the perpendicular distance from the z-axis is r\sin\theta, and the volume element in spherical coordinates is r^2\sin\theta\,dr\,d\theta\,d\phi.

The three integrals separate:

I = \rho \int_0^R r^4\,dr \int_0^{\pi} \sin^3\theta\,d\theta \int_0^{2\pi} d\phi
  • \int_0^R r^4\,dr = R^5/5
  • \int_0^{2\pi} d\phi = 2\pi
  • \int_0^{\pi} \sin^3\theta\,d\theta = \int_0^{\pi}\sin\theta(1 - \cos^2\theta)\,d\theta

Why: substitute u = \cos\theta, du = -\sin\theta\,d\theta. The integral becomes \int_{-1}^{1}(1 - u^2)\,du = [u - u^3/3]_{-1}^{1} = (1 - 1/3) - (-1 + 1/3) = 4/3.

Putting it all together with \rho = 3M/(4\pi R^3):

I = \frac{3M}{4\pi R^3} \cdot \frac{R^5}{5} \cdot \frac{4}{3} \cdot 2\pi = \frac{2MR^2}{5}

The factor 4/3 from the \sin^3\theta integral is what distinguishes the sphere from a cylinder. A cylinder has no \theta variation — its moment of inertia comes only from the radial integral, giving \frac{1}{2}MR^2. The sphere's three-dimensional spread, captured by the \sin^2\theta weighting in the integrand, reduces the coefficient from 1/2 to 2/5.

Where this leads next