Visual walkthrough — Moment of inertia of - rod (about end, centre), disk, ring, sphere (solid, hollow), cylinder

This page is a companion to the parent topic. If a word feels new, we build it here before we use it.

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Step 1 — What does "moment of inertia" even ask?

WHAT. Picture a merry-go-round. A child sitting near the pole is easy to swing around; a child at the rim is hard. Moment of inertia $I$ is a single number that says how hard is this whole object to spin about a chosen line (the axis).

WHY. When something spins at angular speed $\omega$ ("omega" — how many radians it turns per second), a bit of mass at perpendicular distance $r$ from the axis moves along a circle at linear speed $v=\omega r$. Far bits move faster, so they carry more energy and resist spinning more. We need a bookkeeping quantity that weights each bit by how far out it sits.

PICTURE. The red bit is far, the green bit is near. Same $\omega$, but the red bit whips around a big circle.

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Step 2 — Why the formula must contain $r^2$

WHAT. Take one tiny piece of mass $m$ at distance $r$. Its kinetic energy is $\frac{1}{2}m{v}^2$. Substitute $v=\omega r$: $\underset{\text{energy of one bit}}{\underbrace{\frac{1}{2}m{v}^2}}=\frac{1}{2}m(\underset{\text{its speed}}{\underbrace{\omega r}}{)}^2=\frac{1}{2}m{r}^2{\omega }^2$

WHY. Add up every bit of the body. The $\frac{1}{2}$ and ${\omega }^2$ are the same for all bits (rigid body, one $\omega$), so they factor out: $KE=\frac{1}{2}\left(\underset{\text{call this }I}{\underbrace{\sum m{r}^2}}\right){\omega }^2=\frac{1}{2}I{\omega }^2$ The thing in the bracket, $\sum m{r}^2$, is exactly what we name $I$. The square on $r$ was not chosen — it fell out of $v^2=(\omega r{)}^2$. That is why distant mass costs so much: doubling $r$ quadruples its contribution.

PICTURE. Two equal masses; the outer one's bar (its $r^2$ contribution) is four times taller though it is only twice as far.

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Step 3 — The easiest object first: the ring (our building block)

WHAT. A ring is all its mass sitting on one circle of radius $R$. Every single piece is at the same $r=R$.

WHY start here? Because when $r$ is constant it pops straight out of the sum — no calculus needed. The disk will then be made out of many rings, so cracking the ring cracks the disk. $I_{\text{ring}}=\int \underset{\text{same for all}}{\underbrace{R^2}}dm=R^2\int dm=R^{2}M=M{R}^2$ Here $\int dm=M$ just means "all the little pieces add to the total mass $M$."

PICTURE. Every arrow from the axis to the ring has identical length $R$ — that sameness is the whole trick.

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Step 4 — Slice the disk into rings

WHAT. A solid disk of radius $R$ is like a target: nested rings from the centre out. Pick one thin ring at radius $r$ (a value between $0$ and $R$) with a tiny thickness $dr$.

WHY. We already know a ring's contribution is $(\text{its mass})\times r^2$. So if we find the mass of this thin ring, we can add them all up with an integral. This is the Feynman move: reuse the result you already have.

To get the thin ring's mass we spread the disk's mass evenly over its area. Define surface density $\sigma =\frac{M}{\pi R^2}(\text{mass per unit area})$ Unrolled, the thin ring is a strip of length = circumference $2\pi r$ and width $dr$, so its area is $2\pi rdr$ and its mass is $dm=\sigma (2\pi rdr)$

• $2\pi r$ ::: how long the ring is (bigger rings are longer).
• $dr$ ::: how thick the strip is.
• $\sigma$ ::: converts area into mass.

PICTURE. One highlighted ring at radius $r$, thickness $dr$, shown also "unrolled" into a straight strip of length $2\pi r$.

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Step 5 — Add up all the rings (the integral)

WHAT. Each thin ring contributes $r^{2}dm$. Plug in $dm$ from Step 4 and sum from the centre ($r=0$) to the edge ($r=R$): $I={\int }_0^R\underset{\text{ring’s }{r}^2}{\underbrace{r^2}}\underset{\text{ring’s mass}}{\underbrace{(\sigma 2\pi rdr)}}=2\pi \sigma {\int }_0^R{r}^{3}dr$

WHY the $r^3$? Because $r^2\times r=r^3$: the $r^2$ is "how far out," the extra $r$ is because bigger rings are longer (more mass). Now we need the area under $r^3$.

Substitute: $I=2\pi \sigma \cdot \frac{R^4}{4}=\frac{\pi \sigma R^4}{2}$

PICTURE. The bars stack up: each ring's contribution grows like $r^3$, so the outer rings dominate the total area — visualised as a shaded region under the curve $r^3$.

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Step 6 — Put mass back in, get the clean answer

WHAT. Our answer still has $\sigma$ in it. Replace $\sigma =\frac{M}{\pi R^2}$: $I=\frac{\pi R^4}{2}\cdot \frac{M}{\pi R^2}=\frac{M{R}^4}{2R^2}=\frac{1}{2}M{R}^2$ The $\pi$ cancels, and $R^4/R^2=R^2$.

$\boxed{I_{\text{disk}}=\frac{1}{2}M{R}^2}$

WHY the $\frac{1}{2}$ vs the ring's $1$? The ring keeps all its mass at $R$. The disk has plenty of mass at small $r$, which barely resists spinning. Averaging over the whole area, the effective $r^2$ comes out to exactly half of $R^2$. The "$\frac{1}{2}$" is an average, not a coincidence.

PICTURE. Ring vs disk side by side — same $M$, same $R$, but the disk's mass sits inward, so its "resistance bar" is half as tall.

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Step 7 — Edge cases: does the answer still behave?

WHAT / WHY. A formula you trust must survive the extremes. Check three:

1. Length of a cylinder. Stack many disks to make a solid cylinder about its long axis. Every disk gives $\frac{1}{2}dm{R}^2$, and adding them just adds up to $\frac{1}{2}M{R}^2$ — the height $h$ never appears. So a coin and a long rod (same $R$) have the same $I$ about that axis.
2. Shrink to a point ($R\to 0$). Then $I=\frac{1}{2}M(0{)}^2=0$: a point on the axis is trivially easy to spin. ✓
3. All mass at the rim ($r=R$ only). Then it's a ring, and our slicing gives $M{R}^2$, the ceiling from Step 3. The disk's $\frac{1}{2}$ sits below this, as it must, because pulling mass inward can only lower $I$.

PICTURE. Coin and tall cylinder (same $R$) labelled with the identical $\frac{1}{2}M{R}^2$; a shrinking disk vanishing to the axis.

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The one-picture summary

Every step in one frame: mass sits at various $r$, each weighted by $r^2$; rings build the disk; the average lands at $\frac{1}{2}{R}^2$; multiply by $M$.

Recall Feynman retelling — the whole walk in plain words

Spinning something is making its bits run in circles. A bit far from the pole has to run fast, so it fights you hard — and because speed is distance-times-$\omega$, and energy uses speed squared, a bit twice as far fights four times as hard. That squared-distance rule is the entire idea; adding it up over the object is the moment of inertia. The easiest shape is a ring, where every bit is at the same rim distance $R$, giving $M{R}^2$ — the hardest a hoop can be. A disk is just a nest of rings; the inner rings are lazy (small $r$), the outer ones do the work. When you average all their squared distances across the flat area, you get exactly half of $R^2$ — that's where the $\frac{1}{2}$ comes from. Multiply by the total mass and you have $\frac{1}{2}M{R}^2$. Stacking disks into a cylinder changes nothing about the long axis, because you never moved any mass to a new distance. And if you squeezed all the mass out to the rim you'd get the ring back — the ceiling we started from.

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Connections

• Parent topic (Hinglish) — the full table of standard bodies.
• Rotational Kinetic Energy — the $KE=\frac{1}{2}I{\omega }^2$ that forced the $r^2$ in Step 2.
• Parallel Axis Theorem — shift this disk result to an edge axis.
• Perpendicular Axis Theorem — the disk's diameter axis is $\frac{1}{4}M{R}^2$ (half of $\frac{1}{2}$).
• Radius of Gyration — packages $\frac{1}{2}M{R}^2$ as $k=R/\sqrt{2}$.
• Torque and Angular Acceleration — where $I$ is used next, in $\tau =I\alpha$.
• Rolling Motion — this $I$ decides how fast a disk rolls down a ramp.