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

Before we start, one word we keep using: axis. The axis is the imaginary line the body spins around. The definition above tells us the ONLY distance that matters is $r_{\perp }$, the perpendicular drop onto that line — not the distance to the centre, not the distance along the body. If you draw the axis wrong, every $r_{\perp }$ is wrong and every number is wrong. So each example starts by drawing the axis and identifying $r_{\perp }$.

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The scenario matrix

Every problem this topic can throw is one of these case classes. The examples below are tagged with the cell they fill.

Cell	Case class	What makes it tricky	Covered by
A	Single body, standard axis	just pick the formula	Ex 1
B	Axis shifted off the centre	need Parallel Axis Theorem	Ex 2, Ex 6
C	Axis in the plane (diameter)	need Perpendicular Axis Theorem	Ex 3
D	Composite body (add / subtract parts)	$I$ adds; holes subtract	Ex 4, Ex 5
E	Degenerate / limiting input ($R\to 0$, $L\to 0$, point mass)	check the formula doesn't break	Ex 7
F	Zero-distance mass (mass on the axis)	contributes $0$	Ex 4 (hub)
G	Real-world word problem (energy / rolling)	translate words → axis → $I$	Ex 8
H	Exam twist (two bodies, ratio, "which spins easier")	compare, don't just compute	Ex 9

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Ex 1 — Cell A: pure formula, single body

Forecast: Guess the order of magnitude before reading on. Is it closer to $0.01$ or $10\mathrm{kg}{\mathrm{m}}^2$?

What to see: the red circle is the disk; the black dot at the centre is where the axis pierces it (the axis points straight out of the page, toward you). The black arrow labelled $R$ is one radius — notice that a mass element out at the rim has its perpendicular drop $r_{\perp }=R$, while an element halfway in has $r_{\perp }=R/2$. The formula's $\frac{1}{2}$ comes from averaging $r_{\perp }^2$ over all of them.

1. Identify the axis. Look at the figure: the red disk is pierced at its centre by the axis coming out of the page. Each mass element's $r_{\perp }$ is its straight-line distance from that centre dot. Why this step? Cell A only works when the axis is a standard one. Confirming it lets us grab the boxed formula instead of integrating $\int r_{\perp }^{2}dm$ by hand.
2. Pick the formula. For a disk about that axis, the parent note gives $I=\frac{1}{2}M{R}^2$. Why this step? Every bit of mass sits at some $r_{\perp }$ between $0$ and $R$; the $\frac{1}{2}$ is the average of $r_{\perp }^2$ weighted by area — we don't redo that integral, we reuse it.
3. Plug numbers. $I=\frac{1}{2}(2)(0.30{)}^2=\frac{1}{2}(2)(0.09)=0.09\mathrm{kg}{\mathrm{m}}^2$ Why this step? Substituting the given $M$ and $R$ turns the symbolic law into the one number the question actually asks for; squaring $R$ (not doubling it) is essential because $I$ depends on distance squared.

Verify: Units: $\mathrm{kg}\cdot {\mathrm{m}}^2$ ✓. Sanity: a $2\mathrm{kg}$ ring of the same radius would be $M{R}^2=0.18$; the disk is exactly half that because its mass is spread inward — matches "Ring 1, Disk ½". ✓

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Ex 2 — Cell B: axis shifted (parallel axis)

Forecast: Bigger or smaller than the $0.09$ from Ex 1? By how much?

What to see: the small black dot marked "CM" is the old central axis; the black square on the red rim is the NEW axis. The double-headed arrow between them is the gap $d$ — and here it is exactly one radius $R$. Every mass element is now measured perpendicular to the rim axis, which is why every $r_{\perp }$ grows and $I$ increases.

1. Spot that this is NOT a standard axis. The axis no longer passes through the centre of mass, so no boxed formula gives $I$ directly. Why this step? Recognising "off-centre" is the whole trigger for using the Parallel Axis Theorem instead of a table lookup.
2. State the theorem. $I=I_{cm}+M{d}^2$, where $I_{cm}$ is the moment about the parallel axis through the centre of mass, and $d$ is the gap between the two parallel axes. Why this step? It says: shifting an axis away by $d$ always adds $M{d}^2$, because now every mass element's $r_{\perp }$ is larger by the same offset — the theorem is just $\int r_{\perp }^{2}dm$ re-measured from a shifted line.
3. Identify $I_{cm}$ and $d$. The parallel central axis gives $I_{cm}=\frac{1}{2}M{R}^2=0.09$ (from Ex 1). The gap from centre to rim is $d=R=0.30\mathrm{m}$. Why this step? $d$ is the distance between the axes, which here is exactly the radius — see the red segment in the figure.
4. Add. $I=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2=1.5(2)(0.09)=0.27\mathrm{kg}{\mathrm{m}}^2$ Why this step? We combine the two contributions the theorem gave us; because $d=R$, the extra term $M{d}^2$ equals a full $M{R}^2$, doubling on top of the original half — the arithmetic must keep $R$ squared throughout.

Verify: $0.27=3\times 0.09$, i.e. three times the central value. It got bigger, exactly as the theorem promises (you can never make $I$ smaller by moving off the CM). Units ✓.

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Ex 3 — Cell C: axis in the plane (perpendicular axis)

Forecast: For the central perpendicular axis a ring is $M{R}^2=0.25$. About a diameter — same, more, or less?

What to see: the red circle is the ring, the horizontal black arrow is the diameter axis, lying flat in the ring's plane. The dashed vertical line shows the perpendicular drop $r_{\perp }=R$ for the topmost point, while the black dot sitting on the axis itself has $r_{\perp }=0$. Because $r_{\perp }$ varies from $0$ to $R$ around the ring, we cannot say "all mass at $R$" — that is exactly why we need the perpendicular-axis theorem.

1. See why we need a new tool. For the diameter axis, different points on the ring have different perpendicular distances $r_{\perp }$ (points on the axis are at $0$, the top and bottom at $R$). We can't say "all mass at $R$" anymore. Why this step? This distinguishes Cell C from the ring's easy central-axis case, where every $r_{\perp }=R$.
2. Invoke the Perpendicular Axis Theorem. For any flat (planar) body lying in the $x$–$y$ plane, $I_z=I_x+I_y$, where $z$ is perpendicular to the plane and $x,y$ are two in-plane axes through the same point. Why this step? The ring is planar, so this theorem links the easy $z$-axis to the diameter axes without integrating.
3. Use symmetry. The two diameters $x$ and $y$ are identical by symmetry, so $I_x=I_y=I_d$. Then $I_z=2I_d$. Why this step? Symmetry turns one unknown into a factor of 2 — no integration needed.
4. Solve. $I_z=M{R}^2$ (central perpendicular). So $I_d=\frac{1}{2}{I}_z=\frac{1}{2}M{R}^2=\frac{1}{2}(1)(0.5{)}^2=0.125\mathrm{kg}{\mathrm{m}}^2$ Why this step? Rearranging $I_z=2I_d$ gives $I_d$ directly, and substituting the known $I_z=M{R}^2$ with the given numbers produces the answer; halving is legitimate because the two diameters share the whole $I_z$ equally.

Verify: Diameter value ($0.125$) is less than the central value ($0.25$) — right, because half the ring's mass now sits close to the diameter line. Units ✓.

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Ex 4 — Cells D & F: composite body with a hub on the axis

Forecast: Which of the three objects contributes the most? Which contributes nothing?

1. Moments of inertia add. For several masses about the same axis, $I_{\text{total}}=I_1+I_2+\dots$ — this is just $\int r_{\perp }^{2}dm$ (equivalently $\sum m{r}_{\perp }^2$) regrouped by object. Why this step? Because the defining integral is a plain sum over mass, you can compute each piece separately and add.
2. Disk. $I_d=\frac{1}{2}{M}_d{R}^2=\frac{1}{2}(100)(2{)}^2=200\mathrm{kg}{\mathrm{m}}^2$. Why this step? Standard disk formula, axis is its central axis, so every element's $r_{\perp }$ runs $0$ to $R$ exactly as the $\frac{1}{2}M{R}^2$ average assumes.
3. Child (point mass at rim). Treat as a point: $I_c=m{r}_{\perp }^2=50\cdot (2{)}^2=200\mathrm{kg}{\mathrm{m}}^2$. Why this step? A person is small compared to the ride, so the defining $I=\int r_{\perp }^{2}dm$ collapses to a single $m{r}_{\perp }^2$ with $r_{\perp }=R$.
4. Hub-cap ON the axis — Cell F. Its perpendicular distance to the axis is $r_{\perp }=0$, so $I_{\text{hub}}=10\cdot 0^2=0$. Why this step? Mass sitting on the spin line travels no circle at all ($r_{\perp }=0$ in the integral), so it contributes exactly zero. This is the degenerate zero-distance case.
5. Add. $I=200+200+0=400\mathrm{kg}{\mathrm{m}}^2$ Why this step? Step 1 licensed summing the parts, so the total is just the three contributions combined; the hub's zero is kept explicitly to show it genuinely drops out.

Verify: The hub contributes nothing despite weighing $10\mathrm{kg}$ — this is the whole point that "position beats mass" in rotation. Units all $\mathrm{kg}{\mathrm{m}}^2$ ✓.

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Ex 5 — Cell D: subtract a hole (disk with a bite removed)

Forecast: Should the hole make $I$ per kg bigger or smaller than a full disk of the same $R$?

What to see: the red-shaded annulus is the real material; the white circle in the middle (radius $a$) is the missing hole. The two black arrows mark $R$ (outer) and $a$ (inner). Think of it as the big red disk of radius $R$ with a same-density plug of radius $a$ scooped out — that "big disk minus plug" picture is exactly the algebra below.

1. Model as "big disk minus small disk". Imagine filling the hole to make a full disk of radius $R$, then subtracting the plug of radius $a$. Both share the same axis. Why this step? $\int r_{\perp }^{2}dm$ adds and subtracts, so a hole is just a negative disk with the same surface density $\sigma$ as the real material.
2. Work in surface density. Let $\sigma =M/(\pi (R^2-a^2))$ be the density of the real annulus. A full disk of radius $R$ has $I=\frac{1}{2}(\sigma \pi R^2)R^2$; the plug has $I=\frac{1}{2}(\sigma \pi a^2)a^2$. Why this step? Both the filled disk and the plug are made of the same stuff, so the same $\sigma$ describes them; only then can we subtract cleanly.
3. Subtract. $I=\frac{1}{2}\sigma \pi R^4-\frac{1}{2}\sigma \pi a^4=\frac{1}{2}\sigma \pi (R^4-a^4)$ Now substitute $\sigma \pi =M/(R^2-a^2)$: $I=\frac{1}{2}\frac{M}{R^2-a^2}(R^4-a^4)=\frac{1}{2}M\frac{(R^2-a^2)(R^2+a^2)}{R^2-a^2}=\frac{1}{2}M(R^2+a^2)$ Why this step? The factorisation $R^4-a^4=(R^2-a^2)(R^2+a^2)$ makes the annoying denominator cancel, leaving the clean annulus formula $I=\frac{1}{2}M(R^2+a^2)$.
4. Numbers ($R=0.06$, $a=0.015$): $I=\frac{1}{2}(0.020)(0.06^2+0.015^2)=0.010(0.0036+0.000225)=3.825\times 10^{-5}\mathrm{kg}{\mathrm{m}}^2$ Why this step? We substitute the given radii (converted to metres so units come out in $\mathrm{kg}{\mathrm{m}}^2$) into the compact formula from step 3; squaring both radii honours the distance-squared law and the two squared terms simply add.

Verify: If $a\to 0$ (no hole) the formula becomes $\frac{1}{2}M{R}^2$, the full disk — ✓ consistent. If $a\to R$ (all rim, a thin ring) it becomes $\frac{1}{2}M(2R^2)=M{R}^2$, the ring — ✓. Both limits are correct, so the general formula is trustworthy.

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Ex 6 — Cell B: rod about its end, checked two ways

Forecast: The centre value is $\frac{1}{12}M{L}^2$. End value — how many times bigger?

1. Direct formula. $I_{\text{end}}=\frac{1}{3}M{L}^2=\frac{1}{3}(0.6)(1.0{)}^2=0.20\mathrm{kg}{\mathrm{m}}^2$. Why this step? This is the standard end-axis result from the parent derivation; substituting $M$ and $L$ turns it into the requested number.
2. Cross-check with Parallel Axis Theorem. $I_{\text{end}}=I_{cm}+M{d}^2$ with $I_{cm}=\frac{1}{12}M{L}^2$ and $d=L/2$ (centre to end): $I=\frac{1}{12}(0.6)(1{)}^2+0.6{\left(\frac{1}{2}\right)}^2=0.05+0.15=0.20\mathrm{kg}{\mathrm{m}}^2$ Why this step? Getting the same number two independent ways proves both the formula and our axis placement; here $d=L/2$ because the CM sits at the rod's midpoint.

Verify: $\frac{1}{3}:\frac{1}{12}=4:1$, so the end value is $4\times$ the centre value — matches the parent's active-recall fact. ✓

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Ex 7 — Cell E: degenerate & limiting inputs

Forecast: Which of these should give $I=0$, and which should give a finite nonzero answer?

1. (a) Rod shrinking to a point. $I_{\text{end}}=\frac{1}{3}M{L}^2\to \frac{1}{3}M(0{)}^2=0$. Why this step? A point mass sitting on its own pivot has $r_{\perp }=0$ everywhere — no circle to travel, so $I=0$. The formula correctly collapses.
2. (b) Sphere shrinking to a point. $I=\frac{2}{5}M{R}^2\to \frac{2}{5}M(0{)}^2=0$. Same reasoning: all mass at the axis. Why this step? Confirms the sphere formula degenerates to a point mass sensibly ($r_{\perp }\to 0$).
3. (c) All mass at the rim. Ex 5's annulus formula $\frac{1}{2}M(R^2+a^2)$ at $a=R$ gives $\frac{1}{2}M(2R^2)=M{R}^2$ — the ring. Why this step? Pushing every bit of mass out to radius $R$ must reproduce the maximum-$I$ ring; it does.

Verify: (a) and (b) → $0$ (mass on the axis, Cell F logic), (c) → $M{R}^2$ (the ring, the max). All three limits are physically exactly what we'd expect, with no formula blowing up or contradicting itself.

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Ex 8 — Cell G: real-world word problem (energy)

Forecast: Roughly how many joules — a handful, or hundreds?

1. Get $I$. Solid cylinder about its long axis: $I=\frac{1}{2}M{R}^2=\frac{1}{2}(8)(0.20{)}^2=\frac{1}{2}(8)(0.04)=0.16\mathrm{kg}{\mathrm{m}}^2$. Why this step? Energy needs $I$ first; the axis is the long axis, so it's the disk/cylinder formula (length-independent).
2. Apply rotational KE. $KE=\frac{1}{2}I{\omega }^2$ — the rotational twin of $\frac{1}{2}m{v}^2$. Why this step? This formula is the very reason $r_{\perp }^2$ lives inside $I$ (see parent note's "Why $r^2$").
3. Compute. $KE=\frac{1}{2}(0.16)(30{)}^2=\frac{1}{2}(0.16)(900)=72\mathrm{J}$ Why this step? Substituting the $I$ from step 1 and the given $\omega$ into $\frac{1}{2}I{\omega }^2$ yields the stored energy; $\omega$ is squared because energy grows with the square of spin rate.

Verify: Units $\mathrm{kg}{\mathrm{m}}^2\cdot (rad/s{)}^2=\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2=\mathrm{J}$ ✓. $72\mathrm{J}$ is a modest but real store — reasonable for a small flywheel.

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Ex 9 — Cell H: exam twist ("which reaches the bottom first?")

Forecast: Trust your gut before deriving: solid or hollow?

Identify the axis first. As each sphere rolls, it spins about the axis through its own centre of mass, parallel to the ground and perpendicular to the direction of travel. That is the diameter axis, so the relevant $I$ values are the diameter formulas from the parent note: solid $\frac{2}{5}M{R}^2$, hollow $\frac{2}{3}M{R}^2$. (Every $r_{\perp }$ in $\int r_{\perp }^{2}dm$ is measured to that central spin axis.)

1. Write each $I$ as $I=\beta M{R}^2$. Solid sphere: $\beta =\frac{2}{5}=0.4$. Hollow sphere: $\beta =\frac{2}{3}\approx 0.667$. Why this step? Every rolling body's behaviour is controlled by the single pure number $\beta =I/(M{R}^2)$ — the Radius of Gyration squared over $R^2$. Pulling it out lets us treat both spheres with one formula.
2. Split the energy two ways (rolling constraint). Rolling without slipping means the contact point is momentarily still, which forces $v=\omega R$, i.e. $\omega =v/R$. The total kinetic energy is translation plus rotation: $KE=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}M{v}^2+\frac{1}{2}(\beta M{R}^2){\left(\frac{v}{R}\right)}^2=\frac{1}{2}M{v}^2(1+\beta )$ Why this step? This is where the moment of inertia actually bites: the $\frac{1}{2}I{\omega }^2$ term (the parent note's Rotational Kinetic Energy) is energy locked into spinning that cannot go into forward speed. A bigger $\beta$ locks away more.
3. Convert dropped height to speed by energy conservation. After descending a height $h$, gravity has done work $Mgh$, all of it becoming that kinetic energy: $Mgh=\frac{1}{2}M{v}^2(1+\beta )\Rightarrow v^2=\frac{2gh}{1+\beta }$ Why this step? No friction does net work (contact point doesn't slide), so mechanical energy is conserved. This isolates $v$ in terms of $\beta$ — and $M$, $R$ have cancelled, a key exam insight.
4. Get the linear acceleration down the ramp. For motion starting from rest with constant acceleration $a$ over distance $s$ along the incline, $v^2=2as$, and the height dropped is $h=s\sin \theta$. Substituting $v^2$: $2as=\frac{2g(s\sin \theta )}{1+\beta }\Rightarrow a=\frac{g\sin \theta }{1+\beta }$ Why this step? Turning "final speed" into "acceleration" lets us compare who speeds up faster the whole way down, which is what decides the race — not just the finish speed.
5. Plug in both $\beta$ and compare. $a_{\text{solid}}=\frac{g\sin \theta }{1+0.4}=\frac{g\sin \theta }{1.4},a_{\text{hollow}}=\frac{g\sin \theta }{1+0.667}=\frac{g\sin \theta }{1.667}$ The solid sphere has the smaller denominator, hence the larger acceleration: $\frac{a_{\text{solid}}}{a_{\text{hollow}}}=\frac{1+\frac{2}{3}}{1+\frac{2}{5}}=\frac{5/3}{7/5}=\frac{25}{21}\approx 1.19$ Why this step? Same $g$, $\theta$ for both, so the only thing that differs is $\beta$: smaller $\beta$ (mass hoarded near the centre) → bigger $a$.
6. Turn acceleration into descent time. With $s=\frac{1}{2}a{t}^2$ from rest, the time to cover the same ramp length $s$ is $t=\sqrt{2s/a}$, so $t\propto 1/\sqrt{a}\propto \sqrt{1+\beta }$: $\frac{t_{\text{solid}}}{t_{\text{hollow}}}=\sqrt{\frac{1+\frac{2}{5}}{1+\frac{2}{3}}}=\sqrt{\frac{7/5}{5/3}}=\sqrt{\frac{21}{25}}\approx 0.917$ Why this step? This is the actual quantity the question asks for — descent time. The solid sphere's time is about $92\%$ of the hollow one's, i.e. it arrives first.

Answer & interpretation: The solid sphere wins — it accelerates $\approx 19\%$ harder and finishes in $\approx 92\%$ of the hollow sphere's time. Physically, the hollow sphere pushes all its mass out to radius $R$, giving it a larger $\beta$, so a larger share of gravity's energy gets trapped as spin ($\frac{1}{2}I{\omega }^2$) instead of driving it forward. The result is independent of $M$, $R$, and $\theta$ — the classic exam takeaway.

Verify: As a sanity check, a frictionless sliding block has $\beta =0$ and $a=g\sin \theta$, faster than either sphere — correct, since it wastes no energy spinning. And $t_{\text{solid}}/t_{\text{hollow}}<1$ confirms the solid sphere is first. ✓

Recall Which quantity decided the race?

Not mass, not radius, but $\beta =I/(M{R}^2)$ ::: the pure shape fraction. Smaller $\beta$ = mass near the axis = larger acceleration = shorter descent time.

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Connections

• Parallel Axis Theorem — the engine of Ex 2 and Ex 6.
• Perpendicular Axis Theorem — the engine of Ex 3.
• Rotational Kinetic Energy — used in Ex 8 and Ex 9.
• Rolling Motion — the physics behind Ex 9's race.
• Radius of Gyration — the $\beta$ fraction repackaged.
• Torque and Angular Acceleration — how these $I$ values turn forces into spin.