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

WHY / WHAT / HOW

• WHAT is $r$? Always the perpendicular distance to the rotation axis, NOT the distance to the centre.
• WHY $r^2$? Kinetic energy of a particle is $\tfrac12 m v^2 = \tfrac12 m (\omega r)^2$. Summing, $KE = \tfrac12\left(\sum m r^2\right)\omega^2 = \tfrac12 I\omega^2$. The $r^2$ falls out naturally — it is defined so that rotational KE looks like translational KE.
• HOW to compute: pick a mass element $dm$, write $dm$ in terms of linear/area/volume density, express $r$, integrate.

Derivation 1 — Thin Rod about its Centre

Rod length $L$, mass $M$, linear density $\lambda = M/L$. Axis ⟂ to rod through centre.

Take element $dx$ at distance $x$ from centre, so $r=x$, ranging $-L/2$ to $L/2$. $I = \int_{-L/2}^{L/2} x^2 \,\lambda\,dx = \lambda\left[\frac{x^3}{3}\right]_{-L/2}^{L/2} = \lambda\cdot\frac{2}{3}\frac{L^3}{8} = \frac{\lambda L^3}{12}$ Why this step? We used $\int x^2 dx = x^3/3$ and symmetry doubled the half-integral. Now $\lambda L = M$: $\boxed{I_{\text{centre}} = \frac{1}{12}ML^2}$

Derivation 2 — Rod about its End

Now $x$ runs $0 \to L$: $I = \int_0^L x^2 \lambda\,dx = \lambda\frac{L^3}{3} = \frac{1}{3}ML^2$ $\boxed{I_{\text{end}} = \frac{1}{3}ML^2}$ Why bigger? Mass is on average farther from an end axis than from a centre axis. Check with Parallel Axis Theorem: $I_{end} = I_{cm} + Md^2 = \tfrac{1}{12}ML^2 + M(L/2)^2 = \tfrac{1}{12}ML^2 + \tfrac14 ML^2 = \tfrac13 ML^2$. ✓

Derivation 3 — Ring about central axis (⟂ to plane)

Every bit of mass is at the same distance $R$ from the axis. So $r=R$ is constant: $I = \int R^2\,dm = R^2\int dm = MR^2 \quad\Rightarrow\quad \boxed{I_{\text{ring}} = MR^2}$ Why this step? $R$ pulls out of the integral — this is the maximum for a given $M,R$ because all mass sits at the farthest possible point.

Derivation 4 — Disk about central axis

Disk radius $R$, surface density $\sigma=M/(\pi R^2)$. Split into thin rings of radius $r$, thickness $dr$. Each ring has mass $dm = \sigma(2\pi r\,dr)$ and all of it at distance $r$: $I = \int_0^R r^2\,(\sigma 2\pi r)\,dr = 2\pi\sigma\int_0^R r^3\,dr = 2\pi\sigma\frac{R^4}{4} = \frac{\pi\sigma R^4}{2}$ Sub $\sigma=M/(\pi R^2)$: $\boxed{I_{\text{disk}} = \frac{1}{2}MR^2}$ Why this step? We reused the ring result as a building block — Feynman-style decomposition. Disk < ring because the disk's mass is spread inward.

Solid cylinder about its long axis = stack of disks ⇒ same answer $\boxed{\tfrac12 MR^2}$ (length irrelevant).

Derivation 5 — Hollow Sphere (spherical shell)

Mass $M$, radius $R$, surface density $\sigma=M/(4\pi R^2)$. Slice into rings using polar angle $\theta$ from the axis. A ring at angle $\theta$ has:

• radius (distance from axis) $r = R\sin\theta$
• width $R\,d\theta$, circumference $2\pi R\sin\theta$
• mass $dm = \sigma\,(2\pi R\sin\theta)(R\,d\theta)$

$I = \int_0^\pi (R\sin\theta)^2 \,\sigma\,2\pi R^2\sin\theta\,d\theta = 2\pi\sigma R^4\int_0^\pi \sin^3\theta\,d\theta$ Now $\int_0^\pi \sin^3\theta\,d\theta = \frac43$, so $I = 2\pi\sigma R^4\cdot\tfrac43 = \tfrac{8}{3}\pi\sigma R^4$. Sub $\sigma$: $I = \frac{8}{3}\pi R^4\cdot\frac{M}{4\pi R^2} = \frac{2}{3}MR^2 \quad\Rightarrow\quad \boxed{I_{\text{hollow}} = \frac{2}{3}MR^2}$

Derivation 6 — Solid Sphere

Mass $M$, radius $R$, $\rho = M/(\tfrac43\pi R^3)$. Build from thin spherical shells radius $r$, thickness $dr$, shell mass $dm=\rho\,4\pi r^2\,dr$. Each shell contributes (hollow result) $\tfrac23 r^2\,dm$: $I = \int_0^R \frac23 r^2 (\rho 4\pi r^2)\,dr = \frac{8\pi\rho}{3}\int_0^R r^4\,dr = \frac{8\pi\rho}{3}\frac{R^5}{5} = \frac{8\pi\rho R^5}{15}$ Sub $\rho$: $I = \frac{8\pi R^5}{15}\cdot\frac{3M}{4\pi R^3} = \frac{2}{5}MR^2 \quad\Rightarrow\quad \boxed{I_{\text{solid}} = \frac{2}{5}MR^2}$ Why solid < hollow? In a solid sphere, lots of mass sits near the centre ($r<R$), lowering the average $r^2$.

Summary Table (the 20% to memorise)

Common Mistakes (Steel-manned)

Active Recall

Connections

• Parallel Axis Theorem — shifts $I$ off the centre of mass.
• Perpendicular Axis Theorem — relates planar-body axes (disk: $I_z=2I_{diameter}$).
• Rotational Kinetic Energy — origin of the $r^2$.
• Torque and Angular Acceleration — $\tau=I\alpha$ uses these values.
• Rolling Motion — $I$ ratios set rolling acceleration order: sphere > cylinder > ring.
• Radius of Gyration — $I=Mk^2$.

Concept Map

Hinglish (regional understanding)