Moment of Inertia of a Uniform Disc

Consider a solid disc of mass $M$ and radius $R$, and find its moment of inertia about an axis through its center, perpendicular to the plane of the disc. Unlike a ring, a disc's mass is not all at one fixed distance from the axis — but it can be thought of as built up from a great many thin concentric rings, of increasing radius, and the already-known ring result (§5.4.2) can be reused for each of these, then integrated over all of them.

Consider one such elemental ring, of mass $dm$, radius $r$ (where $0\le r\le R$), and small thickness $dr$. Its own moment of inertia is $dI=(dm)r^2$, exactly as for a full ring, just applied to this elemental one. Since the mass is uniformly distributed, the surface mass density (mass per unit area) is $\sigma=\dfrac{M}{\pi R^2}$, and the area of this thin elemental ring is $2\pi r\,dr$ (its circumference $2\pi r$ times its thickness $dr$), so its mass is

Hence $dI=\dfrac{2M}{R^2}r^3\,dr$. Integrating over every elemental ring making up the whole disc, from $r=0$ at the center out to $r=R$ at the rim:

$\boxed{I=\frac{1}{2}MR^2}$ …