Moment of Inertia of a Uniform Ring

Consider a uniform ring of mass $M$ and radius $R$, and find its moment of inertia about an axis through its center, perpendicular to the plane of the ring. Take an infinitesimally small mass element $dm$, of length $dx$ (a tiny arc of the ring's circumference), which — because every point of the ring lies at exactly the same distance $R$ from the central axis — sits at perpendicular distance $R$ from the axis.

The contribution of this element to the moment of inertia is simply $dI=(dm)R^2$. The full length of the ring is its circumference, $2\pi R$. Since the mass is uniformly distributed, the linear mass density is $\lambda=\dfrac{M}{2\pi R}$, so $dm=\lambda\,dx=\dfrac{M}{2\pi R}\,dx$. The moment of inertia of the entire ring is obtained by integrating around the full circumference, from $x=0$ to $x=2\pi R$:

$\boxed{I=MR^2}$ …