Question

Calculate the moment of inertia of a thin ring of mass $m$ and radius $R$ about an axis passing through its centre and perpendicular to the plane of the ring.

Answer 1

Remember that in case of continuous mass distribution, we use the
formula $I=\int \left(dm\right)r^2$ to find out the moment of inertia of the body. $AA$ is the axis about which
rotation of the ring is being considered
Mass of the ring $=M$, circumference of the ring $=2\pi R$.

Consider a small element of the ring at an angle $\theta$ from a particular reference radius. The element subtends and a particular reference radius. The element subtends an angle $d\theta$ at the center.

Length of the element $=Rd\theta$
Mass of the element $=\left(\lambda Rd\theta \right)$
Moment of inertia of the element $=\left(\lambda Rd\theta \right)R^2$
Moment of inertia of the ring $\int^{2 \pi}_{0}(\lambda\ Rd \theta)\ R^{2}=MR^{2}$