Question

A circular lamina of radius $a$ and centre $O$ has a mass per unit area of $k{x}^2$, where $x$ is the distance from $O$ and $k$ is constant. If the mass of the lamina is $M$, find in terms of $M$ and $a$, the moment of inertia of the lamina about an axis through $O$ and perpendicular to the lamina.

Answer 1

Given a circular lamina with centre at O and consider a small part of the lamina as a ring of radius $x$ and thickness $dx$ and mass $dm$.

Area of this element : $A=2\pi dx$

Since mass per unit area is given as $k{x}^2$, mass of the ring can be written as $dm=k{x}^2\times 2\pi xdx=2\pi x^{3}dx$

Moment of inertia about an axis perpendicular to the lamina ,

$\begin{array}{c}I={\int }_0^a{x}^{2}dm={\int }_0^{a}2\pi x^3{x}^{2}dx={\int }_0^{a}2\pi x^{5}dx=2\pi k{{\left[\frac{x^6}{6}\right]}^a}_0\\ I=\frac{\pi k{a}^6}{3}\\ M={\int }_0^{a}2\pi x^{3}dx=\frac{\pi k{a}^4}{2}\\ I=\frac{2M}{a^4}\times \frac{a^6}{3}=\frac{2}{3}M{a}^2\end{array}$