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 a constant. If the mass of the lamina is $M$, in terms of $M$ and $a$ the moment of inertia of the lamina about an axis through $O$ and perpendicular to the lamina is $I=\frac{x}{3}M{a}^2$. Find x.

Answer 1

$M={\int }_0^a\left(dM\right)={\int }_0^a\left(2\pi x\right)dX\left(k{x}^2\right)$
$=\frac{\pi k{a}^4}{2}$
$\therefore \pi k=\frac{2M}{a^4}$
Now, $I={\int }_0^{a}dI={\int }_0^a\left(dM\right)X^2$
$={\int }_0^a\left(2\pi k{x}^{3}dx\right)x^2=\frac{\left(\pi k\right)a^6}{3}$ Substituting the value of $\pi k,$ we get
$I=\frac{2}{3}M{a}^2$