Question

The surface density (mass/area) of a circular disc of radius 𝑎 depends on the distance from the centre as $\rho \left(r\right)=A+Br$. Find its moment of inertia about the line perpendicular to the plane of the disc through its centre.

• A. $2\pi a^3[\frac{A}{4}+\frac{Ba}{5}]$
• B. $2\pi a^4[\frac{A}{8}+\frac{Ba}{7}]$
• C. $2\pi a^4[\frac{A}{4}+\frac{Ba}{5}]$
• D. $2\pi a^4[\frac{A^2}{4}+\frac{Ba}{5}]$

Answer 1

The correct option is C $2\pi a^4[\frac{A}{4}+\frac{Ba}{5}]$

From given,

let the radius is $r$
so, $dI=\int \rho \times 2\times \pi rdr\times r^2$
${\int }_0^{a}dI={\int }_0^a(A+Br)2\times \pi rdr\times r^2=2\pi {\int }_0^a[A{r}^3+B{r}^4]dr=2\pi a^4[\frac{A}{4}+\frac{Ba}{5}]$