Question

Mass per unit area of a circular disc of radius $a$ depends on the distance $r$ from its centre as $\sigma \left(r\right)=A+Br$. The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is

• A. $2\pi a^4(\frac{A}{4}+\frac{aB}{5})$
• B. $\pi a^4(\frac{A}{4}+\frac{aB}{5})$
• C. $2\pi a^4(\frac{Aa}{4}+\frac{B}{5})$
• D. $2\pi a^4(\frac{A}{5}+\frac{Ba}{4})$

Answer 1

The correct option is A $2\pi a^4(\frac{A}{4}+\frac{aB}{5})$

Formula used: $I=M{R}^2$
Let us consider an elemental ring of mass $dm$, radius $r$ and thickness $dr$.

Moment of inertia of the elemental ring
$dI=dm{r}^2$
$dI=\left(\sigma 2\pi rdr\right)r^2$
$dI=\left(2\pi \right(A+Br)r^{3}dr$

Hence,$I=\int dI=2\pi {\int }_0^a(A{r}^3+B{r}^4)dr$

$\Rightarrow I=2\pi a^4(\frac{A}{4}+\frac{Ba}{5})$
Final answer is (a)