Question

The surface density $\left(\frac{mass}{area}\right)$ 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.
• B.
• C.
• D.

Answer 1

The correct option is B

The surface density of a circular disc of radius $a$ depends upon the distance from the centre as

$P\left(r\right)$ = $A+Br$

Therefore the mass of the ring of radius $r$ will be

$\theta$ = $(A+Br)\times 2\pi rdr\times r^2$

Therefore moment of Inertia about the centre will be

= $\underset{0}{\overset{a}{\int }}(A+Br)2\pi r\times dr$ = $\underset{0}{\overset{a}{\int }}2\pi A{r}^{3}dr+\underset{0}{\overset{a}{\int }}2\pi B{r}^{4}dr$

= $2\pi A\left(\frac{r^4}{4}\right)+2\pi B\left(\frac{r^5}{5}\right){|}_0^a$ = $2\pi a^4\left[\right(\frac{A}{4})+(\frac{Ba}{5}\left)\right]$.