Question

The surface mass 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 and passing through its centre

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

Answer 1

The correct option is B

$2\pi a^4[\left(\frac{A}{4}\right)+\left(\frac{Ba}{5}\right)]$

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 moment of inertia of small mass element of the ring of radius $r$ will be

equal to $(A+Br)\times 2\pi rdr\times r^2$

Therefore moment of Inertia about the centre will be

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

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