Question

Answer it fast ........ please ...​

Answer 1

From the circular disc let us consider an elemental hollow disc of inner radius $\sf{r}$ and outer radius $\sf{r+dr.}$

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The area of this hollow disc will be,

$\longrightarrow\sf{dA=\pi(r+dr)^2-\pi r^2}$

Since $\sf{dr}$ is very small, $\sf{(dr)^2}$ can be neglected.

$\longrightarrow\sf{dA=\pi(r^2+2r\,dr)-\pi r^2}$

$\longrightarrow\sf{dA=2\pi r\,dr}$

Let the circular disc be uniform so that the areal density is same for the whole disc and the hollow disc whose mass is $\sf{dM.}$

Given that $\sf{\sigma(r)}$ is areal density. So,

$\longrightarrow\sf{\sigma(r)=\dfrac{dM}{dA}}$

$\longrightarrow\sf{A+Br=\dfrac{dM}{2\pi r\,dr}}$

$\longrightarrow\sf{dM=2\pi(A+Br)r\,dr}$

The hollow disc is just a ring, so its moment of inertia is,

$\longrightarrow\sf{dI=dM\cdot r^2}$

Then moment of inertia of the circular disc is,

$\displaystyle\longrightarrow\sf{I=\int\limits_0^adM\cdot r^2}$

$\displaystyle\longrightarrow\sf{I=\int\limits_0^a2\pi(A+Br)r^3\,dr}$

$\displaystyle\longrightarrow\sf{I=2\pi\int\limits_0^a(A+Br)r^3\,dr}$

$\displaystyle\longrightarrow\sf{I=2\pi\int\limits_0^a(Ar^3+Br^4)\,dr}$

$\displaystyle\longrightarrow\sf{I=2\pi\left[\dfrac{Ar^4}{4}+\dfrac{Br^5}{5}\right]_0^a}$

$\displaystyle\longrightarrow\sf{I=2\pi\left(\dfrac{Aa^4}{4}+\dfrac{Ba^5}{5}\right)}$

$\displaystyle\longrightarrow\underline{\underline{\sf{I=2\pi a^4\left(\dfrac{A}{4}+\dfrac{aB}{5}\right)}}}$