Question

The moment of inertia of a angular disc of external and internal radii $R$ and $R$ and mass $M$ about axis through the center and perpendicular to its plane is-

• A. $\frac{M}{2}(R^2+r^2)$
• B. $M\left(\frac{R^2+r^2}{2}\right)+\frac{M{\ell }^2}{3}$
• C. $M\left(\frac{{\ell }^2+3R^2-3r^2}{12}\right)$
• D. $M\left(\frac{R^2+r^2}{4}\right)+\frac{M{\ell }^2}{12}$

Answer 1

The correct option is A

$\frac{M}{2}(R^2+r^2)$

Mass of this strip is $dM=\frac{M}{\pi (R^2-r^2)}2\pi xdx$

Now moment of inertia of this strip is $dL=dM{x}^2$

$=[\frac{M}{\pi (R^2-r^2)}-2\pi x]x^{2}dx$

$\Rightarrow I={\int }_r^R\frac{2\pi M}{\pi (R^2-r^2)}{x}^{3}dx$

$=\frac{2M}{(R^2-r^2)}{\left[\frac{r^4}{4}\right]}_r^R$

$=\frac{2M}{R^2-r^2}\frac{(R^4-r^4)}{4}$

$=\frac{2M}{4(R^2-r^2)}(R^2+r^2)(R^2-r^2)$

$=\frac{M}{2}(R^2+r^2)$