Question

A disc of radius $R$ has a concentric hole of radius $r(<R)$. Its mass is $m$. Its moment of inertia about an axis through its centre and perpendicular to its plane is

• A. $\frac{1}{2}m(R^2-r^2)$
• B. $\frac{1}{2}m(R^2+r^2)$
• C. $\frac{1}{2}m{(R-r)}^2$
• D. $noneofthese$

Answer 1

The correct option is A $\frac{1}{2}m(R^2-r^2)$

let $I_R$ be the moment of inertia of disc of radius R

let $I_r$ be the moment of inertia of disc of radius r

let $I_{ring}$ be the moment of inertia of ring of width (R-r)

$I_R=I_r+I_{ring}$

$I_{ring}=\frac{m{R}^2}{2}-\frac{m{r}^2}{2}$