Question

A disc of mass m and radius R has a concentric hole of a radius r. 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. $\frac{1}{2}m(R-r^2)$
• C. $\frac{1}{2}m(R^2-r^2)$
• D. $\frac{1}{2}m(R+r^2)$

Answer 1

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

$\therefore$ Moment of inertia of the disc with hole
$I_0=\frac{1}{2}{m}_1{R}^2-\frac{1}{2}{m}_2{r}^2$
$\Rightarrow m_1-m_2=m$ ....(i)Also, $m_1\propto R^2,m_2\propto r^2$
$\therefore \frac{m_1}{m_2}=\frac{R^2}{r^2}$ .....(ii)
$\Rightarrow m_1(1-\frac{r^2}{R^2})=m$
$\therefore m_1=\frac{m{R}^2}{R^2-r^2}$
$m_2=\frac{m{r}^2}{R^2-r^2}$
$I_0=\frac{m}{2(R^2-r^2)}(R^4-r^4)$
$=\frac{m}{2}(R^2+r^2)$