Question

A disc of mass 'm' and radius R has a concentric hole of radius 'r'. Its moment of inertia about an axis through center and normal to its plane is

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

Answer 1

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

$\begin{array}{cc}\text{Let punched disc has a mass 'm'}& \\ I=\int dm{r}^2& =\frac{2\pi m}{\pi (R^2-r^2)}\times {\int }_r^R{r}^{3}dr\\ & =\frac{2m}{4(R^2-r^2}(R^4-r^4)\end{array}$

$\begin{array}{c}I=I^{\prime }-I^{\prime \prime }=\frac{m(R^2+r^2)}{2}\\ \text{Hence option}D\text{is correct}\end{array}$