Question

From a disc of radius R, a concentric circular portion of radius r is cut out so as leave an annular disc of mass M. The moment of inertia of this annular disc about the axis perpendicular to its plane and passing through its center of gravity is

• A. $1/2M(R^2+r^2)$
• B. $1/2M(R^2-r^2)$
• C. $1/2M(R^4+r^4)$
• D. $1/2M(R^4-r^4)$

Answer 1

The correct option is A $1/2M(R^2+r^2)$

$\begin{array}{c}\text{Given, mass of left annular disc}=M.\\ \text{so,}\text{density}=\frac{M}{\pi (R^2-r^2)}\end{array}$

$\begin{array}{c}\text{Now, moment of inertia}\left(I_1\right)\text{of disc of radius}R\\ \text{about central axis}=\frac{m_1{R}^2}{2}(m_1\to \text{its mass})\\ \text{and, moment of inertia}\left(I_2\right)\text{of removed disc}\\ \text{about the central axis}=\frac{m_2{r}^2}{2}(m_2\to \text{its mass)}\end{array}$

$\begin{array}{cc}\therefore \text{moment of inertia of remaluing portion}& =I_1-I_2\\ \text{(I)}& =\frac{1}{2}[M_1{R}^2-M_2{r}^2]\end{array}$

$\begin{array}{c}\text{Now,}{M}_1=\pi R^2\times \frac{M}{\pi (R^2-r^2)}=M\left(\frac{R^2}{R^2-r^2}\right)\\ \text{and,}{m}_2=\pi r^2\times \frac{M}{\pi (R^2-r^2)}=M\left(\frac{r^2}{R^2-r^2}\right)\end{array}$

$\begin{array}{cc}\text{Thus,}I=\frac{1}{2}[\frac{M{R}^4}{(R^2-r^2)}-\frac{M{r}^4}{\left(R^2{r}^2\right)}]& =\frac{M}{2(R^2-r^2)}(R^4-r^4)\\ & =\frac{1}{2}M(R^2+r^2)\end{array}$