Question

A concentric hole of radius R/2 is cut from a thin circular plate of mass M and radius R. The moment of inertia of the remaining plate about its axis will be

• A. $\frac{13}{24}M{R}^2$
• B. $\frac{11}{24}M{R}^2$
• C. $\frac{13}{32}M{R}^2$
• D. $\frac{15}{32}M{R}^2$

Answer 1

The correct option is C $\frac{13}{32}M{R}^2$

$\begin{array}{c}MomentofinertiaofthecutdiscportionaboutO=I_1\\ MomentofinertiaofremainingportionaboutO=I_2\\ M.IofthefulldiscaboutO=I\\ I=I_1+I_2\\ Massofoutdisc=\frac{M}{R^2}\cdot {\left(\frac{R}{2}\right)}^2=\frac{M}{4}\\ I_{1}about{O}^{\prime }=\frac{1}{2}\left(\frac{M}{4}\right){\left(\frac{R}{2}\right)}^2\\ =\frac{M{R}^2}{32}\\ I_{1}aboutO=\frac{M{R}^2}{32}+\frac{M}{4}\left(\frac{R}{4}\right)\\ =\frac{M{R}^2}{32}+\frac{M{R}^2}{16}\\ =\frac{3M{R}^2}{32}\\ I_2=I-I_1\\ =\frac{M{R}^2}{2}-\frac{3M{R}^2}{32}\\ =\frac{13M{R}^2}{32}\end{array}$

Hence,

option $\left(C\right)$ is correct answer.