For simplicity let us use a mathematical trick. We consider...

For simplicity let us use a mathematical trick. We consider the portion of the given disc as the superposition of two. complete discs (without holes), one of positive density and radius $R$ and other of negative density but of same magnitude and radius $R / 2 .$ As (area) $\alpha$ (mass), the respective masses of the considered discs are $(4 m / 3)$ and $(-m / 3)$ respectively, and these masses can be imagined to be situated at their respective centers (C.M). Let us take point $O$ as origin and point $x$ - axis towards right. Obviously the C.M. of the shaded position of given shape lies on the $x$ - axis. Hence the C.M. (C) of the shaded portion is given by $x_{c}=\frac{(-m / 3)(-R / 2)+(4 m / 3) 0}{(-m / 3)+4 m / 3}=\frac{R}{6}$ Thus C.M. of the shape is at a distance $R / 6$ from point $O$ toward $x$ - axis Using parallel axis theorem and bearing in mind that the moment of inertia of a complete homogeneous disc of radius $m_{0}$ and radius $r_{0}$ equals $\frac{1}{2} m_{0} r_{0}^{2}$. The moment of inetia of the small disc of mass $(-m / 3)$ and radius $R / 2$ about the axis passing through point $C$ and perpendicular to the plane of the disc Similarly $$ \begin{gathered} I_{2 C}=\frac{1}{2}\left(-\frac{m}{3}\right)\left(\frac{R}{2}\right)^{2}+\left(-\frac{m}{3}\right)\left(\frac{R}{2}+\frac{R}{6}\right)^{2} \\ =-\frac{m R^{2}}{24}-\frac{4}{27} m R^{2} \\ I_{1 C}=\frac{1}{2}\left(\frac{4 m}{3}\right) R^{2}+\left(\frac{4 m}{3}\right)\left(\frac{R}{6}\right)^{2} \end{gathered} $$ $$ =\frac{2}{3} m R^{2}+\frac{m R^{2}}{27} $$ Thus the sought moment of inertia, $$ I_{C}=I_{1 C}+I_{2 C}=\frac{15}{24} m R^{2}-\frac{3}{27} m R^{2}=\frac{37}{72} m R^{2} $$

For simplicity let us use a mathematical trick. We consider the portion of the given disc as the superposition of two. complete discs (without holes), one of positive density and radius $R$ and other of negative density but of same magnitude and radius $R / 2 .$ As (area) $\alpha$ (mass), the respective masses of the considered discs are $(4 m / 3)$ and $(-m / 3)$ respectively, and these masses can be imagined to be situated at their respective centers (C.M). Let us take point $O$ as origin and point $x$ - axis towards right. Obviously the C.M. of the shaded position of given shape lies on the $x$ - axis. Hence the C.M. (C) of the shaded portion is given by $x_{c}=\frac{(-m / 3)(-R / 2)+(4 m / 3) 0}{(-m / 3)+4 m / 3}=\frac{R}{6}$
Thus C.M. of the shape is at a distance $R / 6$ from point $O$ toward $x$ - axis Using parallel axis theorem and bearing in mind that the moment of inertia of a complete homogeneous disc of radius $m_{0}$ and radius $r_{0}$ equals $\frac{1}{2} m_{0} r_{0}^{2}$. The moment of inetia of the small disc of mass $(-m / 3)$ and radius $R / 2$ about the axis passing through point $C$ and perpendicular to the plane of the disc
Similarly
$$
\begin{gathered}
I_{2 C}=\frac{1}{2}\left(-\frac{m}{3}\right)\left(\frac{R}{2}\right)^{2}+\left(-\frac{m}{3}\right)\left(\frac{R}{2}+\frac{R}{6}\right)^{2} \\
=-\frac{m R^{2}}{24}-\frac{4}{27} m R^{2} \\
I_{1 C}=\frac{1}{2}\left(\frac{4 m}{3}\right) R^{2}+\left(\frac{4 m}{3}\right)\left(\frac{R}{6}\right)^{2}
\end{gathered}
$$
$$
=\frac{2}{3} m R^{2}+\frac{m R^{2}}{27}
$$
Thus the sought moment of inertia,
$$
I_{C}=I_{1 C}+I_{2 C}=\frac{15}{24} m R^{2}-\frac{3}{27} m R^{2}=\frac{37}{72} m R^{2}
$$

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