A circular disc $D_{1}$ of mass $M$ and radius $R$ has two identical $\operatorname{discs} D_{2}$ and…
A circular disc $D_{1}$ of mass $M$ and radius $R$ has two identical $\operatorname{discs} D_{2}$ and $D_{3}$ of the same mass $M$ and radius R attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis OO', passing through the centre of $D_{1}$, as shown in the figure, will be
$\mathrm{MR}^{2}$
$3 \mathrm{MR}^{2}$
$\frac{4}{5} \mathrm{MR}^{2}$
$\frac{2}{3} \mathrm{MR}^{2}$
Solution
Moment of inertia of disc $D_{1}$ about $O O^{\prime}=I_{1}=\frac{M R^{2}}{2}$
M.O.I of $\mathrm{D}_{2}$ about $\mathrm{OO}^{\prime}$ $=I_{2}=\frac{1}{2}\left(\frac{M R^{2}}{2}\right)+M R^{2}=\frac{M R^{2}}{4}+M R^{2}$
M.O.I of $\mathrm{D}_{3}$ about $\mathrm{OO}^{\prime}$ $=I_{3}=\frac{1}{2}\left(\frac{M R^{2}}{2}\right)+M R^{2}=\frac{M R^{2}}{4}+M R^{2}$
so, resultant M.O.I about $\mathrm{OO}^{\prime}$ is $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}$
$
\begin{aligned}
\Rightarrow & I=\frac{M R^{2}}{2}+2\left(\frac{M R^{2}}{4}+M R^{2}\right) \\
&=\frac{M R^{2}}{2}+\frac{M R^{2}}{2}+2 M R^{2}=3 M R^{2}
\end{aligned}
$
