A compound object is formed of a thin rod and a disc attached at the end of the rod. The rod is $0.5 \mathrm{~m}$ long and has mass $2 \mathrm{~kg}$. The disc has mass of $1 \mathrm{~kg}$ and its radius is $20 \mathrm{~cm}$. Find the moment of inertia of the compound object about an axis passing through the free end of the rod and perpendicular to its length.
Q 96.5
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Data : $L=0.5 \mathrm{~m}, R=0.2 \mathrm{~m}, M_{\text {rod }}=2 \mathrm{~kg}, M_{\text {disk }}=1 \mathrm{~kg}$ About a transverse axis through $C M$,
$
I_{\mathrm{CM}, \text { rod }}=\frac{1}{2} M_{\text {rod }} L^2 \quad \text { and } \quad I_{\mathrm{CM}, \text { disk }}=\frac{1}{2} M_{\text {disk }} R^2
$
The MI of the compound object about the given axis,
$
\begin{aligned}
& I_{\text {total }}=I_{\text {rod }}+I_{\text {disk }} \\
& =\left[I_{\mathrm{CM}, \mathrm{rod}}+M_{\mathrm{rod}}\left(\frac{L}{2}\right)^2\right] \\
& +\left[I_{\mathrm{CM}, \text { disc }}+M_{\text {disk }}(L+R)^2\right] \\
& =\left(\frac{1}{12} M_{\text {rod }} L^2+\frac{1}{4} M_{\text {rod }} L^2\right) \\
& +\left[\frac{1}{2} M_{\text {disc }} R^2+M_{\text {disc }}(L+R)^2\right] \\
& =\frac{1}{3} M_{\text {rod }} L^2+M_{\text {disc }}\left[\frac{1}{2} R^2+M_{\text {disk }}(L+R)^2\right] \\
& =\frac{1}{3}(2)(0.5)^2+(1)\left[\frac{1}{2}(0.2)^2+(0.5+0.2)^2\right] \\
& =\frac{1}{6}+\frac{0.04}{2}+0.49=0.167+0.02+0.49 \\
& =0.677 \mathrm{~kg} \cdot \mathrm{m}^2 \\
&
\end{aligned}
$
$
I_{\mathrm{CM}, \text { rod }}=\frac{1}{2} M_{\text {rod }} L^2 \quad \text { and } \quad I_{\mathrm{CM}, \text { disk }}=\frac{1}{2} M_{\text {disk }} R^2
$
The MI of the compound object about the given axis,
$
\begin{aligned}
& I_{\text {total }}=I_{\text {rod }}+I_{\text {disk }} \\
& =\left[I_{\mathrm{CM}, \mathrm{rod}}+M_{\mathrm{rod}}\left(\frac{L}{2}\right)^2\right] \\
& +\left[I_{\mathrm{CM}, \text { disc }}+M_{\text {disk }}(L+R)^2\right] \\
& =\left(\frac{1}{12} M_{\text {rod }} L^2+\frac{1}{4} M_{\text {rod }} L^2\right) \\
& +\left[\frac{1}{2} M_{\text {disc }} R^2+M_{\text {disc }}(L+R)^2\right] \\
& =\frac{1}{3} M_{\text {rod }} L^2+M_{\text {disc }}\left[\frac{1}{2} R^2+M_{\text {disk }}(L+R)^2\right] \\
& =\frac{1}{3}(2)(0.5)^2+(1)\left[\frac{1}{2}(0.2)^2+(0.5+0.2)^2\right] \\
& =\frac{1}{6}+\frac{0.04}{2}+0.49=0.167+0.02+0.49 \\
& =0.677 \mathrm{~kg} \cdot \mathrm{m}^2 \\
&
\end{aligned}
$
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