A thin uniform rod $1 \mathrm{~m}$ long has mass $1 \mathrm{~kg}$. Find its moment of inertia and radius of gyration for rotation about a transverse axis through a point midway between its centre and one end.
Q 96.2
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Data : $M=1 \mathrm{~kg}, \mathrm{~L}=1 \mathrm{~m}$
Let $\mathrm{I}_{\mathrm{CM}}$ and $\mathrm{I}$ be the moments of inertia of the rod about a transverse axis through its centre, and a parallel axis midway between its centre and one end.
Then, $I_{\mathrm{CM}}=\frac{M L^2}{12}$ and $h=\frac{L}{4}$
By the theorem of parallel axis,
$
\begin{aligned}
I & =I_{\mathrm{CM}}+M h^2 \\
& =\frac{M L^2}{12}+\frac{M L^2}{16}=\frac{4 M L^2+3 M L^2}{48} \\
& =\frac{7}{48} M L^2=\frac{7}{48}(1)(1)^2=0.1458 \mathrm{~kg} \cdot \mathrm{m}^2
\end{aligned}
$
If $k$ is the radius of gyration about the parallel axis,
$
\begin{gathered}
I=M k^2 \quad \therefore M k^2=\frac{7}{48} M L^2 \\
\therefore k=\sqrt{\frac{7}{48}} L=\sqrt{\frac{7}{48}} \times 1=0.3818 \mathrm{~m}
\end{gathered}
$
Let $\mathrm{I}_{\mathrm{CM}}$ and $\mathrm{I}$ be the moments of inertia of the rod about a transverse axis through its centre, and a parallel axis midway between its centre and one end.
Then, $I_{\mathrm{CM}}=\frac{M L^2}{12}$ and $h=\frac{L}{4}$
By the theorem of parallel axis,
$
\begin{aligned}
I & =I_{\mathrm{CM}}+M h^2 \\
& =\frac{M L^2}{12}+\frac{M L^2}{16}=\frac{4 M L^2+3 M L^2}{48} \\
& =\frac{7}{48} M L^2=\frac{7}{48}(1)(1)^2=0.1458 \mathrm{~kg} \cdot \mathrm{m}^2
\end{aligned}
$
If $k$ is the radius of gyration about the parallel axis,
$
\begin{gathered}
I=M k^2 \quad \therefore M k^2=\frac{7}{48} M L^2 \\
\therefore k=\sqrt{\frac{7}{48}} L=\sqrt{\frac{7}{48}} \times 1=0.3818 \mathrm{~m}
\end{gathered}
$
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