The radius of gyration of a disc about its transverse symmetry axis is $2 \mathrm{~cm}$. Determine its radius of gyration about a diameter.
Q 96.7
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Data : $\mathrm{k} \mathrm{CM}=2 \mathrm{~cm}$
Let $\mathrm{M}$ and $\mathrm{R}$ be the mass and radius of the disc. Let $\mathrm{I}_{\mathrm{CM}}$ and $\mathrm{K}_{\mathrm{CM}}$ be the $\mathrm{MI}$ and radius of gyration of the disc about its transverse symmetry axis. Let I and $\mathrm{k}$ be the $\mathrm{MI}$ and radius of gyration of the disc about its diameter. Then
$
\begin{aligned}
& I_{\mathrm{CM}}=\frac{M R^2}{2}=M k_{\mathrm{CM}}^2 \text { and } I=\frac{M R^2}{4}=M k^2 \\
& \therefore k_{\mathrm{CM}}^2=\frac{R^2}{2} \text { and } k^2=\frac{R^2}{4} \\
& \therefore \frac{k^2}{k_{\mathrm{CM}}^2}=\frac{R^2}{4} \times \frac{2}{R^2}=\frac{1}{2} \\
& \therefore k^2=\frac{k_{\mathrm{CM}}^2}{2} \quad \therefore k=\frac{k_{\mathrm{CM}}}{\sqrt{2}} \\
& \therefore k=\frac{2}{\sqrt{2}}=\sqrt{2}=1.414 \mathrm{~cm} \\
&
\end{aligned}
$
Let $\mathrm{M}$ and $\mathrm{R}$ be the mass and radius of the disc. Let $\mathrm{I}_{\mathrm{CM}}$ and $\mathrm{K}_{\mathrm{CM}}$ be the $\mathrm{MI}$ and radius of gyration of the disc about its transverse symmetry axis. Let I and $\mathrm{k}$ be the $\mathrm{MI}$ and radius of gyration of the disc about its diameter. Then
$
\begin{aligned}
& I_{\mathrm{CM}}=\frac{M R^2}{2}=M k_{\mathrm{CM}}^2 \text { and } I=\frac{M R^2}{4}=M k^2 \\
& \therefore k_{\mathrm{CM}}^2=\frac{R^2}{2} \text { and } k^2=\frac{R^2}{4} \\
& \therefore \frac{k^2}{k_{\mathrm{CM}}^2}=\frac{R^2}{4} \times \frac{2}{R^2}=\frac{1}{2} \\
& \therefore k^2=\frac{k_{\mathrm{CM}}^2}{2} \quad \therefore k=\frac{k_{\mathrm{CM}}}{\sqrt{2}} \\
& \therefore k=\frac{2}{\sqrt{2}}=\sqrt{2}=1.414 \mathrm{~cm} \\
&
\end{aligned}
$
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