The radius of gyration of a uniform solid sphere of radius $R$ is $\sqrt{\frac{2}{5}} R$ for rotation about its diameter. Show that its radius of gyration for rotation about a tangential axis of rotation is $\sqrt{\frac{7}{5}} R$
Q 89
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Let the mass of the uniform solid sphere of radius R be $\mathrm{M}$. Let $\mathrm{I}_{\mathrm{CM}}$ and $\mathrm{k}_{\mathrm{d}}$ be its $\mathrm{Ml}$ about any diameter and the corresponding radius of gyration, respectively. Then,
$
I_{\mathrm{CM}}=M k_{\mathrm{d}}^2=\frac{2}{5} M R^2 \quad\left(\because k_{\mathrm{d}}=\sqrt{\frac{2}{5}} R, \text { given }\right)
$
Let $\mathrm{I}$ and $\mathrm{k}_{\mathrm{t}}$ be its $\mathrm{Ml}$ about a parallel tangential axis and the corresponding radius of gyration, respectively. Here, $h=R=$ distance between the two axis.
$
\therefore \mathrm{I}=M k_{\mathrm{t}}^2
$
By the theorem of parallel axis,
$
\begin{aligned}
& I=I_{C M}+M h^2 \\
& \therefore M k_{\mathrm{t}}^2=\frac{2}{5} M R^2+M R^2 \quad \therefore k_{\mathrm{t}}^2=\frac{2}{5} R^2+R^2=\frac{7}{5} R^2 \\
& \therefore k_{\mathrm{t}}=\sqrt{\frac{7}{5}} R
\end{aligned}
$
$
I_{\mathrm{CM}}=M k_{\mathrm{d}}^2=\frac{2}{5} M R^2 \quad\left(\because k_{\mathrm{d}}=\sqrt{\frac{2}{5}} R, \text { given }\right)
$
Let $\mathrm{I}$ and $\mathrm{k}_{\mathrm{t}}$ be its $\mathrm{Ml}$ about a parallel tangential axis and the corresponding radius of gyration, respectively. Here, $h=R=$ distance between the two axis.
$
\therefore \mathrm{I}=M k_{\mathrm{t}}^2
$
By the theorem of parallel axis,
$
\begin{aligned}
& I=I_{C M}+M h^2 \\
& \therefore M k_{\mathrm{t}}^2=\frac{2}{5} M R^2+M R^2 \quad \therefore k_{\mathrm{t}}^2=\frac{2}{5} R^2+R^2=\frac{7}{5} R^2 \\
& \therefore k_{\mathrm{t}}=\sqrt{\frac{7}{5}} R
\end{aligned}
$
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