State an expression for the radius of gyration of
(i) a thin ring
(ii) a thin disc, about respective transverse symmetry axis.
OR
Show that for rotation about respective transverse symmetry axis, the radius of gyration of a thin disc is less than that of a thin ring.
(i) a thin ring
(ii) a thin disc, about respective transverse symmetry axis.
OR
Show that for rotation about respective transverse symmetry axis, the radius of gyration of a thin disc is less than that of a thin ring.
Q 70
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(i) The $\mathrm{Ml}$ of the ring about the transverse symmetry axis is
$\mathrm{I}_{\mathrm{CM}}=\mathrm{MR}^2 \ldots(1)$
Radius of gyration : The radius of gyration of the ring about the transverse symmetry axis is $\mathrm{K}=\sqrt{I_{\mathrm{CM}} / M}=\sqrt{R^2}=\mathrm{R}$
(ii) The Ml of the disc about the transverse symmetry axis is
$\mathrm{I}_{\mathrm{CM}}=\frac{1}{2} \mathrm{MR}<^2$..
Radius of gyration : The radius of gyration of the disc for the given rotation axis is
$
k=\sqrt{\frac{I}{M}}=\sqrt{\frac{R^2}{2}}=\frac{R}{\sqrt{2}}
$
Therefore, $k_{\text {disc }}<k_{\text {ring }}$.
$\mathrm{I}_{\mathrm{CM}}=\mathrm{MR}^2 \ldots(1)$
Radius of gyration : The radius of gyration of the ring about the transverse symmetry axis is $\mathrm{K}=\sqrt{I_{\mathrm{CM}} / M}=\sqrt{R^2}=\mathrm{R}$
(ii) The Ml of the disc about the transverse symmetry axis is
$\mathrm{I}_{\mathrm{CM}}=\frac{1}{2} \mathrm{MR}<^2$..
Radius of gyration : The radius of gyration of the disc for the given rotation axis is
$
k=\sqrt{\frac{I}{M}}=\sqrt{\frac{R^2}{2}}=\frac{R}{\sqrt{2}}
$
Therefore, $k_{\text {disc }}<k_{\text {ring }}$.
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