A uniform disc and a hollow right circular cone have the same formula for their M.I., when rotating about their central axes. Why is it so?
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The radius of gyration of a thin ring of radius $Rr$ about its transverse symmetry axis is
$
K _{ r }=\sqrt{I_{ CM } / M_{ r }}=\sqrt{R_{ r }^2}= R _{ r }
$
The radius of gyration of a thin disc of radius $R_d$ about its transverse symmetry axis is
$
k_{ d }=\sqrt{I_{ CM } / M_{ d }}=\sqrt{\frac{M_{ d } R_{ d }^2 / 2}{M_{ d }}}=\frac{1}{\sqrt{2}} R_{ d }
$
Given $k_{ r }=k_{ d }$,
$R_{ r }=\frac{1}{\sqrt{2}} R_{ d }$ or, equivalently, $R_{ d }=\sqrt{2} R_{ r }$
$
K _{ r }=\sqrt{I_{ CM } / M_{ r }}=\sqrt{R_{ r }^2}= R _{ r }
$
The radius of gyration of a thin disc of radius $R_d$ about its transverse symmetry axis is
$
k_{ d }=\sqrt{I_{ CM } / M_{ d }}=\sqrt{\frac{M_{ d } R_{ d }^2 / 2}{M_{ d }}}=\frac{1}{\sqrt{2}} R_{ d }
$
Given $k_{ r }=k_{ d }$,
$R_{ r }=\frac{1}{\sqrt{2}} R_{ d }$ or, equivalently, $R_{ d }=\sqrt{2} R_{ r }$
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