The ratio of radii of gyration of a thin circ | vedclass

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(A) For a thin circular ring of mass $M$ and radius $R$,the moment of inertia about its diameter is $I_{diam} = \frac{1}{2}MR^2$. By the parallel axis

Vedclass · Accepted Answer

$10$

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(A) For a thin circular ring of mass $M$ and radius $R$,the moment of inertia about its diameter is $I_{diam} = \frac{1}{2}MR^2$. By the parallel axis theorem,the moment of inertia about a tangential axis in its plane is $I_{ring} = I_{diam} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$. The radius of gyration $k_{ring}$ is given by $Mk_{ring}^2 = \frac{3}{2}MR^2$,so $k_{ring} = R\sqrt{\frac{3}{2}}$.For a circular disc of mass $M$ and radius $R$,the moment of inertia about its diameter is $I_{diam} = \frac{1}{4}MR^2$. By the parallel axis theorem,the moment of inertia about a tangential axis in its plane is $I_{disc} = I_{diam} + MR^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$. The radius of gyration $k_{disc}$ is given by $Mk_{disc}^2 = \frac{5}{4}MR^2$,so $k_{disc} = R\sqrt{\frac{5}{4}}$.The ratio of radii of gyration is $\frac{k_{ring}}{k_{disc}} = \frac{R\sqrt{3/2}}{R\sqrt{5/4}} = \sqrt{\frac{3}{2} 	imes \frac{4}{5}} = \sqrt{\frac{12}{10}} = \sqrt{\frac{6}{5}}$.To match the form $\sqrt{12}:\sqrt{K}$,we multiply the numerator and denominator by $\sqrt{2}$: $\frac{\sqrt{12}}{\sqrt{10}}$. Thus,$K = 10$.

The ratio of radii of gyration of a thin circular ring and a circular disc of same radius about a tangential axis in their own planes is $\sqrt{12}: \sqrt{K}$. The value of $K$ is

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