Radius of gyration of a thin uniform circular | vedclass

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(A) Let the mass of the disc be $M$ and its radius be $R$.The moment of inertia of a circular disc about an axis passing through its centre and perpen

Vedclass · Accepted Answer

$\sqrt{2}: 1$

Vedclass · Answer

(A) Let the mass of the disc be $M$ and its radius be $R$.The moment of inertia of a circular disc about an axis passing through its centre and perpendicular to its plane is $I_c = \frac{1}{2}MR^2$.The radius of gyration $K_c$ is given by $I_c = MK_c^2$,so $MK_c^2 = \frac{1}{2}MR^2$,which gives $K_c = \frac{R}{\sqrt{2}}$.The moment of inertia of the disc about its diameter is $I_d = \frac{1}{4}MR^2$.The radius of gyration $K_d$ is given by $I_d = MK_d^2$,so $MK_d^2 = \frac{1}{4}MR^2$,which gives $K_d = \frac{R}{2}$.The ratio $K_c : K_d = \frac{R/\sqrt{2}}{R/2} = \frac{2}{\sqrt{2}} = \sqrt{2} : 1$.

Radius of gyration of a thin uniform circular disc about the axis passing through its centre and perpendicular to its plane is $K_{c}$. Radius of gyration of the same disc about a diameter of the disc is $K_d$. The ratio $K_c: K_d$ is

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