Let M and L be the mass and length of a thin | vedclass

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(A) The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to its length is

Vedclass · Accepted Answer

$1: 2$

Vedclass · Answer

(A) The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to its length is $I_1 = \frac{ML^2}{12}$.By definition,$I_1 = MK_1^2$,where $K_1$ is the radius of gyration.So,$MK_1^2 = \frac{ML^2}{12} \implies K_1 = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}}$.The moment of inertia about an axis passing through one end and perpendicular to its length is $I_2 = \frac{ML^2}{3}$.By definition,$I_2 = MK_2^2$,where $K_2$ is the radius of gyration.So,$MK_2^2 = \frac{ML^2}{3} \implies K_2 = \frac{L}{\sqrt{3}}$.The ratio of the radius of gyration in the first case to the second case is $\frac{K_1}{K_2} = \frac{L / (2\sqrt{3})}{L / \sqrt{3}} = \frac{1}{2}$.Thus,the ratio is $1:2$.

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