Radius of gyration of a uniform thin rod of l | vedclass

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

Vedclass · Accepted Answer

$ \frac{L}{\sqrt{12}} $

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(A) The moment of inertia $I$ of a uniform thin rod of mass $M$ and length $L$ about an axis passing through its centre of mass and perpendicular to its length is given by $I = \frac{ML^2}{12}$.By definition,the radius of gyration $K$ is related to the moment of inertia by the equation $I = MK^2$.Equating the two expressions for $I$,we get $MK^2 = \frac{ML^2}{12}$.Canceling $M$ from both sides,we have $K^2 = \frac{L^2}{12}$.Taking the square root of both sides,we find $K = \frac{L}{\sqrt{12}}$.

Radius of gyration of a uniform thin rod of length $L$ about an axis passing normally through its centre of mass is

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