The radius of gyration of a uniform thin rod | vedclass

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Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 the definition of radius of gyration $(K)$,we have $I = MK^2$.Equating the two expressions: $MK^2 = \frac{ML^2}{12}$.Canceling $M$ from both sides: $K^2 = \frac{L^2}{12}$.Taking the square root of both sides: $K = \frac{L}{\sqrt{12}}$.

The 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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