The radius of gyration of a rod of length L a | vedclass

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(C) Let the rod be $AB$ of length $L$ and mass $M$. The center of mass $C$ is at $L/2$ from either end. The axis of rotation passes through a point $N

Vedclass · Accepted Answer

$\frac{L}{3}$

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(C) Let the rod be $AB$ of length $L$ and mass $M$. The center of mass $C$ is at $L/2$ from either end. The axis of rotation passes through a point $N$ at a distance $L/3$ from one end (say $A$).Distance of $N$ from the center $C$ is $d = |L/2 - L/3| = L/6$.Using the parallel axis theorem,the moment of inertia $I$ about the axis passing through $N$ is:$I = I_{CM} + Md^2$$I = \frac{ML^2}{12} + M\left(\frac{L}{6}\right)^2$$I = \frac{ML^2}{12} + \frac{ML^2}{36} = \frac{3ML^2 + ML^2}{36} = \frac{4ML^2}{36} = \frac{ML^2}{9}$The radius of gyration $K$ is defined by $I = MK^2$,so:$MK^2 = \frac{ML^2}{9}$$K^2 = \frac{L^2}{9}$$K = \frac{L}{3}$

The radius of gyration of a rod of length $L$ and mass $M$ about an axis perpendicular to its length and passing through a point at a distance $L/3$ from one of its ends,is :-

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