The radius of gyration of a cylindrical rod a | vedclass

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(A) The moment of inertia $I$ of a uniform rod of mass $m$ and length $\ell$ about an axis passing through its center and perpendicular to its length

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$5$

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(A) The moment of inertia $I$ of a uniform rod of mass $m$ and length $\ell$ about an axis passing through its center and perpendicular to its length is given by $I = \frac{m \ell^{2}}{12}$.By definition,the moment of inertia is also expressed as $I = mk^{2}$,where $k$ is the radius of gyration.Equating the two expressions: $mk^{2} = \frac{m \ell^{2}}{12}$.Solving for $k$: $k^{2} = \frac{\ell^{2}}{12} \Rightarrow k = \frac{\ell}{\sqrt{12}} = \frac{\ell}{2 \sqrt{3}}$.Given $\ell = 10 \sqrt{3} \ m$,we substitute this value:$k = \frac{10 \sqrt{3}}{2 \sqrt{3}} = 5 \ m$.Thus,the radius of gyration is $5 \ m$.

The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through its center is $k$. Given that the length of the rod is $10 \sqrt{3} \ m$,find the value of $k$ in meters.

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