Question

Mass of thin long metal rod is 2 kg and its moment of inertia about an axis perpendicular to the length of rod and passing through its one end is 0.5kgm² . Its radius of gyration is

Moderate

Solution

$I = M K^{2} K = \sqrt{\frac{I}{M}} = \sqrt{\frac{0.5}{2}} = \sqrt{0.25} = 0.5$

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