Question

Four particles each of mass ‘m’ are placed at the corners of a light square frame of side length ' l '. The radius of gyration of the system about an axis perpendicular to the plane of square and passing through its centre is

Moderate

Solution

$I = \sum {mr}^{2} = 4 m {[\frac{l}{\sqrt{2}}]}^{2} = \frac{4 {ml}^{2}}{2} = 2 {ml}^{2}$
Radius of gyration $k = \sqrt{\frac{I}{M}} = \sqrt{\frac{2 {ml}^{2}}{4 m}} = \frac{l}{\sqrt{2}}$

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