Find the radius of gyrational and moment of inertia of a rod of mass `100g` and length `1 m` about an axis passing through its centre and perpendicular to its length.

Explanation

Here, `K = ?, I =?, m = 100 g = 10^(-1) kg, l = 1 m`
About the given axis,
`I = (ml^(2))/(12) = (10^(-1)(1)^(2))/(12) = 0.0083 kg m^(2)`
Also `mK^(2) = (ml^(2))/(12)`
`:. K = (l)/(sqrt(12)) = (1)/(sqrt(12))m = (1)/(3.464) = 0.289 m`