The radius of gyration of a uniform rod of length ℓ, about an axis passing through a point away from the centre of the rod, and perpendicular to it, is

Engineering

Physics

Moment of Inertia

Question

The radius of gyration of a uniform rod of length ℓ, about an axis passing through a point $l\over4$ away from the centre of the rod, and perpendicular to it, is

$\frac{1}{4} ℓ$

$\frac{1}{8} ℓ$

$\sqrt{\frac{7}{48}} ℓ$

$\sqrt{\frac{3}{8}} ℓ$

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$\frac{{Mℓ}^{2}}{12} + M {(\frac{ℓ}{4})}^{2} = {MK}^{2}$

$\frac{ℓ^{2}}{12} + \frac{ℓ^{2}}{16} = K^{2}$

$K = \sqrt{\frac{7}{48}} ℓ$