Question

The radius of gyration of an uniform rod of length l about an axis passing through one of its ends and perpendicular to its length is.

Answer 1

hope it helps you......

Answer 2

Answer:

∴

$Mk^{2} =\frac{ML^{2} }{3}$         k=radius of gyration

$k=\sqrt{L^{2}/3 }$

$k=\frac{L}{\sqrt{3} }$

Explanation:

1. Moment of inetria of an uniform rod of length l about an axis passing through one of its ends is, $I=\frac{ML^{2} }{3}$
2. The moment of inertia of a circular ring about an axis perpendicular to its plane passing through its centre is, $I=MR^{2}$
3. If you take a body with a given moment of inertia around some axis, then melt and recast it into a thin ring with the same mass, such that it has the same value of moment of inertia around the axis through its centre and perpendicular to the plane of the ring, as the original body, the radius of the ring that you get is the radius of gyration.