Determine the radius of gyration of a circular ring of radius r...

The moment of inertia of a circular ring of mass $M$ and radius $r$ about an axis passing through its centre and perpendicular to its plane is $I_{cm} = M r^2$.

Using the parallel axis theorem, the moment of inertia of the ring about an axis passing through a particle on its circumference and perpendicular to its plane is given by:

$I = I_{cm} + M d^2$

Here, the distance between the parallel axes is $d = r$.

$I = M r^2 + M r^2 = 2 M r^2$

The radius of gyration $k$ is defined by $I = M k^2$. Equating the two expressions for $I$:

$M k^2 = 2 M r^2$

$k^2 = 2 r^2$

$k = \sqrt{2}\,r$