Question

Find the radius of gyration of a circular ring of radius $r$ about a line perpendicular to the plane of the ring and passing through one of its particles.

• A.
• B. R
• C.
• D. 2R

Answer 1

The correct option is C

Moment of inertia at the centre and perpendicular to the plane of the ring is $m{R}^2$. So, about a point on the rim of the ring and the axis $\perp$ to the plane of the ring, the moment of inertia

= $m{R}^2+m{R}^2=2m{R}^2$(parallel axis theorem)

$\Rightarrow m{K}^2=2m{R}^2$ ($K$ = radius of the gyration)

$\Rightarrow K=\sqrt{2R^2}=\sqrt{2}R$.