Question

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

Answer 1

The radius of gyration of circular ring is given by a² = 2r²

Explanation:

In view of the principles in the problem ,

The moment of inertia on the ring, which passes through the center.

The point of the rim is therefore perpendicular to the ring axis.

Then,

         $\mathrm{I}=\mathrm{Mr}^{2}$

Now, the ring goes through its particles at a distance r parallel to the axis.

So, the moment of inertia will be,

By the parallel axis theorem,

                                               $I^{\prime}=2 m r^{2}$

Therefore,

In order to obtain the gyration radius :

A new line radius of gyration will be "a ".

Now, equate the expression given for " I'  ".

                                $a^{2}=2 r^{2}$

Thus, the radius of gyration of circular ring is " a² = 2r² ".