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?

Answer 1

Given in the question :-
The moment of inertia perpendicular to the ring , passing though it's centre .
Therefore point of the rim is perpendicular to theplane of the ring.
Then,
I = Mr²

Now, ring passes by its particles at a distance r it will be parallel to the axis .

Therefore, By parallel axis theorm the moment of inertia will be,
$I' = I+mr^{2} =mr^{2} +mr^{2}$
I' = 2mr²

Hence for obtaining the radius of gyration ,for a new line radius of gyration will be a , therefore
$I' = ma^{2}$

Now Equating the given expression for I'
$ma^{2} = 2mr^{2}$
a² = 2r²
$\boxed{a=\sqrt{2r}}$


Hope it Helps.

Answer 2

Moment of inertia at the centre and perpendicular to the plane of the ring.

So, about a point on the rim of the ring and the axis ⊥ to the plane of the ring, the moment of inertia

= mR2 + mR2 = 2mR2 (parallel axis theorem)

mK2 = 2mR2 (K = radius of the gyration)

K = √2R2 = √2 R.