Question

3.
Radius of gyration of a thin circular ring of mass 'm'
and radius 'R about a tangent in the plane of ring​

Answer 1

Answer:√5/2R

Explanation:

The moment of intertia for a particular uniform circular ring about the centroid of axis in the plane of ring is

Ic= 1/4mR²

Using the parallel Axis theorem along the tangent the moment of inertia will be

Iy= Ic+md²

d is distance from centroid to tangent

Hence it is pretty evident that d is not anything else but radius=R

Iy= Ic+mR²

= 1/4mR² + mR²

= 5/4mR²

For radius of gyration

k=√Iy/m

= √1.25mR²/m

= √5/2R