Three rings each of mass m and radius r are so placed that they touch each other. The radius of gyration of the system about the axis as shown in the figure is
Solution:
Moment of Inertia of ring about it's diameter is mR²/2. ( Using Perpendicular axis theorem)
Moment of Inertia about tangential axis in plane of ring will be mR²/2 + mR²= 3/2 mR²
Total moment of inertia about the axis shown in figure
= (3/2 mR² )× 2 + 1/2 mR²= 7/2 mR²
Radius of Gyration is k then 3m×k²= 7/2 mR²
so,
\[ k= \sqrt{\frac{7}{6}}r \]