Question

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 :​

Answer 1

Answer: rt (7/2) r

Explanation:

First calculate  the moment of inertia of the entire system.

The axis passes in such a way that is passes along the centre of one of the circle and along the the tangent for rest 2 rings.

so the total moment of inertia would be mr^2/2 + 2 * 3/2mr^2

which is equal to 7/6 mr^2

we know that it can be equated to mk^2

so, by solving we get k = rt 7/6

:)