How to solve the following moment of inertia problem?

M.I. of each ring about an axis through its center of mass and perp. to its plane = Ma²

Distance of the new axis (z-axis) from the axis through centrer of mass = 2a/√3

M. I. of the ring about the new axis = Ma² + M (4a²/3) = (7/3) Ma²

M. I. of the three rings about the z-axis = 3* (7/3) Ma² = 7Ma²

No, h is not the side of the square. h is the distance between the two axes you are considering. Since one axis is the center of mass and the other is the z-axis, then h is the distance between the center of mass and the origin. That's almost certainly not 1.8 m. The center of mass is somewhere in the middle of the square. In fact I think it IS the center of the square. In order to compute by direct computation, you've got to figure out the distance of each mass from the center of mass and from the z-axis. Then add up the values of mr^2 where r is the appropriate distance.

the moment of inertia (MoI) of each ring about its center=ma^2

distance of center of each from z axis(d)=2a/root 3

by parallel axis theorem -

MoI of a ring about z-axis=ma^2+md^2

=ma^2+m(4/3)a^2

=7a^2/3

so adding the MoI of the three rings algebraically -

we get-

7a^2

guess m correct

hope it helped...

is dis NCERT question