parallel-axis theorem problem?

Question

A thin, uniform rod is bent into a square of side length a.

If the total mass is m, find the moment of inertia about an axis through the center and perpendicular to the plane of the square. Use the parallel-axis theorem.

Thanks for any help!

Answer 1

Each section of the square has mass m/4

A single rod of lenght a rotated through it's center (as if it were a baton held at the center) has I=(m/4)*a^2/12

by the parrallell axis theorm the rod, a/2 from the center will have further inertia=(a/2)^2(m/4)

for a single piece we have m/4a^2/12+m/4a^2/4=ma^2/12.

as there are 4 such pieces the Itotal=ma^2/3

Answer 2

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.