Calculate the moment of inertia of this body

[leynat]

Homework Statement

12 thin, homogenous rods, each of mass m and length l are welded together at the endpoints so they form a cube. Calculate the moment of inertia of this body with respect to an axis through its midpoint.

Relevant Equations

I = I0 + md^2

I use the moment of inertia I = 1/12ml² for an axis perpendicular and passing through the center of mass of a rod.

In a cube built out of 12 rods I have 8 rods at a perpendicular distance l/2 from the axis through the midpoint of a cube. These 8 rods contribute the moment of inertia I₁ = 8(1/12ml² + m(l/2)²) according to the parallel axis theorem:
I = I₀ + md²

What about the 4 remaining vertical rods? They are parallel to the axis passing through the midpoint of a cube. If I consider them cylinders then they contribute 4(1/2mr²+(m(l/2)²) to the overall moment of inertia? And by neglecting r² I get 4(m(l/2)²)?

 

[trurle]

What about the 4 remaining vertical rods? They are parallel to the axis passing through the midpoint of a cube. If I consider them cylinders then they contribute 4(1/2mr²+(m(l/2)²) to the overall moment of inertia? And by neglecting r² I get 4(m(l/2)²)?

A.f.a.i.k., correct. The problem may be in what you did not clearly stated the direction of rotational axis. You can pull a multiple axes through "midpoint of the cube"

 

[leynat]

A.f.a.i.k., correct. The problem may be in what you did not clearly stated the direction of rotational axis. You can pull a multiple axes through "midpoint of the cube"

Thank you. Obviously, I didn't make it clear but I meant the axis perpendicular to the base and passing through the midpoint.

 

[haruspex]

4(1/2mr²+(m(l/2)²)

How far are those 4 rods from the axis?

 

[Orodruin]

A.f.a.i.k., correct. The problem may be in what you did not clearly stated the direction of rotational axis. You can pull a multiple axes through "midpoint of the cube"

This actually does not matter. As can be argued by the symmetry of the cube, the moments of inertia around all axes are the same. You can therefore choose your axis in such a way that the moment of inertia becomes easy to compute.