Calculate the moment of inertia of this body

AI Thread Summary
The moment of inertia for a cube made of 12 rods is calculated using the formula I = 1/12ml² for rods aligned with the axis through the center. Eight rods positioned at a distance of l/2 from the axis contribute I1 = 8(1/12ml² + m(l/2)²) based on the parallel axis theorem. The four remaining vertical rods, considered as cylinders, contribute 4(1/2mr² + m(l/2)²) to the overall moment of inertia, simplifying to 4(m(l/2)²) when neglecting r². The direction of the rotational axis is crucial for accurate calculations, but due to the cube's symmetry, the moments of inertia remain consistent across different axes. This symmetry allows for flexibility in choosing the axis for easier computation.
leynat
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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)?
 
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leynat said:
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"
 
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trurle said:
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.
 
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leynat said:
4(1/2mr2+(m(l/2)2)
How far are those 4 rods from the axis?
 
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trurle said:
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.
 
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