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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 = 1/12ml

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 I

I = I

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(1/2mr

^{2}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 I

_{1}= 8(1/12ml^{2}+ m(l/2)^{2}) according to the parallel axis theorem:I = I

_{0}+ md^{2}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(1/2mr

^{2}+(m(l/2)^{2}) to the overall moment of inertia? And by neglecting r^{2}I get 4(m(l/2)^{2})?
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