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If I have a rigid body consists of three uniform rods, each of mass $m$ and length $2a$, held mutually perpendicular at their midpoints choose a coordinate system with the axes along the rod.

So I will explain how I oriented the rods. I have rod $I_1$ along the y-axis,$I_2$ will be along the z-axis and $I_3$ is along the x-axis my coordinate system is right handed coordinate system.

Now I want to calculate $I_{xx}$

So $I_{xx}$ = $I_{1,xx}$ + $I_{2,xx}$ + $I_{3,xx}$ with $I_{i,xx} = \int( y^2 + z^2 )dm$.

$I_{1,xx}$ we will have z = 0, since it has only y component. $I_{2,xx}$ we will have y = 0, since it has only z component. $I_{3,xx}$ we will have both y and z.

After rearranging we have Hence $I_{xx} = 2 \int (y^2 + z^2) = 2I_{rod,center} = 2/3 ma^2$

Is there anything wrong in my work above?

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  • $\begingroup$ What is an ord? A sketch might help here. Also I suppose you know about the parallel axis theorem. $\endgroup$ – John Alexiou Jun 2 '15 at 22:30
  • $\begingroup$ He means a rod. $\endgroup$ – GCLL Jun 2 '15 at 22:31
  • $\begingroup$ I meant rod yes. $\endgroup$ – Illustionisttt. Jun 2 '15 at 22:36
  • $\begingroup$ well we don't need parrallel axis theorem here since I am using the generalized formula for calculation of moment of inertia @ja72 $\endgroup$ – Illustionisttt. Jun 2 '15 at 22:37
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    $\begingroup$ I'm voting to close this question as off-topic because it is of the "check my work" type. Which are off topic by the rules of the site. $\endgroup$ – Floris Jun 2 '15 at 22:46
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Your calculation is correct. As you mentioned the inertia tensor, you can look at the system from a sligthly different point of view. A rod of length $\ell$ and mass $m$ has an inertia tensor with respect to its center of mass which can be written as

$$I = \frac{1}{12} m \ell^2 \left(I-\hat{n} \otimes \hat {n}\right)$$

where $\hat{n}$ is a versor parallel to it. Note that $\left(I-\hat{n} \otimes \hat {n}\right)$ is a projector in the space perpendicular to $\hat{n}$.

If you add the contributions of the three rods you get

$$I = \frac{1}{12} m \ell^2 \left(3 I-\hat{x} \otimes \hat {x}-\hat{y} \otimes \hat {y}-\hat{z} \otimes \hat {z}\right)$$

which is

$$I = \frac{1}{6} m \ell^2 I$$

because $\hat{x} \otimes \hat {x}+\hat{y} \otimes \hat {y}+\hat{z} \otimes \hat {z}$ is equal to the identity matrix $I$. So the momentum of inertia of the system is the same along any axis, as a consequence of its great simmetry. Setting $\ell=2a$ you obtain your result.

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  • $\begingroup$ versor=vector right? $\endgroup$ – John Alexiou Jun 2 '15 at 23:19
  • $\begingroup$ versor = vector with modulus equal to one $\endgroup$ – GCLL Jun 2 '15 at 23:20

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