# Moment of inertia tensor calculation [closed]

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?

## closed as off-topic by Floris, ja72, Qmechanic♦Jun 3 '15 at 10:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

• What is an ord? A sketch might help here. Also I suppose you know about the parallel axis theorem. – ja72 Jun 2 '15 at 22:30
• He means a rod. – GCLL Jun 2 '15 at 22:31
• I meant rod yes. – Illustionisttt. Jun 2 '15 at 22:36
• well we don't need parrallel axis theorem here since I am using the generalized formula for calculation of moment of inertia @ja72 – Illustionisttt. Jun 2 '15 at 22:37
• 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. – Floris Jun 2 '15 at 22:46

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.

• versor=vector right? – ja72 Jun 2 '15 at 23:19
• versor = vector with modulus equal to one – GCLL Jun 2 '15 at 23:20