Calculating Moment of Inertia Tensor for Rod Along x Axis

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SUMMARY

The moment of inertia tensor for a thin rod of mass M and length L, oriented along the x-axis, is calculated as follows: I_11 = 0, I_22 = (1/3)ML², I_33 = (1/3)ML², with all off-diagonal components (I_12, I_21, I_13, I_31, I_23, I_32) equal to 0. The correct moment of inertia about the center is confirmed to be I = (1/12)ML². This calculation is essential for understanding rotational dynamics in physics.

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  • Ability to apply physical principles to rigid body dynamics
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  • Learn about the parallel axis theorem in rotational dynamics
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Homework Statement



A thin rod has mass M and length L. What is the moment of inertia tensor about the center of mass if placed along the x axis.

Homework Equations



I would write the inertia tensor in component notation, but I don't know how to use Latex.

The Attempt at a Solution



I am getting lost in the inertia tensor equation. For example, when I try to find the I_11 component, plugging in I get:

I_11 = int[rho((x_2^2 -x_1^2)+(x_3^2-x_1^2))dV]

I_11 = iiint[rho(-2x_1^2+x_2^2+x_3^2)dx_1dx_2dx_3]

This is where I am stuck. I do not understand what to use as my various bounds or if this equation is even correct.

Any help would be much appreciated.
 
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Okay, so I think I figured out my problem. I was doing something wrong with the component form.

Here is what I got for my inertia tensor:

I_11 = 0
I_22 = (1/3)ml^2
I_33 = (1/3)ml^2
I_12 = I_21 = 0
I_13 = I_31 = 0
I_23 = I_32 = 0So it looks something like this:

[..0...0.....0...]
[..0...(1/3)ml^2...0...]
[..0...0...(1/3)ml^2]

Does this look correct?
 
it should be 1/12 ml^2 you are doing it about the center
 

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