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## Homework Statement

In Cartesian coordinate system are two point masses of mass [tex]m[/tex] connected with massless rod. Masses are in [tex](0,r,0)[/tex] and [tex](0,-r,0)[/tex].

a) Find the moment of inertia tensor.

b) Points were rotated in OXY plane such that angle between rod and Y-axis is [tex]\vartheta[/tex] (new cooridnates of masses are [tex](-r \sin \vartheta, r \cos \vartheta, 0)[/tex], [tex](r \sin \vartheta, -r \cos \vartheta, 0)[/tex]

## Homework Equations

[tex]\hat{I} = \left(\begin{matrix}

I_{xx} & I_{xy} & I_{xz}\\

I_{yx} & I_{yy} & I_{yz}\\

I_{zx} & I_{zy} & I_{zz}\\

\end{matrix}

\right)[/tex]

[tex]I_{xx} = \sum _{k} m_{k} (y^{2}_{k}+z^{2}_{k}) = \sum _{k} m_{k}(r^{2}_{k} - x^{2}_{k})\![/tex]

[tex]I_{yy} = \sum _{k} m_{k} (z^{2}_{k}+x^{2}_{k}) = \sum _{k} m_{k}(r^{2}_{k} - y^{2}_{k})\![/tex]

[tex]I_{zz} = \sum _{k} m_{k} (x^{2}_{k}+y^{2}_{k}) = \sum _{k} m_{k}(r^{2}_{k} - z^{2}_{k})\![/tex]

[tex]I_{xy} = I_{yx} = - \sum _{k} m_{k} x_{k}y_{k}\![/tex]

[tex]I_{yz} = I_{zy} = - \sum _{k} m_{k} y_{k}z_{k}\![/tex]

[tex]I_{zx} = I_{xz} = - \sum _{k} m_{k} z_{k}x_{k}\![/tex]

## The Attempt at a Solution

a) Using relevant equations I get

[tex]\hat{I} = \left(\begin{matrix} 2 m r^2 & 0 & -2 m r^2 \\ 0 & 4 m r^2 & 0 \\ -2mr^2 & 0 & 2mr^2 \end{matrix}\right)[/tex]

is it correct?

b) I'm not sure, hence I'm asking :) - is it enough (and correct) to calculate:

[tex]\hat{I}' = \hat{I} \left(\begin{matrix} \cos \varphi & \sin \varphi & 0 \\ -\sin \varphi & \cos \varphi & 0 \\ 0 & 0 & 1 \end{matrix}\right)[/tex] or maybe [tex]\hat{I}' = \left(\begin{matrix} \cos \varphi & \sin \varphi & 0 \\ -\sin \varphi & \cos \varphi & 0 \\ 0 & 0 & 1 \end{matrix}\right) \hat{I}[/tex] or sth else?