- #1

- 122

- 44

- Homework Statement
- Show that if ##\rho## is invariant (i.e. stays the same) under a ##\pi##-rotation around the 1 axis, then

- Relevant Equations
- none

Hi,

unfortunately, I am not getting anywhere with the following task

The inertia tensor is as follows

$$\left( \begin{array}{rrr}

I_{11} & I_{12} & I_{13} \\

I_{21} & I_{22} & I_{23} \\

I_{31} & I_{32} & I_{33} \\

\end{array}\right)$$

I had now thought that I could simply rotate the inertia tensor around the angle ##\pi## using the rotation matrix for x, i.e.

$$\left( \begin{array}{rrr}

1 & 0 & 0 \\

0 & cos & -sin \\

0 & sin & cos \\

\end{array}\right)$$

I have received the following

$$\left( \begin{array}{rrr}

I_{11} & I_{12} & I_{13} \\

I_{21} & I_{22} & I_{23} \\

I_{31} & I_{32} & I_{33} \\

\end{array}\right) \left( \begin{array}{rrr}

1 & 0 & 0 \\

0 & -1 & 0 \\

0 & 0 & -1 \\

\end{array}\right)=\left( \begin{array}{rrr}

I_{11} & -I_{12} & -I_{13} \\

I_{21} & -I_{22} & -I_{23} \\

I_{31} & -I_{32} & -I_{33} \\

\end{array}\right)$$

And as you can unfortunately see, that is not the right solution at all. Unfortunately, I don't know what exactly I should calculate here to get the result.

unfortunately, I am not getting anywhere with the following task

The inertia tensor is as follows

$$\left( \begin{array}{rrr}

I_{11} & I_{12} & I_{13} \\

I_{21} & I_{22} & I_{23} \\

I_{31} & I_{32} & I_{33} \\

\end{array}\right)$$

I had now thought that I could simply rotate the inertia tensor around the angle ##\pi## using the rotation matrix for x, i.e.

$$\left( \begin{array}{rrr}

1 & 0 & 0 \\

0 & cos & -sin \\

0 & sin & cos \\

\end{array}\right)$$

I have received the following

$$\left( \begin{array}{rrr}

I_{11} & I_{12} & I_{13} \\

I_{21} & I_{22} & I_{23} \\

I_{31} & I_{32} & I_{33} \\

\end{array}\right) \left( \begin{array}{rrr}

1 & 0 & 0 \\

0 & -1 & 0 \\

0 & 0 & -1 \\

\end{array}\right)=\left( \begin{array}{rrr}

I_{11} & -I_{12} & -I_{13} \\

I_{21} & -I_{22} & -I_{23} \\

I_{31} & -I_{32} & -I_{33} \\

\end{array}\right)$$

And as you can unfortunately see, that is not the right solution at all. Unfortunately, I don't know what exactly I should calculate here to get the result.