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In the following exercice, I get the a different answer than what is asked to show.

A rigid body with a fixed point $o$ is called a top. Consider an orthonormal frame $\{e_1,e_2,e_3\}$ at the point $o$ rigidly attached to the body. The position of the frame itself at time $t$ can be given by an orthogonal matrix $(\alpha_i^j)\;i,j=1,2,3$ composed of the coordiantes of the vectors $\{e_1,e_2,e_3\}$ with respect to some fixed orthonormal frame in space. Thus, the motion of the top corresponds to a mapping $t\mapsto O(t)$ from $\mathbb{R}$ into the group $SO(3)$ of special orthogonal $3\times 3$ matrices.

Show that if $\dot{\bf{r}}(t)$ is the radius vector of a point of a rotating top, then $\dot{\bf{r}}(t)=(O^{-1}\dot{O}\bf{r})(t)$.

My solution: $O^{-1}(t)\bf{r}(t)$ is the coordinate of $\bf{r}$ in the frame $\{e_1,e_2,e_3\}$, thus it is constant. Let $\bf{r}_0=O^{-1}(t)\bf{r}(t)$, then $\mathbf{r}(t)=O(t)\bf{r}_0$. Therefore $\dot{\mathbf{r}}(t)=\dot{O}(t)\mathbf{r}_0=\dot{O}(t)O^{-1}(t)\mathbf{r}(t)$.

Is the conclusion of the exercice wrong? Or is my solution wrong?

Thanks in advance!

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