# Derivative of radius vector of a rotating top

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?