Homework Help: Derivation of non-inertial terms in non-inertial systems using rotation matrices

1. Jan 17, 2010

andresordonez

(I know how to do this without the rotation matrices)
Any suggestion would be much appreciated.

1. The problem statement, all variables and given/known data
Show that the relationship between the forces in the inertial (S') and non-inertial(S) reference frames, with a coordinate transformation given by

$$\vec{r}=R \vec{r'}$$

is:

$$\vec{F'} = \vec{F} + m(\vec{\omega} \times (\vec{\omega} \times \vec{r}) + 2\vec{\omega} \times \vec{v})$$

2. Relevant equations
$$$R = \left( \begin{array}{ccc} \cos(\omega t) & \sin(\omega t) & 0 \\ -\sin(\omega t) & \cos(\omega t) & 0 \\ 0 & 0 & 1 \end{array} \right).$ $R^{-1} = R^t = \left( \begin{array}{ccc} \cos(\omega t) & -\sin(\omega t) & 0 \\ \sin(\omega t) & \cos(\omega t) & 0 \\ 0 & 0 & 1 \end{array} \right)$ $-\vec{\omega} \times (\vec{\omega} \times \vec{r}) = -\omega^2 \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right) \vec{r}$ $\vec{\omega} \times \vec{v} = -\omega \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) \vec{v}$$$

3. The attempt at a solution

$$\ddot{\vec{r}} = R\ddot{\vec{r'}} + 2\dot{R}\dot{\vec{r'}} + \ddot{R}\vec{r'} = \ddot{R}R^{-1}\vec{r} + 2\dot{R}(\dot{R^{-1}}\vec{r}+R^{-1}\dot{\vec{r}}) + R\ddot{\vec{r'}}$$

$$$\ddot{R}R^{-1}\vec{r} = -\omega^2 \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right) \vec{r}$$$
$$$\dot{R}(R^{-1}\vec{r} + R^{-1}\dot{\vec{r}}) = \omega^2 \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right) \vec{r} - \omega \left( \begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0\end{array} \right) \dot{\vec{r}}$$$

then

$$\ddot{\vec{r}} = -\vec{\omega} \times (\vec{\omega} \times \vec{r}) - 2\vec{\omega} \times \dot{\vec{r}} + R\ddot{\vec{r'}}$$
$$R\vec{F'} = \vec{F} + m\vec{\omega}\times(\vec{\omega}\times\vec{r}) + 2m \vec{\omega} \times \dot{\vec{r}}$$

How do I get rid of R??