Derivation of non-inertial terms in non-inertial systems using rotation matrices

[andresordonez]

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

Homework Statement

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:

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

Homework Equations

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

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??

 

[Jhero]

Hello, it seems like you are trying to derive the relationship between forces in inertial and non-inertial reference frames using a coordinate transformation. While your approach is certainly valid, there is an easier way to derive this relationship without involving rotation matrices.

First, let's define the acceleration in the inertial reference frame (S') as \ddot{\vec{r'}}. This can be written in terms of the acceleration in the non-inertial frame (S) as \ddot{\vec{r}} + 2\vec{\omega} \times \dot{\vec{r}} + \vec{\omega} \times (\vec{\omega} \times \vec{r}). This can be easily derived by considering the Coriolis and centrifugal forces that act on a particle in a rotating frame.

Next, we can use the definition of force, \vec{F} = m\ddot{\vec{r}}, to rewrite this equation as \vec{F'} = m(\ddot{\vec{r}} + 2\vec{\omega} \times \dot{\vec{r}} + \vec{\omega} \times (\vec{\omega} \times \vec{r})). This is the desired relationship between forces in the two reference frames.

I hope this helps! Let me know if you have any further questions.