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

  1. Jan 17, 2010 #1
    (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

    [tex] \vec{r}=R \vec{r'} [/tex]

    is:

    [tex] \vec{F'} = \vec{F} + m(\vec{\omega} \times (\vec{\omega} \times \vec{r}) + 2\vec{\omega} \times \vec{v}) [/tex]


    2. Relevant equations
    [tex]
    \[ R = \left( \begin{array}{ccc}
    \cos(\omega t) & \sin(\omega t) & 0 \\
    -\sin(\omega t) & \cos(\omega t) & 0 \\
    0 & 0 & 1 \end{array} \right).\]

    \[ R^{-1} = R^t = \left( \begin{array}{ccc}
    \cos(\omega t) & -\sin(\omega t) & 0 \\
    \sin(\omega t) & \cos(\omega t) & 0 \\
    0 & 0 & 1 \end{array} \right)\]

    \[ -\vec{\omega} \times (\vec{\omega} \times \vec{r}) = -\omega^2 \left( \begin{array}{ccc}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 0 \end{array} \right) \vec{r}\]

    \[ \vec{\omega} \times \vec{v} = -\omega \left( \begin{array}{ccc}
    0 & 1 & 0 \\
    -1 & 0 & 0 \\
    0 & 0 & 0 \end{array} \right) \vec{v}\]
    [/tex]

    3. The attempt at a solution

    [tex]
    \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'}}
    [/tex]

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

    then

    [tex]
    \ddot{\vec{r}} = -\vec{\omega} \times (\vec{\omega} \times \vec{r}) - 2\vec{\omega} \times \dot{\vec{r}} + R\ddot{\vec{r'}}
    [/tex]
    [tex]
    R\vec{F'} = \vec{F} + m\vec{\omega}\times(\vec{\omega}\times\vec{r}) + 2m \vec{\omega} \times \dot{\vec{r}}
    [/tex]

    How do I get rid of R??
     
  2. jcsd
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