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What is a rotating frame

  1. Jul 24, 2014 #1

    Often in physics we need to consider frames of reference that are non-inertial (the Earth spinning on its axis for example). We must therefore see how these rotating reference frames relate to an inertial reference frame.


    [tex]\frac{d^2\mathbf{r}}{dt^2} = \ddot{\mathbf{r}} + 2(\mathbf{\Omega} \times \dot{\mathbf{r}}) +\mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r})[/tex]

    Extended explanation

    Effect on 1st derivatives:

    Consider a rotating frame with an instantaneous angular velocity [itex]\mathbf{\Omega}[/itex]. A unit vector [itex]\mathbf{e}_i[/itex] traces a circle about [itex]\mathbf{\Omega}[/itex] at a rate:

    [tex]\frac{d\mathbf{e}_i}{dt} = \mathbf{\Omega} \times \mathbf{e}_i [/tex]

    A particle will have a position in the rotating frame given by [itex]\mathbf{r} = x_i\mathbf{e}_i[/itex] (where [itex]i[/itex] is summed from 1 to 3) and thus the velocity in an inertial frame is then:

    [tex] \frac{d\mathbf{r}}{dt} = \frac{d}{dt}( x_i\mathbf{e}_i) = \frac{dx_i}{dt}\mathbf{e}_i + x_i\frac{d\mathbf{e}_i}{dt} = \frac{dx_i}{dt}\mathbf{e}_i + x_i(\mathbf{\Omega} \times \mathbf{e}_i)[/tex]

    Example: torque equation:

    For example, in a fixed frame of reference, the equation relating net torque on a body to its rate of change of angular momentum is:

    [tex]\mathbf{\tau}_{net}\ =\ \frac{d}{dt}\left(\tilde{I}\,\mathbf{\Omega}\right)\ =\ \tilde{I}\,\frac{d}{dt}\left(\mathbf{\Omega}\right)\ +\ \frac{d\tilde{I}}{dt}\left(\mathbf{\Omega}\right)[/tex]

    but in a frame rotating with the body, it is:

    [tex]\mathbf{\tau}_{net}\ =\ \frac{d}{dt}\left(\tilde{I}\,\mathbf{\Omega}\right)\ \ +\ \ \mathbf{\Omega}\,\times \left(\tilde{I}\,\mathbf{\Omega}\right)\ =\ \tilde{I}\,\frac{d}{dt}\left(\mathbf{\Omega}\right)\ \ +\ \ \mathbf{\Omega}\,\times \left(\tilde{I}\,\mathbf{\Omega}\right)[/tex]

    Effect on 2nd derivatives:

    The acceleration is then (assuming [itex]\mathbf{\Omega}[/itex] is constant):

    [tex] \frac{d^2\mathbf{r}}{dt^2} = \frac{d^2r}{dt^2}+ 2\frac{dx_i}{dt}(\mathbf{\Omega}\times\mathbf{e}_i) + x_i(\mathbf{\Omega}\times(\mathbf{\Omega}\times \mathbf{e}_i))[/tex]

    Tidying up a bit we have:

    [tex]\frac{d^2\mathbf{r}}{dt^2} = \ddot{\mathbf{r}} + 2(\mathbf{\Omega} \times \dot{\mathbf{r}}) +\mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r})[/tex]

    The [itex] 2(\mathbf{\Omega} \times \dot{\mathbf{r}}) [/itex] term is called the Coriolis acceleration and the [itex]\mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r})[/itex] term is called the centripetal acceleration.

    What we have essentially is:

    Acceleration seen by inertial observer = Acceleration seen by rotating observer + extra terms

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
  2. jcsd
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