- #1
- 19,865
- 10,853
Definition/Summary
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.
Equations
\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})
Extended explanation
Effect on 1st derivatives:
Consider a rotating frame with an instantaneous angular velocity \mathbf{\Omega}. A unit vector \mathbf{e}_i traces a circle about \mathbf{\Omega} at a rate:
\frac{d\mathbf{e}_i}{dt} = \mathbf{\Omega} \times \mathbf{e}_i
A particle will have a position in the rotating frame given by \mathbf{r} = x_i\mathbf{e}_i (where i is summed from 1 to 3) and thus the velocity in an inertial frame is then:
\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)
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:
\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)
but in a frame rotating with the body, it is:
\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)
Effect on 2nd derivatives:
The acceleration is then (assuming \mathbf{\Omega} is constant):
\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))
Tidying up a bit we have:
\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})
The 2(\mathbf{\Omega} \times \dot{\mathbf{r}}) term is called the Coriolis acceleration and the \mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r}) 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!
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.
Equations
\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})
Extended explanation
Effect on 1st derivatives:
Consider a rotating frame with an instantaneous angular velocity \mathbf{\Omega}. A unit vector \mathbf{e}_i traces a circle about \mathbf{\Omega} at a rate:
\frac{d\mathbf{e}_i}{dt} = \mathbf{\Omega} \times \mathbf{e}_i
A particle will have a position in the rotating frame given by \mathbf{r} = x_i\mathbf{e}_i (where i is summed from 1 to 3) and thus the velocity in an inertial frame is then:
\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)
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:
\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)
but in a frame rotating with the body, it is:
\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)
Effect on 2nd derivatives:
The acceleration is then (assuming \mathbf{\Omega} is constant):
\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))
Tidying up a bit we have:
\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})
The 2(\mathbf{\Omega} \times \dot{\mathbf{r}}) term is called the Coriolis acceleration and the \mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{r}) 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!