# What is a rotating frame

1. Jul 24, 2014

### Greg Bernhardt

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

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