Rotating reference frame

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A rotating frame of reference is a special case of a non-inertial reference frame in which the coordinate system is rotating relative to an inertial reference frame. An everyday example of a rotating reference frame is the surface of the Earth.

1 Fictitious forces
2 Relation between positions in the two frames
3 Generalized Derivatives In Rotating Reference Frame
4 Relation between velocities in the two frames
- 4.1 Proof of the formula
5 Relation between accelerations in the two frames

[edit] Fictitious forces

All non-inertial reference frames exhibit fictitious forces. Rotating reference frames are characterized by three fictitious forces

the centrifugal force
the Coriolis force

and, for non-uniformly rotating reference frames,

the Euler force.

Scientists living in a rotating box can measure the speed and direction of their rotation by measuring these fictitious forces. For example, Léon Foucault was able to show the Coriolis force that results from the Earth's rotation using the Foucault pendulum. If the Earth were to rotate a thousand-fold faster (making each day only ~86 seconds long), these fictitious forces could be felt easily by humans, as they are on a spinning carousel.

[edit] Relation between positions in the two frames

To derive these fictitious forces, it's helpful to be able to convert between the coordinates $\left( x^{\prime},y^{\prime},z^{\prime} \right)$ of the rotating reference frame and the coordinates $\left( x, y, z \right)$ of an inertial reference frame with the same origin. If the rotation is about the $z$ axis with an angular velocity $ω$ and the two reference frames coincide at time $t = 0$ , the transformation from rotating coordinates to inertial coordinates can be written

$x = x^{\prime}\ \cos\omega t + y^{\prime}\ \sin\omega t$

$y = y^{\prime}\ \cos\omega t - x^{\prime}\ \sin\omega t$

whereas the reverse transformation is

$x^{\prime} = x\ \cos\left(-\omega t\right) - y\ \sin\left( -\omega t \right)$

$y^{\prime} = y\ \cos\left( -\omega t \right) + x\ \sin\left( -\omega t \right)$

This result can be obtained from a rotation matrix.

[edit] Generalized Derivatives In Rotating Reference Frame

If we have the unit vectors $i, j, k$ representing standard 3 dimensional basis vectors, we can let them rotate because they will remain normalized. If we let them rotate at the speed of $ω$ then each unit vector abides by the following equation:

$\frac{dl}{dt}=\omega \times l$ ,

where $l = {i, j, k}$ . Then if we have a function, $f (t) = f x (t) i + f y (t) j + f z (t) k$ and we want to examine its first dervative we have:

$\frac{df}{dt}=\frac{df_x}{dt}i+\frac{di}{dt}f_x+\frac{df_y}{dt}j+\frac{dj}{dt}f_y+\frac{df_z}{dt}k+\frac{dk}{dt}f_z$

$\frac{df}{dt}=\frac{df_x}{dt}i+\frac{df_y}{dt}j+\frac{df_z}{dt}k+[\omega \times (f_x i + f_y j+f_z k)]$

$\frac{\delta f}{\delta t}+\omega\times f(t)$

Where $\frac{\delta}{\delta t}$ is the rate of change with respect to the rotating coordinate system. That is to say if $f (t)$ is rotating at the same speed as the basis vectors ( $ω$ then $\frac{\delta f}{\delta t}=0$ .

[edit] Relation between velocities in the two frames

A velocity of an object is the time-derivative of the object's position, or

$\mathbf{v} \ \stackrel{\mathrm{def}}{=}\ \frac{d\mathbf{r}}{dt}$

The time derivative of position in a rotating reference frame has two components, one from the time derivative in the inertial reference frame and another from its own rotation. These are related by the equation

$\left( \frac{d}{dt} \right)_{\mathrm{inertial}} = \left( \frac{d}{dt} \right)_{\mathrm{rotating}} + \boldsymbol\omega \times$

where the vector $\boldsymbol\omega$ points along the rotation axis with the magnitude of the angular velocity. Therefore, the velocities in the two reference frames are related by the equation

$\mathbf{v}_{\mathrm{inertial}} \ \stackrel{\mathrm{def}}{=}\ \left( \frac{d\mathbf{r}}{dt} \right)_{\mathrm{inertial}} = \left( \frac{d\mathbf{r}}{dt} \right)_{\mathrm{rotating}} + \boldsymbol\omega \times \mathbf{r} = \mathbf{v}_{\mathrm{rotating}} + \boldsymbol\omega \times \mathbf{r}$

[edit] Proof of the formula

Let's consider a vector a_inertial in the inertial frame of reference, call a_rotating the same vector in the rotating frame of reference. Now, P_t is the position pointed by vector a at time t in the inertial frame of reference, Q is a point which has the same starting position as P₀ (Q₀ = P₀) and rotates according to the inertial frame as if it would appear fixed in the rotating frame.

After a very short time δ t, we have that the vector Q₀ Q_{δ t} is

$\boldsymbol\omega \times \mathbf {a}_{\mathrm{inertial}} \cdot \delta t$

considering some simple vector operations we have

$\overline{P_0 P_{\delta t}} = \mathbf{a}_{\mathrm{inertial}} = \overline{P_0 Q_{\delta t}} + \overline{Q_{\delta t} P_{\delta t}} = \overline{Q_0 Q_{\delta t}} + \overline{Q_{\delta t} P_{\delta t}} = \boldsymbol\omega \times \mathbf{a}_{\mathrm{inertial}} \cdot \delta t + \mathbf{a}_{\mathrm{rotating}}$

differentiating with respect to time we get

$\mathbf{\dot a}_{\mathrm{inertial}} = \boldsymbol \omega \times \mathbf{a}_{\mathrm{inertial}} + \mathbf{\dot a}_{\mathrm{rotating}}$

and observe that

$\boldsymbol \omega \times \mathbf{a}_{\mathrm{rotating}} = \boldsymbol \omega \times \mathbf{a}_{\mathrm{inertial}}$

PLEASE VERIFY THIS LAST STATEMENT

[edit] Relation between accelerations in the two frames

Acceleration is the second time derivative of position, or the first time derivative of velocity

$\mathbf{a}_{\mathrm{inertial}} \ \stackrel{\mathrm{def}}{=}\ \left( \frac{d^{2}\mathbf{r}}{dt^{2}}\right)_{\mathrm{inertial}} = \left( \frac{d\mathbf{v}}{dt} \right)_{\mathrm{inertial}} = \left[ \left( \frac{d}{dt} \right)_{\mathrm{rotating}} + \boldsymbol\omega \times \right] \left[ \left( \frac{d\mathbf{r}}{dt} \right)_{\mathrm{rotating}} + \boldsymbol\omega \times \mathbf{r} \right]$

Carrying out the differentiations and re-arranging some terms yields the acceleration in the rotating reference frame

$\mathbf{a}_{\mathrm{rotating}} = \mathbf{a}_{\mathrm{inertial}} - 2 \boldsymbol\omega \times \mathbf{v}_{\mathrm{rotating}} - \boldsymbol\omega \times (\boldsymbol\omega \times \mathbf{r}) - \frac{d\boldsymbol\omega}{dt} \times \mathbf{r}$

where $\mathbf{a}_{\mathrm{rotating}} \ \stackrel{\mathrm{def}}{=}\ \left( \frac{d^{2}\mathbf{r}}{dt^{2}} \right)_{\mathrm{rotating}}$ is the apparent acceleration in the rotating reference frame.

The three extra terms on the right-hand side result in fictitious forces in the rotating reference frame, i.e., accelerations that result from being in a non-inertial reference frame, rather than from any physical force. Using Newton's second law of motion $F = m a$ , we obtain

the Coriolis force

$\mathbf{F}_{\mathrm{Coriolis}} = -2m \boldsymbol\omega \times \mathbf{v}_{\mathrm{rotating}}$

the centrifugal force

$\mathbf{F}_{\mathrm{centrifugal}} = -m\boldsymbol\omega \times (\boldsymbol\omega \times \mathbf{r})$

and the Euler force

$\mathbf{F}_{\mathrm{Euler}} = -m\frac{d\boldsymbol\omega}{dt} \times \mathbf{r}$

where $m$ is the mass of the object being acted upon by these fictitious forces.

For completeness, the inertial acceleration $\mathbf{a}_{\mathrm{inertial}}$ can be determined from the total physical force $\mathbf{F}_{\mathrm{tot}}$ (i.e., the total force from physical interactions such as electromagnetism) likewise using Newton's second law