Rotating Frame#

Notation and Review#

Because rotational frame are difficult and often not remembered let’s have some quick review:

Angular momentum vector : The angular momentun vector \(\boldsymbol{\omega}\) points towards the direction of rotation (clockwise or counter: CW or CCW). You may also use the direction by right hand rule (direction is towards the thumb as the right hand wraps around the vector either CW or CCW),

$$ \boldsymbol{\omega} = \omega \hat \omega $$

* $\omega$ : Magntiude as the speed of rotation
* $\hat \omega$ : Direction of rotation

As a function of time, $\boldsymbol{\omega}(t)$ may change on either magnitude $\omega$ and/or direction $\hat \omega$

Velocity vector : The velocity vector if you remember has a magnitude of \(v = \omega r\). Because thedirection of the velocity is always perpendicular to the angular velocity vector \(\boldsymbol{\omega}\) and the distance away from the axis \(\mathbf{r}\) we can easily see that therelationship should be a cross product,

\[ \mathbf{v} = \boldsymbol\omega \times \mathbf r \]

Time Derivatives in Rotating Frame#

Consider two frames \(S\) (rotating frame of angular frequency \(\mathbf \Omega\)) and rest frame \(S_0\) with and arbitrary vector \(\mathbf Q\) defined in the \(S\) frame with basis stationary in respect to \(S\)

\[ \mathbf Q = \sum_i{\mathbf Q_i \mathbf e_i} \]

The time derivative of \(\mathbf Q\) along the \(S\) is simply,

\[ \dot{\mathbf Q} = \sum_i{\dot{\mathbf Q_i} \mathbf e_i}\]

The time derivative of \(\mathbf Q\) along the \(S_0\) (denote this as \(\dot {\mathbf Q_0}\)) is then,

\[\begin{split} \dot{\mathbf Q}_0= \sum_i{\dot{Q_i} \mathbf e_i + Q_i \dot{\mathbf e_i}}\\ \dot{\mathbf Q}_0= \dot{\mathbf Q} + \sum{Q_i \dot{\mathbf e_i}}\end{split}\]

The second term has the time derivative of the basis vector \(\dot{\mathbf e_i}\). You may interpret this as a velocity vector along \(\hat e_i\) therefore it must be that,

\[ \dot{\mathbf e_i} = \mathbf\Omega \times \mathbf e_i \]
\[\begin{split} \dot{\mathbf Q}_0= \dot{\mathbf Q} + \sum{Q_i (\mathbf\Omega \times \mathbf e_i)}\\ \boxed{\dot{\mathbf Q}_0= \dot{\mathbf Q} + \mathbf\Omega \times \mathbf Q} \end{split}\]

Newton’s Second Law in Rotating Frame#

To form Newton’s second law we need to define acceleration \(\ddot{\mathbf r}_0\) in the rest frame \(S_0\). Let’s apply the time derivative using the relation of \(\dot{\mathbf Q_0}\) above,

\[ \ddot{\mathbf r}_0 = \overbrace{\frac{d}{dt}\underbrace{\left(\dot{\mathbf r} + \mathbf\Omega \times \mathbf r\right)}_{\mathbf Q'}}^{\dot{\mathbf Q}_0'}\]

Notice that the terms inside the parenthesis is just another vector where we want to take the time derivative with respect to \(S_0\). We once again apply the relation and take the derivative where \(\dot {\mathbf \Omega}_0 = \dot {\mathbf \Omega} = 0\) (constant rotation in both reference frames),

\[ \ddot{\mathbf r}_0 = \ddot {\mathbf r} + 2\mathbf\Omega \times \dot {\mathbf r} + \mathbf\Omega \times \left(\mathbf\Omega \times \mathbf r \right) \]

We may write Newton’s second law in either \(S_0\) frame (\(\mathbf F_0 = m\ddot{\mathbf r_0}\)) or \(S\) frame (\(\mathbf F = m\ddot{\mathbf r}\)). Let’s choose the \(S\) frame because it will result into a force familiar to Earthlings in the rotating Earth.

\[ \underbrace{m\ddot{\mathbf r}_0}_{\mathbf F_0} = 2m\mathbf\Omega \times \dot {\mathbf r} + \underbrace{m\ddot {\mathbf r}}_{\mathbf F} + m\mathbf\Omega \times \left(\mathbf\Omega \times \mathbf r \right)\]

Isolating \(\mathbf F\) gives,

\[\begin{split} \mathbf F = \mathbf{F_0} + \underbrace{2m\dot {\mathbf r} \times \mathbf \Omega}_{F_\text{cor}} + \underbrace{m(\mathbf \Omega \times \mathbf r) \times \mathbf \Omega}_{F_\text{cf}} \\ \boxed{F_\text{cor} = 2m\dot {\mathbf r} \times \mathbf \Omega}\\ \boxed{F_\text{cf} = m(\mathbf \Omega \times \mathbf r) \times \mathbf \Omega} \end{split}\]

  • \(\mathbf F_\text{cor}\) : The coriolis force

  • \(\mathbf F_\text{cf}\) : The centrifugal force notice that we may deduce the equation’s magntiude as something very familiar, the centripetal velocity \(v_\text{cf}\) as \(F_\text{cf} = -mv_\text{cf}\Omega\). This is centrifugal because of the negative sign on the centripetal velocity you’re familiar with.

  • \(\mathbf F\) : The force apprent to observers in the rotating frame \(S\), notice that it is no longer simply \(\mathbf F = m \mathbf a\) but the extra terms which most people call ficticious force