Assume we have an object moving along a path $p_W(t)$ that is described in some fixed reference frame $W$. If we now have a second reference frame $B$ which differs from $W$ by some time varying rotation $R_{BW}(t)$ we can describe the objects path in the reference frame of $B$ then as $p_B(t)=R_{BW}(t)p_W(t)$. If I now want to look at the velocity of this object, we can in $W$ simply compute it as $v_W(t)=\dot{p}_W(t)$. But what is its velocity $v_B(t)$ in the frame $B$? Intuitively I would say it should be $v_B(t)=R_{BW}(t)\dot{p}_W(t)$, however from a mathematical perspective it would rather make sense for it to be $v_B(t)=\dot{R}_{BW}(t)p_W(t)+R_{BW}(t)\dot{p}_W(t)$. Which of those two are correct now, or rather what is the difference between them?

If I try to calculate the above formulas with the following example:

$$p_W(t)= \begin{pmatrix} \text{cos}(t) \\ \text{sin}(t) \end{pmatrix},R_{BW}(t)= \begin{pmatrix} -\text{sin}(t) & \text{cos}(t) \\ -\text{cos}(t) & -\text{sin}(t) \end{pmatrix},$$ so the object has a circular path and the x-Axis of frame $B$ is always along the tangent of this path in the direction of the movement. For the first formula I then get a constant velocity $v_B(t)= \begin{pmatrix} 1 \\ 0 \end{pmatrix}$, and in the second case a constant zero velocity $v_B(t)= \begin{pmatrix} 0 \\ 0 \end{pmatrix}$. Both of them seem correct, the object is indeed moving at constant speed 1 along the x-Axis of frame B. However also from the perspective of frame $B$ the object doesn't seem to move at all, it in fact always stays at the position $ p_B(t)=\begin{pmatrix} 0 \\ -1 \end{pmatrix}$.

So what is now the difference between those two velocities, both of them are described from frame $B$, however one of them actually describes a moving body, the other one a static one? I'm not actually coming from the field of Physics, so I am unfortunately not very familiar with motions in a moving coordinate frame.

  • $\begingroup$ Is $\theta$ a function of time? If so don't forget to include $\frac{d\theta}{dt}$ $\endgroup$ Oct 22 '20 at 9:30
  • 1
    $\begingroup$ I removed the $\theta$ and wrote the variable explicitly as $t$. $\endgroup$ Oct 22 '20 at 9:45

So even if the object was stationary in W it would have velocity in B if the frame is rotating. If frame B rotates relative to W by $\Omega_{BW}$ then

$$ v_B (t) = R_{BW}(t) v_W(t) + \Omega_{BW}\times R_{BW}(t) p_W(t) $$

This comes from the formula for differentiating a vector on a rotating frame

$$ \frac{{\rm d}}{{\rm d}t} \vec{A}_{\rm world} = \mathrm{R}_{\rm frame} \frac{{\rm d}}{{\rm d}t} \vec{A}_{\rm local} + \vec{\Omega}_{\rm frame} \times \mathrm{R}_{\rm frame} \vec{A}_{\rm local} $$

The above is often abbreviated to

$$ \frac{\rm d}{{\rm d}t} \vec{A} = \frac{\partial}{\partial t} \vec{A} + \vec{\Omega} \times \vec{A} $$

where everything is expressed in the non-rotating frame and the partial derivative with time means => how the vector changes if the frame was not rotating.


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