# Kinematics in moving reference frames

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.

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