Wednesday, 29 November 2017

newtonian mechanics - Velocity in a turning reference frame


I often see the relation that $\vec v=\vec v_0+ \vec \omega \times \vec r$ in a turning reference frame, but where does it actually come from and how do I arrive at the acceleration being $$\vec a=\vec a_0+ 2\vec \omega \times\vec v+ \vec \omega \times(\vec \omega \times \vec r)+\dot{\vec \omega} \times \vec r\,\,\text{?}$$


Is there a simple method to see this? All approaches that I saw use some non-intuitive change of differential operators and so on $\left(\frac{d}{dt} \rightarrow \frac{d}{dt}+\vec \omega \times{}\right.$ and so on$\left.\vphantom{\frac{d}{dt}}\right)$.



Answer




I don't think you can do much better than getting your head around the identity $$\frac{d}{dt} \rightarrow \frac{d}{dt}+\vec \omega \times,$$ which holds when the former is applied to vectors. The essential point of the identity is that even if a vector is stationary in one reference frame, it will have some rotational motion in the rotating frame.


It may help to rephrase this in matrix language: for any vector $\vec u$, it reads $$\frac{d}{dt} \begin{pmatrix}u_x\\u_y\\u_z\end{pmatrix} \rightarrow \frac{d}{dt} \begin{pmatrix}u_x\\u_y\\u_z\end{pmatrix} + \begin{pmatrix}0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x\\ -\omega_y & \omega_x & 0\end{pmatrix} \begin{pmatrix}u_x\\u_y\\u_z\end{pmatrix} = \begin{pmatrix} \frac{du_x}{dt} +\omega_y u_z-\omega_z u_y\\ \frac{d u_y}{dt}+\omega_z u_x-\omega_x u_z\\ \frac{du_z}{dt} +\omega_x u_y-\omega_y u_x\end{pmatrix} .$$ Thus, the rate of change of each vector component gets added a linear multiple of the other components, as they "rotate into it".


(For example, if $\vec \omega=\omega \hat{e}_z$, then $$\frac{d}{dt} \begin{pmatrix}u_x\\u_y\\u_z\end{pmatrix} \rightarrow \begin{pmatrix} \frac{du_x}{dt} -\omega u_y\\ \frac{d u_y}{dt}+\omega u_x\\ \frac{du_z}{dt} \end{pmatrix} ,$$ so that $u_x$ and $u_y$ transform into ($\pm$) each other as the frames rotate and the $x$ and $y$ axes rotate into ($\pm$) each other.)


That's the intuition behind the identity. Operationally, it is the easiest to apply (just substitute for $\frac{d}{dt}$), and it gives an unambiguous way to connect rates of change of vector components from one frame to another. What's not to love?


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