For a particle with circular motion, the position vector of the particle can be expressed in the form of polar coordinates as $$\vec{r}=rcos\theta\hat{x}+rsin\theta\hat{y}$$ Thus, the velocity is $$\dot{\vec{r}}=\dot{r}cos\theta\hat{x}-rsin\theta\dot{\theta}\hat{x}+\dot{r}sin\theta\hat{y}+rcos\theta\dot{\theta}\hat{y}$$ Since $r=\sqrt{x^2+y^2}$, we can obtain $$\dot{r}=\frac{\partial r}{\partial x}\dot{x}+\frac{\partial r}{\partial y}\dot{y}=\dot{x}cos\theta+\dot{y}sin\theta$$ Finally, $$\dot{\vec{r}}=(\dot{x}cos\theta cos\theta+\dot{y}sin\theta cos\theta-y\dot{\theta})\hat{x}+(\dot{x}cos\theta sin\theta+\dot{y}sin\theta sin\theta+x\dot{\theta})\hat{y}$$ But, in reality, $$\dot{\vec{r}}=(\dot{x}-y\dot{\theta})\hat{x}+(\dot{y}+x\dot{\theta})\hat{y}$$ What's wrong with this?


Original question: If a particle fixed on a rotating reference frame with angular velocity $\omega\hat{z}$, the velocity of the particle corresponds to the frame would be: $$\dot{\vec{r}}=(\dot{x}-y\dot{\theta})\hat{x}+(\dot{y}+x\dot{\theta})\hat{y}+\dot{z}\hat{z}$$ My modification: If neglecting the motion along $z$, the question will actually become to a 2D question lies in $x-y$ plane. So, according to relativitic, I think we can make the rotating reference frame be static and let the particle rotate around $z$. Can this Equivalent method work? Or why not?

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    $\begingroup$ What you say should happen in reality is wrong as you can immediately see by writing $\vec{r} = x \hat{x}+y \hat{y}$. Its not even consistent with your second equation. $\endgroup$ – user35122 Apr 4 '17 at 6:22
  • $\begingroup$ The eqution, $\dot{\vec{r}}=(\dot{x}-y\dot{\theta})\hat{x}+(\dot{y}+x\dot{\theta})\hat{y}$, is absolutly right. $\endgroup$ – Feynman Apr 4 '17 at 6:36
  • $\begingroup$ R would be constant check en.m.wikipedia.org/wiki/Circular_motion#In_polar_coordinates $\endgroup$ – Utkarsh futous Apr 4 '17 at 8:16
  • $\begingroup$ @Feynman: Where did you find the last equation in your post? Note also that e.g $\partial r/\partial x \neq \cos \theta$. $\endgroup$ – CAF Apr 4 '17 at 9:30
  • $\begingroup$ Wouldn't $\dot{r}=0$ for circular motion? $\endgroup$ – Kyle Kanos Apr 4 '17 at 10:17

Both $r$ and $\theta$ are $x$, $y$ dependence,

\begin{align*} \tan \theta &= \frac{y}{x} \\[4pt] \dot{\theta} \sec^2 \theta &= \frac{x\dot{y}-\dot{x} y}{x^2} \\[4pt] \frac{\dot{\theta} (x^2+y^2)}{x^2} &= \frac{x\dot{y}-\dot{x} y}{x^2} \\[4pt] r^2\dot{\theta} &= x\dot{y}-\dot{x} y \\[4pt] r\dot{\theta} &= \dot{y} \cos \theta-\dot{x} \sin \theta \\[4pt] \mathbf{v} &= \begin{bmatrix} \dot{r} \cos \theta-r\dot{\theta} \sin \theta \\ \dot{r} \sin \theta+r\dot{\theta} \cos \theta \end{bmatrix} \\[4pt] &= \begin{bmatrix} (\dot x \cos \theta+\dot y \sin \theta) \cos \theta- (\dot y \cos \theta-\dot x \sin \theta) \sin \theta \\ (\dot x \cos \theta+\dot y \sin \theta) \sin \theta+ (\dot y \cos \theta-\dot x \sin \theta) \cos \theta \end{bmatrix} \\[4pt] &= \begin{bmatrix} \dot x \\ \dot y \end{bmatrix} \end{align*}


$S$: Inertial frame

$S'$: Rotating frame

$$\mathbf{v} = \mathbf{v}'+\boldsymbol \omega \times \mathbf{r}'$$

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  • $\begingroup$ the question added an edit. $\endgroup$ – Feynman Apr 4 '17 at 11:28
  • $\begingroup$ Answer updated. You should use different denotations for rotating frame and inertial frame. $\endgroup$ – Ng Chung Tak Apr 4 '17 at 17:11

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