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I was studying circular motion and came across a formula for the relative angular velocity between two bodies. I didn't find any kind of demonstration. Is it simply a determined concept, just like the vector product or the angular velocity vector?enter image description here

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    $\begingroup$ The instantaneous relative angular velocity of two objects is just the angular velocity of one of them in the instantaneous inertial reference frame of the other. Think about a Galilean transformation into the frame of $A$ or $B$, and compute the angular velocity there. $\endgroup$
    – G. Smith
    Commented Nov 30, 2019 at 5:01

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take this equation:

$$\vec{v}=\vec{\omega}\times \vec{r}$$

or

$$\vec{v}\times \vec{r}=\left(\vec{\omega}\times \vec{r}\right)\times \vec{r}\tag 1$$ enter image description here

if you write all vectors in components of the S' coordinate system you get:

2D case

$$\vec{r}=r_x\,\hat{\vec{e}}_{x'}$$ $$\vec{v}=v_x\,\hat{\vec{e}}_{x'}+v_y\,\hat{\vec{e}}_{y'}$$ $$\vec{\omega}=\omega_z\,\hat{\vec{e}}_{z'}$$

in (1)

$$(v_x\,\hat{\vec{e}}_{x'}+v_y\,\hat{\vec{e}}_{y'})\times r_x\,\hat{\vec{e}}_{x'}=\left(\omega_z\,\hat{\vec{e}}_{z'}\times r_x\,\hat{\vec{e}}_{x'}\right)\times r_x\,\hat{\vec{e}}_{x'}\tag 2$$

thus:

$$-v_y\,r_x\,\hat{\vec{e}}_{z'}=-\omega_z\,r^2_x\,\hat{\vec{e}}_{z'}$$

or:

$$\omega_z=\frac{v_y}{r_x}\quad, \omega_{AB}= \frac{(V_{AB})_y}{r_{AB}}$$

this is your equation

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  • $\begingroup$ why w pointing in the direction of z? $\endgroup$
    – Helen
    Commented Nov 30, 2019 at 16:47
  • $\begingroup$ In 2D Space you have three degrees of freedom two translations x and y and one rotation about the z axis $\endgroup$
    – Eli
    Commented Nov 30, 2019 at 20:30

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