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Problem:
Consider two particles a and b moving in opposite directions around a circle with angular speed ## \omega ##. At ## t = 0 ## they are both at the point ## \vec{r}=\ell \hat{j} ##. Find the velocity of a relative to b if the radius of the circular path traced out by the particles is ## \ell ##.
Solution:
WLOG, choose b to move counter-clockwise ## \Rightarrow \vec{r_b} = \ell (sin\omega t \ \hat{i} + cos\omega t \ \hat{j}) ##.
Therefore ## \vec{r_a} = \ell (sin(-\omega t) \ \hat{i} + cos(-\omega t) \ \hat{j}) = \ell (-sin\omega t \ \hat{i} + cos\omega t \ \hat{j}) ##.
Thus ## \vec{r_b} - \vec{r_a} = 2 \ell sin\omega t \ \hat{i} ## ; Hence ## \frac{d(\vec{r_b} - \vec{r_a})}{dt}= 2 \omega \ell cos\omega t \ \hat{i} ##.
Correct, yes?
Consider two particles a and b moving in opposite directions around a circle with angular speed ## \omega ##. At ## t = 0 ## they are both at the point ## \vec{r}=\ell \hat{j} ##. Find the velocity of a relative to b if the radius of the circular path traced out by the particles is ## \ell ##.
Solution:
WLOG, choose b to move counter-clockwise ## \Rightarrow \vec{r_b} = \ell (sin\omega t \ \hat{i} + cos\omega t \ \hat{j}) ##.
Therefore ## \vec{r_a} = \ell (sin(-\omega t) \ \hat{i} + cos(-\omega t) \ \hat{j}) = \ell (-sin\omega t \ \hat{i} + cos\omega t \ \hat{j}) ##.
Thus ## \vec{r_b} - \vec{r_a} = 2 \ell sin\omega t \ \hat{i} ## ; Hence ## \frac{d(\vec{r_b} - \vec{r_a})}{dt}= 2 \omega \ell cos\omega t \ \hat{i} ##.
Correct, yes?
