Relative velocities of two particles in circular motion

[wifi]

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?

 

[tiny-tim]

hi wifi!

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?

yes

 

[wifi]

Thanks tiny-tim!