Confusing conceptual question An observer $A$ standing on the circumference of a disc rotating with an uniform angular velocity $\omega = 1$ units , and radius $r=1$ units observes a person $B$ at rest w.r.t ground.Given the $\angle \theta = 30^\circ $ as shown in the figure
Find out the,

*

*relative velocity of $A$ w.r.t $B$

*relative velocity of $B$ w.r.t $A$
My approach , for the first part was that $|V_{A/B}|=rw=1$ units, for the second part i thought since $|V_{B/A}|=|V_{A/B}|=1$, but i found out that the answer to second part is wrong and this formula doesnt work for rotating frames,i also tried this solving with proper maths but i always end up to this conclusion only, maybe i am not hitting the right concept can anyone please help with correct maths, concept for the second part , also how can we write a general expression of $V_{B/A}$ varying with time, the expression i was deriving is as follows $\vec{V_{B/A}}=-(\cos{t}\hat{i}+\sin{t}\hat{j}) $, where centre of circle is the origin. i am really sorry for not typing my work but if users want my work i can share its photo :) .
Edit: All units are in SI system
 A: Let's fix a Cartesian co-ordinate system that is at rest w.r.t the ground and has origin at the centre of the circle. Let's call the co-ordinates of $A$ and $B$ in this co-ordinate system $\vec r_A(t)$ and $\vec r_B(t)$. At time $t=0$ we have
$\vec r_A(0) = (0, -1) \\ \vec r_B(0) = (\sqrt 3, -1)$
At a general time $t$, $\vec r_B(t)$ does not change so $\vec r_B(t) = (\sqrt 3, -1)$, but $A$ has moved around the circle by an angle $\omega t$, so $\vec r_A(t) = (\sin (\omega t), - \cos (\omega t))$.
At time $t$ $A$'s position with respect to $B$ is $\vec r_A(t) - \vec r_B(t)$, and $A$'s velocity with respect to $B$ is
$\displaystyle \frac d {dt} \left( \vec r_A(t) - \vec r_B(t) \right) = \vec v_A(t) - \vec v_B(t) = v_A(t)$
since $v_B(t) = 0$. Similarly, $B$'s position with respect to $A$ is $\vec r_B(t) - \vec r_A(t)$, and $B$'s velocity with respect to $A$ is
$\displaystyle \frac d {dt} \left( \vec r_B(t) - \vec r_A(t) \right) = \vec v_B(t) - \vec v_A(t) = - \vec v_A(t)$
A: I think the problem is that rotating frames don't support the principle of Galilean Invariance. To simplify the issue, get rid of the rotating disk and just consider observers A and B standing motionless, one metre apart. If A chooses an inertial frame of reference ('motionless and irrotational compared to the distant stars), then she will conclude that B is 1 m distant and motionless. But now let B perversely choose a rotating frame with origin at his location and $\omega = 1 $ rad/s. B would claim that A is now rotating around him with velocity $\pi$ m/s. Another observer C could pick another rotating frame with a distant origin and claim that both A and B were moving faster than the speed of light.
That is why inertial rest frames are generally used in classical mechanics. Rotating frames are useful for solving specific problems, but not for kinematics in general. That A is standing on a spinning disk doesn't require him/her to choose a rotating frame of reference (or if they do - to chose it's origin at the axis of the disk).
