Question

Ok now a tough one ! 2 people standing on a rotating disk in angular velocity omega. distance between them is 2d and the center of the disk is between them. one is sliding a box to the other person, the other person catches it after time T. Find the initial velocity vector of the box...

Answer 1

Is that all the information you have? Is d the radius of the disk or some arbitrary distance?

Answer 2

More importantly I suppose, are they both the same distance from the center of the disk ?

Answer 3

Ah well it doesn't really matter if you want to be general about it. Let person A be pushing the box and person B be receiving it. At time t = 0, let them be lined up as shown in the figure. Assuming counterclockwise angular velocity omega, the position of person B is
\[\vec{r}_B(t) = \left< (2d-r_0)\cos(\omega t), (2d-r_0)\sin(\omega t) \right> \]
if it is received at time T, then the position of B is
\[\vec{r}_B(T) = \left< (2d-r_0)\cos(\omega T), (2d-r_0)\sin(\omega T) \right> \]
its initial position was
\[\vec{r}_A(0) = \left< -r_0, 0\right>\]
so the change in position is
\[\Delta \vec{r} = \left< (2d-r_0)\cos(\omega T) + r_0 , (2d-r_0)\sin(\omega T)\right> \]
Assuming no friction, the velocity vector is a constant, so its simply
\[\vec{v} = \frac{\Delta \vec{r}}{T} = \frac{1}{T} \left< (2d-r_0)\cos(\omega T) + r_0 , (2d-r_0)\sin(\omega T)\right>\]
That's about as general as you can make it. If you have more specific conditions to impose you can simplify it from there.