Why is $dr/dt= -V$ 
In the solution, it says we have $dr/dt= -V$ (polar coordinates)
How? I can't see how this can be possible, we know that $r(t)=V/\omega(t)$, and that's it.
 A: The question defines $V$ as the rate at which the string is pulled downward through the ring and consequently the rate at the which the radius changes, $-dr/dt$ (negative as the radius is decreasing with time - the fixed length of string is being pulled down).
This $V$ is different from $\underline{\mathbf{v}}$, the velocity vector of the mass, given by $\underline{\mathbf{v}}=\frac{d\underline{\mathbf{r}}}{dt}$ where $\underline{\mathbf{r}}$ is the position vector (independent of the coordinate system you choose). This is the quantity you rightly define in your second statement, $r(t)=\frac{|\underline{\mathbf{v}}|}{\omega(t)}$.
Overall, you need to be careful between what are scalar quantities ($V$, $r$ and $dr/dt$) and what are vector quantities ($\underline{\mathbf{v}}$ and $\underline{\mathbf{r}}$). Sometimes the vectors will only be bold which can make it difficult to distinguish but the context is key, it is rare to find an error in a textbook as much as we may sometimes wish it so. The devil is in the detail!
A: $r(t)=r(0)-Vt.$ $V$ is not the tangential speed (indeed it's the radial speed).
