Derivative with respect to the radius of a moving particle on a non-frictional surface Basically I was just wondering about the following framing of question:

How can one determine the derivative of the velocity of the particle with respect to the radius when decreasing the force F on the string - (the particle is attached to a string that goes through a hole, as in figure). The surface on which the particle moves, has no friction.

FIGURE (not an inclined plane):

My attempt: So far, I've realized that when decreasing the force F, the radius R will be bigger. I also know that the angular momentum will be preserved. Although, now I'm stuck - so I'm very thankful for every responses/hints to help me solve this problem.
PS. Sorry for my bad english. 
 A: How much of a quantitative answer do you expect? Because it kind of an abstract question. 
First of all, you say "I've realized that when decreasing the force F, the radius R will be bigger." This is not true in a case where the force pulling on the string is bigger than the centrifugal force. The only thing you can be sure of is that $\frac{d^2R}{dt^2} > 0$, but for $\frac{dR}{dt}$, it depends on the initial conditions.
Secondly, you don't have a function for $F(t)$ so it's tough to come up with something numerical. What you know is that there are two forces in the problem, $\vec{F} = -F\hat{r}$ from the string and $\vec{F}_{cen} = \frac{m u_{\theta}^2}{r} \vec{r}$, both radial. Writing the sum of forces, you have
$$
\sum \vec{F} = \left( \frac{mu_{\theta}^2}{r} - F \right) \hat{r} = m\frac{d\vec{u}}{dt} \equiv m \frac{du_r}{dt} \hat{r}
$$
because there are no angular force. So $\frac{du_{\theta}}{dt} = 0$ which as you stated corresponds to conservation of angular momentum. Then you have
$\frac{du_r}{dt} = \left( \frac{mu^2_{\theta}}{r} - F(t) \right)$ which you could try to integrate but $r$ depends on time as well and it because complicated at this point.
I guess that still gives you a clue what to do... I Hope it helps you!
