Why is the radial velocity considered zero? I had recently come across a question which is stated as below:

A disc placed on a large horizontal floor is connected from a vertical cylinder of radius $r$ fixed on the floor with the help of a light inextensible cord of length $l$ as shown in the figure. Coefficient of friction between the disc and the floor is $\mu$. The disc is given a velocity $v$ parallel to the floor and perpendicular to the cord. How long will the disc slide on the floor before it hits the cylinder?

I thought hard for a few days but I couldn't solve it as the mathematics was terrible. Finally, while trying many other things, I tried considering the radial component of velocity to be zero and it worked! I got the answer.
But I am not able to understand the logic behind considering the radial velocity zero. Would someone please help me to understand it?
Edit: The figure is given as below:

 A: Because the cord is inextensible, its tension does no work on the disc (in other words, the disc always moves perpendicular to the cord). Therefore we can model the situation as a linear deceleration under friction. It would be the same if the cylinder's radius were zero and the motion were circular.
A: Let's say the disk has got a non-zero radial velocity. This then has $2$ possibilities. First, the radial velocity is outward along the string and second, the radial velocity is inward along the string.
The First case cannot happen because of the restriction given in the question, the string is inextensible.
For the Second case, if the disk has a velocity inward along the length of the string, the string will slacken after the disk moves a distance $dl$ which will then lead to the tension force, the force exerted by the string and the only force that can provide a angular motion to the disk to instantaneously become zero. So, in this case, the disk will keep moving in the same direction with decreasing speed until the string tightens again to start providing a angular motion.
A: Some commentators also seem to have fallen into the ambiguity of Radial velocity. Radial normally means toward a centre. and a fixed centre. as the string wraps around the cylinder clearly there is no obvious centre as the  motion isn’t circular but some weird spiral and the straight part of string doesn’t continually point back toward a constant common centre. There’s no centre to have radii properly. If the string is unstretchable and doesn’t go slack then however we can say the ball motion is perpendicular to the string along its momentary straight length(which we may try to call radius even though it’s changing centre. Anyway the motion is sort of circular instantaneously about the contact point of string with cylinder which must lead to answer.
A: newest edits in bold.
This is the approximate path consistent with slowing down under friction:
$x(t)=\frac{v_0}{\mu} (1-e^{-\mu t})$
Because this is the solution to $\frac{dx}{dt}=-\mu \frac{dx^2}{dt^2}$.
$ \rightarrow \text{distance traveled along the arclength of the spiral}=\frac{L^2}{2R}$
Therefore,
$\frac{v_0}{\mu} (1-e^{-\mu t}) = \frac{L^2}{2R}$
$\text{Solve for t.}$

New Addition:
$$\begin{eqnarray}
x(\omega t) & = & R \sin(\omega t) + (L - R \omega t) \cos(\omega t)\\
y(\omega t) & = & -R \cos(\omega t) + (L - R \omega t) \sin(\omega t)
\end{eqnarray}$$
$$\begin{eqnarray}
\dot{x}(\omega t) & = & +\omega R \cos(\omega t) - L \omega\sin(\omega t) -R\omega cos(\omega t)+R\omega^2 t sin(\omega t)\\
\dot{y}(\omega t) & = & \omega R \sin(\omega t) + L \omega \cos(\omega t) -R\omega sin(\omega t)-R\omega^2 t cos(\omega t)\\
\end{eqnarray}$$
$$\begin{eqnarray}
\dot{x}(0) & = & 0 \\
\dot{y}(0) & = &  L \omega -R\omega^2 t(0) \\
\end{eqnarray}$$
where $t(0)=0$.
