String and Pillar

There is no friction and gravity in the system. Throughout the process of the 0-mass string wrapping around the circle, energy is conserved and the only energy involved is kinetic of energy of the tiny ball. Let the mass of the ball be $m$ , its initial velocity be $v_0$ and instantaneous velocity at time $t$ be $v$ . Then, we have $\dfrac 12 m v^2 = \dfrac 12 m v_0^2$ $\implies v = v_0$ $\implies L \omega = L_0\omega_0$ , where $L$ and $\omega$ and the instantaneous length and angular velocity of the loose string respectively.

Let the angle wrapped by the string around the circle at time $t$ be $\theta$ . Then, we have:

$\begin{aligned} L \omega & = L_0\omega_0 \\ (L_0 - \theta R) \omega & = L_0 \omega_0 \\ \left(L_0 - \frac {\theta L_0}2 \right) \omega & = L_0 \omega_0 \\ \left(1 - \frac \theta 2 \right) \omega & = \omega_0 \\ \left(1 - \frac \theta 2 \right) \frac {d \theta}{dt} & = \omega_0 \\ \int \left(1 - \frac \theta 2 \right) d \theta & = \int \omega_0 \ dt \\ \theta - \frac {\theta^2} 4 + \color{#3D99F6} C & = \omega_0 t & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ 0 - 0 + C & = 0 & \small \color{#3D99F6} \text{When }t=0, \ \theta = 0 \\ \implies \theta - \frac {\theta^2} 4 & = \omega_0 t & \small \color{#3D99F6} \text{When the string is all wrapped up, }L_0 = \frac {\theta L_0}2 \implies \theta = 2 \\ 2 - \frac {2^2}4 & = 0.25 t \\ \implies t & = \boxed{4} \text{ s} \end{aligned}$

@Trevor Arashiro Wow!!! This is where I feel bad abt making an easy question hard.......I actually found out the parametric equation of the Involute of the circle.....then, integrated it to find the arc length.......and then used time=distance/speed.........!!!! I actually converted this into calculus!!! XD!!! Still....a great Problem!!!