The disc moves on a horizontal floor. The vertical forces (gravity $mg$ and normal reaction $N$) balance each other, so $N = mg$. The only horizontal force acting along the direction of motion is the kinetic friction $f_k$. The tension in the string acts perpendicular to the velocity and provides the necessary centripetal force, doing no work.

The magnitude of the friction force is $f_k = \mu N = \mu mg$.
According to Newton’s second law, the tangential acceleration $a_t$ is: $$ m a_t = -f_k \implies m \frac{dv}{dt} = -\mu mg \implies a_t = -\mu g $$ Integrating this, the velocity at time $t$ is: $$ v(t) = u – \mu g t $$

As the string wraps around the cylinder of radius $r$, the free length of the string decreases. The particle traces the involute of a circle. Let $\phi$ be the angle the string has wrapped. The length of the free string at any instant is $\rho = l – r\phi$.

The small arc length $ds$ traveled by the disc for a small angle of wrap $d\phi$ is given by $ds = \rho d\phi$. The total distance $S$ traveled until the string is completely wound (i.e., when free length becomes 0, so $\phi = l/r$) is: $$ S = \int_{0}^{l/r} (l – r\phi) \, d\phi $$ $$ S = \left[ l\phi – \frac{1}{2}r\phi^2 \right]_0^{l/r} = l\left(\frac{l}{r}\right) – \frac{1}{2}r\left(\frac{l}{r}\right)^2 = \frac{l^2}{r} – \frac{l^2}{2r} = \frac{l^2}{2r} $$

We use the kinematic equation relating displacement, initial velocity, and acceleration: $$ s = ut + \frac{1}{2}at^2 \implies \frac{l^2}{2r} = ut – \frac{1}{2}\mu g t^2 $$ Rearranging this gives a quadratic equation for time $t$: $$ \mu g t^2 – 2ut + \frac{l^2}{r} = 0 $$

We must also check if the disc stops before hitting the cylinder. The stopping distance $S_{stop}$ is found when $v=0$: $$ 0 = u^2 – 2(\mu g) S_{stop} \implies S_{stop} = \frac{u^2}{2\mu g} $$

Case 1: Disc stops before hitting the cylinder

This occurs if $S_{stop} \le S$. $$ \frac{u^2}{2\mu g} \le \frac{l^2}{2r} \implies u^2 \le \frac{\mu g l^2}{r} \implies u \le l\sqrt{\frac{\mu g}{r}} $$ In this case, the time taken is simply the time to stop: $$ 0 = u – \mu g t \implies t = \frac{u}{\mu g} $$

Case 2: Disc hits the cylinder

This occurs if $u > l\sqrt{\frac{\mu g}{r}}$. We solve the quadratic equation derived in Step 3 for $t$: $$ t = \frac{2u \pm \sqrt{4u^2 – 4(\mu g)(l^2/r)}}{2\mu g} = \frac{u \pm \sqrt{u^2 – \frac{\mu g l^2}{r}}}{\mu g} $$ Since the velocity must remain positive ($t < u/\mu g$), we select the smaller root (the minus sign): $$ t = \frac{u}{\mu g} \left( 1 - \sqrt{1 - \frac{\mu g l^2}{u^2 r}} \right) $$