Classical mechanics problem in polar coordinates. * A smooth horizontal table has a vertical post fixed to it which has the form of a circular cylinder of radius $\displaystyle a$. * A light inextensible string is wound around the base of the post ( so that it does not slip ) and its free end of the string is attached to a particle that can slide on the table. * Initially the unwound part of the string is taut and of length $\displaystyle 4a/3$. * The particle is then projected horizontally at right angles to the string so that the string winds itself on to the post. * How long does it take for the particle to hit the post ?. You may make use of the formula $$ \int\sqrt{\, 1 + \phi^{2}\,}\,\mathrm{d}\phi = \frac{1}{2}\,\phi\,\sqrt{\, 1 + \phi^{2}} + \frac{1}{2}\sinh^{-1}\left(\phi\right)\tag A$$ Below is solution recommended by professor, however I would like to make use of the formula $(A)$ suggested above?.

![Figure]( The free part is instantaneously rotating about point C, therefore velocity is perpendicular to this part, and thus, Work done by tension is zero. Energy conservation equation leads to:

$\dfrac{mv^2}{2} = E$

Let $\theta$ be the angle between the direction of the free string at time $t$ and its initial direction, as shown in Figure. Since the length of the free string at time $t$ is $b-a\theta$, it follows that $v=(b-a\theta)\dot\theta$. On using the initial condition $v=u$ when $t=0$, the energy conservation equation becomes:

$(b-a\theta)\dot\theta=u$

Solving differential equation

$$\int_0^{\frac{b}{a}}(b-a\theta)d\theta = u \int_0^{\tau}dt$$

\begin{align}\tau &= \frac{1}{u} \int_0^{\frac{b}{a}}(b-a\theta)d\theta\\\&= \frac{1}{u}[b\theta - \frac{1}{2} a \theta^2]_0^{b/a} \\\&= \frac{b^2}{2au}\end{align}

What sort of solution to be done to make use of the formula (A)?