Rope On Cylinder

Let's consider a very small element of the rope. We will draw the forces acting on it like in the following picture.

The weight of the element can be written as $G=gdm=g\lambda Rd\theta$ . The linear density of the rope $\lambda$ measured in $\frac{kg}{m}$ . Now we write the equations of equilibrium. $\left\{ \begin{matrix} N=g\lambda Rd\theta \sin \theta +T\sin \frac{d\theta }{2}+\left( T+dT \right)\sin \frac{d\theta }{2} \\ \left( T+dT \right)\cos \frac{d\theta }{2}=T\cos \frac{d\theta }{2}+gR\lambda d\theta \cos \theta +\mu N \\ \end{matrix} \right.$ Now we make the assumption that $d\theta$ is very small: $\cos \frac{d\theta }{2}\approx 1$ , $sin \frac{d\theta }{2}\approx \frac{d\theta }{2}$ . Also we will ignore the $d\theta dT$ term in the first equation. $\left\{ \begin{matrix} N=g\lambda Rd\theta \sin \theta +Td\theta \\ dT=gR\lambda d\theta \cos \theta +\mu N \\ \end{matrix} \right.$ Be careful that $\theta$ is not small and cannot be simplified. After substituting the normal force and making some calculations we get: $\frac{dT}{d\theta }-\mu T=gR\lambda \cos \theta +\mu gR\lambda \sin \theta$ This is a first-order nonhomogeneous linear ordinary differential equation with the tension force as a function of angle: $T(\theta )$ . At the initial condition the tension is equal to the weight of the $x$ free rope: $T(0)=g\lambda x$ Solving the the equation we get the tension in the rope as a function of angle and the lenght of the free part $x$ : $T=\frac{g\lambda }{{{\mu }^{2}}+1}\left( {{e}^{\mu \theta }}\left( {{\mu }^{2}}x+2\mu R+x \right)-\left( {{\mu }^{2}}-1 \right)R\sin \theta -2\mu R\cos \theta \right)$ Now we can write the tension at the angle $\pi$ which is equal to the weight of the free part of length $y$ . $T(\pi ,x)=\frac{\lambda g}{{{\mu }^{2}}+1}\left[ {{e}^{\pi \mu }}\left( x{{\mu }^{2}}+2R\mu +x \right)+2R\mu \right]$ We don't forget that the total lenght of the rope is $L$ . We can write 2 conditions: $\left\{ \begin{matrix} \frac{\lambda g}{{{\mu }^{2}}+1}\left[ {{e}^{\pi \mu }}\left( x{{\mu }^{2}}+2R\mu +x \right)+2R\mu \right]=g\lambda y \\ x+y+\pi R=L \\ \end{matrix} \right.$ This is a linear system of 2 equations with 2 unknowns. Solving it we finally get: $\frac{y}{x}=\frac{{{e}^{\mu \pi }}+{{\mu }^{2}}{{e}^{\mu \pi }}}{{{\mu }^{2}}+1}-\frac{2R\mu +4R\mu {{e}^{\mu \pi }}+2R\mu {{e}^{2\mu \pi }}}{\pi R-L{{\mu }^{2}}-L+2R\mu +\pi R{{\mu }^{2}}+2R\mu {{e}^{\mu \pi }}}$ After solving it numerically: $\frac{y}{x}=17.03175$