Tension in a Uniform Thread!

Consider a tiny part of the chain, from stationary condition we have

$\displaystyle T(\theta + \delta \theta)+\lambda R g sin(\theta) \delta \theta=T(\theta)$

$\displaystyle \frac{dT}{d\theta}=-\lambda R g sin(\theta)$

$\displaystyle T=\lambda R g cos(\theta)+C$

we know that T=0 when $\displaystyle \theta = \frac{L}{R}$ so $C=-\lambda R g cos(\theta)$

and, finally, tension in the thread is equal to that in the chain at the highest point

$\displaystyle F=T(\theta=0)=\lambda R g (1-cos(\alpha))=mg \frac{sin^{2}(\frac{\alpha}{2})}{\frac{\alpha}{2}}$

$\theta$ is measured from the top of the cylinder.

$L$ must be equal to $\frac{\pi}{2} \cdot R$ and not $\leqslant \frac{\pi}{2} \cdot R$ as, if $L \leqslant \frac{\pi}{2} \cdot R$ , then we won't be able to make a perfect substitution $\sin(\frac{\pi}{2})\,=\,\frac{L}{R}$ .

The solution is very simple. If you follow the Golden Rule,then the virtual work done by this force T - (dA=Tds) will be equal to the change in the potential energy dU of the small piece of the rope with the mass m ds/L. transfered from the point B to the point A . Hence dA = Tds= dU=mgds R/L (1 - cos L/R). And T= mgR/L (1- cos L/R).