A uniform rope of length $l$ is held motionless on a frictionless hemisphere of radius r with one end of the rope at the top of the hemisphere. The hemisphere is made immobile by gluing it on a horizontal floor. The rope is now released. Immediately after the rope is released

I) Initial tangential acceleration of all elements will be same for the element of mass dm:
$(T+dT)+\left(dm\right)g\sin \theta -T=\left(dm\right)a_t\Rightarrow dT+\left(\frac{m}{l}rd\theta \right)g\sin \theta =\frac{m}{l}rd\theta a_t$
Integrating between 1 and 2

$\Rightarrow \underset{T_1}{\overset{T_2}{\int }}dT+\frac{m}{l}rg\underset{o}{\overset{l/r}{\int }}\sin \theta d\theta =\frac{m}{l}r{a}_t\underset{o}{\overset{l/r}{\int }}d\theta$
$\Rightarrow (T_2-T_1)+\frac{m}{l}rg(1-\cos \frac{l}{r})=\frac{m}{l}r{a}_t\frac{l}{r}$
But at free ends tension is zero
$\therefore T_1=T_2=0 \therefore a_t=\frac{rg}{l}(1-\cos \frac{l}{r})$     
II) Integrating between 1 and  $\theta$
$(T-O)+\frac{m}{l}rg(1-\cos \theta )=\frac{mr}{l}\left[\frac{rg}{l}(1-\cos \frac{l}{r})\right]\theta$
Differentiating w.r.t $\theta$ :
$\frac{dT}{d\theta }+O+\frac{mr}{l}\left[\frac{rg}{l}(1-\cos \frac{l}{r})\right]$
For  $T_{\max \text{'}}\frac{dT}{d\theta }=o$                               $\Rightarrow \sin \theta =\frac{r}{l}(1-\cos \frac{l}{r})$       
$\alpha =90-\theta \Rightarrow \alpha {\cos }^{-1}\left(\frac{r}{\mathcal{l}}(1-\frac{\cos \mathcal{l}}{r})\right)$
D) To prevent slipping 
$F_{friction}=F_{weight}$
$\mu \left(\frac{m}{l}\right)g\left(r\right)\sin \alpha =\left(\frac{m}{l}\right)\left(r\right)(1-\cos \alpha )g$
$\mu =\frac{1-\cos \alpha }{\sin \alpha } [\alpha =\frac{l}{r}]$
$=\tan \left(\frac{\alpha }{2}\right)$