When will the ball lose the hemisphere?

P+N=ma ( m is the masse of the little ball A, a is the acceleration )
Ox:
N-Px = -max;
N-Px= -m ( v^2 ÷ R );
Cos y= Px÷mg (y is the angle at wich A will get seperated from B)
Px=mg cosy;
N - mgcosy = -m( v^2 ÷ R)

From the law of conservation of the mechanical energy :
Emi = Emf;
0 = (1÷2)m( Vf^2) -mgh
Let's find h :
h= R-OH , cosy= OH ÷ R;
h=R-Rcosy = R(1-cosy);
(1÷2)m Vf^2 = mgR(1-cosy); m different from 0;
Vf^2 = 2gR(1-cosy);
N-mgcosy = -(m÷R)2gR(1-cosy);
N= mgcosy -2mg +2mgcosy;
3mgcosy = 2mg;
3cosy=2;
Cosy = 2÷3;
Y=48.18

Let at a height $h$ from the top the ball loses contact. It can be shown then that $l=r(1-\cos\theta)$ where $r$ is the radius of the hemisphere.

The weight $mg$ acting downwards has a component $mg\cos\theta$ radially responsible for the centripetal force,

$\displaystyle \begin{aligned} &mg\cos\theta=\dfrac{mv^2}{r}\\ &v=\sqrt{rg\cos\theta}\end{aligned}$

Since energy remains conserved as no external force acts on the system,

$\displaystyle \begin{aligned} E_{top}&=E_{h} \\ mgr&=mgh+\dfrac{1}{2}mv^2 \\ gr&=gh+\dfrac{1}{2}v^2 \\ gr&=gh+\dfrac{rg\cos\theta}{2} \\ gr&=gh+\dfrac{gh}{2} \\ h&=\dfrac{2}{3}r \end{aligned}$

where we know that $r\cos\theta=h$ .

So the required angle is $\displaystyle \theta=\cos^{-1}\left(\dfrac{h}{r}\right)=\cos^{-1} (\dfrac{2}{3})\approx 48.18^{\circ}$

I would like to propose a harder version of this : What if the ball would have been rolling,not simply sliding ; then what would be the angle and the height ? ( $10/17$ ? )