= Solution
Let $\theta$ be the displacement from the upward vertical and $v$ the <speed> of the ball's <centre of mass>. Its centre follows a circle of radius $R+r$. The <rolling without slipping> constraint is $v=r\omega$, with $\omega$ the ball's spin <angular velocity>. The static contact <friction> does no work on the ball because the instantaneous contact point is stationary. Thus <conservation of energy> gives
$$
Mg(R+r)(1-\cos\theta)
=\frac12Mv^2+\frac12\left(\frac25Mr^2\right)\frac{v^2}{r^2}
=\frac7{10}Mv^2.
$$
Project <Newton's second law> toward the hemisphere's centre. The inward component of gravity is $Mg\cos\theta$ and the <normal force> $N$ points outward, so
$$
Mg\cos\theta-N=\frac{Mv^2}{R+r}.
$$
At contact loss $N=0$. Eliminating $v$ gives the ideal <sphere rolling off a hemisphere> result
$$
\frac{10}{7}(1-\cos\theta_d)=\cos\theta_d,
\qquad
\boxed{\cos\theta_d=\frac{10}{17},\quad\theta_d\simeq53.97^\circ.}
$$
The radius cancels, so replacing $R+r$ by $R$ under the stated small-ball approximation leaves the same answer.
\b[Finite-friction qualification.] This is an ideal rolling-constraint result. It does not describe uninterrupted pure rolling to detachment with any finite <coefficient of static friction>. The tangential <force> and <torque> equations give
$$
M\dot v=Mg\sin\theta-f,\qquad
fr=\frac25Mr^2\frac{\dot v}{r},
$$
so $f=(2/7)Mg\sin\theta$ acts uphill, whereas
$$
N=\frac{Mg}{7}(17\cos\theta-10).
$$
Consequently the required ratio is
$$
\boxed{\frac fN=\frac{2\sin\theta}{17\cos\theta-10}\longrightarrow\infty
\quad\text{as }\theta\to\theta_d^-.}
$$
For a concrete example, a <coefficient of static friction> of one allows rolling only up to about $47.54^\circ$, where $17\cos\theta-10=2\sin\theta$, before sliding begins. A subsequent physical contact-loss angle needs a sliding-friction model. The boxed $53.97^\circ$ answers the source's expressly imposed no-slip idealization.
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