This is a generalization of the question Particle sliding on a sphere when we also have friction given by $F_f = \mu N$.
See the following figure:
Before doing anything, we can imagine what friction will do. It is clear it will reduce the velocity and thus, delaying the "jumping" point from the sphere to a large angle $\theta_J$ (with respect to the vertical). I have tried two approaches:
1) Numerically via Newton's equations
Newton's equations:
$$r: \quad mg \cos(\theta) -N = m \dot{\theta}^2R $$ $$t: \quad mg \sin(\theta) - \mu N = m \ddot{\theta}R, $$ where $R$ is the radius of the sphere.
Substituting one into the other and rearranging I get:
$$\ddot{\theta} = \mu \dot{\theta}^2 + \frac{g}{R} \left( \sin{\theta} - \mu \cos {\theta} \right),$$ I believe this ODE equation does not have an analytical solution (at least that is what WolframAlpha says) so I have integrate it numerically considering $\theta(0) = 0$ and $\dot{\theta}(0) = 0$ as the boundary conditions, as well as requiring the friction force to be smaller than the weight projection in the tangential direction (otherwise the particle won't move).
Then I have computed numerically the moment where the normal $N$ is zero for the first time, that is, for the "jumping" angle $\theta_J$ and velocity $\dot{\theta}_J$ :
$$N(\theta_J, \dot{\theta}_J) = mg \cos \theta_J - mR\dot{\theta_J}^ 2 = 0.$$
From this condition I have been able to extract the "jumping" angle and velocity in function of $\mu$, $\theta_J (\mu)$ and $\dot{\theta_J}(\mu)$. Suprisingly, after all this "complex" computations I got what seems linear relationships for $\theta_J (\mu)$ and $\dot{\theta_J}(\mu)$ (for $m=1$, $R=1$, $g = 9.8$, $\theta_0 = 0.57º$ and
$\dot{\theta}_0 = 1.5 \text{ rad/s}$ ):
For the first plot, the linear fit is given by:
$$ \theta_J = a\mu + b $$
for $a=21.07 \pm 0.3$ and $b=42.0$. The interceipt $b$ is the value expected for $\theta_J (\mu = 0)$, which is given by (from the problem without friction):
$${\theta_J}(\mu = 0) = \cos^{-1} \left(\frac{2}{3} + \frac{{\dot{\theta}_0}^2R}{3g}\right) = 42.0.$$
This linear behaviour stands out as the ODE is highly nonlinear! So I suspect there might be a simpler way to compute the "jumping" points which avoids computing the full trajectory and makes use of the "jumping" condition given by $N(\theta_J, \dot{\theta}_J) = 0$ to get (perhaps) an analytical linear solution for $\theta_J (\mu)$ and $\dot{\theta_J}(\mu)$?
2) Via Energy conservation Following the non-friction case, the easier way to compute it is considering energy conservation to find a relationship between the "jumping" velocity and angle $\dot{\theta_J} (\theta_J)$. I have tried it but I get stuck. I started considering the loss of mechanical energy is the work of the friction force: $$\Delta E_m = W_{F_f}$$ Now, the left hand side is: $$\Delta E_m = mgh_J - \frac{1}{2} m v_J^2 = mgR(1- \cos \theta_J) - \frac{1}{2} m R^2 \dot{\theta_J}^2, $$ where $v_J$ is the linear jumping velocity and $h_J$ the height from the top of the sphere where the ball "jumps".
And the right hand side: $$ W_{F_f} = \int_0^{\theta_J} F_f R d\theta = R m \mu \left( \int_0^{\theta_J} g \cos \theta d\theta - \int_0^{\theta_J} R \dot{\theta}^2 d\theta \right) = R m \mu \left( g \sin \theta_J - R \int_0^{\theta_J} \dot{\theta}^2 d\theta \right), $$ where at the second equality I have substituted the expression for the normal $N(\theta, \dot{\theta})$. The question is how to approach this last integral... I substituted $\dot{\theta}^2$ from the 2nd order ODE from the previous approach and managed to integrate both terms and obtained the same expression as in the left hand side, basically finding that $$ 1 = 1,$$ which is not very useful.
QUESTION: Any ideas or approaches to solve this problem with an apparent simple solution? Is it true that $\theta_J (\mu)$ and $\dot{\theta_J}(\mu)$ follow linear relationships or is it just an artifact of the range studied and farther out the functions have different behaviours?
Comment: I have to say too I am using as initial conditions $\theta_0 = 0.57º$ and $\dot{\theta}_0 = 1.5 \text{ rad/s}$ so the particle does not get stuck before moving. That is because it will only move if, initially (which is the worse case), $W_t > F_f$, for $W_t$ the tangential weight. This implies there is a condition for a maximum $\mu$ given $\theta_0$ and $\dot{\theta}_0$: $$ \mu \leq \mu_\text{max} \equiv \frac{g \sin \theta_0}{g \cos \theta_0 - \dot{\theta}_0^ 2 R}.$$ For the initial conditions taken, $\mu_\text{max} = 0.0129$.
