I have the following problem:
A particle of mass $m$ is at rest at the top of a semispherical surface of radius $R$. After an impulse, it starts to slide along a meridian with speed $v_0$. The surface has dynamic friction coefficient $\mu$. At what angle of colatitude does the particle lose contact with the surface?
I know how to solve the problem in the case $\mu=0$, but in this case things seem to be more complicated. These is my attempt at a solution.
Let $\theta$ be the colatitude, $N$ the surface´s reaction. By $\vec F=m\vec a$ we get:$$N=m(g \cos \theta-R\dot \theta ^2),$$ and $$R\ddot\theta=g(\sin \theta-\mu \cos \theta)+\mu R \dot \theta ^2.$$ In the case $\mu = 0$ I would multiply last equation by $\dot \theta$ to obtain the angular velocity in function of the angle (which is equivalent to use the kinetic energy theorem) and then set $N=0$ in the first equation.
It´s easily seen that this strategy doesn´t work for the general case. Is it possible to solve these differential equations analytically?