General motion of a cone on an inclined surface 
Suppose that a solid cone is placed horizontally on an inclined surface and is initially at rest. How will the cone move when it starts motion due to its weight?

I know that its motion depends on the incline angle and also on the friction coefficient of the surface (as I observed by doing some experiments), but I can't establish the relation between them. Can anyone help me?
 A: 
"How will the cone move when it starts motion due to its weight"
to answer this equation you have to obtain the equations of motion.
you have 3 generalized coordinates

*

*rotate about then z -axis angle $\psi$

*incline plane x position $s_x$

*incline plane y position $s_y$
thus you get :
Kinetic energy
$$T=\frac 1 2\,{{\it \dot{s}_x}}^{2}m+\frac 1 2\,{{\it \dot{s}_y}}^{2}m+\frac 1 2\,{\dot{\psi}}^{2}{{\it J_z}
}^{2}-\tau_z\,\psi
$$
Where $\tau_z$ is the torque due to the friction forces and $J_z$ is the inertia of the cone z component.
Potential  energy
$$U=m\,g \left( \sin \left( \alpha \right) {\it s_y}+{\it s_z} \right) $$
Where $s_z$ is the z component of the  COM.
Thus: The EOM's:
$$m\,\ddot{s}_x+F_{Rx}=0$$
$$m\,\ddot{s}_y+F_{Ry}+m\,g\,\sin(\alpha)=0$$
$$J_z^2\,\ddot{\psi}+\tau_z=0$$
with $\tau_z=F_{Ry}\,y_s\quad $ and $F_{Rx}=\mu\,N\,,F_{Ry}=\mu\,N\quad , N=m\,g\,\cos(\alpha)$
According to the EOM's the cone will move "diagonal" on the incline plane and rotate about the cone z- axis
A: For a solid cone the COM is $\frac{h}{4}$ above its base
Along the incline we can write the following equations.
Forces:
$$Mg\sin\theta+f=Ma$$
Torque:
$$\frac{fh}{4} = I \alpha$$
In this case $I = \frac{3}{5}m(\frac{r^2}{4}+h^2)$
Rolling:
$$\alpha = \frac{a}{h/4}$$ 
Since we have 3 equations and 3 unknowns $(f, a, \alpha)$  the system can be solved.
