Rotation and value of the derivative of an angle If I have the following equation of motion for an axis:
$$
\mathbf{e}_{\phi} \quad  \colon \quad m\left(r\ddot{\phi}\sin\theta +2\dot{r}\dot{\phi}\sin\theta\right)=0
$$
Can I then conclude from the above that phi is constant  and phi dot = 0 since there is no motion in phi?
Because I know that if initially there is NO acceleration in phi so my motion in phi is either at rest or a uniform one (constant speed) but since we are here at the very start there can't have been a force setting my motion from rest to a uniform one and thus there is NO motion in phi thus phi is constant and its first and second derivatives are null.
Or what is missing in my deduction.
to put in context here is a diagram and the forces acting on my mass:

\begin{align}
\mathbf{e}_{r} \quad & \colon \qquad m\left(\ddot{r}-r\dot{\phi}^{2}\sin^{2}\theta\right)=mg\cos\theta-k\left(r-l_{0}\right)\\
\mathbf{e}_{\theta} \quad & \colon \qquad m\left(-r\dot{\phi}^{2}\sin\theta\cos\theta\right)=-mg\sin\theta+N\\
\mathbf{e}_{\phi} \quad & \colon \qquad m\left(r\ddot{\phi}\sin\theta +2\dot{r}\dot{\phi}\sin\theta\right)=0
\end{align}
 A: Dividing by $m$ and multiplying by $r \sin \theta$, we have
\begin{equation}
(r \sin \theta) (r \ddot{\phi}  \sin \theta+ 2 \dot{r} \dot{\phi} \sin \theta) = 0 \quad \Leftrightarrow \quad \frac{d}{dt} \left( r^2 \dot{\phi} \sin^2 \theta \right) = 0.
\end{equation}
Note that $r^2 \dot{\phi} \sin^2 \theta$ can be recognized as the $z$-component of the angular momentum, which means that your $e_\phi$ equation is simply a restatement of angular momentum conservation.  This makes sense, since you have rotational symmetry about the $z$-axis, and so by Noether's theorem there will be a conserved quantity in the system associated with this symmetry.
With the equation in this form, it is also pretty obvious that $\dot{\phi}$ will be a non-zero constant if and only if $r$ is also a constant.  If angular momentum is to be conserved, then the mass must be moving faster when it is closer to the vertex of the cone, and slower when it is farther away.
As to your logical deduction, I'm having trouble understanding it.  You seem to be arguing that the object can't be in motion initially (in any direction—nothing in your argument is special to the $\phi$ direction so far as I can tell.)  If that's true, then yes, $\dot{\phi} = 0$ is a valid solution, and the angular momentum of the mass will be constant (and zero) throughout the mass's motion.  But it's also possible to set the mass in motion with a non-zero initial velocity, and in this more general case, $\dot{\phi}$ will not be zero (and, as mentioned above, will not be constant unless $r$ is a constant as well.) 
