A particle is sliding in a cone maintaining contact with it all the time. Analyzing the situation in spherical coordinates I was able to find that in the total acceleration of the particle there will be a component along $\hat{\theta}$ ,where $\theta$ is the polar angle(here $\theta=\alpha$ ),the equation is
$\ddot{r}=\left(\ddot{r}-r \sin ^{2}\theta\dot{\phi}^{2}\right) \hat{r}$+$(2 \dot{r} \dot{\phi} \sin \theta+r sin\theta \ddot{\phi}) \hat{\phi}-r \dot{\phi}^{2} \sin\theta\cos \theta \hat{\theta}$
But looking it intuitively I feel that there should be no component along $\hat{\theta}$, because any component along $\hat{\theta}$ would pull the particle away from the contact with the cone.
Why is my intuition wrong?
Acceleration in the direction of the $\hat\theta$ vector does not mean that the $\theta$ coordinate component changes, because the coordinate system itself curves.
For instance, pure circular movement around the vertical axis (which is one form of movement along the cone) would mean acceleration horizontally in toward the axis, which is partly the direction of $-\hat\theta$, partly the direction of $-r$ (assuming the vertex of the cone is at the origin).