TimJ said:
I am not sure if I understand correctly. Did you ment doing this:
<br />
m \frac{d \textbf{v}}{dt}=m v \frac{d \textbf{v}}{ds} =m v^2 r \left(\frac{d\phi}{ds} \left(\hat{\mathbf{\phi}} \cos \theta \frac{d\theta}{ds}+\hat{\mathbf{\theta}}\frac{d \theta'}{ds} \right)+\left (\hat{\mathbf{\phi}} \sin\theta+\hat{\mathbf{\theta}} \theta' \right) \frac{d^2\phi}{ds^2} \right)<br />
Sort of, but the spherical unit vectors themselves will also depend on position/arclength, so for example,
\frac{d}{ds}\left(r\theta'\dot{\phi}\mathbf{\hat{\theta}}\right)=\frac{d}{ds}\left(r\theta'v\frac{d\phi}{ds}\mathbf{\hat{\theta}}\right)=r\frac{d\theta'}{ds}v\frac{d\phi}{ds}\mathbf{\hat{\theta}}+r\theta'\frac{dv}{ds}\frac{d\phi}{ds}\mathbf{\hat{\theta}}+r\theta'v\frac{d^2\phi}{ds^2}\mathbf{\hat{\theta}}+r\theta'v\frac{d\phi}{ds}\frac{d\mathbf{\hat{\theta}}}{ds}
=r\left[v\theta''\left(\frac{d\phi}{ds}\right)^2+\theta'\left(\frac{dv}{ds}\right)\left(\frac{d\phi}{ds}\right)+v\theta'\left(\frac{d^2\phi}{ds^2}\right)\right]\mathbf{\hat{\theta}}+rv\theta'\left(\frac{d\phi}{ds}\right)\left[\frac{d}{ds}\left(\cos\theta\cos\phi\mathbf{\hat{x}}+\cos\theta\sin\phi\mathbf{\hat{y}}-\sin\theta\mathbf{\hat{z}}\right)\right]
=r\left[v\theta''\left(\frac{d\phi}{ds}\right)^2+\theta'\left(\frac{dv}{ds}\right)\left(\frac{d\phi}{ds}\right)+v\theta'\left(\frac{d^2\phi}{ds^2}\right)\right]\mathbf{\hat{\theta}}
+rv\theta'\left(\frac{d\phi}{ds}\right)\left[-\theta'\frac{d\phi}{ds}\sin\theta\cos\phi\mathbf{\hat{x}}-\frac{d\phi}{ds}\cos\theta\sin\phi\mathbf{\hat{x}}-\theta'\frac{d\phi}{ds}\sin\theta\sin\phi\mathbf{\hat{y}}+\frac{d\phi}{ds}\cos\theta\cos\phi\mathbf{\hat{y}}-\theta'\frac{d\phi}{ds}\cos\theta\mathbf{\hat{z}}\right]
=r\left[v\theta''\left(\frac{d\phi}{ds}\right)^2+\theta'\left(\frac{dv}{ds}\right)\left(\frac{d\phi}{ds}\right)+v\theta'\left(\frac{d^2\phi}{ds^2}\right)\right]\mathbf{\hat{\theta}}+rv\theta'\left(\frac{d\phi}{ds}\right)^2\left[-\theta'\mathbf{\hat{r}}+\cos\theta\mathbf{\hat{\phi}}\right]
The part that confuses me most are the unit vectors \mathbf{\hat{\theta}} and \mathbf{\hat{\phi}}.
The idea is that when you compare the r, \theta and \phi components of v\frac{d\textbf{v}}{ds} with the \theta and \phi components of \textbf{F}_g+\textbf{F}_f you will get 3 simultaneous differential equations where everything is expressed in spherical coordinates.
The 3 DEs you got by comparing Cartesian components had a mix of Cartesian coordinates (x,y,z,v_x etc.) and spherical coordinates (\theta, \phi etc) all of which will have some dependence on s, making it very unclear as to how to solve them!
At least when you compare spherical components like this, you get everything in terms of spherical coordinates.
However, by looking at the DEs you get doing this, you can see it is still very difficult to solve them!:sad:
Luckily

, this:
Because the tangental component looks in the direction of the movement and because so doing I get \frac{d \phi}{ds}
gives me an idea to simplify things considerably...
Recall that the acceleration along a curve can be written \textbf{a}=\frac{d\textbf{v}}{dt}=\frac{dv}{dt}\textbf{T}+v^2\kappa\textbf{N}_c where \kappa is the curvature and \textbf{N}_c is the unit normal to the curve (not the normal to the surface).
Looking at the acceleration in this form, it should be clear that any changes in the speed of the particle will be due entirely to the tangential component of the applied force!
That means,
m\frac{dv}{dt}=\textbf{F}_g\cdot\textbf{T}+\textbf{F}_f\cdot\textbf{T}=mgr\sin\theta\theta' \frac{d \phi}{ds}-\mu mg\cos\theta
Which is much simpler than solving 3 coupled DEs!