TJ

Hi. How is acceleration [math]a[/math] written in spherical coordinates, so that [math]a[/math] is not dependant from time but from the line element [math]ds=\sqrt{dr^2+r^2d\theta^2+r^2\sin^2\theta^2d\phi^2}[/math] So I would like to have [math]\frac{d}{ds}[/math] instead of [math]\frac{d}{dt}[/math] in the next equation:

What is the speed at any point of a path [math] \theta = \theta(\phi)[/math] on a rough sphere. The forces that affect the moving of a point particle are the force of gravity and the force of friction. I tried this way but I am not sure if the normal vector is right: [math] \mathbf{T}=\frac{dx}{ds}\mathbf{i}+\frac{dy}{ds}\mathbf{j}+\frac{dz}{ds}\mathbf{k} [/math] [math] \mathbf{N}=(-\frac{dz}{ds}-\frac{dy}{ds})\mathbf{i}+(-\frac{dz}{ds}+\frac{dx}{ds})\mathbf{j}+(\frac{dx}{ ds}+\frac{dy}{ds})\mathbf{k} [/math] [math] F_g = -mg \mathbf{k} [/math] [math] F_n=|\textbf{F}_g\cdot\textbf{N}|=|-mg (\frac{dx}{ds} +\frac{dy}{ds}) \textbf{k}|=mg (\frac{dx}{ds} +\frac{dy}{ds}) [/math] [math] \textbf{F}_{friction}=-\mu F_n\textbf{T}=-\mu mg (\frac{dx}{ds} +\frac{dy}{ds}) \textbf{T} [/math] [math] ma=F_g -F_{riction} [/math] [math] \frac{1}{2}\frac{d(v^2)}{ds}=g \frac{dz}{ds}-\mu g (\frac{dx}{ds} +\frac{dy}{ds}) [/math] After integrating and converting to spherical coordinates i get: [math] v=\sqrt{2g((\cos{\theta_0}-\cos \theta)-\mu((\cos \phi_0 \sin \theta_0 - \cos \phi \ \sin \theta)+(\sin \phi_0 \sin \theta_0 - \sin \phi \sin \theta)) )} [/math] What do you think about the formulas about. Have I done it wrong? I think it would be better to go in spherical coordinates but I didn't know how. Merged post follows: Consecutive posts mergedI found how to solve it.