Applied Mathematics: Spherical Polar Coordinates and Newton's Second Law I've been attempting this question but can't seem to find a solution.
Question:
A particle of mass $m$ moves under the influence of a force which, in spherical polar coordinates, only acts in the radial direction. The force is therefore of the form
$$\boldsymbol{F} =f\boldsymbol{e}_r $$
where $f$ is specified. Hence, using Newton's second law of motion, show that the quantity
$$r^2\dot{\phi}\sin^2\theta$$
is constant during the motion. (where $\dot{\phi}$ is the derivative of $\phi$).
Where $\boldsymbol{e}_r = \sin{\theta}\cos{\phi}\space \boldsymbol{i} +\sin{\theta}\sin{\phi}\space \boldsymbol{j} + \cos{\theta}\space \boldsymbol{k}$
 A: The acceleration in spherical coordinates is given by
$$\eqalign{
  & {\bf{a}} = \left( {\ddot r - r{{\dot \theta }^2} - r{{\dot \varphi }^2}{{\sin }^2}\theta } \right){{\bf{e}}_r}  \cr 
  & \,\,\,\, + \left( {r\ddot \theta  + 2\dot r\dot \varphi  - r{{\dot \varphi }^2}\sin \theta \cos \theta } \right){{\bf{e}}_\theta }  \cr 
  & \,\,\,\, + \left( {r\ddot \varphi \sin \theta  + 2\dot r\dot \varphi \sin \theta  + 2r\dot \theta \dot \varphi \cos \theta } \right){{\bf{e}}_\phi } \cr} $$
According to the newtons second law ${\bf{F}}=m{\bf{a}}$ and that ${\bf{F}}=f{\bf{e}}_r$, acceleration component in $\phi$ direction should be zero
$$ {r\ddot \phi \sin\theta + 2\dot r\dot \phi \sin \theta  + 2r\dot \theta \dot \phi \cos \theta } =0$$
multiply by $r\sin\theta$ to get
$$\eqalign{
  & \,\,\,\,\, {r^2}\left( {\ddot \phi } \right){\sin ^2}\theta  + \left( {2\dot rr} \right)\dot \phi {\sin ^2}\theta  + {r^2}\dot \phi \left( {2\dot \theta \sin \theta \cos \theta } \right)  \cr 
  &  = {d \over {dt}}\left( {{r^2}\dot \phi {{\sin }^2}\theta } \right) = 0 \cr} $$
and finally 
$${{r^2}\dot \phi {{\sin }^2}\theta }=c$$
