A particle P has polar coordinates (r, theta). The movement of P is described as the vectorial sum of forces,

F1=-mk(sq)r
F2=-2mbv

(a) write the equations of movement
(b) suppose that the second derivative of theta is 0, determine r(t), theta(t) and thus deduce r(theta), and the speed of the particle P.

I'm not sure about the notation you're using - I can't tell which direction your force vectors are acting. However, when working in polar coordinates remember (memorize!) that:


\vec v = \dot r \hat {e_r} + r \dot {\theta} \hat {e_{\theta}} \\
\vec a = (\ddot r - r \dot \theta ^2) \hat {e_r} + (r \ddot \theta + 2 \dot r \dot \theta) \hat {e_{\theta}}


where \hat {e_r} \text and \hat {e_{\theta}} are the unit vectors in the radial and tangential directions, respectively. Using these equations, and the fact that the forces can be divided into radial and tangential components so that you have two equations in the form \vec F = m \vec a should allow you to get the answers.

Yup I've got the answer. Thanks!

@ebaines, Can You elaborate on how you arrived at those equations?

Wow - dragging up a post over two years old! But OK - see: Polar coordinate system - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Polar_coordinate_system)
About two thirds of the way down you'll find these two equations:


\frac {d \vec{r}} {dt} = \dot {r} \hat {r} + r \dot {\theta} \hat {\theta} \\

\frac {d^2 \vec{r}} {dt^2} = (\ddot r - r {\dot {\theta}}^2) \hat {r} + (r \ddot \theta + 2 \dot {r} \dot {\theta}) \hat {\theta}


The notation used is slightly different, in that here \hat {r} is the unit vector in the radial direction and \hat {\theta} is the unit vector in the tangential direction. If you're studying kinematics in polar coordinates the derivation of this should be in your text book.

@ebaines ,never mind.. I figured it out moments after posting my question.. I don't think this site has a "delete question" button anywhere.. thanks anyway.
for the curious /people too lazy to remember it: here's a simple way to do it:
let A point P(r,angle w) can be transformed to cartesian coordinates in the following way:
unit vector along r: (cos(w),sint(w)) : equation 1
and unit vector along t: (-sin(w),cos(w)) :equation 2
deriving equations 1 and 2 wrt time give relations with each other in terms of their derivatives.
velocity = d(vector r)/dt and acceleration is d(vector velocity)/dt.
where vector r=r*(equation 1)