# Work in rectangular and polar coordinates: result is not the same?

1. Oct 7, 2006

### SNES

I'm a little puzzled with the work of a force considering rectangular and polar coordinates.

In rectangular coordinates we have:

(1) [F]=(Fx + Fy [j]) is the force vector.
(2) [dr]=(dx + dy [j]) is small displacement.

Then the work is:
(3) W = [F] [dr] = (Fx dx + Fy dy)

In polar coordinates, we have:
(4) [F] = F [r1] + phi [theta1]
(5) [dr] = dr [r2] + r dtheta [theta2]

In this case the work is:
(6) W = [F] [dr] = F dr cos(phi - r dtheta)

Now I'd like to get the last result starting from equation (3), considering:

(7) Fx=F cos(phi)
(8) Fy=F sin(phi)

(9) x=r cos(theta) and dx=dr cos(theta) - r sin(theta) dtheta
(10) y=r sin(theta) and dy=dr sin(theta) + r cos(theta) dtheta

After inserting these quantities in equation (3), I get:

(11) W = F dr cos(phi-theta) + F r dtheta sin(phi-theta)

My question is: the result in equation (11) is the same of the one given by equation (6)?

Thank you!
SNES

2. Oct 7, 2006

### Staff: Mentor

Those don't look right.

$\vec{F}\,=\,F_r(r,\theta)\hat{r}\,+\,F_\theta(r,\theta)\,\vec{\theta}$

$d\vec{r}\,=\,dr\hat{r}\,+\,rd\theta\,\vec{\theta}$.

Last edited: Oct 7, 2006