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

  1. Oct 7, 2006 #1
    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. jcsd
  3. Oct 7, 2006 #2

    Astronuc

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    Staff Emeritus
    Science Advisor

    Those don't look right.


    [itex]\vec{F}\,=\,F_r(r,\theta)\hat{r}\,+\,F_\theta(r,\theta)\,\vec{\theta}[/itex]

    [itex]d\vec{r}\,=\,dr\hat{r}\,+\,rd\theta\,\vec{\theta}[/itex].
     
    Last edited: Oct 7, 2006
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