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Finding work done in spherical coordinates

  1. Dec 13, 2011 #1
    1. Find the work done by the force F=r3*cos2[itex]\varphi[/itex]*sin[itex]\varphi[/itex]*[itex]\hat{r}[/itex] + r3*cos[itex]\varphi[/itex]*cos(2[itex]\varphi[/itex]) [itex]\hat{\varphi}[/itex]
    from the point (0,0,0) to (2,0,0)

    2. Relevant equations
    Work=[itex]\int[/itex] F*dr
    where dr= dr[itex]\hat{r}[/itex] + rd[itex]\varphi[/itex][itex]\hat{\varphi}[/itex]


    3. The attempt at a solution

    When muliplying the line element, dr, by the force, F, I come up with
    [itex]\int[/itex] r3*cos2[itex]\varphi[/itex]*sin[itex]\varphi[/itex] dr +[itex]\int[/itex] r4*cos[itex]\varphi[/itex]*cos(2[itex]\varphi[/itex]) d[itex]\varphi[/itex]

    I believe the r goes from 0 to 2, and there is no change in [itex]\varphi[/itex]

    I end up with 4*cos[itex]^{2}[/itex][itex]\varphi[/itex]*sin[itex]\varphi[/itex]
    but then when I plug in 0 for [itex]\varphi[/itex], the answer ends up being zero, which I have a hard time believing since it moves from 0 to 2.
    Any help is greatly appreciated
    Thank You in advance.
     
    Last edited: Dec 13, 2011
  2. jcsd
  3. Dec 13, 2011 #2

    Simon Bridge

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    What is the force along [itex]\varphi = 0[/itex]?
    (This should simplify your line integral.)
     
  4. Dec 13, 2011 #3
    Check the force along the path you are given (because it's really a line integral through a vector field), and it should become fairly simple to see why that is. Notice that your psi component didn't really change in the integral that mattered.
     
  5. Dec 14, 2011 #4
    Thanks for the replies
    I believe you are saying that the force along [itex]\varphi[/itex]=0 is just zero along the r component, which is the only component that matters, since there is no motion in the other two coordinates. Since my force is actually zero at [itex]\varphi[/itex]=0, it doesn't matter that I went from (0,0,0) to (2,0,0), since no force means no work.

    Am I interpreting your comments correctly?
    Thanks for the feedback. Always appreciated.
     
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