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Work done along a semicircle

  1. Dec 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the work done along the semicircle (x-1)2+y2=1 from the point (0,0,0) to (2,0,0)
    When F=r3cos2φsinφ[itex]\hat{r}[/itex] + r3*cosφ*cos(2φ) [itex]\hat{φ}[/itex]


    2. Relevant equations
    Work
    W=∫F*dr
    where dr=dr= dr[itex]\hat{r}[/itex] + rsinφdφ[itex]\hat{φ}[/itex]



    3. The attempt at a solution

    I convert the equation of the line (x-1)2+y2=1 to cylindrical coordinates.
    Where x=rsinφ
    and y=rcosφ
    so (x-1)2+y2=1
    eventually becomes r=2cosφ

    now i plug in this r value into the force vector given, which gives me F=8cos5φsinφ[itex]\hat{r}[/itex] + 8*cos4φ*cos(2φ) [itex]\hat{φ}[/itex]

    W=∫F*dr

    So now i dot product this with the dr, which is now dr= dr[itex]\hat{r}[/itex] + 2cosφ
    sinφdφ[itex]\hat{φ}[/itex] due to the r value I found.

    The dot product comes out to be

    ∫8cos5φsinφdr + 16*cos5φ*cos(2φ)sinφdφ

    and because r=2cosφ, dr=-2sinφ dφ

    so this is once again plugged in to the integral of the dot product, creating

    ∫-16cos5φsin2φdφ + 16*cos5φ*cos(2φ)sinφdφ

    where I integrate from pi/2 to 0, since it ends at x=2 and y=0, but begins at x=0 and y=0

    This gives me -4/35, which does not seem to be a correct answer
    Any mistakes?
    Any help is appreciated
    Thank You in advance
     
  2. jcsd
  3. Dec 15, 2011 #2

    ehild

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    The equation is not correct.The second terms has to be rdφ[itex]\hat{φ}[/itex].
    And make the vector [itex]\vec{dr}[/itex] different from dr, change of the coordinate r.


    ehild
     
  4. Dec 15, 2011 #3
    For d[itex]\vec{r}[/itex] I actually left out the second term, which is the elevation angle, since it is constant. I guess I should have clarified that I am using this in spherical coordinates, but I did produce a typo It should be
    dr[itex]\hat{r}[/itex]+rsinθd[itex]\varphi[/itex][itex]\hat{\varphi}[/itex].
    Which of course just simplifies to
    dr[itex]\hat{r}[/itex]+rd[itex]\varphi[/itex][itex]\hat{\varphi}[/itex]
    because θ constantly equals pi/2
    So you can ignore the lone sin at the end of the force vector, since it goes to one. Yet it still goes to a negative value unfortunately.
     
  5. Dec 15, 2011 #4

    ehild

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    The integration path is not clear: You can reach from (0.0,0) to (2,0,0) along the upper semicircle or the bottom one, yielding integrals of opposite signs.

    ehild
     
  6. Dec 15, 2011 #5
    My apologies, we must go along the circle where y>0
     
  7. Dec 15, 2011 #6

    ehild

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    The integral is negative then.

    ehild
     
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