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Homework Help: Prove the work done by F is zero for a curve on a sphere

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that if an object moves along any smooth simple curve C that lies on the sphere [tex]x^2 + y^2 + z^2 = a^2[/tex] in the force field [tex]F(x,y,z) = f(x,y,z)(xi + yj +zk)[/tex] where [tex]f[/tex] is a continuous function, then the work done by [tex]F[/tex] is zero.


    2. Relevant equations

    3. The attempt at a solution

    I tried to show the curl of F was zero but realized that since f can be anything it'd be impossible to show that the curl was zero, atleast i think so :confused:. Not really sure how else to approach this problem. Looking for a hint in the right direction.
     
  2. jcsd
  3. Feb 3, 2010 #2
    Off the top of my head i'd say to do it in generality you'd be looking at something like Greens theorem or Stokes theorem (one of them deals with conservative fields can't remember which)
     
  4. Feb 3, 2010 #3
    We've yet to learn Stokes Theorem, though we have learned Green's theorem. But I don't see how Green's theorem would apply here as it is just a specific case of Stokes Theorem for two dimension (correct?).
     
  5. Feb 3, 2010 #4

    LCKurtz

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    At a point on the surface of the sphere, think about what direction your force field points.
     
  6. Feb 3, 2010 #5
    Doesn't that depend on the function f(x,y,z) ? since the force field will change depending on what that function is.
     
  7. Feb 3, 2010 #6

    LCKurtz

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    f(x,y,z) is a scalar. The direction is going to be parallel to the vector, so...
     
  8. Feb 3, 2010 #7
    At any point the force would be the vector (f(x,y,z)x , f(x,y,z)y, f(x,y,z)z). Hmm the light bulb still hasn't gone on :(
     
  9. Feb 3, 2010 #8

    LCKurtz

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    If f was constant what direction would your force field be? Remember f can only change the length of the vector or reverse its direction.
     
  10. Feb 3, 2010 #9
    The force would be going straight outwards, wouldn't it? with the length modified by f(x,y,z). which would mean it was perpendicular to the curve right?
     
  11. Feb 3, 2010 #10

    LCKurtz

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    Bingo! Well, almost bingo. It might be pointed inward if f was negative at a point. All the scalar f can do is vary the length but the force must always point inward or outward. Now think about what direction you are moving along your surface curve in comparison to the force.
     
  12. Feb 3, 2010 #11
    Well if the force is perpendicular at one point on the surface curve, and the surface is sphere wouldn't the force always be perpendicular to the curve regardless of which curve is chosen.
     
  13. Feb 3, 2010 #12

    LCKurtz

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    Aahhyup! So?
     
  14. Feb 3, 2010 #13
    Well the work done by F would be zero then since it doesn't effect the "particle" when moving from point a on the curve to point b.

    I now understand it intuitively. Thank you :) How would i go about proving this though. Would i need to parametrize a curve along the sphere and show that it is perpendicular to the force vector?
     
  15. Feb 3, 2010 #14

    LCKurtz

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    Aye, there's the rub. The pesky equations. What you have is a curve on the sphere:

    [tex]\vec R(t) = \langle x(t),y(t),z(t)\rangle[/tex]

    with

    [tex]|\vec R(t)| = 1[/tex]

    and a force that can be written as

    [tex]\vec F = f(\vec R)\vec R[/itex]

    And you need to show

    [tex]W = \int_C \vec F \cdot d\vec R = 0[/tex]

    I'm going leave you to think about that now. Express it in terms of t and see if you can figure out how to get that = 0. You obviously need to use

    [tex]|\vec R(t)| = 1[/tex]

    somehow in your argument, eh? Good luck. Sack time here.
     
  16. Feb 3, 2010 #15
    Thanks for the help. Good night :)
     
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