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Equality between centrifugal force and V

  1. Sep 24, 2008 #1
    Hi everyone,

    Today in a question I saw an equality like this but I couldn't figure out how they are equal? Can anyonu help me please?

    mv^2/ r = | dV/dr |
     
  2. jcsd
  3. Sep 24, 2008 #2
    What do the various symbols in this equality represent?
     
  4. Sep 24, 2008 #3
    I can't figure it out either, probably because they are not equal. If the equation was: m*v^2/r = dv/dr*m then they would be equal.
     
  5. Sep 24, 2008 #4
    In the left side of the equation is the known centrifugal force. m--mass,v--velocity,r--radius,on the right side V is the potential,r--radius
     
  6. Sep 24, 2008 #5
    Are you working at a level where you can relate forces to gradients of potential?
     
  7. Sep 24, 2008 #6

    tiny-tim

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    Hi soul! :smile:

    Could that be mv^2/ r = | dU/dr |, where U is the gravitational or electric potential?
     
  8. Sep 24, 2008 #7
    Yes, tiny-tim.it is. I think I am not good enough to express myself. Thank you.
     
  9. Sep 24, 2008 #8
    Writing mv^2/ r = | dV/dr | is like writing ma=F (which you might be more familiar with?). Does this help?
     
  10. Sep 24, 2008 #9
    Could you explain it in a more detailed way, please,JimChampion?
     
  11. Sep 24, 2008 #10
    F=ma is a common way of expressing Newton's second law of motion for an object with constant mass m.

    The mv^2/ r = | dV/dr | is equivalent to writing ma=F because

    v^2/r is the centripetal acceleration

    |dV/dr| is the magnitude of the centripetal force (given by the gradient of the potential V)
     
  12. Sep 24, 2008 #11
    I don't really know what level of explanation you're after!

    When I did A-level physics (in the UK, 1994) knowing about potential and how it relates to force was part of the course. I now teach A-level physics, and this is no longer part of the course: anything with calculus is avoided and probably wouldn't be encountered unless you studied an undergraduate physics course.
     
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