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Use energy conservation to determine the forces

  1. Mar 5, 2007 #1
    1. The problem statement, all variables and given/known data

    Consider an isolated system comprising two particles of masses m1 and m2, whose position vectors, in an inertial frame, are x1 and x2 and whose velocity vectors are v1 and v2. The interaction of the particles may be described by an energy function

    E = 1/2 m1v1^2 + 1/2 m2v2 ^2 + U(x1, x2).

    (a) Suppose that U = −k/r2, where k is a positive constant. Use energy conservation to determine the forces acting between the particles.


    2. Relevant equations

    E = 1/2 m1v1^2 + 1/2 m2v2 ^2 + U(x1, x2). (1)

    U = −k/r2, (2)

    3. The attempt at a solution

    substituting U = −k/r2 into (1) gives

    E = 1/2 m1v1^2 + 1/2 m2v2 ^2 −k/r2

    then dont know what to do to determine the forces ,
    i would think that gravtional forces are acting between the particles.
     
  2. jcsd
  3. Mar 5, 2007 #2

    Dick

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    What does r2 mean? r usually denotes distance from x1 to x2. So r2=r^2 or just r? In either case, the force is related to the derivative of the potential energy function U with respect to distance. Why is this post titled 'relativity'?
     
  4. Mar 5, 2007 #3
    r2 means rsquared which should be r^2
    relativity, momentum forces...
    well the quesiton was longer , and other parts involved relativty, if u want me to post the rest of the quesiton let me know.
     
  5. Mar 5, 2007 #4

    Dick

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    If the potential is k/r^2 then it's not gravity. And you're using nonrelativistic expressions for kinetic energy. Maybe you'd better post the full question.
     
  6. Mar 5, 2007 #5
    (i)Which aspect of Newtonian relativity requires
    U to depend only on the separation vector x1 − x2?

    (ii) Which further aspect of Newtonian relativity requires U to depend only on the magnitude r = |x1 − x2| of the separation vector?

    and part (iii) is given as question (a)
     
  7. Mar 5, 2007 #6

    Dick

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    Ahhhh! NEWTONIAN relativity. Did you figure out (i) and (ii)? For (iii) I'll reiterate, the force is the negative derivative of the potential. What is the direction of the force?
     
  8. Mar 5, 2007 #7
    no couldnt do (i) and (ii).

    force will act towards the particles
     
  9. Mar 5, 2007 #8

    Dick

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    What ARE the principles of Newtonian relativity?
     
  10. Mar 5, 2007 #9
    Newtonian relativity-The special principle of relativity states that physical laws should be the same in all inertial reference frames, but that they may vary across non-inertial ones
     
  11. Mar 5, 2007 #10

    Dick

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    So in particular you should be able to translate and rotate positions without affecting the physics (such as forces or potentials). What does this tell you about (i) and (ii)?
     
    Last edited: Mar 5, 2007
  12. Mar 5, 2007 #11
    U= -Gm1m2/ x1-x2

    (i)Which aspect of Newtonian relativity requires
    U to depend only on the separation vector x1 − x2?

    mechanics.
     
  13. Mar 5, 2007 #12

    Dick

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    If you translate the system in space, x1 and x2 change but their difference remains constant, right? Is that important?
     
  14. Mar 5, 2007 #13
    yes it is important..well is it gravtional (netwonain relatiity that is reuqired)
     
  15. Mar 5, 2007 #14

    Dick

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    Is that a question? Can you answer what principle for (i)?
     
  16. Mar 5, 2007 #15
    (i)the principle of conservation of energy
     
  17. Mar 5, 2007 #16

    Dick

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    The principle is translational invariance - a special case of frame invariance. Likewise, what would rotational invariance have to do with (ii)?
     
  18. Mar 5, 2007 #17
    (ii) Which further aspect of Newtonian relativity requires U to depend only on the magnitude r = |x1 − x2| of the separation vector?


    rotational invariance-

    a vector quantity is rotationally invariant if its value remains the same under a rotation of its input vectors. Both the dot product and the cross product are rotationally invariant, while vector addition and scalar multiplication, in general, are not.
     
  19. Mar 5, 2007 #18

    Dick

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    Basically, yes. x1-x2 is not rotationally invariant but |x1-x2| is. Now, what about your force. The latest version of the potential you sent looks more like -k/r than -k/r^2. Which is it and which force law does that lead to?
     
  20. Mar 6, 2007 #19
    U = −k/r^2 this is the potenetial that is given in the question.

    Newtons force law of gravition:

    F= Gm1m2/ r^2
     
  21. Mar 6, 2007 #20

    Dick

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    Fine. But if the potential is -k/r^2 then what is the force? (Don't repeat 'gravitation').
     
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