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Repulsive central force

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data

    1. The problem statement, all variables and given/known data
    A particle of mass m moves under action of a repulsive central force Fr=Cr-3 with constant C greater than 0. At a very large distance from the centre of the force, the partcle has kinetic energy K and its impact parameter is b. Use conservation of energy and angular momentum to show that the closest m comes to the centre of force is rmin=[tex]\sqrt{b^2+ C/2K}[/tex]



    2. Relevant equations
    L=mbv0
    E=.5mvr2+L2/2mr2+V(r)



    3. The attempt at a solution
    So far, I said that since the object is far away its total energy is K
    I think V(r) = -3C/r2, but I'm not sure.
    And at rmin, E==.5mvrmin2+(mbv)2/2mr2-3C/rmin2

    I don't know what to do next.
     
  2. jcsd
  3. Oct 22, 2009 #2

    Andrew Mason

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    Your nomenclature is confusing. Do you mean: [itex]F(r) = Cr^{-3}[/itex] ??

    AM
     
  4. Oct 22, 2009 #3
    Yeah, thats what i meant
    I didnt realize it was like that when i copied it
     
  5. Oct 23, 2009 #4

    Andrew Mason

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    Conservation of angular momentum:

    [tex]\vec L = m \vec v \times \vec r[/tex]

    so:

    (1) [tex]|L| = mv_0rsin\theta = mv_0r(b/r) = mv_0b[/tex]

    and:

    (2) [tex]U(r) + \frac{1}{2}mv^2 = constant = K[/tex]

    So far, you have this correct. [I will use U for potential energy since it is confusing to use v for speed and potential.]

    Since F = -dU/dr, -U is the antiderivative of F

    [itex]F = Cr^{-3}[/itex] so [itex]U = \frac{1}{2}Cr^{-2}[/itex]


    Therefore, from (2)

    (3) [tex]\frac{1}{2}Cr^{-2} + \frac{1}{2}mv^2 = K[/tex]

    Next, you have to find v at minimum r. At that point, what is L (hint: what the angle between v and r?)? You should be able to express v at minimum r in terms of v_0, r and b and then K, r and b. Substitute that into (3) and you should get your answer.


    AM
     
    Last edited: Oct 23, 2009
  6. Oct 23, 2009 #5
    ok so the angle between r and v at rmin should be 90?
    so,
    [tex]|L| = mv_0r = mv_0b[/tex]
    sp rmin=bv/v0?
     
  7. Oct 23, 2009 #6

    Andrew Mason

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    The angle is 90. But your expression for L is not correct. Why are you using v_0 at minimum r? [itex]L \ne mv_0r[/itex]. v_0 is the speed at the beginning when U = 0.

    AM
     
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