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Force as a function of position

  1. Jul 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m is acted on by a force F(x) = -k/x2. The particle is released from rest a distance b from the origin of the attractive force F(x). Show that the time to reach the origin is given by t = [tex]\pi[/tex](mb3/8k)1/2.

    2. Relevant equations

    F(x) = -k/x2
    F = ma = m dv/dt = m dv/dx dx/dt = m v dv/dx

    3. The attempt at a solution

    m v dv = -k/x2

    v dv = -k/mx2 dx

    [tex]\int[/tex] v dv = -k/m [tex]\int[/tex] x-2 dx

    Integrating the left from zero to v and the right side from b to zero:

    1/2 v2 = an integral that diverges???!!!
     
  2. jcsd
  3. Jul 23, 2010 #2

    rock.freak667

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    Actually, v=0 at x=b if you read the part that says "he particle is released from rest a distance b from the origin of the attractive force F(x)".
     
  4. Jul 23, 2010 #3
    Something doesn't seem right about this problem. Perhaps I am not understanding it correctly..
    But your result would actually seem to make sense to me. I mean, what is the force at the origin?
    F(0) = -k / 0^2
    as you approach the origin the force toward the origin becomes infinite....
    How are you supposed to find how long it takes a particle to reach the origin under infinite acceleration?
    I mean unless it works out nicely because the distance becomes infinitesimal as the acceleration becomes infinite and maybe it works....

    But, there is something else, what if "b" is negative, then the force is always away from the origin, and therefore it is impossible to solve for the case where b is negative.
     
  5. Jul 23, 2010 #4

    rock.freak667

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    I am guessing it is similar to gravitation

    F=GMm/r2
     
  6. Jul 23, 2010 #5

    vela

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    Remember, you're after the time, not the velocity. Integrate with these limits instead:

    [tex]\int_0^{v(x)} v\,dv = -\int_b^x \frac{k}{mx^2}\,dx[/tex]

    to get v(x), the velocity as a function of position. Then you need to do one more integration to find the time to reach the origin. The velocity diverges, but the time is finite.
     
  7. Jul 23, 2010 #6
    Thanks for the responses. The particle is in a conservative force field so the energy is the sum of the potential and the kineticenergy and is a constant. So its potential energy at rest a distance b from the origin will equal its kinetic energy as it reaches the origin.

    To find the potential, F(x) = -dV/dx where V is the potential energy.

    -dV = -kx-2 dx

    Integrating the right hand integral from infinity to b:

    V = -k/b

    1/2 mv2 = k/b

    So we get v = (2k/mb)1/2

    dx/dt = (2k/mb)1/2

    dt = (mb/2k)1/2 dx

    Integrating the right integral from 0 to b:

    t = (mb3/2k)1/2 which, unfortunately , is still off by a factor of [tex]\pi[/tex]/2.
     
  8. Jul 24, 2010 #7

    vela

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    What exactly is v? It's the velocity of the particle when?
    Note that your velocity is a constant, which means the particle isn't accelerating, which, in turn, means there's no net force on the particle. This is obviously wrong.
     
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