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Homework Help: Mechanics Question

  1. Sep 9, 2007 #1
    1. The problem statement, all variables and given/known data
    Hi, I'd just like to know if I'm on the right track or a hint or something to help. The problem is: "A particle of mass m is released from rest a distance b from a fixed origin of force that attracts the particle according to the inverse square law:

    F(x) = -kx^(-2). Show that the time required for the particle to reach the origin is
    [tex] \pi \sqrt{ \frac{mb^3}{8k}} [/tex]

    2. Relevant equations
    F = ma, dV/dx = -F

    3. The attempt at a solution
    Initially, I rather hoped to solve the ODE [tex] m \ddot{x} + kx^{-2} = 0 [/tex]
    However, this was nonlinear and I didn't know how to solve it analytically.

    So, I decided to look at the potential. I know the negative derivative of the potential with respect to x is the force. So, I separated and integrated and got
    [tex] V(x) = \int kx^{-2} dx = \frac{-1}{x} + K = \frac{mv^2}{2} [/tex]. I know that when x = b, my velocity is zero (released from rest), so I get the K, the constant of integration, should be 1/b.

    So, my idea now was to solve for velocity as a function of x. I then could call velocity dx/dt, separate and integrate hopefully. I then could set my x(t) = 0 and solve for t hopefully. This, however, did not work well. I got

    [tex] v = \sqrt{ \frac{2k}{m} * ( 1/b - 1/x)) }[/tex] . Separation of variables (calling v = dx/dt) yielded the recipriocal of that fraction being integrated, which I was unaware of how to do, and did not look simple.

    I'm pretty sure I'm approaching this problem in a far too difficult manner. Any hints or suggestions would be appreciated, I don't want a full solution, just a hint in the right direction. Thank you.
    Last edited: Sep 9, 2007
  2. jcsd
  3. Sep 9, 2007 #2


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    Homework Helper

    Note that you forgot the k factor when you did the integral.

    I think you're almost there... use separation of variables in the last part as you were going to... I think a variable substitution will allow you to solve that integral. Might be a trigonometric substitution since the answer has a [tex]\pi[/tex].
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