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Homework Help: Variable acceleration problem.

  1. Jun 19, 2003 #1
    A particle of mass m is released from rest at a distance b from a fixed origin of force that attracts the particle according to the inverse square law:

    F = -kx^-2

    Show that the time required for the particle to reach the origin is:

    [pi](mb^3/8k)^1/2

    I have no idea where the pi came from.

    This is what I've done.

    F=dp/dt
    dp/dt=mdv/dt
    dv/dt=d^2x/dt^2
    m*d^2x/dt^2=m*dv/dt
    m*d^2x/dt^2=-k/x^2
    m*x^2*d^2x=-k*dt^2
    m[inte][inte]x^2dxdx=-k[inte][inte]dtdt

    After solving the double integrals and pluging in the constants I get.

    t = sqrt[-1/(6k)*m(16b^4)]

    I'm going to be really embarrased if my calculus is wrong.

    So am I doing wrong. I still have no idea where a pi comes from!

    Thanks

    Frank
     
  2. jcsd
  3. Jun 19, 2003 #2

    Tom Mattson

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    Here's your problem. d2x/dt2 cannot be separated as dxdx/dtdt.

    You'll have much better luck by noting the following:

    a=dv/dt
    a=(dv/dx)(dx/dt)
    a=(dv/dx)(v)

    or

    a=v(dv/dx)

    Now, try that with your force:

    -k/x2=mv(dv/dx)
    (-k/x2)dx=mvdv

    You can get x in terms of v, and then simply note that v=dx/dt and integrate again to find the time. Give it a shot and let me know if you need more help.

    edit: fixed an omission
     
    Last edited: Jun 19, 2003
  4. Jun 19, 2003 #3
    I get -b*sqrt(2)/(3*sqrt(k/(b*m)) = sqrt(2*b^3*m/9k)

    I still don't see where a pi could come from.
     
    Last edited: Jun 19, 2003
  5. Jun 19, 2003 #4

    Tom Mattson

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    Can you show your work? That will help a lot!

    Here's how I started:

    -(k/x2)dx=mvdv

    (sorry, I accidentally left the 'm' out in my last post, but I have fixed it)

    -k[inte]bx(1/x2)dx=m[inte]0vvdv
    (k/x)-(k/b)=(1/2)mv2

    Are you with me so far? If so, can you show me how you did the second integral?
     
  6. Jun 19, 2003 #5
    Except for up to there. I made an algebriac mistake.

    k/x - k/b != k/(x-b)

    Let me start over. Give me a couple minutes.

    Thanks for your help so far.
     
  7. Jun 19, 2003 #6
    Okay cool.

    The integral ins't too prety so I let me Ti-89 solve it.

    I get t = -Sin([oo])b^3/2*sqrt(2m)[pi]/4sqrt(k)

    Which is = [pi]sqrt[Sin([oo])*b^3m/8k]

    What the heck does Sin([oo]) mean?

    Thanks. I just need to slow down. Stop making dumb mistakes!

    I need to stop forgetting that chain rule property.

    Thanks a lot!

    Frank
     
  8. Jun 19, 2003 #7

    Tom Mattson

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    Uh-oh.

    It sin([oo]) is indeterminate, because it keeps oscillating. That means you can't assign a value to it. Me smells a mistake here.
     
  9. Jun 19, 2003 #8
    I'm it's the software in my calculator.

    I get a headache looking at this integral:

    sqrt[m/2][inte]dx/sqrt[k/x-k/b]

    I'm sure it's not as hard as it looks. I can drag up my old Calculus text, or look for a tabulated solution, to solve it another way.
     
  10. Jun 19, 2003 #9

    Tom Mattson

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    I would try to work it into a form that matches an integral in a table. My guess is that it will turn out to be an inverse trig function, which is how you will get a π out of it.
     
  11. Jun 19, 2003 #10
    I've been playing with it for half an hour, I'm not getting anywhere. It's beyond my mathematical abilities. Everything I do I still get an indeterminate solution.

    If b>0, which it is, then I believe we get a solution.
     
  12. Jun 19, 2003 #11

    Tom Mattson

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    Without knowing what you've done, I really can't say much on this.

    Physically, you should get a solution, because this force is the same as a point mass or a point charge. Mathematically however, there seems to be a problem, because you encounter an integral like this:

    [inte]b0((x/(b-x))1/2dx

    whose integrand diverges at the lower boundary. Still, there should be a way to handle this improper integral.
     
  13. Jun 19, 2003 #12
    That's the form I found.

    If x is under the following condition:

    0<=x<b

    The solution is fine.
     
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