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Equation of Motion

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data
    A mass m has speed v0 at the origin and coasts along the x axis in a medium with force F(v). Use the chain rule of differentiation to write the equation of motion in the separated form m*v*dv/F(v)=dx.


    2. Relevant equations
    F(v)= -c(v^3/2)


    3. The attempt at a solution
    So far, i drew a diagram to show what was happening. I know that F(v) is the drag force. I'm really just confused about what the equation of motion is. I know that it is going to have a velocity and mass, I think.
    Can any one help?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 7, 2008 #2

    CompuChip

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    The equation of motion is just Newtons law,
    [tex]F(v) = m a[/tex]
    where a is the acceleration
    [tex]a = \frac{dv}{dt} = \frac{d^2s}{dt^2}[/tex].
     
  4. Dec 7, 2008 #3
    Okay, that is what i have done so far, is this right,
    I have m*v0= -c*v3/2.
    then i am suppose to differentiate this using the chain rule which my book is calling the v*dv/dx rule to get a separated form m*v*dv/F(x)=dx. But i don't know what to differentiate with respect to, v or x, but my equation doesn't have an x. So maybe my equation is wrong, I am not quite sure.
     
  5. Dec 7, 2008 #4

    CompuChip

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    How do you get the right hand side? Don't use any specific form for F(v), just the equation I gave you.
    Then, in m*a, replace a = dv/dt by something which looks like what you want (try dv/dt = dv/dt * something), using the chain rule.
     
  6. Dec 7, 2008 #5
    So even though the drag force = F(v) i don't sub that part in?
     
  7. Dec 7, 2008 #6

    CompuChip

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    No, because there is also an F(v) and not c and v^(3/2) in your final answer.
     
  8. Dec 7, 2008 #7
    so in the end when i use the chain rule should I end up with the drag force equation? Sorry I am still a little confused.
     
  9. Dec 7, 2008 #8

    CompuChip

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    No, they want you to show how you can go from
    F(v) = m a
    to
    F(v) = m v (dv/dx)
    using the chain rule, and then rewrite this to
    m*v*dv/F(v)=dx.

    The point being that the last line is a differential equation for v as a function of x with separated variables (all the v's on one side, all the x's on the other) so you can presumably solve it more easily than solving
    m x''(t) = F(x'(t))
    for x as a function of t.
     
  10. Dec 7, 2008 #9
    ok, so understand what i was suppose to do, and got it to work, so was there any reason why they told me what the drag force was?
     
  11. Dec 7, 2008 #10

    CompuChip

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    Probably in the next question they are going to ask you to solve the equation by integrating both sides :smile:
     
  12. Dec 7, 2008 #11
    they do ask me that, so do i substitute that in now for F(v)
     
  13. Dec 7, 2008 #12
    never mind i totally got it to work! thank you for all of the help!
     
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