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Velocity relation

  1. Mar 9, 2005 #1

    Wiz

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    I have a relation for velocity as v= sqrt [(1-x)/x)]
    I need to find the time the particle takes to reach x= 0.25
    Initially it is at x=1
    I had problems integrating furthur...to find time
    Can ya help??..
    Wiz
     
  2. jcsd
  3. Mar 9, 2005 #2
    I did this integral on my calulator and it's gross. Integration by parts perhaps? See what you can do with that..

    [tex]\int \sqrt{\frac{1-x}{x}} dx = \int (\sqrt{(1-x)}) * (\sqrt{x})^{-1} dx[/tex]


    Jameson
     
    Last edited: Mar 9, 2005
  4. Mar 9, 2005 #3

    dextercioby

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    From the definition of velocity
    [tex] v(t)=:\frac{dx}{dt} [/tex] Separation of variables and corresponding limits of integration give:

    [tex] \int_{0}^{\tau} dt=\int_{1}^{0.25} \frac{dx}{\sqrt{\frac{1-x}{x}}} [/tex]

    The integral in the first member is really simple,it is the time you need,denoted by me with " \tau"...

    For the integral in the RHS,use 2 substitutions

    First
    [tex] x=t^{2} [/tex]
    The second
    [tex] t=\sin y [/tex]

    Pay attention to:
    1.Signs.
    2.Transformations of the limits of integration.

    Daniel.



    EDIT:I'm getting a "tau" smaller than 0.It's weird.Are u sure the problem is stated correctly...?
     
    Last edited: Mar 9, 2005
  5. Mar 10, 2005 #4

    Wiz

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    prob

    I'm not sure......here's the original prob from which i derived tht velocity relation.

    There is a particle initially on x = 1
    It is acted upon by a force which varies as F(x)= -0.5 * x^-2
    What is the time it takes to reach x= 0.25

    Thanks,
    Wiz
     
  6. Mar 10, 2005 #5

    Wiz

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    forgot

    I forgot.....the mass of the particle is 10^-2 kg.....
     
  7. Mar 10, 2005 #6

    dextercioby

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    That's another problem,man,totally different.

    The advice is the same.Separate variables & integrate with corresponding limits.(EDIT 1:It doesn't work,it's second order).

    Daniel.

    EDIT 2:See next post for further comments.
     
    Last edited: Mar 10, 2005
  8. Mar 10, 2005 #7

    dextercioby

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    Nope,i'm afraid this problem (the second u posted) is not completely integrable

    It requires solving the ODE

    [tex] \frac{d^{2}x(t)}{dt^{2}}=-|C|\frac{1}{x^{2}(t)} [/tex]

    ,but subject only to the initial condition:

    [tex] x(0)=1 [/tex]

    Unfortunately,the closure of the Cauchy problem requires 2 independent initial conditions to determine the 2 coefficients (<============ a II-nd order ODE).

    Daniel.

    P.S.Look it up again.
     
  9. Mar 10, 2005 #8

    Wiz

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    The prob i posted is the same,
    All i did was....i found out acc. from the force..
    Then integrated it and i found velocity......the constant in tht case was c=-1

    Btw i hv mentioned tht the particle is initially at x=1
    The prob is same as given in the book,
    This is an iit-jee question meant for students who hv cleared the 12th grade...and i am in 11th rite now,
    Thanks for the help but i think u are getting a bit too advanced for me,,,

    Wiz
     
  10. Mar 10, 2005 #9

    dextercioby

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    Could you please post your work.I'm really curious of what you've done...

    Daniel.
     
  11. Mar 10, 2005 #10

    Wiz

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    Oh,my god.......i am sooooo sorry
    the correct eq is F(x)= -k * 0.5 * x^-2
    where k= 10^-2 Nm^2
    the mass of the body is 10^-2 kg
    Now itz 100% correct.

    Wht i did was, i got a= -0.5 * x^-2 as k and m cancelled out
    so i got dv/dt=(-0.5 * x^-2)
    hence dv/dx*dx/dt=(-0.5 * x^-2)
    hence dv*v=(-0.5 * x^-2)dx
    integrating i got c=-1 and simplification led to the vel eq i posted....
    Cud u plz check it now?
     
    Last edited: Mar 10, 2005
  12. Mar 10, 2005 #11

    dextercioby

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    It looks correct.(In general,the constant are less worrying).

    Daniel.

    P.S.Can u uniquely determine v=v(x)...?
     
  13. Mar 10, 2005 #12
    You can be sure. I also got: v= sqrt [(1-x)/x)].
    It's time just to integrate. Substitution sqrt [(1-x)/x)]=t leads to rather simple integral.
     
  14. Mar 10, 2005 #13

    dextercioby

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    Are u sure it's not a

    [tex] \int a(x) \ dx=\int v \ dv [/tex]

    Plug the values & integrate.The RHS is simple

    [tex] \int v \ dv =\frac{1}{2}v^{2}+C [/tex]



    Daniel.
     
    Last edited: Mar 10, 2005
  15. Mar 10, 2005 #14
    I hope we shall see it after integration. There must be "tau">0. If we'll have opposite we can just change the sign. Am i right?
     
  16. Mar 10, 2005 #15
    Sorry. i didn't notice edition. I think You are right in both posts
     
  17. Mar 10, 2005 #16

    dextercioby

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    The acceleration is force divided by mass:

    [tex] a(x)=\frac{F(x)}{m}=\frac{-5\cdot 10^{-3} x^{-2}}{10^{-2}}=-0.5\cdot x^{-2} [/tex]

    Therefore:

    [tex] \int a(x) \ dx=-0.5\int x^{-2} \ dx =0.5 x^{-1} [/tex]

    What initial condition does the velocity satisfy...?

    Daniel.
     
  18. Mar 12, 2005 #17

    Wiz

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    Hey i solved the prob,
    The fact we forgot was tht since the body is moving towards the orgin due to -ve value of acc, the correct relation wud be v= - sqrt[(1-x)/x] ----(1)
    Now put x = sin^2($)
    hence dx/d$ = sin2$
    hence dx = sin2$*d$ --------(2)

    By (1),
    - dx/sqrt[(1-x)/x] = dt
    => -dx /cot($) = dt
    By (2)

    dt = - sin2$*d$/cot($)
    => dt = -2sin^2($)*d$
    Now integrate lhs from 0 to T and rhs from pi/2 to pi/6
    So the final ans I got was
    T= [pi/3 + sqrt(3)/4 ]
    which happens to be correct :smile:

    I hope I havent made any typing mistakes this time..
    Thanks for replying....
     
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