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Chain sliding off a table, find the function a(t)

  1. Nov 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A chain lies on a frictionless table at rest, half off the edge, and half on. As soon as it is let go, it begins accelerating due to gravity only. Determine the acceleration of the chain as a function of time. The mass is m, gravity is g, and the length of the chain is L.

    2. Relevant equations
    F=ma
    K=(1/2)mvv
    Ug=mgh
    W=integral(F*dr)
    a=dv/dt , v=dd/dt

    3. The attempt at a solution
    Let x=the length of chain hanging vertically. Therefore x starts at L/2, and ends at L, the min and max accelerations are g/2 and g.
    I figured out that a=x/L*g
    I then tried to use energy, with my zero line level with the table, so I was dealing with negative values. I would get mgL/8 =-xxm/(2L)+mvv/2. Solving for v gives a messy square root, and I don't see how time fits into the equation.

    I know that the function will be piecewise because once the acceleration reaches g, it will remain there. I just need to find the function as it grows from g/2 to g.
     
  2. jcsd
  3. Nov 8, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi packers7! Welcome to PF! :smile:

    Hint: x/L*g = a = dv/dt = dv/dx times … ? :smile:
     
  4. Nov 8, 2008 #3
    oops, my energy equation is wrong, the first term, to the left of the = should be negative. And the first term to the right of the = should also have *g.

    dv/dt=dv/dx*dx/dt
    So a=dv/dx*dx/dt.
    Taking the differential of my energy equation, simplified, yields dv/dx=x/(L*v). Should I substitute v with the energy equation, solved for v (sqrt of stuff)?
    I tried using a=dv/dx*dx/dt. Which goes to x*g/L=x/(L*v)*dx/dt. That leads to dx/dt=v*g
    Am I on the right track?
     
  5. Nov 8, 2008 #4

    tiny-tim

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    What happened to dv? :confused:

    Start again:

    you can do it two ways …

    i] use Newton's second law F = ma and a = dv/dx to get a differential equation for v and x

    ii] just use KE + PE = constant, to get an ordinary equation for v and x.

    They come to the same thing in the end, but try both of them, just for practice. :smile:
     
  6. Nov 8, 2008 #5
    sorry, I don't think I'm making any progress. I can't figure out when t comes into the answer.

    I tried using F=m*dv/dx
    m*a=m*dv/dx
    a=dv/dx
    x*g/L*dx=dv
    integrate
    x*x*g/(2*L)+C=v
    C=-L*g/8
    I can't use v=distance/time since that's avg velocity.

    With energy, I tried
    K+Ug=constant
    .5*m*v*v+(x/L)*m*g*(-x/2)=-m/2*g*(-L/4)
    .5*v*v-x*x*g/(2*L)=-g*L/8
    Do I differentiate here?
    v*dv-x*g/L*dx=0
    v*dv=x*g/L*dx
    dv/dx=x*g/(L*v)
    I feel like I'm at a dead end :uhh:

    doesn't a=dv/dt, not dv/dx? x is not time, I made it a length that changes
     
    Last edited: Nov 8, 2008
  7. Nov 8, 2008 #6

    tiny-tim

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    That's right, except a isn't dv/dx, is it?

    No! v = dx/dt … so you integrate! :smile:
     
  8. Nov 8, 2008 #7
    then I'd get: .5[(dx/dt)^2]=x*x*g/(2*L)-g*L/8
    (dx/dt)^2=x*x*g/L-g*L/4
    (dx/dt)^2=(4*x*x*g-g*L*L)/(4*L)
    move x so it's with dx
    [(dx)^2]/(4*x*x*g-g*L*L)=4*L*[(dt)^2]
    integrate
    (dx)(-[tanh^-1[2*x/L]]/(2*g*L)=(dt)4*L*t
    integrate again
    -(x*[tanh^-1[2*x/L]]+(L/4)ln(L*L-4*x*x))/(2*g*L)=L*t*t/2
    ???
    this seems like it's too complicated to be right...
     
  9. Nov 8, 2008 #8

    tiny-tim

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    (have a square-root: √ and a squared: ² smile:)

    Why are you integrating twice?

    And where did tanh-1 come from? :confused:

    v = dx/dt = a√(x² - b²) …
     
  10. Nov 9, 2008 #9
    So if I take the square root of (dx/dt)^2=(4*x*x*g-g*L*L)/(4*L)
    I get dx/dt=√[(4x²g-gL²)/(4L)]
    dx*√[4L/(4x²g-gL²)]=dt , integrate
    ln(2gx+g√[4x²-L²])/√[g] + C = t
    test x=L/2, t=0 yields C=-ln(gL)/√[g]
    solve for x
    x=L(1+e^(2t√[g]))/(4*e^(t√[g]))
    plug in to a=xg/L
    a=g/(4*e^(t√[g]))+g*e^(t√[g])/4
    is that correct? i tested t=0, and a=g/2, which is right!
    a=g/2*cosh(t√[g])
    therefore at t=ln(2+√[3])/√[g], a=g ?
     
  11. Nov 9, 2008 #10

    tiny-tim

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    Hi packers7! :smile:
    No … where do you get ln from? :confused:

    Hint: to make the integration easier, rewrite it as

    (√(L/g))dx/√(x² - L²/4) = dt :smile:
     
  12. Nov 9, 2008 #11
    I'm plugging in the equation into http://integrals.wolfram.com/index.jsp
    using what you have, Sqrt[L/g]*Sqrt[1/(x*x-L*L/4)]
    integrates into √[L/g]*ln(2x+√[4x²-L²]) (+C)
    i checked by taking the derivative of that, and it works.
    am I making a mathematical mistake?
     
  13. Nov 9, 2008 #12

    tiny-tim

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    ok … bizarrely, that is right :rolleyes:

    but that gives you t as a function of x, while the question asks for x as a function of t …

    how are you going to "invert" something so complicated? :cry:

    Hint: to solve dx/√(x² - b²) the easy way, make a substitution! :smile:
     
  14. Nov 9, 2008 #13
    my logic is that if i solve that integrated equation for x, then i can plug that in for a=xg/L, getting an equation that gives acceleration as a function of time
    is that what you're trying to tell me to do? x gives the amount of chain hanging vertically, not the acceleration.
     
  15. Nov 10, 2008 #14

    tiny-tim

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    ok … show me!
     
  16. Nov 10, 2008 #15
    i've already done it in an earlier post...

    √[L/g]*ln(2x+√[4x²-L²]) (+C)=t
    C=-√[L/g]*ln(L) (by using x=L/2 and t=0)
    then solving for x yields x=(L²+L²*e^(2t√[g/L]))/(4L*e^(t√[g/L])
    x=L*(1+e^(2t√[g/L]))/(4e^(t√[g/L])), which equals (L/2)*cosh(t√[g/L])

    a=xg/L
    substitute x=(L/2)*cosh(t√[g/L])
    a=(g/2)*cosh(t√[g/L])

    (at t=0, a=g/2 and if a=g, t=√[L/g]*ln(2+√[3]))
    :biggrin:
     
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