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Rope on a Table

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data

    A flexible rope of length 1.0 m slides from a frictionless table top. The rope is initially released from rest with 30 cm hanging over the edge of the table. Find the time at which the end of the rope left on the table will reach the end of the table (basically when the length of the rope hanging of the table will be the whole 1 m. This is problem 9-21 in Thornton/Marion's Classical Dynamics).


    2. Relevant equations

    Let L be the total length of the rope and x be the portion hanging off the table.


    3. The attempt at a solution

    I started off with [tex]dp/dt = mg x/L[/tex]

    because only the part hanging off the table is accelerated due to gravity.

    I then rewrote that as [tex]dp=mgx/L \:dt[/tex]. Since the initial momentum is zero [tex]dp=p(t)-p(0)=p(t)=mdx/dt \:dt[/tex] where t is just some later time.

    So I'm stuck at [tex]dx/dt=gx/L \:dt[/tex]. I can integrate over position but I'll have this (dt)^2 term that I don't know what to do with.
     
  2. jcsd
  3. Oct 3, 2009 #2
    we know that the rope will gain mass as it falls so we will have to do this in terms of mass density which will be represented with [tex]\sigma[/tex]
    force of gravity=mg=( [tex]\sigma[/tex] x) [tex]\cdot[/tex] x [note a dot equals the time derivative]
    momentum= [tex]\sigma[/tex] L [tex]\cdot[/tex] x
    F=ma=( [tex]\sigma[/tex] x)g= [tex]\sigma[/tex] L [tex]\cdot[/tex] [tex]\cdot[/tex] x
    now integrate and you will be able to get your length of rope as a function of time.
     
  4. Oct 3, 2009 #3
    hi there
    I have a solution for your problem but I am not sure it is right. Anyway, here it is.
    I started with the relation W = (integral)F dx
    Work done equals change in kinetic energy and initial KE is zero. Plugging in W = 0.5mv(squared) and F = (x/L)mg and integrating it, I got Lv(squared) = gx(squared)
    Using v = v(initial)+at and equating x and 0.7L because the question is about when the rope travels 0.7 m that is on the table, I finally got: [tex]\ t = 0.7 \sqrt{\frac{L}{g}}[/tex].
    Frankly, I was shocked by this equation's similarity with the simple harmonic motion. =D
    Thanks
     
  5. Oct 3, 2009 #4
    I am really sorry, I did a big mistake. I assumed the mass to be constant. Mine is flawed
     
  6. Oct 4, 2009 #5
    That's going to involve solving a second order linear ODE. I was thinking that you can go through this problem with only separation of variables.
     
  7. Oct 4, 2009 #6
    How about using the conservation of energy?

    [tex]\begin{array}{rcl}U&=&\int\limits_0^x -\mu g h{\rm d}h \\ &=&-\frac{\mu g} 2 x^2\end{array}[/tex]

    and

    [tex]K=\frac 1 2 \mu L v^2^[/tex]

    and we know that:

    [tex]\Delta (U+K)=0[/tex]

    hence

    [tex]v=\sqrt{\frac g L} \sqrt{x^2-x_0^2^}} \Leftrightarrow \frac{{\rm d} x}{\sqrt{x^2-x_0^2}}=\sqrt{\frac g L} {\rm d}t[/tex]

    [tex]\fbox{t=\sqrt{\frac L g} \cosh^{-1} \left(\frac x {x_0}\right)}[/tex]


    ([tex]\mu=\frac m L[/tex], [tex]L[/tex]=1m, [tex]x_0[/tex]=30 cm)
     
    Last edited: Oct 4, 2009
  8. Oct 4, 2009 #7
    That looks correct but I'm just wondering what you're motivation was for integrating the potential energy?
     
  9. Oct 4, 2009 #8
    Potential energy varies with height. You need to calculate the potential energy of a small segment at a given height first and then integrate that expression over the length of the rope.
     
  10. Oct 4, 2009 #9
    So you've chosen the tabletop as your zero reference point?
     
  11. Oct 4, 2009 #10
    Yes.
     
  12. Oct 5, 2009 #11
    I get the same thing by just using F = m.a

    the weight of the piece thats over the edge is [tex] \frac {m g x} {L} [/tex], but
    the entire rope has to move so the mass wil be just m
    (x = the length of the rope over the edge)

    this getst you [tex] \frac {m g x} {L} = m \frac {d^2x} {dt^2} [/tex] or

    [tex] \frac {d^2x} {dt^2} = \frac {g} {L} x [/tex]

    this is a very easy linear equation with constant coefficients.

    the solution of is [tex] C_1 e^{\sqrt {\frac{g}{L}} t} + C_2 e^{-\sqrt \frac{g}{L} t} [/tex]

    using what we now about x(0) you get [tex] C_1 + C_2 = x(0) [/tex]

    and [tex] x'(0) = 0 = C_1 - C_2 [/tex] so [tex] C_1 = C_2 = x(0)/2 [/tex]

    the solution now becomes [tex] x = x(0) \frac { e^{\sqrt {\frac{g}{L}} t} + e^{-\sqrt \frac{g}{L} t} } {2} = x(0) cosh(\sqrt {\frac{g}{L}} t) [/tex]

    and solving that gets you

    [tex] t=\sqrt{\frac L g} \cosh^{-1} \left(\frac x {x_0}\right) [/tex]
     
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