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Rope falling through a hole on a frictionless table

  1. Mar 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A rope with uniform density ##\lambda=\frac{m}{L}## is placed on a frictionless table with an initial length ##y_0## hanging through the hole. Derive a differential equation for the position of the bottom of the rope and then using this solve for the time required for the rope to leave the table.

    2. Relevant equations
    ##p=mv##
    ##F=\frac{dp}{dt}##
    ##p=\dot{m}v+m\dot{v}##

    3. The attempt at a solution

    I'm currently stuck with determining which differential equation I've derived is the correct one (although they only differ slightly).

    The first method is just by analyzing the motion in the y direction. The acceleration will be given by ##ma=\lambda gy## where ##m## is the mass of the entire rope and ##y## (a function of time) is the length of the rope that's fallen through the hole. We can rewrite ##m## as ##\lambda L## from which we obtain: ##a=\frac{gy}{L}## or written more conveniently $$\ddot{y}-\frac{g}{L}y=0$$

    The second method is the way my professor suggested to go about solving it. We consider the forces in both the ##x## and ##y## directions. To do this I considered the mass on the table ##m_T## and the mass falling off the table ##m_F## separately. Defining the ##x## and ##y## axis so the movement on the x and y axis will be positive, in the x direction we have $$F_x=\frac{dp}{dt}=\dot{m_T}v+m_T\dot{v}\hspace{5mm} \mbox{and} \hspace{5mm} F_y=\dot{m_F}v+m_F\dot{v}$$ Then using the relation ##\dot{m_T}=-\dot{m_F}## we can express the first equation as ##F_x=-\dot{m_T}v+m_T\dot{v}##. Now the same force ##\lambda gy## will act on both of the sections therefore $$F_x+F_y=2\lambda gy=\dot{v}(m_F+m_T)$$ Now using ##m_F=y\lambda## and ##m_T=\lambda(L-y)## we get ##2\lambda gy=\dot{v}\lambda L## which after cancelling terms and changing to a more convenient form: $$\ddot{y}-\frac{2g}{L}y=0$$

    I feel like something is definitely off about the second method particularly with how the force is acting on the masses but I just can't see what.
     
  2. jcsd
  3. Mar 16, 2016 #2

    Andrew Mason

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    Why not stick to the first method? It looks fine to me. The second method is unnecessarily complicated because all parts of the rope accelerate at the same rate whether it is vertical or horizontal. But the force on the rope depends only on the amount of rope hanging down (ie. y).

    AM
     
  4. Mar 16, 2016 #3
    Yea I agree with what you've said I just thought it was weird that it's almost the same equation but not quite.
     
  5. Mar 17, 2016 #4

    Andrew Mason

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    There is only one force on the entire rope and it is λgy. The net force on the part hanging down is λgy - T where T is the tension in the rope.

    AM
     
  6. Mar 17, 2016 #5
    Ah thanks that makes sense!
     
  7. Mar 17, 2016 #6

    haruspex

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    It is not stated whether the rope on the table is laid out straight or in a random heap. It makes quite a difference.
     
  8. Mar 17, 2016 #7
    I suppose I could've included a picture, in this case it is laid out straight on a table.
     
  9. Mar 17, 2016 #8

    haruspex

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    Ok.
     
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