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Thornton and marion problem: systems of particles

  1. Apr 4, 2006 #1
    the problem appears to be simple:

    "a flexible rope of length 1 meter slides from a frictionless table top. the rope is initially released from rest with 30cm hanging over the edge of the table. find the time at which the left end of the rope reaches the edge of the table."

    (thornton and marion, 5th edition, chapter 9, problem 21, page 381.)

    i have attempted to solve this problem in a variety of ways:

    method 1) breaking up the rope into two sections and applying newton's laws.

    i find the mass of each section as a function of time, find the momentum of each side, find the time derivative of the momentum of each side and use

    dP1/dt = M1 g - T,
    dP2/dt = T, where T is the tension in the rope, P1 and M1 are the momentum and mass, respectively, for the part of the rope that is hanging off of the side, and P2 is the momentum of the part of the rope that is still on the table.

    all said and done, i got

    ddx/dtdt = (x+0.3)*g.
    (that's the second derivative of x with respect to time on the left hand side.)

    solving the diff eq, i get two exponentials and a constant term, but this leads to a transcendental equation, so i figured it was time to try something else.

    2) same precedure, but i replaced the second derivative of x with dx/dt * d(dx/dt)/dx, using the chain rule.

    but this set up another messy differential equation, since i have to take a square root, then divide by the right hand side.

    3) conservation of energy. i found the mass and center of mass of the hanging portion in its intial position and then calculated the potential energy. then i found the mass and center of mass of the hanging portion after 'x' more meters of rope fell over the side.

    so U_before = U_after + K, where K is the kinetic energy.

    i got

    dx/dt = (gx(x+0.6)/(x+0.3))^(1/2).

    this also is a mess! :mad:

    what am i missing that makes this problem soluble? :confused:
  2. jcsd
  3. Apr 4, 2006 #2

    Physics Monkey

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    Hi Brad,

    I would simply set up Newton's equations, although the energy method is very simple too. I have a couple of questions for you to help get you started. First, what is x? As you have things set up now, x must be dimensionless in order that you can add .3 to it, but then you have the second derivative of x having the units as g. Also, shouldn't the force vanish when the rope doesn't hang over the edge at all? Does your equation of motion reflect this?

    Try to make some progress with these hints, and we'll go from there.
  4. Apr 4, 2006 #3
    0.3 has the units of meters. (0.3 meters hang over the edge.) x is the additional length of the rope that falls over the edge. ie, i want to find the time it takes for x to go from 0m to 0.7m (since the total length of rope is 1 m.)

    the force acting on the rope is the force due to gravity, it seems, so i doubt that it would vanish once all of the rope is over the edge (unless that's how gravity works! :tongue2: ).

    looking over somewhat similar problems, i don't think conservation of energy would work, since the other problems don't conserve energy! (what makes me mad: the other problems only ask for VELOCITIES! :mad: :mad: :mad: )

    another technique:

    M dv/dt + dM/dt v = Mg, where M was as before, and v is dx/dt.

    i got v = ((x^2 + 0.6x)g)^(1/2).

    doing that integration looks to me like it's a nightmare, so i used my calculator and got that the time was 0.5986 seconds. anyone concur?
  5. Apr 4, 2006 #4

    Physics Monkey

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    Good, I just wanted to make sure you had everything straight. You are still missing a factor of length somewhere in your differential equation.

    To solve the differential equation, make a change of variables from x to x + .3 and then integrate.

    Just out of curiosity, what problems are you referring to that don't conserve energy?
  6. Apr 4, 2006 #5

    good call with the units; i forgot to divide out a term with units lenght on the right hand side at one point.

    except... after doing that, i just get the free-fall result

    v = (2gx)^(1/2), which is not reassuring. :mad:

    the problems that don't conserve energy include hanging a chain from a roof at one point and holding up the other side and then dropping it. another is when a rope falls through a hole. (ie, example 9-2 and problem 9-15 in thornton and marion, fifth edition.)

    i think i'm just going to have to wait until my prof's office hours tomorrow. :/
  7. Apr 4, 2006 #6

    Physics Monkey

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    How did you get the free fall result? Your differential equation is d^2 x/dt^2 = (x + .3 L) g/L (I put all the units in just for clarity). Now, if you make a transformation from from x to y = x+.3L, then d^2y/dt^2 = d^2x/dt^2 and the differential equation for y is really simple. It does not give the free fall result.

    Edit: Also, regarding the energy conservation issue, it's a little bit tricky with these macroscopic objects. In the first case you mentioned, for example, energy is clearly not conserved because one side of the rope is fixed. Also, a coiled rope is different from an uncoiled rope. Conservation of energy happens to be valid in this problem.
    Last edited: Apr 4, 2006
  8. Apr 5, 2006 #7
    yeah, that was the way to solve it!

    i spent an hour in my prof's office working through this and another problem. man, this assignment sucked, not least of all because i had all of three days to do it. :mad:

  9. Apr 5, 2006 #8

    Physics Monkey

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    You're welcome. Glad you got it.
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