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Tougher than it looks

  1. Oct 27, 2005 #1
    Source: PhysicsBowl
    A chain of length 2d and mass 2m is placed over a frictionless table such that half of the rope is on the table; the other half dangling, and it is just starting to accelerate. Find its acceleration as a function of time.

    I couldn't get this one.

    I tried net force = ma
    (m+d)g = 2ma; where d represents the mass of the part of the chain that has slipped. But I can't get it in terms of t. Any help?
     
  2. jcsd
  3. Oct 27, 2005 #2

    daniel_i_l

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    If I understood correctly the only thing falling is the chain like this:
    [" ******
    ---------* ||
    --------|* \/
    --------|* "]
    * = chain
    and the "arrow" is the direction that it is falling.
    Hint: all mass falls with the same acceleration(on earth).
     
  4. Oct 27, 2005 #3
    I know, but I can't solve for a in terms of t.
     
  5. Oct 28, 2005 #4

    daniel_i_l

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    If there's no friction then the acceleration is constant (same a for every t);
     
  6. Oct 28, 2005 #5
    As more of the chain is sliding off the table, the remaining chain will experience a larger force pulling it off teh table. Acceleration is not constant
     
  7. Oct 28, 2005 #6

    daniel_i_l

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    Yes there will be a larger FORCE pulling it of the table but there will also be more mass to pull. Since F=ma the more massive the body is the more force you need to accelerate it. you can also see this mathematicly:
    F=ma , in this case the force is mg so:
    mg = ma /m
    g=a
    As you can see, the mass is irrelevant since the more mass you add to the force (mg) the more you need to accelerate it (ma) so the masses cancel out. If there was friction then it would be different cause you'd have more force to overcome the friction and then you'd speed up.
     
  8. Oct 28, 2005 #7

    Tide

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    No, the acceleration is not constant. If x is the length of chain hanging over the edge then the acceleration is

    [tex]\frac {gx}{2L}[/tex]

    The force acting on the chain is the weight of the hanging chain but it's the entire chain that is being accelerated.
     
  9. Oct 28, 2005 #8

    daniel_i_l

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    You totally right, I was only looking at the mass hanging over the end:(
     
  10. Oct 28, 2005 #9
    Were you able to solve this? I tried solving it but I can't. I set up a few relationships, but I think this problem requires some calculus and I'm really weak with that. Can someone help further please?
     
  11. Oct 28, 2005 #10

    Tide

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    The solution works out to [itex]x = \frac {L}{2} \cosh \left( \sqrt {\frac {g}{2L}} t\right)[/itex].
     
  12. Oct 29, 2005 #11
    how did you do it?
     
  13. Oct 29, 2005 #12

    Tide

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    There are two ways to do it. The easier way is to write down an equation expressing energy conservation. It relates the speed of the mass to the length of overhanging mass. Once you have that, solve for v and, since v = dx/dt you can integrate to find x(t).

    The other way is to start with the second derivative of x equals gx/2L. It's a little messier but if you're into differential equations you should be able to handle it.
     
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