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Modern Classical Mechanics

  1. Aug 31, 2006 #1
    I have three questions. I've worked on these for several hours and havn't come up with very much at all. Any help would be greatly appreciated.

    1.a perfectly flexible cable has length l. Initially, the cable is at rest, with a length xnaught of it hanging vertically over the edge of a table. Neglecting friction compute the length hanging over the edge after a time t. Assume that the sections of the cable remain straight during the motion.

    (I am assuming the cable (l) is laying horizontaly on top of the table, then xnaught is hanging from the corner downward.)

    For this problem the correct answer is x=xnaught*cosh(sqrt(g/l)*t).
    How do you arrive at this answer?

    2.A particle of mass m, initially at rest, moves on a horizontal line subject to a force F(t)=a*e^(-b*t). Find its position and velocity as a function of time.

    Correct answers are:
    x(t)=xnaught+ (a/(b*m))*[t-1/b(1-e^(-b*t))]

    3. A massless spring of rest length l and spring constant k has a mass m attached to one end. The system is set on a table with the mass vertically above the spring.

    a) What is the new equilibrium height of the mass above the table?
    b) The spring is compressed a distance c below the new equilibrium point and released. Find the motion of the mass assuming the free end of the spring remains in contact with the table.
    c) Find the critical compression for which the spring will break contact with the table.

    (I don't have the answers to this problem. I just needed to check mine.)
  2. jcsd
  3. Aug 31, 2006 #2


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    Hey Nick,

    According to the rules of this forum, you have to post your attempts at solving the problem in order to get help.
  4. Sep 1, 2006 #3
    Sorry, I appologize for not reading the rules of the forum beforehand.

    for the first problem I drew a FBD of the table and the cable. I was reading about catenary's and see that it's almost the same (one end is just flat, which is where the square root comes from inside the cosh() I think). for a catenary "y=a*cosh(x/a)"
    I'm kind of lost as to where to begin integrating from to arrive at the correct answer. Maybe I'm looking at it the wrong way. Thanks.
  5. Sep 1, 2006 #4
    well... for the first one, you can probably do the energy equation. set reference on the table...

    the sum of energy at one point should equal total energy.

    potential energy can be calculated using center of mass.
  6. Sep 1, 2006 #5
    For the first one, it's best to find the acceleration of the system in terms of the length [tex]x[/tex] of cable hanging over the edge. That should be simple enough (Perhaps assign the cable a linear density [tex]\lambda[/tex]). Do you have such an equation yet? Next, do a little integrating and you'll end up with the answer (that's the tricky part). The "cosh" has nothing to do with a catenary. It's just a simplification of exponentials which result from natural logs which come from integrations.
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