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Chain on table - calculate work

  1. Nov 22, 2007 #1
    1. A chain is help on a frictionless table with one-fourth of its length hanging over the edge. If the chain has length L and mass M, how much work is required to pull the hanging part back on the table?

    i dont think i could have W = mgL/4
    will the work change each time a new chain ring is on the table?
    is there work from the horizontal chain if the table is frictionless?


    2. A small solid marble of mass m and radius r rolls without slipping along the loop the loop track having been released from rest somewhere on the straight section of the track.
    a) from what minimum height above the the bottom of the track must the marble be released in order that it just stay on the track at the top of the loop? (the radius of the loop is R , assume R>>r )
    b) If the marble is released from height 6R above the bottom of the track, what is the horizontal component of the force acting on it when it is at height R in the loop.

    Ui + Ki = Uf + Kf
    mgh = mg2R
    h = 2R
    same as height it reaches
    how can I account for friction?
     
  2. jcsd
  3. Nov 22, 2007 #2
    I'm a bit short on time, but here's a tip on question #1:

    Since the forces are all conservative, the work done is equal in magnitude to the change in potential energy of the chain. The most straightforward way to think about it (neglecting comments about "center of mass" and so forth), is to discretize the hanging part of the chain into small elements of length dy and mass dm=M/L*dy. The change of the potential energy for such an element at point y, once the chain is pulled completely up, is:

    [tex]dE(y) = dm \times g \times y = \frac{M g y dy}{L} [/tex]

    You must now integrate on the dangling part of the chain, i.e., from 0 to -L/4:

    [tex]\Delta E = \int_{-L/4}^0 dE(y) = \frac{Mg}{32}[/tex] (minus signs not withstanding)

    This makes sense if you think about the dangling chain as a small, independent chain of length L/4 and mass M/4, with center of mass at -L/8 (half of L/4, the length). Pulling up this center off mass to table height requires work equal to:

    W = mass of dangling chain * height of center of mass * g , or

    [tex]W = \frac{M}{4} \times \frac{L}{8} \times g[/tex]

    which is the same answer.

    --------
    Assaf
    Physically Incorrect
     
  4. Nov 22, 2007 #3

    Doc Al

    User Avatar

    Staff: Mentor

    re: A small solid marble

    First figure out how fast the marble needs to be going at the top of the loop in order to maintain contact. (Hint: Use Newton's 2nd law, recognizing that the motion is circular.) Then use conservation of energy, but don't neglect rotational KE. (It's rolling without slipping.)
     
  5. Nov 22, 2007 #4
    so

    F=ma
    centripedal force
    Fc = mv^2/r
    vcm = ωr
    Fc = m(ωr)^2/r
    Fc = mrω^2 = ma
    ω = sqrt(a/r)
    since a = g
    ω = sqrt(g/r)

    v = vcm

    K total = (1/2)Mv^2 + (1/2)Iω^2
    K total = (1/2)M(ωr)^2 + (1/2)Iω^2
    K total = (1/2)M([sqrt(g/r)]r)^2 + (1/2)I[sqrt(g/r)]^2
    K total = (1/2)Mgr + (1/2)Ig/r
    K total = (1/2)Mgr + (1/2)Ig/r
    I of a sphere = (2/5)Mr^2
    K total = (1/2)Mgr + (1/2)((2/5)Mr^2)g/r
    K total = (1/2)Mgr + (1/5)Mgr
    K total = (7/10)Mgr

    Etotal = Utotal - Ktotal
    Mgh = (7/10)Mgr
    h = (7/10)r

    is this correct for the first part?

    thanks
     
  6. Nov 23, 2007 #5

    Doc Al

    User Avatar

    Staff: Mentor

    marble a)

    OK, you found the speed (and ω) of the marble at the top of the loop. Good!

    (I would have left it as: gR = mv^2, but what you did was just fine.)

    Here you found the total KE of the marble at the top of the loop. Good!

    Almost. You found the height above the top of the loop, but they wanted the height above the bottom of the loop.
     
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