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Magnitude of Acceleration of Masses

  1. Feb 7, 2010 #1
    1. There's a pulley with a larger mass (m1) on one side, and a smaller mass (m2) on the other side. What is the magnitude of acceleration of the masses?



    2. I think this is going to use conservation of energy equations: U = mgy and K = (1/2)mv^2



    3. I began by setting up these equations:

    D = delta
    f = final i = initial

    DUm1 + DKm1 = 0
    DUm2 + DKm2 = 0
    DUm1 + DKm1 = DUm2 + DKm2

    (Um1f - Um1i) + (Km1f - Km1i) = (Um2f - Um2i) + (Km2f - Km2i)
    (0 - Um1i) + (Km1f - 0) = (Um2f - Um2i) + (Km2f - 0)
    (-m1gy1i) + ((1/2)m1v1^2 = (m2gy2f - m2gy2i) + ((1/2)m2v2^2

    Now I think -v1 = v2, and so v1^2 = v2^2

    I can move a couple things around:

    m1((1/2)v^2 - gy1i) = m2((1/2)v^2 + gy2f - gy2i)

    But I'm not sure this gets me anywhere. Any thoughts? Thank you again for your help.
     
    Last edited: Feb 7, 2010
  2. jcsd
  3. Feb 7, 2010 #2

    ideasrule

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    You don't need to use the conservation of energy. Since both mass and acceleration was mentioned in the problem, you should immediately think of F=ma.

    Draw a free-body diagram for both masses, then write out Newton's second law for both masses.
     
  4. Feb 7, 2010 #3
    Does this problem assume no rotational influence on the answer? If so, use [tex]F=ma[/tex] to generate an equation of acceleration for each mass. You will have two unknowns: tension and acceleration, which can be solved with your two equations. The key is to realize that each mass's acceleration has the same magnitude as the other mass, because you cannot create nor destroy rope, and the rope is assumed not flexible. Therefore, when one mass goes down 1 meter, the other must go up 1 meter.

    If you experience problems deriving the equation, many websites go over this exact pulley problem:
    http://en.wikipedia.org/wiki/Atwood_machine
    http://hyperphysics.phy-astr.gsu.edu/hbase/Atwd.html
     
  5. Feb 7, 2010 #4
    My professor started the semester with Conservation of energy, and I've heard from other students that's not the way most classes go. With that in mind, we have yet to encounter any equation that looks like F = ma or tension. Or force at all for that matter.

    Thanks for the links, I'm going to read them over right now.
     
  6. Feb 7, 2010 #5

    PhanthomJay

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    If you must use conservation of energy, your equation is wrong. You must set the change in energy of the system of both masses equal to zero, not individually equal to zero. You should also assume for simplicity that both masses are lined up at the start (no initial PE for each).
     
  7. Feb 7, 2010 #6
    He had us do a problem yesterday with the techniques he'd like us to use for this problem I think.

    He wanted us to show that acceleration = -g. We did this like this:

    dv/dt = (dv/dK)(dK/dU)(dU/dy)(dy/dt)

    a = (1/mv) (-1) (mg) (v)

    a = -g

    Using this system, I came up with:

    a = (dv/dK) (dK/dU) (g(m1-m2)) (v)

    (dU/dy) represents weight and (dy/dt) is velocity
    I don't remember what dv/dk and dK/dU represent.

    Based on those articles, the answer is a = g(m1-m2)/(m1+m2), but I don't know what dV/DK and dK/dU represent.

    Am I on the right track?
     
  8. Feb 7, 2010 #7

    PhanthomJay

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    I have never seen such a roundabout way of solving acceleration; I have no comment on it. Either you use Newton's 2nd law, as others have previously noted,or, since you haven't been taught it , use conservatuion of energy:
    Delta K1 + Delta U1 + Delta K2 + Delta U2 = 0. You were on the right track before, with those couple of adjustments necessary. Then you need to use kinematic equations to solve for the acceleration. A bit of long way around to solve the problem.
     
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