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Algebra of pulley-mass energy equation

  1. Nov 24, 2014 #1
    1. The problem statement, all variables and given/known data

    A 0.20-kg block and a 0.25-kg block are connected to each other by a string draped over a pulley that is a solid disc of inertia 0.50 kg and radius 0.10 m. When released, the 0.25-kg block is 0.30 m off the ground. What speed does this block have when it hits the ground?

    2. Relevant equations

    ΔE = 0
    Ei = Ef (closed system)
    E = K + UG + Eth + Es
    UiG = Ktransf + Krotf
    Ktrans = ½mv2
    Krot = ½Iω2
    ISolid Cylinder = ½mr2
    ω = v/r

    3. The attempt at a solution

    First, I step up the energy equation and substitute in my energy expressions. I've used the subscript "1" to represent the 0.20-kg mass, the subscript "2" to represent the 0.25-kg mass, and the subscript "p" to represent the pulley.

    U1iG + U2iG = U1fG + U2fG + K1f + K2f + Kpf

    m1gh1i + m2gh2i = m1gh1f + m2gh2f + ½m1v12 + ½m2v22 + ½Iω2

    I know what the next step is, but I'm not sure how to get there algebraically.

    I should end up with:

    (m1 - m2)gΔh = ½(m1 + m2)vf2 + ¼mpvf2

    I understand that v1, v2, and vp are equal, so no problem there. I see that how we substitute ½mr2 for I, and v/r for ω. I don't understand how we end up with (m1 - m2)gΔh. I understand that we pull the common variable g out, and I can kind of understand how the h-values become Δh. However, it looks like this when I set it up:

    m1gh1i + m2gh2i - m1gh1f - m2gh2f = ½m1v12 + ½m2v22 + ½Iω2

    Which becomes:

    g(m1h1i + m2h2i - m1h1f - m2h2f) = ½m1v12 + ½m2v22 + ½Iω2

    Any help is greatly appreciated.
  2. jcsd
  3. Nov 25, 2014 #2


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    Homework Helper

    The length of the rope does not change. If the 0.25 kg block descends by 0.3 m, the other block rises by 0.3 m.
    (h1f-h1i= h2i-h2f)
  4. Nov 25, 2014 #3
    Thanks for the reply, ehild.

    I initially responded with another question, but I think I understand now.
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