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Translational velocity help

  1. Dec 15, 2008 #1
    1. The problem statement, all variables and given/known data
    A disk is released from rest from the top of an incline. The bottom of the incline is a vertical distance h=15m below the top. The wheel rolls without slipping. The moment of inertia of the disk is given by I=1/2MR2. What is the translational velocity of the disk at the bottom of the incline? (use g=9.8m/s2)


    2. Relevant equations
    PE= mgy and KE= 1/2mv^2 and ω=v/r


    3. The attempt at a solution
    I ended up with 17.1 m/s but it keeps telling me its wrong
     
  2. jcsd
  3. Dec 15, 2008 #2

    hage567

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    Homework Helper

    Are you using rotational kinetic energy anywhere in your calculations? Notice that you were given the moment of inertia of a disk.
     
  4. Dec 16, 2008 #3
    Ki = 0
    Kf = translation kinetic energy + rotational kinetic energy = 1/2 Mv^2 + 1/2 I w^2
    In the above, substitute I = 1/2 MR^2 and w = v/R to find Kf in terms of M and v.

    Ui = Mgy
    Uf = 0
    Kf + Uf = Ki + Ui
    Substitute values and solve for v.
     
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