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Sledgehammer on pivot, using inertia, center of mass, finding speed

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data
    A sledgehammer with a mass of 2.60kg is connected to a frictionless pivot at the tip of its handle. The distance from the pivot to the center of mass is rcm=0.610m, and the moment of inertia about the center of mass is Icm=0.04kg*m2. If the hammer is released from rest at an angle of Θ=53.0° such that H=0.487m, what is the speed of the center of mass when it passes through horizontal?

    2. Relevant equations
    Thin rod, about end I=1/3M*rcm=0.529 kg*m2
    Then some torque equations I don't know which to use, if any
    and α=τnet/I
    And eventually I'll have to use either
    vtan=ω*r or v=√arad*r
    360°=2π radians

    3. The attempt at a solution
    I think Fgrav runs along H. ϕ=143° so Ftan=Fgrav*sin(-143°) where Fgrav=m*g.
    I can also get I=Icm+M*rcm2 but then I get lost and don't know where to go with all the info. If I get α, I don't know how I would use rotational kinematics since there is no time given. Any clues to get going would help. Thanks
  2. jcsd
  3. Nov 10, 2008 #2
    How about conservation of energy here? That way you do not have to deal with any forces.


    Also, why did you calculate the mass moment of inertia if it was given to you in the problem statement?
  4. Nov 10, 2008 #3
    That's good advice
  5. Nov 10, 2008 #4
    You know that the initial velocity is=0 so KE1 is 0. It has only PE1.

    At the horizontal, PE2=0, so you can find the velocity.

    (If you do not understand why PE2=0, please ask and I will explain)
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