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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?
    [​IMG]

    2. Relevant equations
    Thin rod, about end I=1/3M*rcm=0.529 kg*m2
    I=Icm+Md2
    Then some torque equations I don't know which to use, if any
    τ=r*F*sinϕ
    τ=m*r2
    τgrav=-M*g*rcm
    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.

    PE1+KE1=PE2+KE2

    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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