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Mass falls on lever, catapulting mass on other end

  1. Dec 2, 2012 #1
    1. The problem statement, all variables and given/known data
    You have constructed a simple catapult which basically consists of a 4 meter long rod of negligible mass attached to a fulcrum point offset from the center of mass by 1.5 meters. You decide to launch a m2 = 0.100 kg mass from the long end of the rod by dropping a m1 = 20.0 kg ball of putty from a height of h = 5 m onto the short side of the rod. The putty sticks to the rod when it hits it, causing the rod to rotate about the fulcrum until the short end of the rod hits the ground. At this point the small mass is launched into the air at an angle of 45 degrees. You may model the mass of putty and the small block as point particles. How far away from the launch point does the mass land when it hits the ground?

    m1 = 20.0 kg
    m2 = 0.100 kg
    r1 = .5 m
    r2 = 3.5 m
    L = 4 m
    h = 5 m
    θ = 45°

    2. Relevant equations



    3. The attempt at a solution
    In my solution, I assumed that no Kinetic Energy was lost during the collision and that the linear Kinetic Energy just translated into Rotational Kinetic Energy. So the rotational kinetic energy of mass 1 as it starts to rotate is equal to its change in GPE from falling. That was the part I was most unsure of.

    I then used energy conservation to solve for ωf:

    KA0 + UA0 + KB0 + UB0 = KAf + UAf + KBf + UBf
    where A refers to mass 1, B refers to mass 2, t0 is the time of impact, and tf is the time of release

    KAf + KBf = KA0 + UA0 + UB0 - UBf

    Abstract away the known energies: let C = KA0 + UA0 + UB0 - UBf
    KAf + KBf = C

    .5mArA2ω2 + .5mBrB2ω2 = C

    ω = √(2C/(mArA2 + mBrB2))
     
  2. jcsd
  3. Dec 3, 2012 #2

    TSny

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    The collision of the putty with the stick will probably be inelastic, so you don't want to assume that kinetic energy is conserved during the collision.

    However, there's another quantity that you can argue is conserved during the collision.
     
  4. Dec 3, 2012 #3
    It's momentum, right? vf = (m0 * v0)/mf
    But what is mf?
     
  5. Dec 3, 2012 #4
    Oh I think I get it now. mvr = (Ia + Ib)w
    So once I find the angular velocity right after impact, the rest of my equations hold, right?
    Total energy = Kinetic Energy + Gravitational Potential Energy?
     
  6. Dec 3, 2012 #5

    TSny

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    Yes. Good!
     
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