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Rotational Kinematics: Falling Rod

  1. Jan 8, 2012 #1
    1. A uniform rigid rod of length L and mass M is attached to a frictionless pivot point at one end. The rod is initially held completely horizontal and is released from rest. What is the linear and angular velocities at the instant the rod is in the vertical position?


    2. -ΔU = ΔK

    3. -ΔU = ΔK(trans) + ΔK(rotat)
    MgΔh = .5MV^2 + .5Iω^2 and I = (1/3)ML^2 since it is a rod pivoted at the end
    MgΔh = .5MV^2 + .5((1/3)ML^2)ω^2 and Lω = V -------> V = (ω/L)
    2gΔh = V^2 + (1/3)V^2
    2gΔh = (4/3)V^2
    V = ((3/2)gΔh) ^ (1/2) and ω = V/L = (((3/2)gΔh)^(1/2))/L

    Is this correct? It then says to confirm your answer by using two photogates at the vertical position. The info for that problem is:

    Length of the rod: 1m
    Time it takes to get to the vertical position: 0.4419s
    Time it takes to pass between the photogates: 0.0032s
    Distance between the photogates: 0.016m
    Mass of the rod (doesn't matter): 92.4g
     
  2. jcsd
  3. Jan 8, 2012 #2

    ehild

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    I do not think your derivation is correct. First, you did not specified what Δh and V are.
    If you consider the motion of the rod with respect to the pivot then it is rotation about a fixed axis. KE=Iω2, I = 1/3 ML2.

    ehild
     
  4. Jan 8, 2012 #3
    Sorry for not being clearer:
    Δh = L since h(i) = 0 and h(f) = L
    and V is the linear velocity at the vertical position
     
    Last edited: Jan 8, 2012
  5. Jan 9, 2012 #4

    ehild

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    The change of the potential energy of the uniform rigid rod is not mgL. Think: the piece at the pivot does not change height, while the height of the opposite end changes by L.

    Linear velocity of what piece of the rod do you speak about ?

    ehild
     
  6. Jan 9, 2012 #5
    HINT: 1.consider center of mass OF THE ROD ?
    {--- BODY HAS NO TRANS. KINETIC ENERGY}
     
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