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Falling rod with a mass on either end

  1. Feb 26, 2008 #1
    [SOLVED] Falling rod with a mass on either end

    1. The problem statement, all variables and given/known data

    A wooden rod of negligible mass and length 85.0cm is pivoted about a horizontal axis through its center. A white rat with mass 0.520kg clings to one end of the stick, and a mouse with mass 0.250kg clings to the other end. The system is released from rest with the rod horizontal.

    If the animals can manage to hold on, what are their speeds as the rod swings through a vertical position?

    2. Relevant equations

    (K1a + K1b) + (U1a + U1b) = (K2a + K2b) + (U2a + U2b)

    U = mgR, R = 1/2d of the rod
    K1 = 0
    U2 = 0

    3. The attempt at a solution

    (magR + mbgR) = 1/2v^2(ma + mb)

    gR (ma+mb) = 1/2v^2 (ma+mb)

    2gr = v^2, v = [tex]\sqrt{2gr}[/tex]

    v = 2.9m/s wrong

    not sure what I'm doing wrong here.

    any help is appreciated.
    Last edited: Feb 26, 2008
  2. jcsd
  3. Feb 26, 2008 #2


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    The rat goes down, losing potential energy and the mouse goes up gaining potential energy. The potential energy changes don't add. They subtract. They have different signs.
  4. Feb 26, 2008 #3
    hmmm, so you're saying it would be:

    magR - mbgR = 1/2mav^2 + 1/2mbv^2?

    and from here would'nt it be:

    gR(ma-mb) = 1/2v^2(ma+mb)

    [tex]\sqrt{2gr((ma-mb)/(ma+mb))}[/tex] = v?
  5. Feb 26, 2008 #4


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    I think so. In fact, I know so. Does that work?
    Last edited: Feb 26, 2008
  6. Feb 26, 2008 #5
    yes it did! wow I was stuck on that one for hours and I was actually thinking of changing the signs, but when I did it I changed the signs of both the potential and kinetic energies so they (the masses) both cancelled out anyways, thanks for pointing out the difference; thanks again!
    Last edited: Feb 26, 2008
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