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Falling Velocity Relative to Weight

  1. Feb 26, 2009 #1
    1. The problem statement, all variables and given/known data

    A wooden rod of negligible mass and length 80.0 cm is pivoted about a horizontal axis through its center. A white rat with mass 0.490 kg clings to one end of the stick, and a mouse with mass 0.240 kg 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

    free fall acceleration = 9.80 m/s^2.

    3. The attempt at a solution

    I would love to attempt at the solution, but I haven't even the slightest clue where to start, I'm quite confused :(
  2. jcsd
  3. Feb 26, 2009 #2


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    I would think about conservation of energy. The total kinetic energy of the unfortunate mammals is equal to the change in gravitational potential energy.
  4. Feb 26, 2009 #3
    I'm not making too much headway on this problem..

    I've got

    PE = change in mgh and
    KE = change in 1/2mvf^2 - 1/2mvi^2

    PE = 5.7232, assuming m = 0.490 kg + .240 kg, and g = 9.8, and h = .80m

    but I don't really know where to go from there. and is that PE even correct? or do I have to do it separately for each animal, and add them? ahh, so confused
  5. Feb 26, 2009 #4


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    Ok, so mouse goes up, rat goes down. PE=mgh. One h is positive and one is negative. Since one goes up and one goes down. Hence, don't add the masses. Calculate each one separately. Initial KE is zero.
  6. Feb 26, 2009 #5
    Alright, i'm still a little confused but I'm getting there, I think..

    PE(mouse) = (9.8m/s)(.24kg)(0.8m) = 1.8816
    PE(rat) = (9.8)(.49)(-0.8) = -3.8416

    change in PE = 5.7232

    5.7232 = 1/2 mv^2

    v = 3.38?

    Which is still wrong, so.. I'm doing something wrong hehe
  7. Feb 26, 2009 #6


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    You have to ADD the PE's. If the mouse and the rat had the same mass there would be no change in PE.
  8. Feb 26, 2009 #7
    Ok, so..

    Total PE = -1.96

    -1.96 = (1/2) (0.49 + 0.24)(v^2)

    Assume the negative sign is negligible?

    v = 2.3?
  9. Feb 26, 2009 #8


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    The minus sign isn't negligible. You just have to think about what it means relative to your choice of the sign of h. But yes, roughly 2.3m/s.
  10. Feb 26, 2009 #9
    The website is telling me 2.3 is incorrect.. I re-calculated, and it's coming up 2.3172, rounded to the tenths is 2.3..

    Is there anything else we might not be accounting for?
  11. Feb 27, 2009 #10


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    I don't think so. I worked it out again and agree with your number. Did you put correct units on it?
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