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Rotational Kinetic Energy: Pole Balanced Vertically

  1. Nov 26, 2013 #1
    1. The problem statement, all variables and given/known data

    A 1.80m long pole is balanced vertically with its tip on the ground. It starts to fall and its lower end does not slip. What will be the speed of the upper end of the pole just before it hits the ground? [hint: Use conservation of energy]

    l=1.890m


    2. Relevant equations

    mgh=1/2I(w)^2
    v=wr

    3. The attempt at a solution

    So I know how to do the problem but I don't quite understand it.

    The original method I used to solve the problem was:

    mgh=1/2I(w)^2
    mgh=(1/2)(1/3m(l^2))(w^2)
    *masses cancel to get:
    gh=(1/6)(ml^2)(w^2)
    Solve for w:
    sqrt((6gh)/l^2)=w

    However this got me the wrong answer and I realized instead of plugging in h, you must use l/2. Why is this? Is there an equation or something I am missing?

    Correct Solution:

    sqrt((6g(l/2))/l^2)=w
    sqrt((6(9.8m/s^2)(1.8m/2))/(1.8^2)=w
    w=4.04rad/s
    v=rw
    = 1.8m(4.04rad/s)
    =7.27m/s

    Thanks!
     
  2. jcsd
  3. Nov 26, 2013 #2
    Here you should consider the center of mass of the falling rod.It is located at the geometric center of the rod.
    Welcome to PF!
     
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