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Homework Help: PLEASE HELP - Rotational Kinetic Energy

  1. Nov 7, 2005 #1
    Please help me get this, it's due tonigh tonight on webassign. The question is:

    A 3.30 m long pole is balanced vertically on its tip. It is given a tiny push. What will be the speed of the upper end of the pole just before it hits the ground? Assume the lower end does not slip.

    I know to use MgH = 1/2 MV^2 + 1/2 Iw^2

    but when you plug in for w, what do you use? w is rotational velocity and you have to know the time frame, and it doesn't give you that. how do you do this problem??
  2. jcsd
  3. Nov 7, 2005 #2
    My editing caused double posted for some reason. :\
  4. Nov 7, 2005 #3
    Since there is no slippage, all energy is rotational (PE = RE.) Your moment of inertia in this case is [itex]\frac{1}{3}mL^2[/itex], where "L" is the length of the pole. You can find the potential energy when the pole is standing up (mgh) but remember that the center of mass is at h = L/2. And remember that [itex]v = r\omega[/itex]

    Oh, and let me guess, Physics - Giancoli :tongue:
    Last edited: Nov 7, 2005
  5. Nov 7, 2005 #4
    haha yeah its giancoli. thanks so much for your help, but i still have one question though. you you are plugging in v/r for w, what do you use for r?
  6. Nov 7, 2005 #5
    The length of the pole because it asks for the speed of the top end of the pole which is 3.30m from the pivot point.
    Last edited: Nov 7, 2005
  7. Nov 7, 2005 #6
    hey thanks i got it
  8. Nov 7, 2005 #7
    No Problem :smile:
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