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Kinetic energy of free fall

  1. Oct 28, 2006 #1
    An airplane propeller is 2.08 m in length (from tip to tip) with mass 117 kg. The propeller is rotating at 2400 rev/min about an axis through its center. Its rotational kinetic energy is 1.33x10^6 J. If it were not rotating, how far would it have to drop in free fall to acquire the same kinetic energy?

    I tried using this equation:

    K1 (becomes 0) +U1 = K2+U2 (becomes 0)

    since the object would start from rest, K1=0. I will call the height of the object y=h, and the point at which it reaches required kinetic energy will be at y=0

    U1 = K2
    mgh = 1.33x10^6 J
    h=(1.33x10^6 J)/(mg)

    But that answer is wrong. The computer doesn't give me the right answer, but it says that one is wrong. Any thoughts? Thank you.

  2. jcsd
  3. Oct 28, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks right to me. What was the exact answer that you put in?
  4. Oct 28, 2006 #3
    I put in 1159 m (1.33x10^6 J)/(117g). It usually leaves room for errors up to 10%.
  5. Oct 28, 2006 #4
    OOHH!!! Nevermind..it wanted the answer in km and I put it in m. Stupid me!
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