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Kinetic energy of free fall

  • Thread starter rleung3
  • Start date
  • #1
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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.

Ryan
 

Answers and Replies

  • #2
Doc Al
Mentor
44,874
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Looks right to me. What was the exact answer that you put in?
 
  • #3
18
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I put in 1159 m (1.33x10^6 J)/(117g). It usually leaves room for errors up to 10%.
 
  • #4
18
0
OOHH!!! Nevermind..it wanted the answer in km and I put it in m. Stupid me!
 

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