- #1
rleung3
- 18
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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
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