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Rotational Kinetic Energy of an airplane propeller

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data
    An airplane propeller is 2.02 m in length (from tip to tip) with mass 127 kg and is rotating at 2300 rpm about an axis through its center. You can model the propeller as a slender rod.
    a)What is its rotational kinetic energy?
    b)Suppose that, due to weight constraints, you had to reduce the propeller's mass to 75.0\% of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?


    2. Relevant equations
    K=0.5 (mr^2) w^2
    (m***, radius, w angular speed)


    3. The attempt at a solution
    a)
    2300 RPM = 38.33 RPS = 76.66 radians/s
    2300RPM=240 rad/s

    KE=0.5 (127) (1.01)^2 (240)^2 = 3731117.76 J

    b)
    new mass = 95.25kg

    KE = 0.5(95.25)(1.01)^2 (w)^2

    2 KE /(95.25)(1.01)^2) = (w)^2

    [2 KE /(95.25)(1.01)^2)]^0.5 = w


    but im sure my original KE is wrong so..

    thanks for the help
     
    Last edited: Dec 8, 2008
  2. jcsd
  3. Dec 8, 2008 #2

    LowlyPion

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  4. Dec 8, 2008 #3
    2300RPM=240 rad/s

    KE=0.5 (1/12)M L^2 (w)^2 ?
    giving

    1/24 x 127 x 2.02^2 x 240^2 = 1243705.9
    ?
     
  5. Dec 8, 2008 #4

    LowlyPion

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    Homework Helper

    I get 241 for ω.

    I didn't calculate the rest out, but that looks like the right method now.
     
  6. Dec 9, 2008 #5
    bingo
    thanks lots and lots
     
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