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Moment of Inertia

  1. Dec 1, 2007 #1
    Helicopter's main rotor consists of 4 4m long individual blades. One end of each of the four blades is attached to a central motor unit. Assuming that each of the blades has a uniform mass per unit length of 7.5 kg/m calculate the moment of inertia of the main rotor.
    blades are set to rotate at 6000rotations/min. Calculate the total angular momentum of the main rotor.

    moment of inertia of a rod about one end is (Ml^2)/3
    so I did (7.5kg/m x 4m)(4^2)/3
    = 160 what units?

    4x160 since there are four blades? = 640

    6000rotations/min Angular momentum given by L=Iw =640 x (6000/60) = 64000 which seems too big!

    could someone tell me if this is right
    thanks
     
  2. jcsd
  3. Dec 1, 2007 #2
    It's more or less right. You can get the units of I by dimensional analysis.
    The only mistake I can see right now is in your calculation of w- check what you've done there.
     
  4. Dec 2, 2007 #3
    should it be 6000*2PI / 60 ?
     
  5. Dec 2, 2007 #4
    captainjack,

    I came up with an answer of 128,000 J*s. You've done everything right so far, except in your first post you didn't calculate angular velocity correctly, but you have in your second post. Plug everything in and you should arrive at 128,000 (kg * m^2)/s or J*s
     
  6. Dec 2, 2007 #5
    Sorry but how?
    Moment of Inertia =640
    L = I w
    w = 6000*2PI/60
    L = 640 * (6000*2PI)/60
    = 402123
    Am I doing something really silly?
     
  7. Dec 2, 2007 #6
    What the heck? I just redid the calculation and got the same answer you just got. It must be 402,123, I must've made a mistake earlier. But I do believe that is the correct answer.
     
  8. Dec 2, 2007 #7
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