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Roational Mechanics of Helicopter Blades

  1. Oct 13, 2011 #1
    1. The problem statement, all variables and given/known data

    A uniform helicopter rotor blade is 8.82 m long, has a mass of 108 kg, and is attached to the rotor axle by a single bolt. (a) What is the magnitude of the force on the bolt from the axle when the rotor is turning at 302 rev/min? (Hint: For this calculation the blade can be considered to be a point mass at its center of mass. Why?) (b) Calculate the torque that must be applied to the rotor to bring it to full speed from rest in 7.14 s. Ignore air resistance. (The blade cannot be considered to be a point mass for this calculation. Why not? Assume the mass distribution of a uniform thin rod.) (c) How much work does the torque do on the blade for the blade to reach a speed of 302 rev/min?

    2. Relevant equations

    F=mv[itex]^{2}[/itex]/r

    F= mr[itex]\omega^{2}[/itex]

    Since it's a uniform rod;

    I=[itex]\frac{1mL^{2}}{12}[/itex]
    [itex]T[/itex] = I[itex]\alpha[/itex]
    [itex]\alpha[/itex] = [itex]\omega[/itex]/t
    W=.5I[itex]\omega^{2}[/itex]

    3. The attempt at a solution

    Using F= mr[itex]\omega^{2}[/itex]
    I changed 302rev/min to 31.625 rad/s
    plugging in m= 108kg r = 4.41m

    F= 4.76 * 10^5
    -->This answer is wrong.


    [itex]\alpha[/itex] = [itex]\omega[/itex]/t
    [itex]\alpha[/itex] ~ 4.429 rad/s[itex]^{2}[/itex]

    I=[itex]\frac{mL^{2}}{12}[/itex]
    [itex]T[/itex] = I[itex]\alpha[/itex]
    [itex]T[/itex] = 775.58 Nm
    --> This answer is also wrong

    W=.5I[itex]\omega^{2}[/itex]
    W=87530 J
    --> Also wrong

    Any help would be greatly appreciated, thanks :)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 13, 2011 #2
    This question is still getting the better of me if anyone could help.
     
  4. Oct 14, 2011 #3

    NascentOxygen

    User Avatar

    Staff: Mentor

    I can't see any mistake, other than units of F have been omitted.

    I believe moment of inertia for a rod rotated about its end is I = (m L2) /3

    http://en.wikipedia.org/wiki/List_of_moments_of_inertia" [Broken]
     
    Last edited by a moderator: May 5, 2017
  5. Oct 14, 2011 #4
    Right, rotated about it's end the moment of inertia is I = (m L2) /3
    I was assuming that the rotational axis is in the center of the blade though, leading to my equation.
    Using I = (m L2) /3 actually leads to correct answers. Thanks ^^
     
  6. Oct 14, 2011 #5
    Scratch that.... I get the correct answers according to the book, but using the numbers from online I get the wrong answers.... makes no sense to me
     
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