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[Mechanics] Friction decelerating a helicopter rotor assembly

  1. May 31, 2017 #1
    1. The problem statement, all variables and given/known data
    A Chinook helicopter’s rotor blades and hub have a top speed of 300 revs/min and a combined mass of 300 kg.
    On a maintenance test the blade assembly is allowed to stop without applying the brake, in this condition the blades take 48 seconds to come to a standstill. The effective radius off the rotor is 6.8 m.

    I need to calculate the following:

    a. The deceleration of the rotor blades, assuming a constant rate of deceleration.

    b. The number of turns the rotor assembly will make.

    c. The frictional resistance at this deceleration.

    2. Relevant equations

    3. The attempt at a solution
    I have completed the first two parts:

    300 rev/min * (2pi)/60 = 31.416 rad/s

    Acceleration = (ω2 - ω1)/(t2 - t1) = (0-31.416)/(48) = -0.6545 rad/s^(2)

    ϴ = ω1 * t + 1/2 * a * t^(2) = (31.416*48)+(0.5x(-0.6545)*48^(2)) = 753.984 rads

    753.984/(2pi) = 120 revolutions

    For the final part, all I have came up with so far is that the friction will be acting downwards as a result of gravity, F = mg = 300*9.81 = 2943N

    I'm not sure if this is the right thing to do though, as the question asks the friction 'at this deceleration', so I guess I need an equation that contains acceleration in it.

    Thank you for any help!

    EDIT: Do I need to use the equation:

    Frictional torque = angular acceleration * Inertia?
     
    Last edited: May 31, 2017
  2. jcsd
  3. May 31, 2017 #2

    berkeman

    User Avatar

    Staff: Mentor

    Yes, or something similar. The force downward due to gravity would not enter into the calculation, IMO.
     
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