CATAGORY: Mechanics
QUESTION: A helicoptor rotor blade can be considered  a long thin rod,  If
each of the three rotor blades is 3.37 m long and has a mass of of 160 kg,
calculate the moment of inertia of the three rotor blades about the axis
of
rotation.  How much torque must the motor apply to bring the blade up to a
speed of 5.o rev/s in 8.0 s/

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The angular acceleration alpha is given by

             Change in rev/s  X 2 pi
(1)  alpha = ------------------------                     
                   time

              (5.0 rev/s - 0.0 rev/s) x 2 pi
           = -------------------------------
                       8.0s
                       
           = 3.93 rad/s^2
           
The torque is given by

(2)   Torque = I alpha

Where I is the rotational moment of inertia of the propeller.  For a
rod pivoted about its center

(3)     I = 1/12 M L^2

Where M is the mass and L is the length.  Here M = 160kg and L =3.37m
so

(4)     I = 151 kg m^2


Substitution into (2) gives

(5)   Torque = 151 kg m^2  X  3.93 rad/s^2 

             = 593 Nm                      

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       *                 *
       * Torque = 593 Nm *
       *                 *
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