Question

Let us treat a helicopter rotor blade as a long thin rod, as shown in the figure. (Figure 1) . If each of the three rotor helicopter blades is 3.75 m long and has a mass of 135 kg, calculate the moment of inertia of the three rotor blades about the axis of rotation. b- How much torque must the motor apply to bring the blades from rest up to a speed of 5.9 rev/s in 10 s ?

Answer 1

The moments of inertia are listed on and a long thin rod through its end is:
I = ¹/₃ML²
so each rotor has a moment of inertia of:
I = ¹/₃(135 kg)(3.75 m)²  = 632.81 kgm²
Three rotors would have three times this moment:
I = (632.81 kgm²)3 = 1898.43kgm²

Now we need to solve a kinematics question:
w_o = 0
w = (5.9 Revolutions/s)(2p radians/revolution) = 37.052 rad/s
t = 10 s

Now apply:
w = w_o + at
a = 3.70 rad/s/s

And now find the torque using F = ma:
t = Ia
t = (1898.43 kgm²)(3.70 rad/s/s) =7024.191 Nm of torque