Let us treat a helicopter rotor blade as a long thin rod. a. if each of the SIX

Question

Let us treat a helicopter rotor blade as a long thin rod.

a. if each of the SIX blades is 3.75 m long and has a mass of 135kg, calculate the moment of inertia of the six rotor blades about the axis of rotation.

b. how much torque must the motor apply to bring the blades from rest to a speed of 8000rpm in 12.0s.

c. how much torque must the tail rotor apply to the helicopter's center of mass to stabalize the helicopter if it is taking off of a frictionless surface.

d. what is the power output of the main rotor engine during acceleration

e. through what angular displacement did the blades turn as they accelerated

Explanation / Answer

part a:

as axis of rotation is at the edge of the rod, moment of inertia=(1/3)*mass*length^2

=(1/3)*135*3.75^2

=632.8125 kg.m^2

hence moment of all six blades=6*632.8125=3.797*10^3 kg.m^2

part b:

as we know that, torque=moment of inertia*angular acceleration

here angulat acceleration=(final angular speed - initial angular speed)/time

=(8000 rpm-0)/12

=8000*2*pi radians/(60 seconds*12)

=69.8132 rad/s^2

hence toruqe needed=632.8125*69.8132=4.418*10^4 N.m

part c:

for rotational stabiltiy,the tail rotor must apply equal but in opposite direction torque

hence magnitude of torque require=4.418*10^4 N.m

part d:

poqwe output=total work done/time

=0.5*moment of inertia*angular speed^2/time

=0.5*632.8125*(8000*2*pi/60)^2/12

=1.8506*10^7 W

part e:

angular displacement=initial angular speed*time+0.5*angular acceleration*time^2

=0.5*((8000*2*pi/60)/12)*12^2

=5.0265*10^3 rad

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