A helicopter has two blades?

Question

each of which has a mass of 240 kg and can be approximated as a thin rod of length R = 6.2 m.

a) What is the total moment of inertia of the two blades about the axis of rotation?

(b) Determine the rotational kinetic energy of the spinning blades.

Sorry, didn't see this. The blades are rotating at an angular speed of 46 rad/s.

Answer 1

a) Moment of inertia of each blade is ML^2/3 (see link below). So total moment of inertal is I = 2*ML^2/3 = 2*240*6.2^2/3 = 6160
b) Rotational kinetic energy is I*w^2/2 = 6160*46^2/2 = 6.4*10^6 J

In my previous answer I assumed that blades are suspended at the middle and considered I = ML^2/12, but as Alexander pointed blades most probably are fixed at the end.

Answer 2

Helicopter Two Blades

Answer 3

T = Ia; where T is torque, I is moment of inertia, and a is angular acceleration. T also = F r; where F is force applied a distance r from the axis of rotation. Here, r = R = 6.2 m.

Thus, I = T/a = FR/a and you have the moment of inertia if you know the force applied to the blade tips and the blades are accelerating at a.

a) But given you failed to mention any blade acceleration, or any torque on the rotors or force on the blade tips, I cannot be caculated. Torque exists if and only if there is angular acceleration and you failed to mention any.

b) Further, as you failed to give the angular velocity (w) in KE = 1/2 I w^2; where w = at, the angular velocity after accelerating t time, we cannot find rotational kinetic energy.

Oops, my bad, Alex is absolutely correct. In my I = FR/a, I should have continued on with I = FR/a = MaR/a = MR; where M is the effective mass of a "thin rod" (gotten by integration), which is 1/3 mR as Alex noted.

Answer 4

Moement of inertia of each blade about its end, where it is attached to the axis:
I = 1/3 mR²