The raduis of rotor R = 9m
Mass of helicoper is M = 10 tonnes + 5 x (m = 120 kg blages)
Velocity of the tips of the blades is v = 300 m/s
http://www.sa-transport.co.za/aircraft/military/mi-24_hind_d_ajk.JPG
What is the angle of the cone formed by the rotor?
2007-09-14 10:01:05 · 4 answers · asked by Alexander 6 in Science & Mathematics ➔ Physics
g = 10 m/s2
assume that blades have uniform cross-section
2007-09-14 10:01:43 · update #1
Alan:
contrary to what most people think the blades are (almost) freely hinged to the axle in vertical and to some degree in tangetial direction.
The blades simply cannot support the full weight of the helicopter by shear bending. Note they can barely support their own weight when not rotating on the ground.
The helicoper is hanged on the blades by some other force. The key is to realize exactly what force.
2007-09-14 11:06:38 · update #2
Geeze...normally I can at least get a grasp of how you might solve physics questions, even if I can't solve them, but this one's a lulu. Speaking as a civil engineer, I'd need to know the properties of the rotor blades - cross sectional dimensions and modulus of elasticity. I know how much force is on each blade, then I could calculate the deflection using the section and material properties.
Added: Of course! You know the mass of the blade, and the angular speed, so you can calculate the tension at the base of the blade. Then it's a vector problem with the sum of the vertical component of the tension in all the blades equalling the weight of the helicopter.
Further added: Alright, I took a hack at it. The tension in the blade = centripedal force in the blade.
Fc = mv^2/r
m = 120 kg; v = 150 m/s (at mid length); r = 4.5 m (at mid length)
Fc = 120(150^2)/4.5 = 600 kN
Mheli = 10,000 kg
Wheli = 100 kN or 20 kN/blade (Note: I'm not sure if I should include the weight of the blades here - I want to say I shouldn't)
You have a right triangle with the blade tension as the hypotenuse, of length 600kN; vertical leg must be 20 kN.
sin theta = (20/600); theta = 1.91° above the horizontal, so the angle of the cone is about 176.2°.
2007-09-14 10:50:58 · answer #1 · answered by alan_has_bean 4 · 2⤊ 0⤋
Centripetal force Fc applied as a torque about the hinge equals the torque due to gravity force Fg. Assume the blade hinges are at the center of the hub. For each particle of the rotor at distance r from the hinge and with mass dm, Fc perpendicular to the blade Fc(perp) = sin(theta)*dm*w^2*r*cos(theta) and Fg(perp) = gM*dm/m*cos(theta). So sin(theta) = gM/(m*w^2*R/2). But w^2 = (v/(R*cos(theta)))^2 thus sin(theta) = 2R*cos^2(theta)*gM/(m*v^2).
To be continued, maybe by someone else...
Edit: Alan's formulation for Fc is more succinct than mine and differs only in lacking the cos^2(theta) factor. I believe that's necessary to resolve the horizontal cent. force to blade tension and to account for velocity v presumably being that of the deflected, not relaxed, tips. Of course at the small angle being considered, it's a tiny difference. However you should include the blade weight; they're part of the total weight. Using a total weight of 10600 kg changes theta to 2.02 deg for a cone angle of 175.96 deg.
2007-09-14 20:29:13 · answer #2 · answered by kirchwey 7 · 2⤊ 0⤋
Hey, that's an Mi-24. Those things can fly without even turning the main rotor.
http://www.youtube.com/watch?v=E8ftEKI0cLg
just kidding
Rotating image:
http://www.combataircraft.com/aircraft/hmi24_r.gif
2007-09-14 17:27:32 · answer #3 · answered by Anonymous · 0⤊ 0⤋
http://youtube.com/watch?v=BUfoRDh3gcE
2007-09-16 14:48:52 · answer #4 · answered by (Ω) 3 · 0⤊ 0⤋