Roational Mechanics of Helicopter Blades

In summary, a uniform helicopter rotor blade is 8.82 m long, has a mass of 108 kg, and is attached to the rotor axle by a single bolt. The magnitude of the force on the bolt from the axle when the rotor is turning at 302 rev/min is F= (mr/omega^2) and the torque that must be applied to the rotor to bring it to full speed from rest in 7.14 s is T= I\alpha. For the blade to reach a speed of 302 rev/min, Work is done by the torque on the blade equal to .5*I\alpha.
  • #1
Zyxer22
16
0

Homework Statement



A uniform helicopter rotor blade is 8.82 m long, has a mass of 108 kg, and is attached to the rotor axle by a single bolt. (a) What is the magnitude of the force on the bolt from the axle when the rotor is turning at 302 rev/min? (Hint: For this calculation the blade can be considered to be a point mass at its center of mass. Why?) (b) Calculate the torque that must be applied to the rotor to bring it to full speed from rest in 7.14 s. Ignore air resistance. (The blade cannot be considered to be a point mass for this calculation. Why not? Assume the mass distribution of a uniform thin rod.) (c) How much work does the torque do on the blade for the blade to reach a speed of 302 rev/min?

Homework Equations



F=mv[itex]^{2}[/itex]/r

F= mr[itex]\omega^{2}[/itex]

Since it's a uniform rod;

I=[itex]\frac{1mL^{2}}{12}[/itex]
[itex]T[/itex] = I[itex]\alpha[/itex]
[itex]\alpha[/itex] = [itex]\omega[/itex]/t
W=.5I[itex]\omega^{2}[/itex]

The Attempt at a Solution



Using F= mr[itex]\omega^{2}[/itex]
I changed 302rev/min to 31.625 rad/s
plugging in m= 108kg r = 4.41m

F= 4.76 * 10^5
-->This answer is wrong.


[itex]\alpha[/itex] = [itex]\omega[/itex]/t
[itex]\alpha[/itex] ~ 4.429 rad/s[itex]^{2}[/itex]

I=[itex]\frac{mL^{2}}{12}[/itex]
[itex]T[/itex] = I[itex]\alpha[/itex]
[itex]T[/itex] = 775.58 Nm
--> This answer is also wrong

W=.5I[itex]\omega^{2}[/itex]
W=87530 J
--> Also wrong

Any help would be greatly appreciated, thanks :)
 
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  • #2
This question is still getting the better of me if anyone could help.
 
  • #3
Zyxer22 said:
Using F= mr[itex]\omega^{2}[/itex]
I changed 302rev/min to 31.625 rad/s
plugging in m= 108kg r = 4.41m

F= 4.76 * 10^5
-->This answer is wrong.
I can't see any mistake, other than units of F have been omitted.

[itex]\alpha[/itex] = [itex]\omega[/itex]/t
[itex]\alpha[/itex] ~ 4.429 rad/s[itex]^{2}[/itex]

I=[itex]\frac{mL^{2}}{12}[/itex]
[itex]T[/itex] = I[itex]\alpha[/itex]
[itex]T[/itex] = 775.58 Nm
--> This answer is also wrong
I believe moment of inertia for a rod rotated about its end is I = (m L2) /3

http://en.wikipedia.org/wiki/List_of_moments_of_inertia"
 
Last edited by a moderator:
  • #4
Right, rotated about it's end the moment of inertia is I = (m L2) /3
I was assuming that the rotational axis is in the center of the blade though, leading to my equation.
Using I = (m L2) /3 actually leads to correct answers. Thanks ^^
 
  • #5
Scratch that... I get the correct answers according to the book, but using the numbers from online I get the wrong answers... makes no sense to me
 

1. What is the purpose of rotational mechanics in helicopter blades?

The purpose of rotational mechanics in helicopter blades is to generate lift and control the movement of the helicopter. The blades are designed to rotate at different speeds and angles, creating a difference in air pressure that allows the helicopter to fly.

2. How do helicopter blades create lift?

Helicopter blades create lift through the principle of Bernoulli's equation, which states that as the speed of a fluid (in this case, air) increases, its pressure decreases. As the blades rotate, they create a difference in air pressure, with lower pressure on the top of the blade and higher pressure on the bottom. This pressure difference generates lift, allowing the helicopter to fly.

3. How do helicopter blades control the movement of the helicopter?

The pitch of the blades, or the angle at which they are positioned, can be adjusted to control the direction and speed of the helicopter. By changing the pitch, the amount of lift generated by the blades can be altered, allowing the helicopter to move up, down, forward, or backward.

4. What factors affect the performance of helicopter blades?

The performance of helicopter blades can be affected by various factors, including the shape and design of the blades, the speed of rotation, the angle of attack, and the weight and balance of the helicopter. Environmental factors such as air density, temperature, and wind speed can also affect the performance of the blades.

5. How do scientists improve the design of helicopter blades?

Scientists use advanced computational simulations and wind tunnel experiments to study the performance of different blade designs and make improvements. They also consider factors such as material strength, weight, and cost when designing and testing new blade designs for optimal performance and safety.

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