Angular Momentum In A Helicopter

  • Thread starter godtripp
  • Start date
  • #1
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So we have a helicopter with a blade that has the moment of inertia "I" that is "2R" long and is spinning at an angular velocity [tex]\omega[/tex].

To stabilize the aircraft, the tail ruder must exert an equal an opposite force.

Can someone point me in the direction of how to calculate the force needed?

I'm thinking conservation of angular momentum, so I want to say

F=I[tex]\omega[/tex].

But I don't believe that's correct.
So I started thinking energy.

F=1/2I[tex]\omega[/tex].^2

Which still seems funny.


It's a problem I invented to test my understanding and prepare for the final, so if I'm missing a variable that would make this calculation easier let me know.

Thank you!
 

Answers and Replies

  • #2
472
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Hi there,

I am not an expert, and never really gotten close to a chopper, but I thought that they had more than one blade (3 or 4), placed opposite from each other. If my thoughts are correct, then the momentum of each blade should counteract the others.

Cheers
 
  • #3
54
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Those are dual rotor helicopters, which have several varieties of how they can be aligned. However single rotor helicopters traditionally use the description above.

I did a little bit of wiki-research before I considered the physics.
http://en.wikipedia.org/wiki/Helicopter_rotor
The descriptions are rather brief and informative if you're interested.
 
  • #4
54
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Ok so here's my thought L(Helicopter)=I(rotor)*Omega(rotor)

taking the derivative with respect to time on each side I get Torque=I*alpha(rotor)

So F*R=I*alpha

So F=(I*alpha)/R

is that correct?
 

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