Moment of Inertia of helicopter blades.

[eptheta]

I have built a model of rotating blades of a helicopter and want to mathematically calculate it's moment of inertia. I am unsure on how to do that.
Its a simple model:
a square wooden base with uniform thickness and 4 equally long blades made of styrofoam from the midpoints of the 4 sides of the square.
The rotational axis is through the centre of the square

Lets just say i know the density, mass and thickness of the base and the blades, how do i get moment of inertia for a specific length of blades ?

Thank you

 

[rcgldr]

You could approximate the rotor blades as rods rotated at the end.

Angular inertia = m L² / 3

http://en.wikipedia.org/wiki/List_of_moments_of_inertia

Note that real helicopter blades are hinge mounted at the base to reduce stress on the main rotor head.

 

[eptheta]

What if i have something like this:
http://img706.imageshack.us/img706/418/moin.png
I have those extra triangles at the center to deal with. can i just get the MOI of the triangles and add it to the MOI of the rectangular blades ?
yeah, i know I'm approximating a LOT, but it'll have to do.
Can anyone help ? Can anyone show me the derivation of it as well ?

 

[pgardn]

I have built a model of rotating blades of a helicopter and want to mathematically calculate it's moment of inertia. I am unsure on how to do that.
Its a simple model:
a square wooden base with uniform thickness and 4 equally long blades made of styrofoam from the midpoints of the 4 sides of the square.
The rotational axis is through the centre of the square

Lets just say i know the density, mass and thickness of the base and the blades, how do i get moment of inertia for a specific length of blades ?

Thank you

The moment of inertia of a rectangular plate might get you even closer depending how you want to set it up. Its usually given in textbooks. I personally don't like to do the integration on some of these figures.

The moment of inertia of a rectangular plate is 1/12m(a^2 + b^2) where a is the width and b the length of the plate. Note that this is the moment of inertia taken from the center of mass of the plate.

I see your picture. So a and b are b and l...

So really one plate would have length 2l +b and width b... and there are two of them.

 

[eptheta]

The moment of inertia of a rectangular plate might get you even closer depending how you want to set it up. Its usually given in textbooks. I personally don't like to do the integration on some of these figures.

The moment of inertia of a rectangular plate is 1/12m(a^2 + b^2) where a is the width and b the length of the plate. Note that this is the moment of inertia taken from the center of mass of the plate.

I see your picture. So a and b are b and l...

So really one plate would have length 2l +b and width b... and there are two of them.

From what you've said, if i take two rectangular plates, then after adding the two MOIs i' have an extra square in between. And i don't think i can just subtract that...

Also, do you know the derivation of 1/12 M(a²+b²) or a link to it ? I can't find it in my textbook.

I'm not sure if this is correct, but i tried my best to do the integration based on the baterials densities. I havn't added limits yet, but can you verify this for me ?:For the triangle in the middle for each blade:
{σ : density of triangle}
m=1/2 r2hσ
r2=2m/hσ
therefore I=∫2m/hσ
=m2/hσ dm=dx*b*h*rho where dx is a small distance on the wing.
{ρ: density of blade}
therefore ∫r² dm
substituting for dm:
=∫x²*b*h*ρ dx
=1/3 b*h*ρ*x³

The dimensions are right [ML²] but I'm really not too sure...
Thanks

 

[pgardn]

From what you've said, if i take two rectangular plates, then after adding the two MOIs i' have an extra square in between. And i don't think i can just subtract that...

Also, do you know the derivation of 1/12 M(a²+b²) or a link to it ? I can't find it in my textbook.

I'm not sure if this is correct, but i tried my best to do the integration based on the baterials densities. I havn't added limits yet, but can you verify this for me ?:


For the triangle in the middle for each blade:
{σ : density of triangle}
m=1/2 r2hσ
r2=2m/hσ
therefore I=∫2m/hσ
=m2/hσ


dm=dx*b*h*rho where dx is a small distance on the wing.
{ρ: density of blade}
therefore ∫r² dm
substituting for dm:
=∫x²*b*h*ρ dx
=1/3 b*h*ρ*x³

The dimensions are right [ML²] but I'm really not too sure...
Thanks

You don't have to worry about the middle square its taken care of if you use the sides I gave you.

And I am going to admit I don't like doing the math on these. So here is an old discussion on this board.
https://www.physicsforums.com/showthread.php?t=57119

 

[eptheta]

Really ? So i just multiply it by 2 ? i.e I= 1/6 M(a²+b²)