Re: Centrifugal force of a helicopter blade

Well, you may already know this, but the first thing I want to address is 
the common misuse of centrifugal force.  Centrifugal force 
is an imaginary force that arises when you consider noninertial frames (an 
inertial frame would be one of a stationary observer).  
This concept can be best understood with an example - take a rock on a 
string. 

When you swing the rock  in a circle, an observer in the inertial frame 
(like the person swinging the rock), would see a centripetal 
force (of F=ma) of the string acting on the rock to cause it to follow the 
line of the circle....and not fly off on a tangent.  However 
to a person in the same reference frame as the swinging rock (don't ask me 
how the person got there!   :-)   ) sees the rock at rest 
and therefore has to invent a force to counteract the tension on the string 
and make all the the forces balance out.- the centrifugal 
force.  (This is the same force that seems to pull you to the side when you 
make a sharp turn in a car......in reality, you are just 
lacking a force to pull you along the circle of the turn with the car.) 

For your answer, I will calculate the centripetal force.  This may all seem 
nitpickey, but it is good to understand the true nature of 
the forces in a system. 

::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

Now to your real question........I thought about this question for a while 
(mostly because most of my general physics knowledge 
seems to have seeped from my brain) and I seem to have an answer for you.  
Applying what I have just explained above, I will 
attempt to calculate the centripetal force at the hub of a helicopter 
blade. 

If you just consider one side of the helicopter blade at a time, you can 
break down the system to something that can be solved with 
some simple calculus and physics equations (I sincerely hope you know 
calculus.....if not, email me back and I will try a simpler 
way of explaining it).  First of all, the acceleration for any particle (we 
will start small) moving in a circle can be described by the 
the equation: 

        a = -(v^2)/r  (where 'v' is the particle's velocity, and 'r' is the 
radius of the particle from the centre of motion) 

Combine this with Newton's Second Law (or F=ma) and you get the centripetal 
force acting on the particle: 

        F = (mv^2)/r     (where 'm' is the mass of the particle). 

(Note: the acceleration equation was negative because the motion is 
directed inward; the force is also inward, but we will not 
bother carrying the negative sign) 

This last equation is what we need to solve your question.  The only 
problem now is that we have a radius to deal with.  First of all, 
sub in the following equivalent equation for 'v': 

    v = (RPM*2*pi*r)/(60 s/min) 

    where,    RPM = the revolutions per minute of the blade 
                  pi = 3.14 
                  r = radius of the part being calculated from the hub 
                  60 s/min = a conversion factor to get the result in terms 
of seconds 

The equation you now have (after substitution and reduction) is:

     F = 4*m*r*(pi*RPM/60)^2 

If you integrate over the length of the one half of the helicopter blade 
(i.e. from the hub to one of the tips - we will call the limits 'a' 
and 'b'), you can derive a general equation for the centripetal force: 

    F = 2m*(pi*RPM / 60 s/min)^2 * (b^2 - a^2) 

The above equation, as promised, should give you an expression to describe 
the centripetal force of one blade of a helicopter (from 
the centre to the tip) on the hub - just sub in the limits of the start of 
the blade and the tip for 'a' and 'b' respectively, sub in for the 
mass of the blade and solve.  The opposite blade will have the same force, 
but in the opposite direction. 

::::::::::::::::::::::::::::::::::::::::::::::::::::

I hope this answers your question for you.  It was hard trying to express 
these calculus and physics equations on the computer 
without resorting graphics.  If you need any further explanation of this 
answer, please feel free to email me.  You might also want 
to refer to a good physics text for further explaination....I listed the 
one I used below. 
 

 Kurt Frost 
 kfrost@sympatico.ca 
 
 

REFERENCES: 

Benson, H.  1991.  University Physics.  New York: John Wiley & Sons, Inc.