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It is well known that helicopters use the concept of centrifugal stiffening. That is, there is a centrifugal force that acts against the motion of the blade due to the thrust and thereby reducing flapping angle $\beta$. Now, in order to use the analysis of centrifugal force, we must be in a rotating frame of reference.

Now, however, when the thrust force of a blade element is determined the velocity that is used always contains the $\Omega r$ term where $\Omega$ is the rotational velocity. If we consult wikipedia then we find $\vec{v_i} = \vec{v_r} + \vec{\Omega}\times \vec{r} $, where subscripts $_i$ and $_r$ denote inertial and rotating reference frame, respectively.

This leads me to think that the $\Omega r$ term in fig 3.4 appears since we are in an inertial frame. But, the forces resulting from fig 3.4 are used in fig 4.12 which is a rotating reference frame.

How is this possible, and where am I making a mistake?

Blade section of a helicopter

Forces acting on blade element in rotating frame of reference

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1 Answer 1

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You go from inertial frame (the helicopter frame reference system) to the rotating frame (the one in the blade), via D'Alembert approach, where time derivative of momentum becomes apparent inertia forces. As an example,

$m \mathbf{a} = \mathbf{F} \qquad \rightarrow \qquad \mathbf{0} = - m \mathbf{a} + \mathbf{F} = \mathbf{F}^{in}+ \mathbf{F} $.

Here, what you interpret as centrifugal force in the frame rotating with the blade, it's just the opposite of the centripetal force that makes a blade rotate around the axis, if you look at it from the inertial reference frame, the one on the helicopter fuselage, here.

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  • $\begingroup$ the unit of CF is $~\frac{[kg\,m^2]}{s^2}~$ this is not the force unit? $\endgroup$
    – Eli
    Aug 31, 2022 at 9:42
  • $\begingroup$ The elementary centrifugal force of an elementary section of the blade $dr$ is $dF_c = dM r \Omega^2 = (m dr) r \Omega^2$, where $m$ is the mass density per unit length and thus it has physical unit in the SI $kg/m$, and the elementary force has the right physical dimension, i.e. $\frac{kg}{m} \, m \, m \frac{1}{ s^2} = \frac{kg \,m}{s^2} = N$. $\endgroup$
    – basics
    Aug 31, 2022 at 9:50
  • $\begingroup$ o.k the unit of m is $~\frac{[kg]}{m}~$ $\endgroup$
    – Eli
    Aug 31, 2022 at 10:04
  • $\begingroup$ Good, if you're ok with the answer, please accept it, so that the community knows that is a satisfactory answer to your question $\endgroup$
    – basics
    Aug 31, 2022 at 10:08
  • $\begingroup$ So basically, the lift force on the blade element is conserved when going from inertial to rotational frame of reference. The only thing that is changed is the acceleration in the respective frame. This is because the acceleration is caused by the force, and the force does essentialy not care about the reference frame. Is this true? $\endgroup$
    – lWindy
    Aug 31, 2022 at 10:18

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