Sadly! But yeah, great lecture!

Thank you for your feedback. It's very encouraging to hear.

Yeah couldn’t agree more :)

I'm assuming you want this relative to the CG again... First construct the Rayleigh's Dissipation function, R, where R = (1/2)c(ḣ_cg + eα̇)² + (1/2)c_t(α̇)² This looks very similar to the potential energy expression, except that the k's become c's and the displacements become velocities. Substituting this into the extended form of Lagrange's equation (take a look at: kzfaq.info/get/bejne/hLxnlsdenLjeXX0.html ) will yield a damping matrix that looks like [c] = [ [c, ce], [ce, (ce²+c_t)] ]

@@Freeball99 Yes, I wanted at CG. Thanks a lot for the reply.

The choice of coordinates is somewhat arbitrary. I could have modeled this problem that way. My rule-of-thumb is that when dealing with gravity, generally use x-y coordinates (which makes one less prone to careless errors). But, to be clear, I could have used x-y coordinates for this problem too. My suggestion is to try it both ways and compare the results.

Got it - thanks. When you wrote the velocity of the CG, you just included the vertical component: h dot - e*alpha dot. Is this because the CG velocity in the horizontal direction as it pitches is negligible? If my understanding is correct, KE in general plane motion is 0.5*m*V^2 + 0.5*J*omega^2, where V is a vector (with two components in 2-D).

@@ironpine5815 Yes, the displacement in the horizontal direction is negligible due to the assumption of small displacements (and small rotations). In this problem, I assumed small displacements from the start (probably should have made that clearer). For the pendulum problem I never made the small displacement assumption.

@@Freeball99 Awesome - thanks for the clarification and prompt replies!

There is no mention of any external force in the problem. This is a free vibration problem.

This is a function of the coordinate system I have used. I chose the h-direction to be positive downwards and the α-direction to be positive when the airfoil pitches upward. As a result, the C.G. moves in the NEGATIVE h-direction when α is positive - as drawn. As a result, we subtract the pitch effect. IT SHOULD BE NOTED HOWEVER, that for a typical, real airfoil, the C.G. is usually AFT of the elastic axis (so e would be negative based on my coordinates). In this case, the sign would be flipped and the pitch effect would be added.

Maybe one day!

EDIT: I realize that I gave you bad information here. I wasn't thinking clearly...if you write the EOM's about the CG (we would then assume that h describes the plunge of the CG rather than the elastic axis), then the velocities are decoupled and so the mass matrix would become diagonal, however the stiffness matrix would then become coupled. Your energies would become T = (1/2)m(dh/dt)² + (1/2)J_CG(dα/dt)² V = (1/2)K(h + eα)² + (1/2)K_t(α)² Plugging these into Lagrange's equations will give you the EOMs that you are after. [m] = [ [m , 0], [0, J_CG] ] and [k] = [ [K, Ke ], [Ke, (Ke² + K_t)] ]

@@Freeball99 Thank you very much for the reply.

SHORT ANSWER: You are not told that gravity is present in the problem, so the potential energy associated with the mass moving up and down is due to the potential energy in the spring only. LONGER ANSWER: If gravity were present in the problem, then you could still ignore the change in potential energy due to the motion of the mass up and down (translation) IF YOU ASSUME that the mass is oscillating about it's equilibrium point then you can ignore gravitational effects. I have explained this effect in this video (kzfaq.info/get/bejne/itaKgLegkrLNl2g.html). If gravity were present in the problem, you would, however, need to include the rotational potential due to gravity (i.e. gravitational force on the mass will produce a moment which cannot be ignored). The rotational component of this basically acts like a pendulum. In the case of a pendulum, you CANNOT ignore gravity.

@@Freeball99 That makes perfect sense, thank you. I may try the problem with a gravity component just to compare the behaviour.

@@McTreestump That is definitely a worthwhile exercise. What you should find is that by including gravity, the x location is increased by a constant offset x0 = F/k. NOTE: Gravity WILL have a effect on the torsional/rotational part of the system (since this is a pendulum motion). The effects on the rotational motion on the airfoil due to gravity CANNOT be ignored by considering oscillations about an equilibrium position (since it changes the effective stiffness properties of the torsional system - much like a pendulum). Going through this exercise should make this all obvious.

@@Freeball99 Hi, great video and explanation! I know it has been a while since these comments but I just wanted to point out that in this case since the equilibrium position is close to horizontal (because of the torsional spring kt), the gravity term CAN be ignored even for rotational case. You are right though that it shows up in the pendulum but I think that is because the pendulum is vertical at its equilibrium position and hence for small oscillations the effect of gravity is like stiffness. Let me know if I misunderstood something.

The motion of the cg with respect to O was already captured in the velocity expression (the - eα_dot term). This is what give rise to the me^2 expression which is the moment of inertia of the mass about point O.

m is the mass of the airfoil (which is assumed to be concentrated at the c.g.)

It is certainly my intention to get to flutter eventually. I have a video showing a very simple example of aeroelastic instability using a 1DOF model (kzfaq.info/get/bejne/l7FhZdOrz6yah6s.html). Flutter analysis, however, requires at minimum a 2D model. To perform a full 2D work up will require a lot of math. I will make a video for it eventually when I can figure out the best way to keep the video (reasonably) short.

@@Freeball99 a video about fluttering would be great Sir