I am trying to wrap my head around the energy terms that arise from the following discrete systems vibration problem:

Wing Vibration Problem

The potential energy terms seem simple enough. I include the elastic potential of the spring with stiffness $k$ and the torsional stiffness of the wing $k_T$:

$$ V = \frac{1}{2}kz^{2} + \frac{1}{2}k_T \theta^{2} $$

The kinetic energy terms of the wing is something I'm uncertain about. I visualise the wing in much the same way as a rolling wheel - I include the translational motion of the entire wing in the z direction due to the spring (down being +ve) and the rotational motion of the wing about some point to the right of $G$ a distance $e$ away. The parallel axis theorem must be used here as the rotation does not occur about the center of gravity $G$:

$$ T = \frac{1}{2}(I+me)\dot\theta^{2} + \frac{1}{2}m\dot z^{2} $$

However, my kinetic energy terms seem incomplete since the Langrange equations do not match what is provided in the mass matrix in (a). What am I missing? I thought about including the tangential velocity of the wing mass would make sense but the rotational energy of the system is already accounted for by the $\frac{1}{2}(I+me)\dot\theta^{2}$ term. What have I not taken into account?


1 Answer 1


For the kinetic energy, do the analysis in the center of mass frame. The key observation is that the vertical height of the center of mass is $$ z_\mathrm{cm} = z - e \sin\theta \; . $$ Thus, for small oscillations, the translation velocity is approximately $\dot{z}-e\dot{\theta}$. We can then write the total kinetic energy as the sum of translational and rotational energy: $$ T = \underbrace{\frac{1}{2} I \dot{\theta}^2}_{\text{rotation KE}} + \underbrace{\frac{1}{2} m \left( \dot{z} - e \dot{\theta} \right)^2}_{\text{translation KE}} $$ I think this gives the required answer.

  • $\begingroup$ This does indeed give me the correct answer. I'm still just a little confused about how doing the analysis in the center of mass frame does not require the use of the parallel axis theorem since the point of rotation is not aligned with the center of mass. Or maybe I misinterpreted the question and "effective moment of inertia $I$ about $G$" already takes this into account? $\endgroup$
    – user155876
    Nov 7, 2017 at 4:38
  • $\begingroup$ Well, one must to do the analysis in the CM frame, or else the rotational and translational energies are not separable. When you do the analysis in the CM frame, you measure the rotation of the wing about its center of mass, so there is no need for the parallel axis theorem. The angle of rotation about the CM is still $\theta$. And the translational velocity of the CM is $\dot{z}_\mathrm{cm}$. $\endgroup$
    – jcandy
    Nov 7, 2017 at 6:01

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