5
$\begingroup$

So I have a question regarding this system:

System

It is supposed to be a simple model of an aircraft with the fuselage idealized as a concentrated mass $M_0$ and the wings modeled as rigid bars carrying weights at the end of mass $M$. The flexibility of the wings are modeled with springs with torsional stiffness $k_t$.

The fuselage is free to translate in the $x$ direction while the small mass rotates about the larger mass in the $\theta$ direction. One can derive the equations of motion of the system with Lagrange's equation with $x$ and $\theta$ as the generalized coordinates.

One way to derive the kinetic energy of the small mass ($K_m$) is to obtain the velocity of it ($\dot{x}_m$) as a function of the velocity of the larger mass ($\dot{x}$) and the angular velocity ($\dot{\theta}$).

$$ \dot{x}_m = \dot{x} + l\dot{\theta}\\ K_m = \frac{1}{2}M\dot{x}_m^2 = \frac{1}{2}M(\dot{x}^2 + 2l\dot{x}\dot{\theta} + l^2\dot{\theta}^2) $$

However, if one were to use the definition of the kinetic energy of a rigid body $B$ rotating about a moving point $C$ in inertia frame $F$ (see note 1 below), which is:

$$ K_B = \frac{1}{2}m_B\dot{x}_C^2 + \frac{1}{2}I_B\dot{\theta}_B^2 $$

Applied to this question, the kinetic energy of the small mass is:

$$ K_m = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}I_m\dot{\theta}^2 = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}(Ml^2)\dot{\theta}^2 = \frac{1}{2}M(\dot{x}^2 + l^2\dot{\theta}^2) $$

It seems that the kinetic energy derived by the 2 methods results in 2 different expressions. The correct answer provided by my school is the former. But what is missing in the second approach that leads to the difference?

Note 1: This definition comes from "Introduction to Structural Dynamics and Aeroelasticity" from Hodges and Pierce. The definition can also be commonly found in other physics textbooks like "Introduction to Classical Mechanics" by D. Morin.

$\endgroup$
  • 1
    $\begingroup$ Hi Joel, and welcome to Physics Stack Exchange! Excellent question :-) (the kind I'd point other posters to as an example of what a good question should look like!) To stave off some possible objections: where did your second expression ($K_B$) come from? $\endgroup$ – David Z Oct 3 '15 at 12:39
  • $\begingroup$ Hi, thanks! I've updated the question. Hope to get some replies soon :) $\endgroup$ – Joel Oct 3 '15 at 13:00
  • $\begingroup$ Question for the asker: do the equation of motion differ? Related question on the site: physics.stackexchange.com/q/50075 $\endgroup$ – dmckee Oct 3 '15 at 17:40
  • $\begingroup$ @dmckee I don't think so since the difference term is not a total derivative. By the way, to the OP, it seems to me the position of the point $M$ is $x_{M}=x+l\sin\theta$ and so your expression are in the small $\theta$ limit. I suppose the fuselage $M_{0}$ to be a point. The small angle approximation is obviously fine, but it is clear also that your expression for the centrifugal force $I_{B}\dot{\theta_{B}}^{2}/2$ is in fact correct for all angles (am I correct ?). So it's even more puzzling. Be aware that not all your notations are defined unfortunately: all the B,C variables, ... $\endgroup$ – FraSchelle Oct 7 '15 at 11:09
  • $\begingroup$ Ah ok I think I start understanding the problem. More certainly none of the above kinetic energy is correct, since the problem is not Galilean invariant due to the rotation, and so you have to add non-inertial effect by hand. The simplest way to find something you might call a kinetic energy would be to write the equation of motion, then to infer a Lagrangian and/or Hamiltonian from them, and to identify the kinetic energy if you can. $\endgroup$ – FraSchelle Oct 7 '15 at 11:36
1
$\begingroup$

The inconsistency arises because the norm of the velocity vector is computed incorrectly.

The formula for the kinetic energy is $T=\frac{1}{2}m\cdot \overrightarrow{{v}}^2$.

Now for small values of $\theta$ the movement of the mass M is collinear with the direction of the x-axis. In that case $\overrightarrow{v}=[v_x, v_y, v_z] = [\dot{x}+l\dot{\theta}, 0, 0]$ is a good approximation and the OP's first formula for $K_m$ is correct in that approximation.

For the second form of the OP's $K_m$ formula they add $\dot{x}$ and $l\dot{\theta}$ quadratically, which is correct only if the two respective vectors are orthogonal to each other. For instance it would be approximately correct at $\theta\approx \pi/2$.

For the problem at hand, the first $K_m$ formula is correct.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.