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I want to derive equation of motion for the system shown in picture. How do I choose a generalized coordinate in order to calculate kinetic and potential energy of the system? I need the aforementioned quantities to implement Lagrange equations for determining the equation of motion. How many DOFs does the system has?

enter image description here

Gert
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user3127040
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    What are we looking at? K1 and K2 are springs? What are c1 and c2? What is m-bar? Can the object rotate as well as vibrate in the x- and y-directions? –  Dec 13 '15 at 19:32
  • It looks like a system with three degrees of freedom - two translation and one rotation. Assuming there is no motion perpendicular to the plane of the paper. – Floris Dec 13 '15 at 20:18
  • @SamuelWeir K1 and K2 are springs, c1 and c2 are dashpots. mbar denotes mass per unit length... well, it might, I'm not sure. I don't know how many generalized coordinate is needed to describe the motion of this system. one rotational or two translational or maybe both of them simultaneously?... – user3127040 Dec 13 '15 at 20:21
  • Dashpots lose energy, no? – Gert Dec 13 '15 at 20:26
  • @Floris Three independent DOFs? – user3127040 Dec 13 '15 at 20:36
  • Yes - $x, y$ and $\theta$. – Floris Dec 13 '15 at 20:37
  • @Gert it does not matter here. damping forces considered as non-conservative forces and appear in the right hand side of the Lagrange equations. – user3127040 Dec 13 '15 at 20:44

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Consider the net forces $F_x$, $F_y$ and moments $M_A$ at a point A, let's say the corner of the beam

pic

now consider the center of mass at a point C located at $(c_x,c_y)$ relative to A. This location changes in each time frame because it is measured from an inertial reference frame (and not in the body frame).

The three degree of freedom of the body can be the position of A $(x_A,y_A)$ and the orientation angle $\theta$. The equations of motion from a Newtonian point of view are

$$ \begin{align} F_x & = m \ddot{x}_A - m c_y \ddot{\theta} - m c_x \dot{\theta}^2 \\ F_y & = m \ddot{y}_A + m c_x \ddot{\theta} - m c_y \dot{\theta}^2 \\ M_A &= (I_C+ m (c_x^2+c_y^2)) \ddot{\theta} + m (c_x \ddot{y}_A - c_y \ddot{x}_A ) \end{align} $$

Where $m$ is the total mass, and $I_C$ is the mass moment of inertia about the center of mass. (See Derivation of Newton-Euler equations of motion for a derivation of the equations of motion not on the center of mass in 3D).

What you are interested in the motion due to the net forces at each frame which is

$$\begin{pmatrix} \ddot{x}_A \\ \ddot{y}_A \\ \ddot{\theta} \end{pmatrix} = \begin{vmatrix} \frac{1}{m} + \frac{c_y^2}{I_C} & - \frac{c_x c_y}{I_C} & \frac{c_y}{I_C} \\ - \frac{c_x c_y}{I_C} & \frac{1}{m} + \frac{c_x^2}{I_C} & -\frac{c_x}{I_C} \\ \frac{c_y}{I_C} & - \frac{c_x}{I_C} & \frac{1}{I_C} \end{vmatrix} \begin{pmatrix} F_x \\ F_y \\ M_A \end{pmatrix} + \begin{pmatrix} c_x \dot{\theta}^2 \\ c_y \dot{\theta}^2 \\ 0 \end{pmatrix} $$

The last part that is needed is to expressed the forces as a function of the degrees of freedom. For small movements you can estimate

$$ \begin{pmatrix} F_x = -k_2 x_A - c_1 (\dot{x}_A-L \dot{\theta}) \\ F_y = -k_1 y_A - c_2 (\dot{y}_A) - m g \\ M_A = - c_1 L (\dot{x}_A-L \dot{\theta}) - c_x m g \end{pmatrix}$$

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John Alexiou
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