Question 4.7.2: The support of an airfoil section frequently studied in a wi......

Applying Newton’s law of motion along the y-direction and with the FBD shown in Figure 4E2b, and remembering that for small oscillation sin θ ≈ θ, one has

Summing moments about G, it gives

where  J_{G}  is the moment of inertia about G, J_{0} =   J_{G}   +    me²  with   J_{0}  being the moment of inertia about O.

The above two required equations may be expressed in matrix form as

\begin{bmatrix} m  &  – me \\   0 &   J_{0}  –   me² \end{bmatrix}  \begin{pmatrix}   \ddot{y} \\  \ddot{θ} \end{pmatrix}   +    \begin{bmatrix} k   &   0 \\   ek   &   k_{T}   \end{bmatrix}  \begin{pmatrix} y \\  θ \end{pmatrix}   =    \begin{pmatrix} 0 \\ 0 \end{pmatrix}   .

With reference to this matrix equation it is clear that if the center of gravity G coincides with the point of support O such that e =  0, the translational dof is uncoupled from the rotational dof. Additionally, if e ≠ 0 and one sums all moments about O instead of G, one can show that the resulting equations of motion in matrix form are

\begin{bmatrix} m  &  – me \\   –   me  &  J_{0} \end{bmatrix}  \begin{pmatrix}   \ddot{y} \\  \ddot{θ} \end{pmatrix}   +    \begin{bmatrix} k   &   0 \\   0   &   k_{T}   \end{bmatrix}  \begin{pmatrix} y \\  θ \end{pmatrix}   =    \begin{pmatrix} 0 \\ 0 \end{pmatrix}   .

In this equation the mass and stiffness matrices are symmetric. This is called a system with dynamic coupling because the mass matrix is not diagonal.