Question 4.7.4: Consider again the system of Example 4.7.1 and assume that t......

The dissipation function given by equation (4.150) becomes

F=\frac{1}{2} \sum_{r=1}^n \sum_{s=1}^n c_{r s} \dot{q}_r \dot{q}_s              (4.150)

F=\frac{1}{2}\left[c_1 \dot{q}_1^2+c_2\left(r \dot{q}_2-\dot{q}_1\right)^2\right]

Substitution into equation (4.151) yields the generalized forces

Q_j=-\frac{\partial F}{\partial \dot{q}_j}, \text { for each } j=1,2, \ldots, n            (4.151)

Adding the moment as indicated in Example 4.7.1, the second generalized force becomes

Q_2=M(t)-c_2 r^2 \dot{q}_2+r c_2 \dot{q}_1

Next, using T and U as given in Example 4.7.1, recalculate the equations of motion using equation (4.144) to get, for i=1 :

\frac{d}{d t}\left(\frac{\partial T}{\partial \dot{q}_i}\right)-\frac{\partial T}{\partial q_i}+\frac{\partial U}{\partial q_i}=Q_i \quad i=1,2, \ldots, n               (4.144)

and for i=2 :

Combining the expressions for i=1 and i=2 yields the matrix form of the equations of motion:

\left[\begin{array}{cc} m & 0 \\ 0 & J \end{array}\right] \ddot{ \mathrm{x} }(t)+\left[\begin{array}{cc} c_1+c_2 & -r c_2 \\ -r c_2 & r^2 c_2 \end{array}\right] \dot{ \mathrm{x} }(t)+\left[\begin{array}{cc} k_1+k_2 & -r k_2 \\ -r k_2 & r^2 k_2 \end{array}\right] \mathrm{x} (t)=\left[\begin{array}{c} 0 \\ M(t) \end{array}\right]