Question 4.7.1: Derive the equations of motion of the system of Figure 4.18 ......

The motion of this system can be described by the two coordinates x and θ, so a good choice of generalized coordinates is q_1(t) = x(t)\text{ and }q_2(t) = θ(t). The kinetic energy becomes

T=\frac{1}{2} m \dot{q}_1^2+\frac{1}{2} J \dot{q}_2^2

The potential energy becomes

U=\frac{1}{2} k_1 q_1^2+\frac{1}{2} k_2\left(r q_2-q_1\right)^2

Here Q_1=0 and Q_2=M(t). Using equation (4.145) yields, 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}=0 \quad i=1,2, \ldots, n         (4.145)

\frac{d}{d t}\left(m \dot{q}_1+0\right)-0+k_1 q_1+k_2\left(r q_2-q_1\right)(-1)=0

or

m \ddot{q}_1+\left(k_1+k_2\right) q_1-k_2 r q_2=0            (4.147)

Similarly, for i=2, equation (4.144) yields

\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)

J \ddot{q}_2+k_2 r^2 q_2-k_2 r q_1=M(t)       (4.148)

Combining equations (4.147) and (4.148) into matrix form yields

\left[\begin{array}{cc} m & 0 \\ 0 & J \end{array}\right] \ddot{ \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]           (4.149)

Here the vector x(t) is

\mathrm{x} (t)=\left[\begin{array}{l} q_1(t) \\ q_2(t) \end{array}\right]=\left[\begin{array}{c} x(t) \\ \theta(t) \end{array}\right]