Question 4.7.4: The inverted pendulum hinged at O of a carriage shown in Fig......

Let x and θ be the horizontal displacement of the pivot O on the carriage and angular displacement or rotation of the pendulum about O, respectively.
At the center of gravity G the radial acceleration towards O is a_{r}    =  \ddot{ℓ}   −  ℓ \dot{θ}²  =  −  ℓ  \dot{θ}² , since the length of the pendulum is constant.
The transversal acceleration at G as indicated in the FBD is a_{θ} = ℓ \ddot{θ}  − 2 \dot{ℓ} \dot{θ}  = ℓ\ddot{θ},  since the length of the pendulum is constant.
With reference to the FBD in Figure 4E4b and applying Newton’s law of motion in the horizontal direction, one has
(M+m) \ddot{x}  +  m (ℓ\ddot{θ} cosθ)  −  m( ℓ\dot{θ}²sinθ )  =  \sum{F_{x}}  =  f(t) −  c\dot{x}.

Re-arranging, one obtains

(M+m) \ddot{x}  +  m (ℓ cosθ)\ddot{θ}  −  m( ℓsinθ ) \dot{θ}²   +  c\dot{x}  =  f(t) .            (i)

Taking moments about O, with clockwise moments being positive,

( \frac{mℓ²}{3}  +   mℓ²)  \ddot{θ}   +   m \ddot{x} (ℓ cosθ)  =   –  c_{r}  \dot{θ}   +   mg  ( ℓsinθ )  .

Re-arranging, one may expressed this in a more familiar form as

( \frac{4  mℓ²}{3} )  \ddot{θ}   +   m (ℓ cosθ)\ddot{x}   +  c_{r}  \dot{θ}   =   (mg ℓ)sinθ .                                (ii)

Equations (i) and (ii) are highly nonlinear. However, if a small oscillation is assumed such that sinθ≈θ, cosθ≈1, and \dot{θ}² can be disregarded, then the two equations of motion reduce to

(M+m) \ddot{x}   +  mℓ  \ddot{θ}   +  c\dot{x}  =   f(t) .                                (iii)

mℓ  \ddot{x}  +   ( \frac{4  mℓ²}{3} )  \ddot{θ}   +  c_{r}  \dot{θ}   –  (mg ℓ)  θ  =   0  .                                          (iv)

Writing in matrix form, one has

\begin{bmatrix}M  +  m &  mℓ \\ mℓ   &  \frac{4  mℓ²}{3} \end{bmatrix}  \begin{pmatrix} \ddot{x} \\ \ddot{θ}\end{pmatrix}   +    \begin{pmatrix}c & 0 \\ 0 & c_{r} \end{pmatrix}      \begin{pmatrix} \dot{x} \\ \dot{θ}\end{pmatrix}  +  \begin{bmatrix}0 & 0 \\ 0 & –  mgℓ\end{bmatrix}  \begin{pmatrix} x\\ θ \end{pmatrix}   =  \begin{pmatrix} f(t)\\ 0 \end{pmatrix}   .                              (v)

This system is said to possess dynamic coupling because the off-diagonal terms in the inertia matrix are not zero. However, it is not a 2-dof oscillatory system in the usual sense. Owing to the fact that the diagonal term associated with θ is negative, this system is unstable. Furthermore, it may be appropriate to point out that the dimensions of x and θ are different. Therefore, if numerical integration techniques are to be applied to the solution of Equation (v), it may be more convenient to rewrite it as

\begin{bmatrix}M  +  m &  m \\ m   &  \frac{4  m}{3} \end{bmatrix}  \begin{pmatrix} \ddot{x} \\ \ddot{θ}ℓ\end{pmatrix}   +    \begin{pmatrix}c & 0 \\ 0 & c_{r}/ℓ² \end{pmatrix}      \begin{pmatrix} \dot{x} \\ \dot{θ}ℓ\end{pmatrix}  +  \begin{bmatrix}0 & 0 \\ 0 & –  \frac{mg}{ℓ}\end{bmatrix}  \begin{pmatrix} x\\ θℓ \end{pmatrix}   =  \begin{pmatrix} f(t)\\ 0 \end{pmatrix}   .                              (vi)

Now, the elements in the displacement, velocity, and acceleration vectors have identical dimensions.