Question 11.9: A system of rigid bodies in its static equilibrium position ...
A system of rigid bodies in its static equilibrium position is shown in Fig. 11.3. The homogeneous wheel \mathscr{B} of radius a is free to rotate in a smooth bearing about Q, and the block of mass m is supported by a linear spring of stiffness k_2 attached to an inextensible cable that wraps around the wheel. The other end of the cable is fastened to a linear spring of modulus k_1 fixed to the machine foundation . There is no slip between the cable and the wheel when the block is displaced vertically and released . Derive the equations of motion for the system.
Because the elastic spring response is linear, we may consider the motion about the prestretched static equilibrium position of the system . In consequence, gravity has no further influence on the motion. This system has two degrees of freedom characterized by the independent generalized coordinates \left(q_1, q_2\right)=(x, \phi) measured from the equilibrium state shown in Fig. 11.3. Due to the no slip and inextensibility constraints, the extension \tilde{x} the foundation spring is \tilde{x}=a \phi. The smooth bearing reaction force is workless, and the remaining forces that act on the system are conservative . Therefore, relative to the static equilibrium state, the Lagrangian function for the system is given by
L=T-V=\frac{1}{2} m \dot{x}^2 + \frac{1}{2} I \dot{\phi}^2 – \left[\frac{1}{2} k_1 a^2 \phi^2 + \frac{1}{2} k_2(x – a \phi)^2\right]. (11.67a)
where I is the moment of inertia of the wheel about its principal axis at Q. Notice in passing that the total kinetic energy in (11.67a) has the form (11.24) in which \left[M_{j k}\right]=\operatorname{diag}[m, I]. With (11.66) in mind, we first determine
T=\frac{1}{2} M_{j k}\left(q_r\right) \dot{q}_j \dot{q}_k (11.24)
\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_r}\right) – \frac{\partial L}{\partial q_r}=0, \quad r=1,2, \ldots, n (11.66)
\frac{\partial L}{\partial \dot{x}}=m \dot{x}, \quad \frac{\partial L}{\partial x}=-k_2(x – a \phi), (11.67b)
\frac{\partial L}{\partial \dot{\phi}}=I \dot{\phi}, \quad \frac{\partial L}{\partial \phi}=-k_1 a^2 \phi + a k_2(x – a \phi), (11.67c)
and thereby obtain the following coupled pair of linear differential equations of motion for the system:
m \ddot{x} + k_2 x – k_2 a \phi=0, \quad I \ddot{\phi} + \left(k_1 + k_2\right) a^2 \phi – k_2 a x=0 . (11.67d)
The structure of these equations is similar to (11.40f) studied earlier.
m_1 \ddot{x}_1 + \left(k_1 + k_2\right) x_1 – k_2 x_2=0, \quad m_2 \ddot{x}_2 + \left(k_1 + k_2\right) x_2 – k_2 x_1=0 . (11.40f)