Question 6.14: A pendulum bob B attached to a rigid rod of negligible mass ...
A pendulum bob B attached to a rigid rod of negligible mass and length l is suspended from a smooth movable support at O that oscillates about the natural undeformed state of the spring so that x(t)=x_O \sin \Omega t in Fig. 6.19. Apply equation (6.80) to derive the equation of motion for the bob.
\mathbf{M}_O=\dot{\mathbf{h}}_O + \mathbf{v}_O \times \mathbf{p} (6.80)
The forces that act on the pendulum bob B are shown in the free body diagram in Fig. 6.19. Notice that the tension T in the rod at B is directed through the moving point O. Moreover, the spring force and normal reaction force of the smooth supporting surface also are directed through O; but these forces do not act on B, so they hold no direct importance in its equation of motion. Consequently, the moment about the point O of the forces that act on B at \mathbf{x}_B=\ell \mathbf{e}_r in the cylindrical system shown in Fig. 6.19 is given by
\mathbf{M}_O=\mathbf{x}_B \times \mathbf{W}=-\ell W \sin \phi \mathbf{k}. (6.81a)
The absolute velocity of B is determined by \mathbf{v}_B=\mathbf{v}_O + \boldsymbol{\omega} \times \mathbf{x}_B, in which \omega=\dot{\phi} \mathbf{k} and \mathbf{v}_O=\dot{x} \mathbf{i}=x_O \Omega \cos \Omega t \mathbf{i}=v_O \mathbf{i}. Thus,
\mathbf{v}_B=v_O \mathbf{i} + \ell \dot{\phi} \mathbf{e}_\phi, \quad \text { with } \quad v_O=x_O \Omega \cos \Omega t. (6.81b)
With the linear momentum \mathbf{p}=m \mathbf{v}_B and use of (6.81b) , we find
\mathbf{v}_O \times \mathbf{p}=v_O \mathbf{i} \times m \ell \dot{\phi} \mathbf{e}_\phi=m v_O \dot{\phi} \ell \sin \phi \mathbf{k}. (6.81c)
The moment of momentum about O is given by \mathbf{h}_O=\mathbf{x}_B \times \mathbf{p}=m \ell\left(v_O \cos \phi + \right. \ell \dot{\phi}) \mathbf{k}, and its time rate of change is
\dot{\mathbf{h}}_O=m \ell\left(a_O \cos \phi – v_O \dot{\phi} \sin \phi + \ell \ddot{\phi}\right) \mathbf{k} (6.81d)
in which a_O=\dot{v}_O=-x_O \Omega^2 \sin \Omega t. Substituting (6.81a), (6.8Ic), and (6.81d) into (6.80), we find -\ell W \sin \phi \mathbf{k}=m \ell\left(-x_O \Omega^2 \sin \Omega t \cos \phi + \ell \ddot{\phi}\right) \mathbf{k}. Hence, with W = mg, the equation of motion for the bob may be written as
\ddot{\phi} + p^2 \sin \phi=\frac{x_O \Omega^2}{\ell} \cos \phi \sin \Omega t, (6.81e)
where p^2=g / \ell. The solution of (6.81e) for small \phi(t) is discussed later in our study of mechanical vibrations. (See Example 6.15 , page 161.)