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## Q. 6.14

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) ## Verified Solution

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.)