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Chapter 6

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)

Screenshot 2022-10-10 154611

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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.)