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)

Question

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)

[latex]\mathbf{M}_O=\dot{\mathbf{h}}_O + \mathbf{v}_O imes \mathbf{p}[/latex] (6.80)

Accepted Answer

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 [latex]\mathbf{x}_B=\ell \mathbf{e}_r[/latex] in the cylindrical system shown in Fig. 6.19 is given by

[latex]\mathbf{M}_O=\mathbf{x}_B imes \mathbf{W}=-\ell W \sin \phi \mathbf{k}[/latex]. (6.81a)

The absolute velocity of B is determined by [latex]\mathbf{v}_B=\mathbf{v}_O + \boldsymbol{\omega} imes \mathbf{x}_B[/latex], in which [latex]\omega=\dot{\phi} \mathbf{k}[/latex] and [latex]\mathbf{v}_O=\dot{x} \mathbf{i}=x_O \Omega \cos \Omega t \mathbf{i}=v_O \mathbf{i}[/latex]. Thus,

[latex]\mathbf{v}_B=v_O \mathbf{i} + \ell \dot{\phi} \mathbf{e}_\phi, \quad ext { with } \quad v_O=x_O \Omega \cos \Omega t[/latex]. (6.81b)

With the linear momentum [latex]\mathbf{p}=m \mathbf{v}_B[/latex] and use of (6.81b) , we find

[latex]\mathbf{v}_O imes \mathbf{p}=v_O \mathbf{i} imes m \ell \dot{\phi} \mathbf{e}_\phi=m v_O \dot{\phi} \ell \sin \phi \mathbf{k}[/latex]. (6.81c)

The moment of momentum about O is given by [latex]\mathbf{h}_O=\mathbf{x}_B imes \mathbf{p}=m \ell\left(v_O \cos \phi + ight. \ell \dot{\phi}) \mathbf{k}[/latex], and its time rate of change is

[latex]\dot{\mathbf{h}}_O=m \ell\left(a_O \cos \phi - v_O \dot{\phi} \sin \phi + \ell \ddot{\phi} ight) \mathbf{k}[/latex] (6.81d)

in which [latex]a_O=\dot{v}_O=-x_O \Omega^2 \sin \Omega t[/latex]. Substituting (6.81a), (6.8Ic), and (6.81d) into (6.80), we find [latex]-\ell W \sin \phi \mathbf{k}=m \ell\left(-x_O \Omega^2 \sin \Omega t \cos \phi + \ell \ddot{\phi} ight) \mathbf{k}[/latex]. Hence, with W = mg, the equation of motion for the bob may be written as

[latex]\ddot{\phi} + p^2 \sin \phi=\frac{x_O \Omega^2}{\ell} \cos \phi \sin \Omega t[/latex], (6.81e)

where [latex]p^2=g / \ell[/latex]. The solution of (6.81e) for small [latex]\phi(t)[/latex] is discussed later in our study of mechanical vibrations. (See Example 6.15 , page 161.)

Question 6.14: A pendulum bob B attached to a rigid rod of negligible mass ...

Related Answered Questions

Determine the Coriolis deflection of a projectile fired eastward at latitude λ. Derive the classical relations for the motion and the range when the Earth’s rotation is neglected. ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

A relativistic particle P, initially at rest at the origin in frame ψ, is moving along a straight line under a constant force Fo. Determine the relativistic speed and the distance traveled by P as functions of time. ...

Verified Answer:

A particle P in an inertial reference frame has an initial velocity vo at the place xo, and subsequently moves under the influence of a force that is proportional to the time and acts in a fixed direction e. Find the position and velocity of P at time t . ...

Verified Answer:

Verified Answer:

Verified Answer: