Question 10.13: A physical pendulum shown in Fig. 10.15 swings in its vertic...
A physical pendulum shown in Fig. 10.15 swings in its vertical plane of symmetry about a smooth, fixed axle at Q. The pendulum is released from rest at the placement \theta_0 with angular speed \omega_0. (i) Apply the work—energy principle to derive the equation of motion of the pendulum. Confirm the solution by application of Euler’s equation. (ii) Describe the general solution, find the small amplitude circular frequency of the oscillation, and describe a simple pendulum having the same period. (iii) What is the mechanical power expended over the interval [0, t] as a function of θ?
Solution of (i). The idea here is to obtain the work—energy equation for a pivoted rigid body, and then derive from this first integral the differential equation of motion for θ(t). The center of gravity G of the pendulum is at a distance \ell from Q, and hence the rotational work done by the torque in turning the pendulum about the smooth hinge Q is determined by (10.118) in which \boldsymbol{\omega}(t)=\dot{\theta}(t) \mathbf{k} and \mathbf{M}_Q(\mathscr{B}, t)=M_1 \mathbf{i} + M_2 \mathbf{j} + M_3 \mathbf{k}-m g \ell \sin \theta \mathbf{k}, where M_1 and M_2 are unknown bearing reaction torques, which are workless. Because the hinge is smooth, the bearing reaction component M_3=0 and the support reaction force R is workless. Hence, with \theta(0)=\theta_0 at t_o=0,
\mathscr{W}(\mathscr{B}, t)=\int_{t_0}^t \mathbf{M}_Q(\mathscr{B}, t) \cdot \omega(t) d t=\Delta K(\mathscr{B}, t) . (10.118)
\mathscr{W}=\int_0^t – m g \ell \sin \theta \dot{\theta} d t=-m g \ell \int_{\phi_0}^\theta \sin \theta d \theta . (10.122a)
This yields the rotational work done by the applied loads as
\mathscr{W}=m g \ell\left(\cos \theta – \cos \theta_0\right). (10. 122b)
Notice that this is just the gravitational work done: \mathscr{W}_g=m g h=m g \ell(\cos \theta – \cos {\theta}_0).
We choose a body frame \varphi=\left\{Q ; \mathbf{i}_k\right\} in the vertical plane of symmetry, as shown in Fig. 10.15, so that k is a fixed principal axis with I_{31}^Q=I_{32}^Q=0. Then the first relation in (10.75) gives \mathbf{h}_{r Q}=I_Q \omega=I \dot{\theta} \mathbf{k}, and hence the rotational kinetic energy for the body is
\mathbf{h}_{r Q}=I_{33} \omega(t) \mathbf{e}_3=I_Q \boldsymbol{\omega}, \quad \mathbf{M}_Q=I_{33} \dot{\omega} \mathbf{e}_3=I_Q \dot{\boldsymbol{\omega}} (10.75)
K(\mathscr{B}, t)=\frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{h}_r Q=\frac{1}{2} I \dot{\theta}^2 . (10. 122c)
With \dot{\theta}(0)=\omega_0 at t_o=0, the change in the kinetic energy is \Delta K(\mathscr{B}, t)=\frac{1}{2} I\left[\dot{\theta}^2 – \omega_0^2\right]; and, with (10.122b), the work-energy principle (10.118) yields
m g \ell\left(\cos \theta – \cos \theta_0\right)=\frac{1}{2} I\left[\dot{\theta}^2 – \omega_0^2\right] . (10.122d)
The equation for the motion \theta(t) of the physical pendulum follows by differentiation of (10.122d) with respect to θ:
I \ddot{\theta} + m g \ell \sin \theta=0. (10.122e)
This result is readily confirmed by use of Euler’s law through the second relation in (10.75) . This yields
\mathbf{M}_Q(\mathscr{B}, t)=I_Q \dot{\omega}(t)=I \ddot{\theta} \mathbf{k}, (10.122f)
where \mathbf{M}_Q(\mathscr{B}, t)=M_1 \mathbf{i} + M_2 \mathbf{j} – m g \ell \sin \theta \mathbf{k}. We thus recover (10.122e), and find also that the reaction torques vanish: M_1=M_2=0. These are the torque s identified in (10.72). They vanish because k is a principal axis.
M_1 \equiv \mathbf{M}_Q \cdot \mathbf{e}_1=I_{13} \dot{\omega} – I_{23} \omega^2, \quad M_2 \equiv \mathbf{M}_Q \cdot \mathbf{e}_2=I_{23} \dot{\omega} + I_{13} \omega^2 (10.72)
Solution of (ii). Equation (10.122e) has the same form as (6.67b) for the simple pendulum. Therefore, its exact solution may be expressed in terms of an elliptic integral of the first kind or in terms of a Jacobian elliptic function, as shown in Section 7.10. By (10.122e), the small amplitude motion is described by \ddot{\theta} + p^2 \theta=0 with circular frequency p=(m g \ell / I)^{1 / 2} and period \tau =2 \pi(I / m g \ell)^{1 / 2}. The length \ell_s of a simple pendulum having the same period is \ell_s=I / m \ell=R^2 / \ell, R denoting the radius of gyration.
\ddot{\theta} + p^2 \sin \theta=0, \quad T=m \ell\left(\dot{\theta}^2 + p^2 \cos \theta\right) (6.67b)
Solution of (iii). The mechanical (rotational) power expended is given by (10.121). Hence, by (10.122b), \mathscr{P}(\mathscr{B}, t)=d \mathscr{W}(\mathscr{B}, t) / d t=-m g \ell \dot{\theta} \sin \theta. Alternatively, by (10.122c), \mathscr{P}(\mathscr{B}, t)=d K(\mathscr{B}, t) / d t=I \dot{\theta} \ddot{\theta} and use of (10.122e) yields the same result. So, the power expended during [0, t] is given by \Delta \mathscr{P} \equiv \mathscr{P}(\mathscr{B}, t) – \mathscr{P}(\mathscr{B}, 0)=-m g \ell\left(\dot{\theta} \sin \theta – \omega_0 \sin \theta_0\right); and with (10.122d), we thus find, as a function of \theta(t), that
\Delta \mathscr{P}=m g \ell\left(\omega_0 \sin \theta_0 \mp \sin \theta \sqrt{2 p^2\left(\cos \theta – \cos \theta_0\right) + \omega_0^2}\right), (10.122g)
where the sign is chosen accordingly as \dot{\theta} increasing (-) or decreasing (+) in time.