Question 10.15: Apply the energy principle to derive the first integral of t...
Apply the energy principle to derive the first integral of the equation of motion for the physical pendulum in Fig. 10.15, page 467 .
The bearing reaction force R at the smooth support Q in Fig. 10.15 is workless; the bearing reaction torque \mu_Q \equiv M_1 \mathbf{i}+M_2 \mathbf{j}, because there is no rotation of the body about these directions, also is workless, in fact \boldsymbol{\mu}_Q=\mathbf{0}; and the gravitational force W is conservative with potential energy V=m g \ell(1 – \cos{\theta}). Clearly, the system is conservative and (10.132) holds .The rotational kinetic energy is given by K_{r Q}=\frac{1}{2} I \omega^2, where I \equiv I_{33}^Q With \omega=\dot{\theta}, (10.132) yields the first integral of the equation of motion for the physical pendulum:
K(\mathscr{B}, t)+V(\mathscr{B})=E, \text { a constant. } (10.132)
For initial data \theta(0)=\theta_0 and \dot{\theta}(0)=\omega_0, the constant E=\frac{1}{2} I \omega_0^2 + m g \ell(1 – \cos{\theta}_0), and the last result then agrees with (10.122d).
m g \ell\left(\cos \theta – \cos \theta_0\right)=\frac{1}{2} I\left[\dot{\theta}^2 – \omega_0^2\right] (10.122d)