Question 10.16: Apply the energy method to derive the first integral of the ...
Apply the energy method to derive the first integral of the equation of motion of the rod described in Fig. 10.11, page 447.
The resultant force R exerted on the rod by the smooth hinge bearing is workless, and the gravitational force on the rod is conservative with total potential energy V=\frac{1}{2} m g \ell(1 – \cos \theta). The body frame 2=\{Q ; \mathbf{i}, \mathbf{j}, \mathbf{k}\} is a principal reference frame at Q. Therefore, with the aid of (10.80a) and (10.80c) in (10.102), the total kinetic energy of the rod relative to Q is K=K_{r Q}=\frac{1}{6} m \ell^2\left(\dot{\theta}^2 + \Omega^2 \sin ^2 \theta\right).
\boldsymbol{\omega}=-\Omega(\sin \theta \mathbf{i} + \cos \theta \mathbf{j}) + \dot{\theta} \mathbf{k} (10.80a)
\mathbf{I}_Q=\frac{1}{3} m \ell^2\left(\mathbf{i}_{11} + \mathbf{i}_{33}\right) \text {, } (10.80c)
The smooth hinge bearing exerts no torque about the hinge axis , the k direction. But the component \mu_1 of the bearing reaction torque exerted on the rod about the i-axis does work on the rod to control its spin. Therefore, we shall need to apply the general energy principle (10.131). The total bearing reaction torque exerted on the rod, from (10.80e), is \mathbf{M}_Q^B=\mu_1 \mathbf{i}=-\frac{2}{3} m \ell^2 \Omega \dot{\theta} \cos \theta \mathbf{i}. Then with (10.80a) and \theta_0=\theta(0) \text { at } t=0, (10.114) for a fixed point Q gives
\mathscr{W}(\mathscr{B}, t)=\mathscr{W}_t(\mathscr{B}, t)+\mathscr{W}_r(\mathscr{B}, t) (10.114)
\mathscr{W}_N=\int_0^t \mathbf{M}_Q^B \cdot \omega d t=\frac{2}{3} m \ell^2 \Omega^2 \int_{\theta_0}^\theta \cos \theta \sin \theta d \theta=\frac{1}{3} m \ell^2 \Omega^2\left(\sin ^2 \theta-\sin ^2 \theta_0\right) (10.133a)
Recall the foregoing kinetic and potential energy functions to form \left.\Delta \mathscr{E}\right|_0 ^t=\left.\Delta K\right|_0 ^t + \left.\Delta V\right|_0 ^t. Then with (10.133a), the general energy principle (10.131) yields
\left.\frac{1}{6} m \ell^2\left(\dot{\theta}^2 + \Omega^2 \sin ^2 \theta\right)\right|_0 ^t + \left.\frac{1}{2} m g \ell(1 – \cos \theta)\right|_0 ^t=\frac{1}{3} m \ell^2 \Omega^2\left(\sin ^2 \theta-\sin ^2 \theta_0\right) (10.133b)
\Delta \mathscr{E}=\mathscr{W}_N(\mathscr{B}, t) (10.131)
wherein \left.\Delta \mathscr{E}\right|_0 ^t \equiv \mathscr{E}(t)-\mathscr{E}(0), subject to the initial data \dot{\theta}(0)=0 \text { and } \theta(0)=\theta_0. We thus obtain the first integral of Euler’s equation of motion for the rod:
\dot{\theta}^2=\Omega^2\left(\sin ^2 \theta – \sin ^2 \theta_0\right) + 3 \frac{g}{\ell}\left(\cos \theta – \cos \theta_0\right) (10.133c)
This is the same as (10.80g) obtained by direct integration of the equation of motion (10.80f) . Relative to the vertical shaft, (10.133c) determines the angular speed of the rod as a function of θ.
\dot{\theta}=\pm \sqrt{\Omega^2\left(\cos ^2 \theta_0 – \cos ^2 \theta\right)+3 \frac{g}{\ell}\left(\cos \theta – \cos \theta_0\right)} (10.80g)
\ddot{\theta}+\left(p^2-\Omega^2 \cos \theta\right) \sin \theta=0, \quad p^2 \equiv \frac{3 g}{2 \ell} (10.80f)
\dot{\theta}^2=\Omega^2\left(\sin ^2 \theta – \sin ^2 \theta_0\right) + 3 \frac{g}{\ell}\left(\cos \theta-\cos \theta_0\right) (10.133c)