Question 10.5: Apply the foregoing results to find the moment of momentum r...
Apply the foregoing results to find the moment of momentum relative to Q for the connecting rod of the simple machine in Fig. 10.6, page 425.
The rod frame 2=\{Q ; \mathbf{i}, \mathbf{j}, \mathbf{k}\} in Fig. 10.6 is a principal reference frame for which \hat{\mathbf{e}}_k \equiv \mathbf{i}_k and whose total angular velocity referred to frame 2 is given in (10.41b). Hence, \omega_1=\omega_2=0, \omega_3=\dot{\beta} + \Omega are the principal components \hat{\omega}_k \equiv \omega_k. The principal components of the moment of inertia about the rod’s end point Q, bearing in mind the basis directions, may be read from (9.46d) or from Fig. D.7 of Appendix D.
I_H(\mathscr{B})=\frac{m\ell^2}{3}(i_{11}+i_{22}) (9.46d)
For a uniform thin rod of mass m and length \ell, however, it is instructive to show directly that I_{22}^Q=I_{33}^Q=\int_{\mathscr{B}} r^2 d m=(m / \ell) \int_0^{\ell} r^2 d r, so
\boldsymbol{\omega}=(\dot{\beta} + \Omega) \mathbf{k} (10.41b)
\mathbf{M}_Q(\mathscr{B}, t)=\dot{\mathbf{h}}_{r Q}(\mathscr{B}, t)+\mathbf{r}^*(\mathscr{B}, t) \times m(\mathscr{B}) \mathbf{a}_Q (9.46)
I_{22}^Q=I_{33}^Q=\frac{1}{3} m \ell^2. (10.63a)
Thus, (10.62) yields \mathbf{h}_{r Q}=I_{33}^Q \omega_3 \mathbf{i}_3, that is,
\mathbf{h}_{r Q}=\frac{m \ell^2}{3}(\dot{\beta} + \Omega) \mathbf{k}. (10.63b)
This is the same as (10.41d) obtained earlier by direct integration. Let the reader show that the nontrivial principal moments of inertia about the center of mass are I_{22}^C=I_{33}^C=m \ell^2 /12, and thus confirm that the mom \mathbf{h}_{r C}=\left(m \ell^2 / 12\right)(\dot{\beta} + \Omega) \mathbf{k} and thus confirm that the moment of momentum relative to the center of mass is \mathbf{h}_{r c}=\left(m \ell^2 / 12\right)(\dot{\beta} + \Omega) \mathbf{k}, as shown differently in (10.50d).
\mathbf{h}_{r Q}(\mathscr{B}, t)=\frac{m \ell^2}{3}(\dot{\beta} + \Omega) \mathbf{k} (10.41d)
\mathbf{h}_{r C}=\sigma(\dot{\beta}+\Omega) \int_{-\ell / 2}^{\ell / 2} \rho^2 d \rho \mathbf{k}=m(\dot{\beta} + \Omega) \frac{\ell^2}{12} \mathbf{k} (10.50d)
