Question 10.11: (i) Apply (10.95) to determine the total kinetic energy of t...

Solution of (i). Since Q in Fig. 10.6 is a moving base point, the total  kinetic energy of the rod is obtained from (10.95). With   \boldsymbol{\omega}=(\dot{\beta}  +  \Omega) \mathbf{k}  by (10.41b) and  \mathbf{h}_{r Q}  in (10.41d), (10.99) yields the kinetic energy of the rod relative to Q:

\mathbf{h}_{r Q}(\mathscr{B}, t)=\frac{m \ell^2}{3}(\dot{\beta}  +  \Omega) \mathbf{k}.                  (10.41d)

\boldsymbol{\omega}=(\dot{\beta}  +  \Omega) \mathbf{k}                 (10.41b)

K_{r Q}(\mathscr{B}, t)=\frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{h}_{r Q}=\frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{I}_Q \boldsymbol{\omega} .                    (10.99)

K_{r Q}(\mathscr{B}, t)=\frac{1}{2} \omega \cdot \mathbf{h}_{r Q}=\frac{m \ell^2}{6}(\dot{\beta}  +  \Omega)^2                   (10.103a)

The same result also follows easily from (10.102). With   \dot{\mathbf{r}}^*=\boldsymbol{\omega} \times \mathbf{r}^*=\frac{\ell}{2}(\dot{\beta}  +  \Omega) \mathbf{j}  and use of (10.41e) and (10.103a), (10.95) yields the total kinetic energy of the connecting rod:

K_{r Q}=\frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{I}_Q \boldsymbol{\omega}=\frac{1}{2} \hat{I}_{11}^Q \hat{\omega}_1^2  +  \frac{1}{2} \hat{I}_{22}^Q \hat{\omega}_2^2  +  \frac{1}{2} \hat{I}_{33}^Q \hat{\omega}_3^2.                  (10.102)

\mathbf{v}_Q=\alpha \Omega \mathbf{b}                  (10.41e)

K(\mathscr{B}, t)=\frac{1}{2} m\left[a^2 \Omega^2  +  \ell a \Omega(\dot{\beta}  +  \Omega) \cos \beta  +  \frac{\ell^2}{3}(\dot{\beta}  +  \Omega)^2\right].                (10.103b)

Solution of (ii). The solution based on (10.94) is simpler to  construct. The kinetic energy of the center of mass of the rod is given by (10.91). Thus, recalling relations introduced above , we obtain

K^*(\mathscr{B}, t) \equiv \frac{1}{2} m(\mathscr{B}) \mathbf{v}^*(\mathscr{B}, t) \cdot \mathbf{v}^*(\mathscr{B}, t)                         (10.91)

\mathbf{v}^*=\mathbf{v}_Q  +  \boldsymbol{\omega} \times \mathbf{r}^*=a \Omega \mathbf{b}  +  \frac{\ell}{2}(\dot{\beta}  +  \Omega) \mathbf{j},                 (10.103c)

and hence

K^*(\mathscr{B}, t)=\frac{1}{2} m\left[a^2 \Omega^2  +  a \Omega \ell \cos \beta(\dot{\beta}  +  \Omega)  +  \frac{\ell^2}{4}(\dot{\beta}  +  \Omega)^2\right] .                         (10.103d)

With the aid of (9.28) referred to the principal basis at the center of mass of the rod, (10.102) yields easily the kinetic energy relative to C,

\mathbf{I}_C=\frac{m \ell^2}{12}\left(\mathbf{i}_{22}^*  +  \mathbf{i}_{33}^*\right)                    (9.28)

K_{r C}(\mathscr{B}, t)=\frac{m \ell^2}{24}(\dot{\beta}  +  \Omega)^2.                  (10.103e)

Now (10.94) leads again to (10.103b). The reader will appreciate the simplicity of the mnemonic rule (10.94) and the consequent result (10.101) for the moving center of mass, compared with (10.95) for an arbitrary moving point Q.

K(\mathscr{B}, t)=\frac{1}{2} m \mathbf{v}^* \cdot \mathbf{v}^*  +  \frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{I}_C \boldsymbol{\omega} .                      (10.101)