Question 10.12: At an instant of interest to, the center of mass of a rigid ...

The total kinetic energy of the body at the moment of concern may be found from (10.101). An easy calculation gives   K^*=\frac{1}{2} m \mathbf{v}^* \cdot \mathbf{v}^*=6250 \mathrm{~N} \cdot \mathrm{m},  the kinetic energy of the center of mass. To find   K_{r C}=\frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{h}_C  in (10.100), bearing in mind that  \mathbf{h}_C  is given, we must first find   \boldsymbol{\omega}=\omega_k \mathbf{e}_k  such that   \mathbf{h}_C=\mathbf{I}_C \boldsymbol{\omega}.  For the assigned moment of inertia tensor, the moment of momentum relative to C is given by   \mathbf{h}_C=\left(25 \omega_1  –  15 \omega_2\right) \mathbf{e}_1  +  \left(-15 \omega_1  +  25 \omega_2\right) \mathbf{e}_2  +  30 \omega_3 \mathbf{e}_3.  Equating these scalar components with those given at  t_o,  we find three equations for the components   \omega_k :

K_{r C}=\frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{h}_C=\frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{I}_C \boldsymbol{\omega} .                      (10.100)

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)

25 \omega_1  –  15 \omega_2=1100, \quad  –  15 \omega_1+25 \omega_2=-500, \quad 30 \omega_3=600.

These yield   \boldsymbol{\omega}=50 \mathbf{e}_1  +  10 \mathbf{e}_2  +  20 \mathbf{e}_3  rad/sec . Hence, with the  assigned value for   \mathbf{h}_C,  the kinetic energy relative to the center of mass is   K_{r C}=\frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{h}_C=31,000  N . m. The total kinetic energy of the body in the inertial frame now follows from (10.101 ) or (10.94):  K=37,250  \mathrm{~N} \cdot \mathrm{m}.

K(\mathscr{B}, t)=K^*(\mathscr{B}, t)  +  K_{r C}(\mathscr{B}, t)                     (10.94)