At an instant of interest to, the center of mass of a rigid body of mass m = 20 kg has an intrinsic velocity v* = 25t m/sec, the moment of momentum about the center of mass is hc = 1100e1 - 500e2+ 600e3 kg . m²/ sec, and the moment of inertia tensor is Ic = 25(e11 + e22) - 15(e12 + e21) + 30e33kg

Question

At an instant of interest to, the center of mass of a rigid body of mass m = 20 kg has an intrinsic velocity v* = 25t m/sec, the moment of momentum about the center of mass is [latex]\mathbf{h}_C=1100 \mathbf{e}_1 - 500 \mathbf{e}_2 + 600 \mathbf{e}_3[/latex] kg . m²/ sec, and the moment of inertia tensor is [latex]\mathbf{I}_C=25 \left(\mathbf{e}_{11} + \mathbf{e}_{22}\right) - 15\left(\mathbf{e}_{12} + \mathbf{e}_{21}\right) + 30 \mathbf{e}_{33}[/latex] kg .m², referred to a body reference frame [latex]\varphi=\left\{C ; \mathbf{e}_k\right\}[/latex]. Find the total kinetic energy of the body.

Accepted Answer

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

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

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

[latex]25 \omega_1 - 15 \omega_2=1100, \quad - 15 \omega_1+25 \omega_2=-500, \quad 30 \omega_3=600[/latex].

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

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

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

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