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11.6: Derivation of kinetic energy of a rigid body

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    103617
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    Derivation: Kinetic energy of a rigid body

    We now derive the kinetic energy \(T_{C}\) of a rigid body \(C\), by summing the kinetic energies of all point masses \(i\) inside the rigid body.

    \[\begin{align} T_{C}= & \sum_{i} \frac{1}{2} m_{i}\left|\overrightarrow{\boldsymbol{v}}_{i}\right|^{2} \tag{11.31} \label{11.31}\\[4pt] = & \frac{1}{2} \sum_{i} m_{i}\left(\overrightarrow{\boldsymbol{v}}_{G}+\overrightarrow{\boldsymbol{\omega}}_{C} \times \overrightarrow{\boldsymbol{r}}_{i / G}\right) \cdot\left(\overrightarrow{\boldsymbol{v}}_{G}+\overrightarrow{\boldsymbol{\omega}}_{C} \times \overrightarrow{\boldsymbol{r}}_{i / G}\right) \tag{11.32} \label{11.32}\\[4pt] = & \frac{1}{2} \sum_{i} m_{i}\left|\overrightarrow{\boldsymbol{v}}_{G}\right|^{2}+\frac{1}{2} \sum_{i} m_{i}\left|\overrightarrow{\boldsymbol{\omega}}_{C} \times \overrightarrow{\boldsymbol{r}}_{i / G}\right|^{2} \\[4pt] & +\frac{1}{2} \sum_{i} 2 m_{i} \overrightarrow{\boldsymbol{v}}_{G} \cdot\left(\overrightarrow{\boldsymbol{\omega}}_{C} \times \overrightarrow{\boldsymbol{r}}_{i / G}\right) \tag{11.33} \label{11.33}\\[4pt] = & \frac{1}{2} \sum_{i} m_{i}\left|\overrightarrow{\boldsymbol{v}}_{G}\right|^{2}+\frac{1}{2} \sum_{i} m_{i}\left|\overrightarrow{\boldsymbol{\omega}}_{C} \times \overrightarrow{\boldsymbol{r}}_{i / G}\right|^{2} \\[4pt] & +\left(\sum_{i} m_{i} \overrightarrow{\boldsymbol{r}}_{i / G}\right) \cdot\left(\overrightarrow{\boldsymbol{v}}_{G} \times \overrightarrow{\boldsymbol{\omega}}_{C}\right) \tag{11.34} \label{11.34}\\[4pt] = & T_{C, \text { trans }}+T_{C, \text { rot }} \tag{11.35} \label{11.35}\\[4pt] T_{C, \text { trans }}= & \frac{1}{2} m_{\mathrm{tot}}\left|\overrightarrow{\boldsymbol{v}}_{G}\right|^{2} \tag{11.36} \label{11.36}\\[4pt] T_{C, \text { rot }}= & \frac{1}{2} \sum_{i} m_{i}\left|\overrightarrow{\boldsymbol{\omega}}_{C} \times \overrightarrow{\boldsymbol{r}}_{i / G}\right|^{2} \tag{11.37} \label{11.37}\end{align}\]

    where the kinematic equation \(\overrightarrow{\boldsymbol{v}}_{i}=\overrightarrow{\boldsymbol{v}}_{G}+\overrightarrow{\boldsymbol{\omega}}_{C} \times \overrightarrow{\boldsymbol{r}}_{i / G}\) and the vector identity \(\overrightarrow{\boldsymbol{a}} \cdot(\overrightarrow{\boldsymbol{b}} \times \overrightarrow{\boldsymbol{c}})=\overrightarrow{\boldsymbol{c}} \cdot(\overrightarrow{\boldsymbol{a}} \times \overrightarrow{\boldsymbol{b}})\) were used.

    For 2D planar motion we obtain:

    \[\begin{align} T_{C, \mathrm{rot}, 2 \mathrm{D}} & =\frac{1}{2} \sum_{i} m_{i}\left|\omega_{C} \hat{\boldsymbol{k}} \times \rho_{i / G} \hat{\boldsymbol{\rho}}_{G}\right|^{2} \tag{11.38} \label{11.38}\\[4pt] & =\frac{1}{2}\left(\sum_{i=I_{i, z z}} m_{i} \rho_{i / G}^{2}\right) \omega_{C}^{2} \tag{11.39} \label{11.39}\\[4pt] & =\frac{1}{2} I_{G, z z} \omega_{C}^{2} \tag{11.40} \label{11.40}\end{align}\]

    In summary:

    \[T_{C, 2 D}=T_{C, \text { trans }}+T_{C, \mathrm{rot}, 2 D}=\frac{1}{2} m_{\mathrm{tot}}\left|\overrightarrow{\boldsymbol{v}}_{G}\right|^{2}+\frac{1}{2} I_{G, z z} \omega_{C}^{2} \tag{11.41} \label{11.41}\]


    This page titled 11.6: Derivation of kinetic energy of a rigid body is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Peter G. Steeneken via source content that was edited to the style and standards of the LibreTexts platform.