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10.2: Linear Momentum in a Moving Frame

  • Page ID
    47281
    • Franz S. Hover & Michael S. Triantafyllou
    • Massachusetts Institute of Technology via MIT OpenCourseWare
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    The expression for total velocity may be inserted into the summed linear momentum equation to give

    \begin{align} \sum_{i=1}^N \vec{F}_i \,\, &= \,\, \sum_{i=1}^N \dfrac{d}{dt} (m_i (\vec{v}_o + \vec{\omega} \times \vec{r}_i)) \\[4pt] &= \,\, m \dfrac{\partial \vec{v}_o}{\partial t} + \dfrac{d}{dt} \left[ \vec{\omega} \times \sum_{i=1}^N m_i \vec{r}_i \right] , \end{align}where \(m = \displaystyle\sum_{i=1}^N m_i\), and \(\vec{v}_i = \vec{v}_o + \vec{\omega} \times \vec{r}_i\). Further defining the center of gravity vector \(\vec{r}_G\) such that

    \[ m \vec{r}_G \, = \, \sum_{i=1}^N m_i \vec{r}_i, \]

    we have \[ \sum_{i=1}^N \vec{F}_i \, = \, m \dfrac{\partial \vec{v}_o}{\partial t} + m \dfrac{d}{dt} (\vec{\omega} \times \vec{r}_G). \]

    Using the expansion for total derivative again, the complete vector equation in body coordinates is

    \[ \vec{F} \, = \, \sum_{i=1} N \, = \, m \left( \dfrac{\partial \vec{v}_o}{\partial t} + \vec{\omega} \times \vec{v}_o + \dfrac{d \vec{\omega}}{dt} \times \vec{r}_G + \vec{\omega} \times (\vec{\omega} \times \vec{r}_G) \right). \]

    Now we list some conventions that will be used from here on:

    \begin{align*} \vec{v}_o \,\, &= \,\, \{ u, \, v, \, w \} \quad \text{(body-referenced velocity)} \\[4pt] \vec{r}_G \,\, &= \,\, \{ x_G, \, y_G, \, z_G \} \quad \text{(body-referenced location of center of mass)} \\[4pt] \vec{\omega} \,\, &= \,\, \{ p, \, q, \, r \} \quad \text{(rotation vector, in body coordinates)} \\[4pt] \vec{F} \,\, &= \,\, \{ X, \, Y, \, Z \} \quad \text{(external force, body coordinates)}. \end{align*}

    The last term in the previous equation simplifies using the vector triple product identity \[ \vec{\omega} \times (\vec{\omega} \times \vec{r}_G) \, = \, (\vec{\omega} \cdot \vec{r}_G) \vec{\omega} - (\vec{\omega} \cdot \vec{\omega}) \vec{r}_G, \]and the resulting three linear momentum equations are

    \begin{align} X \,\, &= \,\, m \left[ \dfrac{\partial u}{\partial t} + qw - rv + \dfrac{dq}{dt} z_G - \dfrac{dr}{dt} y_G + (q y_G + r z_G)p - (q^2 + r^2)x_G \right] \\[4pt] Y \,\, &= \,\, m \left[ \dfrac{\partial v}{\partial t} + ru - pw + \dfrac{dr}{dt} x_G - \dfrac{dp}{dt} z_G + (r z_G + p x_G)q - (r^2 + p^2) y_G \right] \\[4pt] Z \,\, &= \,\, m \left[ \dfrac{\partial w}{\partial t} + pv - qu + \dfrac{dp}{dt} y_G - \dfrac{dq}{dt} x_G + (p x_G + q y_G)r - (p^2 + q^2) z_G \right]. \end{align}

    Note that about half of the terms here are due to the mass center being in a different location than the reference frame origin, i.e., \(\vec{r}_G \neq \vec{0}\).


    This page titled 10.2: Linear Momentum in a Moving Frame is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Franz S. Hover & Michael S. Triantafyllou (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.