Free Books

Linear Momentum of the Center of Mass

Consider a system of $ N$ point-masses $ m_i$, each traveling with vector velocity $ \underline{v}_i$, and not necessarily rigidly attached to each other. Then the total momentum of the system is


$\displaystyle \underline{p}\eqsp \sum_{i=1}^N m_i \underline{v}_i
\eqsp \sum_{...
...ine{x}_i \right)
\eqsp M \frac{d}{dt} \underline{x}_c
\isdef M \underline{v}_c
$

where $ M=\sum m_i$ denotes the total mass, and $ \underline{v}_c$ is the velocity of the center of mass. Thus, the momentum $ \underline{p}$ of any collection of masses $ m_i$ (including rigid bodies) equals the total mass $ M$ times the velocity of the center-of-mass.
Next Section:
Whoops, No Angular Momentum!
Previous Section:
Literature Relevant to Brasses