PX155 - E6 - system of particles II

as before, consider a body as a system of particles subject to internal and external forces
total force on the particle :

\vec{F} = {\vec{F}}_{i} + \sum_{j \neq i} {\vec{F}}_{j i} = \frac{d {\vec{p}}_{i}}{d t}

if positions ${\vec{r}}_{i}$ are measured from $O$ , then the torque about $O$ :

{\vec{τ}}_{i} = {\vec{r}}_{i} \times ({\vec{F}}_{i} + \sum_{j \neq i} {\vec{F}}_{j i}) = \frac{d {\vec{L}}_{i}}{d t}

total torque

\vec{τ} = \sum_{i} {\vec{r}}_{i} \times ({\vec{F}}_{i} + \sum_{j \neq i} {\vec{F}}_{j i}) = \sum_{i} {\vec{r}}_{i} \times {\vec{F}}_{i} + \sum_{i} \sum_{j \neq i} {\vec{r}}_{i} \times F_{j i} = \sum_{i} \frac{d {\vec{L}}_{i}}{d t}

$\sum_{i} \sum_{j \neq i} {\vec{r}}_{i} \times F_{j i} = 0$ as for central forces, acting along the line between particles, PX155 - A3 - newton's third law: internal force term cancels out
- both $F_{21}$ and $F_{12}$ have some perpendicular distance from the pivot, so, generate equal and opposite torques

\vec{τ} = \sum_{i} ({\vec{r}}_{i} \times {\vec{F}}_{i}) = \frac{d}{d t} \sum_{i} {\vec{L}}_{i} = \frac{d \vec{L}}{d t}

total torque from external forces on a composite body equals the rate of change of its total angular momentum:

\vec{τ} = \frac{d \vec{L}}{d t}