12.11.1 Angular Momentum of a Rigid Body

Equation 12.49 is the angular momentum of a single particle. The angular momentum of a rigid body composed of particles with individual angular momenta $\vec{L}_{1}$ , $\vec{L}_{2}$ , $\vec{L}_{3}, \dots$ is the vector sum

\vec{L} = {\vec{L}}_{1} + {\vec{L}}_{2} + {\vec{L}}_{3} + \dots = \sum_{i}^{} {\vec{L}}_{i}

(12.53)

We can combine Equations 12.52 and 12.53 to find the rate of change of the system’s angular momentum:

\frac{d \vec{L}}{d t} = \sum_{i}^{} \frac{d {\vec{L}}_{i}}{d t} = \sum_{i}^{} {\vec{τ}}_{i} = {\vec{τ}}_{net}

(12.54)

Because any internal forces are action/reaction pairs of forces, acting with the same strength in opposite directions, the net torque due to internal forces is zero. Thus the only forces that contribute to the net torque are external forces exerted on the system by the environment.

For a system of particles, the rate of change of the system’s angular momentum is the net torque on the system. Equation 12.54 is analogous to the Chapter 11 result $d \vec{P} / d t = \vec{F}_{net}$ , which says that the rate of change of a system’s total linear momentum is the net force on the system.