Torque and angular momentum for a system of particles

To get the total angular momentum of a system of particles about a given point we need to add vectorially the angular momenta of individual particles. Thus, for a system of n particles,

$\mathbf{L}=\mathbf{l_1}+\mathbf{l_2}+...+\mathbf{l_n}=\Sigma_{i = 1}^{n}\mathbf{l_i}$

The angular momentum of the i th particle is given by

$\mathbf{l_i}=\mathbf{r_i} \times \mathbf{p_i}$

where $\mathbf{r_i}$  is the position vector of the i th particle with respect to a given origin and $\mathbf{p} = (m_i \mathbf{v_i})$ is the linear momentum of the particle. (The particle has mass mi and velocity $\mathbf{v_i}$) We may write the total angular momentum of a system of particles as

$\mathbf{L}=\Sigma \mathbf{l_i}=\Sigma_{i} \mathbf{r_i} \times \mathbf{p_i}$

This is a generalisation of the definition of angular momentum (Eq. 7.25a) for a single particle to a system of particles. 

Using Eqs. (7.23) and (7.25b), we get

$\frac{d\mathbf{L}}{dt}=\frac{d}{dt}(\Sigma \mathbf{l_i})=\Sigma_{i} \frac{d\mathbf{l}}{dt} =\Sigma \tau_i$

where τ$_i$ is the torque acting on the $i^{th}$ particle;

τ$_i$ = $\mathbf{r_i} \times \mathbf{F_i}$

The force $\mathbf{F_i}$ on the $i^{th}$ particle is the vector sum of external forces $\mathbf{F_i^{ext}}$ acting on the particle and the internal forces $\mathbf{F_i^{int}}$ exerted on it by the other particles of the system. We may therefore separate the contribution of the external and the internal forces to the total torque

τ = $\Sigma_{i}\tau=\Sigma_{i}\mathbf{r_i}\times \mathbf{F_i}$

τ = τ$_{ext}$ + τ$_{int}$

where τ$_{ext}$ = $\Sigma_{i}\mathbf{r_i}\times \mathbf{F_i^{ext}}$

and τ$_{int}$ = $\Sigma_{i}\mathbf{r_i}\times \mathbf{F_i^{int}}$

We shall assume not only Newton’s third law of motion, i.e. the forces between any two particles of the system are equal and opposite, but also that these forces are directed along the line joining the two particles. In this case the contribution of the internal forces to the total torque on the system is zero, since the torque resulting from each action reaction pair of forces is zero. We thus have, τ$_{int}$ = 0 and therefore τ = τ$_{ext}$.

Since τ = $\Sigma \tau_i$, it follows from Eq. (7.28a) that

$\frac{d\mathbf{L}}{dt}=0$

Thus, the time rate of the total angular momentum of a system of particles about a point (taken as the origin of our frame of reference) is equal to the sum of the external torques (i.e. the torques due to external forces) acting on the system taken about the same point. Eq. (7.28 b) is the generalisation of the single particle case of Eq. (7.23) to a system of particles. Note that when we have only one particle, there are no internal forces or torques. Eq.(7.28 b) is the rotational analogue of

$\frac{d\mathbf{p}}{dt}=\mathbf{F_{ext}}$

Note that like Eq.(7.17), Eq.(7.28b) holds good for any system of particles, whether it is a rigid body or its individual particles have all kinds of internal motion.