Relation between angular momentum and total external torque:

Let us consider \( \vec{F_i} \) be the resultant external force acting on the ith particle of a system of n particles and \( \vec{F_{ij}} \) be the internal force on the ith particle due to the jth particle.

So the resultant internal force on the ith particle due to all other particle is \( \displaystyle{\sum_{j=1}^{n}\vec{F_{ij}}} \), where \( i\neq{j} \).

If \( \vec{p_i} \) be the linear momentum of the ith particle of mass \( m_i \) moving with velocity \( \vec{v_i} \) and \( \vec{r_i} \) be the position vector of ith particle then the total force on the ith particle is given by,

\( \vec{F_i}+\displaystyle{\sum_{j=1}^{n}\vec{F_{ij}}}=\displaystyle{\frac{d}{dt}\vec{p_i}}=\frac{d}{dt}(m_i\vec{v_i})\tag{1} \)

Multiplying both side by \( (\vec{r_i}\times) \) we get,

\( \vec{r_i}\times\vec{F_i}+\displaystyle{\sum_{j=1}^{n}\vec{r_i}\times\vec{F_{ij}}}=\vec{r_i}\times\displaystyle{\frac{d}{dt}\vec{p_i}}=\vec{r_i}\times\frac{d}{dt}(m_i\vec{v_i}) \)

or, \( \vec{r_i}\times\vec{F_i}+\displaystyle{\sum_{j=1}^{n}\vec{r_i}\times\vec{F_{ij}}}=\frac{d}{dt}\left[m_i(\vec{r_i}\times\vec{v_i})\right] \tag{2}\)

Summing over \( i \) in equation (2) we get,

\( \displaystyle{\sum_{i=1}^{n}}\vec{r_i}\times\vec{F_i}+\displaystyle{\sum_{i=1}^{n}}\displaystyle{\sum_{j=1}^{n}\vec{r_i}\times\vec{F_{ij}}}=\frac{d}{dt}\left[\displaystyle{\sum_{i=1}^{n}}m_i(\vec{r_i}\times\vec{v_i})\right]\tag{3} \)

Now double sum in equation (3) is composed of terms such as

\( \vec{r_i}\times\vec{F_{ij}}+\vec{r_j}\times\vec{F_{ji}}\\=\vec{r_i}\times\vec{F_{ij}}-\vec{r_j}\times\vec{F_{ij}}\\=(\vec{r_i}-\vec{r_j})\times\vec{F_{ij}}\\=0 \)

Since the internal force between the particles are central forces i.e., \( \vec{F_{ij}} \) has the same direction as that of \( (\vec{r_i}-\vec{r_j}) \).

Therefore, \( \displaystyle{\sum_{i=1}^{n}}\displaystyle{\sum_{j=1}^{n}\vec{r_i}\times\vec{F_{ij}}}=0 \)

Now equation (3) will be,

\( \displaystyle{\sum_{i=1}^{n}}\vec{r_i}\times\vec{F_i}=\frac{d}{dt}\left[\displaystyle{\sum_{i=1}^{n}}m_i(\vec{r_i}\times\vec{v_i})\right] \)

or, \( \displaystyle{\vec{\tau}=\frac{d\vec{L}}{dt}} \)

Where \( \vec{\tau}=\displaystyle{\sum_{i=1}^{n}}\vec{r_i}\times\vec{F_i} \) is the total external torque acting on the system of n particles. and \( \vec{L}=\displaystyle{\sum_{i=1}^{n}}m_i(\vec{r_i}\times\vec{v_i}) \) is the total angular momentum of the system of particles.

So the total external torque acting on a system of particles is equal to the time rate of change of total angular momentum of the system of particles, provided that the internal forces between the particles are central forces.