König's theorem (kinetics)

The first part expresses the angular momentum of a system as the sum of the angular momentum of the centre of mass and the angular momentum applied to the particles relative to the center of mass. [1]

${\displaystyle \displaystyle {\vec {L}}={\vec {r}}_{CoM}\times \sum \limits _{i}m_{i}{\vec {v}}_{CoM}+{\vec {L}}'={\vec {L}}_{CoM}+{\vec {L}}'}$

Considering an inertial reference frame with origin O, the angular momentum of the system can be defined as:

${\displaystyle {\vec {L}}=\sum \limits _{i}({\vec {r}}_{i}\times m_{i}{\vec {v}}_{i})}$

The position of a single particle can be expressed as:

${\displaystyle {\vec {r}}_{i}={\vec {r}}_{CoM}+{\vec {r}}'_{i}}$

And so we can define the velocity of a single particle:

${\displaystyle {\vec {v}}_{i}={\vec {v}}_{CoM}+{\vec {v}}'_{i}}$

The first equation becomes:

${\displaystyle {\vec {L}}=\sum \limits _{i}({\vec {r}}_{CoM}+{\vec {r}}'_{i})\times m_{i}({\vec {v}}_{CoM}+{\vec {v}}'_{i})}$

${\displaystyle {\vec {L}}=\sum \limits _{i}{\vec {r}}'_{i}\times m_{i}{\vec {v}}'_{i}+\left(\sum \limits _{i}m_{i}{\vec {r}}'_{i}\right)\times {\vec {v}}_{CoM}+{\vec {r}}_{CoM}\times \sum \limits _{i}m_{i}{\vec {v}}'_{i}+\sum \limits _{i}{\vec {r}}_{CoM}\times m_{i}{\vec {v}}_{CoM}}$

But the following terms are equal to zero:

${\displaystyle \sum \limits _{i}m_{i}{\vec {r}}'_{i}=0}$

${\displaystyle \sum \limits _{i}m_{i}{\vec {v}}'_{i}=0}$

So we prove that:

${\displaystyle {\vec {L}}=\sum \limits _{i}{\vec {r}}'_{i}\times m_{i}{\vec {v}}'_{i}+M{\vec {r}}_{CoM}\times {\vec {v}}_{CoM}}$

where M is the total mass of the system.

The second part expresses the kinetic energy of a system of particles in terms of the velocities of the individual particles and the centre of mass.

Specifically, it states that the kinetic energy of a system of particles is the sum of the kinetic energy associated to the movement of the center of mass and the kinetic energy associated to the movement of the particles relative to the center of mass.[2]

${\displaystyle K=K'+K_{\text{CoM}}}$

The total kinetic energy of the system is:

${\displaystyle K=\sum _{i}{\frac {1}{2}}m_{i}v_{i}^{2}}$

Like we did in the first part, we substitute the velocity:

${\displaystyle K=\sum _{i}{\frac {1}{2}}m_{i}({\bar {v}}'_{i}+{\bar {v}}_{\text{CoM}})^{2}}$

${\displaystyle K=\sum _{i}{\frac {1}{2}}m_{i}({\bar {v}}'_{i}+{\bar {v}}_{\text{CoM}})\cdot ({\bar {v}}'_{i}+{\bar {v}}_{\text{CoM}})=\sum _{i}{\frac {1}{2}}m_{i}{v'_{i}}^{2}+{\bar {v}}_{\text{CoM}}\cdot \sum _{i}m_{i}{\bar {v}}'_{i}+\sum _{i}{\frac {1}{2}}m_{i}v_{\text{CoM}}^{2}}$

We know that ${\displaystyle {\bar {v}}_{CoM}\cdot \sum _{i}m_{i}{\bar {v}}'_{i}=0,}$ so if we define:

${\displaystyle K'=\sum _{i}{\frac {1}{2}}m_{i}{v'_{i}}^{2}}$

${\displaystyle K_{\text{CoM}}=\sum _{i}{\frac {1}{2}}m_{i}v_{\text{CoM}}^{2}={\frac {1}{2}}Mv_{\text{CoM}}^{2}}$

we're left with:

${\displaystyle K=K'+K_{\text{CoM}}}$