The Angular Momentum of a System of Particles — Question 3

Question 3

Cross terms vanish only when the reference is the mass center

For a particle system write $\vec r_i=\vec R_C+\vec\rho_i$ and $\vec v_i=\vec V_C+\vec u_i$, where $C$ is the center of mass.

1. Derive the orbital-plus-internal decomposition of total angular momentum.

2. Identify the identities that remove the cross terms and explain why an arbitrary reference point does not automatically do so.

Question 3 - Solution

Cross terms vanish only when the reference is the mass center

(a) Derivation and calculation

Substitute the given decompositions and expand: $$\vec L_O=\sum(\vec R_C+\vec\rho_i)\times m_i(\vec V_C+\vec u_i).$$ The four terms are $$\begin{aligned} \vec L_O={}&\vec R_C\times M\vec V_C+\vec R_C\times\sum m_i\vec u_i\\ &+\left(\sum m_i\vec\rho_i\right)\times\vec V_C+\sum\vec\rho_i\times m_i\vec u_i. \end{aligned}$$

(b) Result and interpretation

The mass-center definition and its derivative imply $$\sum m_i\vec\rho_i=0,\qquad\sum m_i\vec u_i=0.$$ Both cross terms therefore vanish: $$\boxed{\vec L_O=\vec R_C\times M\vec V_C+\vec L_C},\qquad \vec L_C=\sum\vec\rho_i\times m_i\vec u_i.$$ An arbitrary reference point does not generally satisfy the two mass-weighted zero-sum identities.

Physical check

The result separates center orbital motion from the angular momentum of motion relative to the center.