Derive total angular momentum for the collection of particles

Question

So I was trying to figure out how the total angular momentum for the collection of particles is derived. Here in Wikipedia https://en.wikipedia.org/wiki/Angular_momentum#Center_of_mass is the derivation, but there is this one part that I do not understand. How does the term

$$ \sum_ir_i\times m_iV $$

vanish? I also know that $\sum_i m_ir_i=0$, but how does that help making the term vanish?

See also https://physics.stackexchange.com/a/499323/392 – John Alexiou Oct 04 '19 at 13:37 — John Alexiou, Oct 04 '19 at 13:37

score 1 · Accepted Answer · answered Oct 04 '19 at 10:27

Rearrange the above equation as $\sum_i m_ir_i \times V$. Notice that this is the same as $m_i$ is a scalar. Now, $V$ is a constant vector in the above equation. We know that $\sum_i m_ir_i$ is $0$. So multiplying both sides by a constant vector is also $0$.

Derive total angular momentum for the collection of particles

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