# Total Angular Momentum of system of particles

In my mechanics book while working out the total angular momentum's formula analogous to the center of mass of the system

from the fig it concludes $$r_i=R+r_i'$$ Taking derivative $$v_i=V+v_i'$$ formula for angular momentum $$l=\sum r_i \times m_iv_i$$

now putting the value of $$v_i$$ and $$r_i$$ we get this $$l= \sum R \times m_iv + \sum r_i' \times m_iv_i' + {\sum m_ir_i'} \times v + R \times \frac{d\sum m_ir_i'}{dt}.$$

Now in the book it says $$\sum m_ir_i'$$ is defines as the radius vector of the center of mass in the very coordinate system whose origin is the center of mass therefore it is a null vector. How can $$\sum m_ir_i'$$ can be the null vector is it shifting the whole coordinate system to COM given in the figure if thats the case it still does not make sense to me

The definition of the center of mass is

$$\sum_i m_i r_i = \sum_i m_i R = m\, R$$

$$R = \frac{ \sum_i m_i r_i }{\sum_i m_i }$$

But since $$r_i = R + r'_i$$, the above is also

$$\require{cancel} \sum_i m_i r_i = \sum_i m_i (R+r'_i) = \sum_i m_i R + \cancel{ \sum_i m_i r'_i }$$

So the definition of the CoM, implies that $$\sum_i m_i r'_i =0$$.

Note that the derivative of the above is $$\sum_i m_i v'_i =0$$ must also be true which is used in the derivation of angular momentum.

As a result, you can state that At every instant, only the vector $$R$$ tracks the center of mass. and the vector $$r'_i$$ cancels out of the calculation. Essentially this is the definition of center of mass.

• Does that mean if i calculate COM from $r_i$ point than the COM vector will be a null vector as COM of a body or system of bodies cannot shift. Commented Jul 23, 2019 at 17:40
• Sorry, I don't understand the question. You can interpret $\sum_i m_i r'_i =0$ as measuring the center of mass relative to $R$. If the sum is zero then $R$ is the center of mass location. Commented Jul 23, 2019 at 18:14