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

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


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – tanuj23199
    Commented Jul 23, 2019 at 17:40
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jul 23, 2019 at 18:14

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