The term $$\sum_i m_ir_i$$ vanishes when the origin is defined as the center of mass (CM) of a rigid body. This is due to the definition of the CM, where the sum of the positions of mass elements relative to the CM equals zero. The position vector $$\vec R$$ represents the CM relative to an arbitrary origin, while $$\vec {r_i}'$$ denotes the position of each mass element $$dm_i$$ relative to the CM. The relationship $$\vec X = \frac{\sum_i m_i\vec r_i}{\sum_i m_i}$$ confirms that the CM is the point where the weighted average of the positions results in zero when calculated from the CM's perspective.
PREREQUISITESStudents of physics, mechanical engineers, and anyone studying dynamics and kinematics of rigid bodies will benefit from this discussion.
I think the origin is at an arbitrary position. Vector ##\vec R## is the position of the CM of a rigid body relative to the origin. This looks like part of the proof that the angular momentum of a rigid body is the sum of the angular momentum of the CM (orbital) and about the CM (spin). The primed coordinate ##\vec {r_i}'## is the location of element ##dm_i## relative to the CM, i.e. ##\vec {r_i}'=\vec r-\vec R##. Otherwise yes, the sums are zero by definition of the CM.PeroK said:
Define ##\vec R## as the position of the CM relative to an arbitrary origin and ##\vec r## and ##\vec {r_i}'## respectively as the position of ##m_i## relaitve to the origin and relative to the CM. Then ##\vec r_i = \vec R+\vec {r_i}'## so you can write the position of the CM relative to the origin, $$\vec X=\frac{\sum_i m_i\vec r_i}{\sum_i m_i}=\frac{\sum_{i} m_i(\vec R+\vec {r_i}')}{\sum{_i} m_i}=\vec R+\frac{\sum{_i} m_i\vec {r_i}'}{\sum{_i} m_i}.$$ By definition, ##\vec X = \vec R##. The only way for this to be true is to have ##\dfrac{\sum_i m_i\vec {r_i}'}{\sum_i m_i}=0.##Istiakshovon said:
