Cross products vanish Classical Mechanics

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The discussion centers on the vanishing of cross terms in the context of an object moving in the x-y plane, specifically referencing equation 8.9. The key reason for this is the definition of the center of mass, which simplifies the calculations in the center of mass frame of reference. Participants suggest examining equations 5.56 and 5.58 for further clarity, emphasizing that in this frame, the position vector r' becomes zero. A question arises regarding the integral of the cross product of r' and v', prompting a deeper exploration of the mathematical implications. Understanding these concepts is crucial for grasping the mechanics involved in the problem.
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As seen in the picture, this question is about an object moving in the x-y plane. But I don't get why in equation 8.9 the cross terms vanish? If anyone can help me, that would be really nice.
 

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... they vanish due to the definition of "center of mass".
The text explains with maths in the passage right below eq8.9.
Have you tried working out the cross terms for some example?
 
The book answers your question already. You need to think about the center of mass in the center of mass frame of reference. Also the author suggests hou look at equation 5.56
 
Equation 5.58 is just R_cm = sum m_i r_i / M.

I get why r' is zero in the frame of the center of mass. But if that's the case, why is the integral of the cross product of r' and v' not equal to zero also?
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
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