Many particle system - problem with cross product

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SUMMARY

The discussion focuses on the properties of a many particle system, specifically addressing the equation R × ∑(m_i ̇r_i) = 0. Here, R represents the center of mass, r_i is the position vector of the ith particle relative to the center of mass, and ̇r_i denotes the time derivative of r_i. The conclusion drawn is that the sum of the momenta of the particles, when taken with respect to the center of mass, results in zero due to the definition of center of mass, which states that the total momentum of a system of particles is conserved.

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  • Understanding of vector calculus, specifically cross products.
  • Familiarity with the concept of center of mass in physics.
  • Knowledge of particle dynamics and momentum conservation principles.
  • Basic proficiency in mathematical notation and summation notation (∑).
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This discussion is beneficial for physics students, researchers in classical mechanics, and anyone studying particle systems and their dynamics.

lavster
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in a many particle system we have a center of mass R and position vector of the ith particle with respect to centre of mass is r_i. hence the position vector measured from the origin is R_i=R+r_i.

why does [tex]R\times\sum (m_i \dot{r}_i) =0[/tex], where [tex]\dot{r}_i[/tex] denotes the rate of change wrt time and m_i is the mass of the ith particle?

thanks
 
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hi lavster! :smile:

(have a sigma: ∑ and try using the X2 tag just above the Reply box :wink:)

because by definition of centre of mass,

∑ miri' = (∑ miri)' = 0' :wink:
 

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