Proof of angular momentum conservation

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SUMMARY

The discussion focuses on the conservation of angular momentum as presented in the text "Introduction to Lagrangian and Hamiltonian Mechanics" from NTNU. The user seeks clarification on the differentiation of the angular momentum expression at equation (1.12), specifically questioning the absence of the term d/dt(r) x p. Additionally, they inquire about the derivation of the relationship ri X Fji = 1/2(rij X Fji) and its implications for torque calculations. The conclusion reached is that the sum Σij rij x Fji equals zero, leading to the understanding that L dot equals torque.

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  • Understanding of Lagrangian mechanics
  • Familiarity with Hamiltonian mechanics
  • Knowledge of vector calculus, particularly cross products
  • Basic principles of angular momentum and torque
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KT KIM
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This is from text [Introduction to Lagrangian and Hamiltonian Mechanics] on NTNU opencourse.
Annnnd... I don't use english as my primary language, so sorry for poor sentences.

I can't get two things in here.

First, at (1.12) I can't understand how L dot derivated like that.
Since I know differentiation of cross product should be done like

d/dt(AxB)=d/dt(A) x B + A x d/dt(B)

then, at (1.12), why it doesn't have the terms of d/dt(r) x p ?
I think it only has the terms of r x d/dt(p)

Second, I can't get how
upload2.png

were derived by using (1.13), How could ri X Fji = 1/2(rij X Fji)
is possible?

These might be dumb questions, but please help me.
 
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KT KIM said:
en, at (1.12), why it doesn't have the terms of d/dt(r) x p ?
##\boldsymbol p_i=m_i\dot{\boldsymbol r}_i,\quad \dot{\boldsymbol r}_i\times\dot{\boldsymbol r}_i=0##
by the way ##\sum_{ij}{\boldsymbol r}_{i}\times\boldsymbol F_{ji}=0##
 
Thank you for clear explanation. Got the first one.

Yes Σij rij x Fji = 0 so eventually it makes L dot = tau (torque)
But, what I want know is the mathematical manuever that makes ri X Fji = 1/2(rij X Fji)
 
do the calculation for two particles directly and everything will be clear
 

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