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Angular Momentum 
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#1
May412, 01:36 PM

P: 2

Considering a Rigid Body/Angular Momentum/Torque
We know that Torque(ext) = dL/dt Now with respect to stationary point S: L(s, cm) = Ʃ(ρi x mivi) and that dL(cm)/dt = Ʃτ(ext, CM) Now with respect to ANY point, P, that is accelerating: L(s,p) = L(cm) + ρ(cm) x Mv(cm) And hence, Ʃτ(ext, p) = dL(rel_p)/dt + ρ(cm) x Ma(p) Ʃτ(ext, p) = dL(rel_cm)/dt + ρ(cm) x Ma(cm) Can someone explain to me why this happens?: Why this happens? : Ʃτ(ext, p) = dL(rel_p)/dt + ρ(cm) x Ma(p) Thanks 


#2
May412, 04:46 PM

Homework
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Thanks
P: 9,784

If you would take the trouble to describe all your variables you've a better chance of a response.



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