Angular Momentum


by naestibill
Tags: angular, momentum
naestibill
naestibill is offline
#1
May4-12, 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
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haruspex
haruspex is offline
#2
May4-12, 04:46 PM
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P: 9,154
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