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Angular momentum in cartesian coordinates (Lagrangian)
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[QUOTE="JulienB, post: 5485804, member: 574060"] Actually I just found in other pages that apparently for many people the ##r## refers to the ##\rho## I was mentioning. Then my ##x_i## and ##\dot{x}_i## stay the same as before (except with ##\rho## instead of ##r##), and I get those expression for ##\vec{L}##: ##L_x = m[\dot{z}y - z(\dot{\rho} \sin \varphi + \rho \dot{\varphi} \cos \varphi)]## ##L_y = m[z(\dot{\rho} \cos \varphi - \rho \dot{\varphi} \sin \varphi) - \dot{z}x]## ##L_z = mr^2 \dot{\varphi}## And the magnitude is very long: ##|\vec{L}| = m \sqrt{\dot{z}^2\rho^2 - 2 z\dot{z} \bigg[x\big(\dot{\rho} \cos \varphi - \rho \dot{\varphi} \sin \varphi\big) + y\big(\dot{\rho} \sin \varphi + \rho \dot{\varphi} \cos \varphi\big) \bigg] + z^2 (\dot{\rho}^2 + \rho^2 \dot{\varphi}^2) + \rho^4 \dot{\varphi}^2}## Could that be? I really thought it was going to be something of the form ##I \omega##, maybe I cannot see it because of the cylindrical coordinates. EDIT: I just saw I posted in "Advanced Physics" by mistake. I hope it's not too bad, sorry for that.Julien. [/QUOTE]
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Angular momentum in cartesian coordinates (Lagrangian)
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