Understanding the Derivative of Angular Velocity in Lagrangian Mechanics

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SUMMARY

The discussion centers on the derivative of angular velocity in the context of Lagrangian mechanics, specifically addressing the expression for the time derivative of the square of velocity, represented as d/dt(ẋ²). Participants clarify that the correct application of the chain rule leads to the conclusion that this derivative equals 2ẋẋ̇, not simply 2ẋ. The conversation emphasizes the importance of understanding the notation used in Lagrangian mechanics, particularly the distinction between generalized coordinates and their derivatives.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with calculus, specifically derivatives and the chain rule
  • Knowledge of generalized coordinates and their notation
  • Basic concepts of angular velocity and its derivatives
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  • Study the application of the chain rule in calculus
  • Explore the principles of Lagrangian mechanics in detail
  • Learn about generalized coordinates and their significance in physics
  • Investigate the relationship between angular velocity and linear velocity in mechanics
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Physics students, mechanical engineers, and anyone studying advanced mechanics who seeks to deepen their understanding of Lagrangian dynamics and the mathematical tools used in this field.

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\underbrace{d}_{dt} (\dot{x})^{2} = 2\ddot{x} ? or is it just 2\dot{x} ?
 
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welcome to pf!

hi blueberrypies! welcome to pf! :smile:

neither :redface: …

use the chain rule :wink:
 


nevermind. I'm going to work on this more.
 
Alright, I still haven't figured this out so if anyone can help point out what I'm missing I would greatly appreciate it.

My book has this
attachment.php?attachmentid=31450&stc=1&d=1295467586.png



But I don't get this. Shouldn't it be
attachment.php?attachmentid=31451&stc=1&d=1295467716.png
 

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no, that would be ∂L/∂t

for ∂L/∂θ', remember that θ' is just a symbol, you can replace it by x (or anything), and forget the ' :wink:
 
tiny-tim said:
no, that would be ∂L/∂t

for ∂L/∂θ', remember that θ' is just a symbol, you can replace it by x (or anything), and forget the ' :wink:

Ohhh!

I see what you are saying :) Thanks!
 

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