How Does Goldstein Derive the Kinetic Energy Equation in Classical Mechanics?

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Discussion Overview

The discussion revolves around the derivation of the kinetic energy equation in classical mechanics as presented by Goldstein. Participants are examining the mathematical steps involved, particularly focusing on the application of the chain rule in differentiation.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • Some participants inquire about the expression md²r/dt² · dr/dt and its equivalence to the time derivative of (1/2 m (dr/dt)².
  • Several participants request explicit workings out of the differentiation process.
  • There are questions regarding the application of the chain rule to the time derivative of (dr/dt)².
  • One participant expresses confusion about the left-hand side of the equation, suggesting it should be straightforward.
  • Another participant references an external link to a discussion on the derivative of kinetic energy, indicating it contains relevant information.
  • One participant acknowledges difficulties with the rest of Goldstein's material, suggesting it may be complex.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the interpretation of the left-hand side of the equation or the application of the chain rule, indicating ongoing confusion and differing viewpoints.

Contextual Notes

Some participants express uncertainty about the correctness of their interpretations and calculations, particularly regarding the left-hand side of the equation and its relationship to the right-hand side.

bengeof
how is md^2r/dt^2 . dr/dt = d/dt (1/2 m (dr/dt)^2 )

Thank You
 
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bengeof said:
how is md^2r/dt^2 . dr/dt = d/dt (1/2 m (dr/dt)^2 )

Thank You

Chain rule.
 
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Likes   Reactions: Ibix
Can you work it out for me explicitly ?
 
Can you state the chain rule? Can you apply it to the time derivative of ##(dr/dt)^2##?
 
Ibix said:
Can you state the chain rule? Can you apply it to the time derivative of ##(dr/dt)^2##?

2dr/dt . . is that right ?
 
bengeof said:
I was able to work out the right hand side. But the left hand side is my problem. .

That makes no sense. The LHS is the answer to the derivative on the RHS. There's nothing to work out on the left.

bengeof said:
2dr/dt . . is that right ?

The stackexchange link I gave you is all about why that is not the answer!
 
okay Got it. Thank you so much !
 
  • #10
The rest of Goldstein may present difficulties.
 

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