Conservation of Energy in Mechanics for Point Mass

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SUMMARY

The discussion centers on the conservation of energy in mechanics, specifically the transition from the equation mv . dv/dt = -mgv . ez to d/dt (1/2 mv . v) = -mg dz/dt. The key point is understanding how the derivative d/dt can be factored out of the left-hand side, which involves recognizing that v . ez equals dz/dt. Additionally, the factor of 1/2 arises from the derivative of the kinetic energy term 1/2 mv². This simplification relies on foundational calculus principles, particularly the chain rule.

PREREQUISITES
  • Understanding of basic mechanics and the concept of point mass.
  • Familiarity with calculus, specifically derivatives and the chain rule.
  • Knowledge of vector notation and operations, particularly in physics.
  • Basic grasp of energy conservation principles in physics.
NEXT STEPS
  • Study the chain rule in calculus to understand derivative manipulation.
  • Review kinetic energy concepts and their derivation in mechanics.
  • Explore vector calculus, focusing on dot products and their physical interpretations.
  • Examine the principles of conservation of energy in various mechanical systems.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in deepening their understanding of mechanics, particularly in relation to energy conservation and calculus applications in physics.

Bucho
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Reading "Atmospheric Thermodynamics" I'm stumped almost as soon as I've started. I've probably bitten off more than I can chew and this also might even be more of a math question than a physics one but where I'm stuck is where they "simplify" from:

mv . dv/dt = -mgv . ez (where ez is a unit vector on the z-axis and the dots signify scalar multiplication by v)

to

d/dt 1/2mv . v = d/dt 1/2 mv2 = -mg dz/dtWhat I do get is that on the right we have v . ez = dz/dt, since velocity is the derivative of displacement (z) with respect to time.

What I don't get is the operation on the left by which the d/dt is just magically pulled out, leaving the v behind to work its own kind of magic on the other v. The text simply says "This equation can be simplified:" so I guess the authors presume a reader of more skill than I currently have in terms of playing with derivatives. What is it about a derivative that allows that d/dt to be stripped of its v and yanked out in front like that? And where the heck does that 1/2 appear from?
 
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It is just the derivative
[tex]\dfrac{dv^2}{dt} = 2 v \dfrac{dv}{dt}[/tex]
Which comes from the rule
[tex]\dfrac{d f(t)^\alpha}{dt} = \alpha f(t)^{\alpha-1} \dfrac{df(t)}{dt}[/tex]
 
Thanks Matteo, that's somewhat familiar but I'm so rusty on this stuff it's not funny.
 

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