Difference between revisions of "Forest UCM Energy TimeDepPE"

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(Created page with " Time dependent force. What happens if you have a time dependent force that still manages to satisfy :<math>\vec \nabla \times \vec F = 0</math>? Because of the above, and St…")
 
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Time dependent force.
 
Time dependent force.
  
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:<math>\Delta T = W = \int \vec F \cdot d \vec r</math>
 
:<math>\Delta T = W = \int \vec F \cdot d \vec r</math>
 +
 +
or
 +
 +
:<math>d T = \frac{dT}{dt} dt = (m \vec \dot v \cdot v) dt  = \vec F \cdot d \vec r</math>
  
 
The for a potential defined as  
 
The for a potential defined as  
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:<math>U(r,t) = - \int \vec {F}(r,t) \cdot d \vec r</math>
 
:<math>U(r,t) = - \int \vec {F}(r,t) \cdot d \vec r</math>
  
you would be left with
+
or
:<math>\Delta T = - U(r,t) </math>
+
 
 +
:<math>dU(r,t) = \frac{\partial}{\partial x} dx +\frac{\partial}{\partial y} dy +\frac{\partial}{\partial z} dz +\frac{\partial}{\partial t} dt </math>
 +
:<math>= - \vec F \cdot d \vec r +  \frac{\partial}{\partial t} dt </math>
 +
:<math>= - dT +  \frac{\partial}{\partial t} dt </math>
  
 +
or
 +
:<math>dT + dU = \frac{\partial}{\partial t} dt  </math>
  
  

Revision as of 15:39, 24 September 2014

Time dependent force.

What happens if you have a time dependent force that still manages to satisfy

[math]\vec \nabla \times \vec F = 0[/math]?

Because of the above, and Stoke's Theorem , you would be able to find a close loop where zero work is done at some given time.

If we consider the work energy theorem

[math]\Delta T = W = \int \vec F \cdot d \vec r[/math]

or

[math]d T = \frac{dT}{dt} dt = (m \vec \dot v \cdot v) dt = \vec F \cdot d \vec r[/math]

The for a potential defined as

[math]U(r,t) = - \int \vec {F}(r,t) \cdot d \vec r[/math]

or

[math]dU(r,t) = \frac{\partial}{\partial x} dx +\frac{\partial}{\partial y} dy +\frac{\partial}{\partial z} dz +\frac{\partial}{\partial t} dt [/math]
[math]= - \vec F \cdot d \vec r + \frac{\partial}{\partial t} dt [/math]
[math]= - dT + \frac{\partial}{\partial t} dt [/math]

or

[math]dT + dU = \frac{\partial}{\partial t} dt [/math]



Forest_UCM_Energy#Time_Dependent_PE