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> | ||
− | + | or | |
− | :<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
- ?
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
or
The for a potential defined as
or
or