Forest UCM Energy TimeDepPE - Revision history

Foretony: /* Example */

2014-09-24T15:54:57Z

Example

Foretony at 15:45, 24 September 2014

2014-09-24T15:45:33Z

Foretony at 15:44, 24 September 2014

2014-09-24T15:44:44Z

Foretony at 15:44, 24 September 2014

2014-09-24T15:44:08Z

Foretony at 15:43, 24 September 2014

2014-09-24T15:43:23Z

Foretony at 15:42, 24 September 2014

2014-09-24T15:42:52Z

Foretony at 15:42, 24 September 2014

2014-09-24T15:42:26Z

Foretony at 15:39, 24 September 2014

2014-09-24T15:39:58Z

Foretony: Created page with " Time dependent force. What happens if you have a time dependent force that still manages to satisfy : $\vec \nabla \times \vec F = 0$ ? Because of the above, and St…"

2014-09-24T15:32:46Z

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…"

New page

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>

The for a potential defined as

:<math>U(r,t) = - \int \vec {F}(r,t) \cdot d \vec r</math>

you would be left with
:<math>\Delta T = - U(r,t) </math>

[[Forest_UCM_Energy#Time_Dependent_PE]]

@@ Line 37: / Line 37: @@
 ==Example==
-Leaky charge
+Charge escaping from a charged conducting sphere into the air.
 [[Forest_UCM_Energy#Time_Dependent_PE]]

← Older revision		Revision as of 15:45, 24 September 2014
Line 34:		Line 34:

	Mechanical Energy is only conserved if the potential is not time dependent.		Mechanical Energy is only conserved if the potential is not time dependent.
		+
		+	==Example==
		+
		+	Leaky charge

	[[Forest_UCM_Energy#Time_Dependent_PE]]		[[Forest_UCM_Energy#Time_Dependent_PE]]

← Older revision		Revision as of 15:44, 24 September 2014
Line 32:		Line 32:
	:<math>d(T + U) = \frac{\partial U}{\partial t} dt \ne </math>constant		:<math>d(T + U) = \frac{\partial U}{\partial t} dt \ne </math>constant

		+
		+	Mechanical Energy is only conserved if the potential is not time dependent.

	[[Forest_UCM_Energy#Time_Dependent_PE]]		[[Forest_UCM_Energy#Time_Dependent_PE]]

← Older revision		Revision as of 15:44, 24 September 2014
Line 28:		Line 28:
	:<math>dT + dU = \frac{\partial U}{\partial t} dt </math>		:<math>dT + dU = \frac{\partial U}{\partial t} dt </math>

		+	or

		+	:<math>d(T + U) = \frac{\partial U}{\partial t} dt \ne </math>constant


	[[Forest_UCM_Energy#Time_Dependent_PE]]		[[Forest_UCM_Energy#Time_Dependent_PE]]

@@ Line 17: / Line 17: @@
 If a potential U for the force exists such that
-:<math>\vec F = - \vec \nabla U(r,t) \cdt d \vec r</math>
+:<math>\vec F = - \vec \nabla U(r,t) \cdot d \vec r</math>
 or

@@ Line 3: / Line 3: @@
 What happens if you have a time dependent force that still manages to satisfy
-:<math>\vec \nabla \times \vec F = 0</math>?
+:<math>\vec \nabla \times \vec {F}(\vec r, t) = 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.

@@ Line 15: / Line 15: @@
 :<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
+If a potential U for the force exists such that
-:<math>U(r,t) = - \int \vec {F}(r,t) \cdot d \vec r</math>
+:<math>\vec F = - \vec \nabla U(r,t) \cdt 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>dU(r,t) = \frac{\partial U}{\partial x} dx +\frac{\partial U}{\partial y} dy +\frac{\partial U}{\partial z} dz +\frac{\partial U}{\partial t} dt </math>
-:<math>= - \vec F \cdot d \vec r +  \frac{\partial}{\partial t} dt </math>
+:<math>= - \vec F \cdot d \vec r +  \frac{\partial U}{\partial t} dt </math>
-:<math>= - dT +  \frac{\partial}{\partial t} dt </math>
+:<math>= - dT +  \frac{\partial U}{\partial t} dt </math>
 or
-:<math>dT + dU = \frac{\partial}{\partial t} dt  </math>
+:<math>dT + dU = \frac{\partial U}{\partial t} dt  </math>

@@ Line 1: / Line 1: @@
 Time dependent force.
@@ Line 11: / Line 10: @@
 :<math>\Delta T = W = \int \vec F \cdot d \vec r</math>
 The for a potential defined as
@@ Line 16: / Line 19: @@
 :<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>

Forest UCM Energy TimeDepPE - Revision history

Foretony: /* Example */

Foretony at 15:45, 24 September 2014

Foretony at 15:44, 24 September 2014

Foretony at 15:44, 24 September 2014

Foretony at 15:43, 24 September 2014

Foretony at 15:42, 24 September 2014

Foretony at 15:42, 24 September 2014

Foretony at 15:39, 24 September 2014

Foretony: Created page with " Time dependent force. What happens if you have a time dependent force that still manages to satisfy :\vec \nabla \times \vec F = 0? Because of the above, and St…"

Foretony: Created page with " Time dependent force. What happens if you have a time dependent force that still manages to satisfy : $\vec \nabla \times \vec F = 0$ ? Because of the above, and St…"