Difference between revisions of "Forest UCM EnergyIntPart"

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=Translational invariance=
 
=Translational invariance=
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Consider two particles that interact via a conservative force <math>\vec{F}</math>
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Let <math>\vec{r}_1</math> identify the location of object 1 from an arbitrary reference point and <math>\vec{r}_2</math> locate the second object.
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The vector that points from object 2 to object 1 may be written as
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:<math>\vec r = \vec{r}_1 - \vec{r}_2</math>
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The distance between the two object is given as the length of the above vector
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If the force is a central force
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:<math>\vec F = \frac{k}{r^3} \vec r = \frac{k}{\left | r_1 - r_2 \right |^3}  \left ( \vec{r}_1 - \vec{r}_2 \right )</math>
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;Notice
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:;The interparticle force is independent of the coordinate system's position, only the difference betweenthe positions matters
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If object 2 was fixed so it is not accelerating and we place the origin of the coordinate system on object 2
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Then the force is that of a single object
  
 
=One potential for Both Particles=
 
=One potential for Both Particles=
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==Both forces from same potential==
 
==Both forces from same potential==
  
just take appropriate derivative
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If the above force is conservative then a potential exists such that
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:<math>\vec{F}_{12} = - \vec{\nabla}_1 U(\vec{r}_1) </math>
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::<math> = - \vec{\nabla}_1 U(\vec{r}_1) = - \left( \hat i \frac{\partial}{\partial x_1} +  \hat j \frac{\partial}{\partial y_1} +  \hat k \frac{\partial}{\partial z_1} \right ) U(\vec{r}_1)</math>
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:<math>= - \vec{\nabla}_1 U(\vec{r}_1 - \vec{r}_2) </math>
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Newton's 3rd law requries that
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:<math>\vec{F}_{21} = - \vec{F}_{12} = - \left (- \vec{\nabla}_1 U(\vec{r}_1)  \right ) = \vec{\nabla}_1 U(\vec{r}_1 - \vec{r}_2) </math>
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:: <math>- \left (- \vec{\nabla}_2 U(\vec{r}_2)  \right ) = -\vec{\nabla}_2 U(\vec{r}_1 - \vec{r}_2)</math>
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or
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:<math> \vec{\nabla}_1 U(\vec{r}_1 - \vec{r}_2)  = -\vec{\nabla}_2 U(\vec{r}_1 - \vec{r}_2)</math>
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;You can find the net external force on a body in the system once you have the potential for the system
  
 
==Total work given by one potential==
 
==Total work given by one potential==
  
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The total work is the sum of
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:the work done by <math>\vec{F}_{12}</math> as object 1 moves through <math>d\vec{r}_1</math>
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plus
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:the work done by <math>\vec{F}_{21}</math> as object 1 moves through <math>d\vec{r}_2</math>
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This NET work can be determine by taking the derivative of the potential energy
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Proof:
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: <math>W_{\mbox{total}} = d \vec{r}_1 \cdot \vec{F}_{12} + d \vec{r}_2 \cdot \vec{F}_{21} </math>
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::<math>= d \vec{r}_1 \cdot \vec{F}_{12} + d \vec{r}_2 \cdot (-\vec{F}_{12}) </math>
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::<math>= \left ( d \vec{r}_1  - d \vec{r}_2 \right ) \cdot \vec{F}_{12} </math>
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::<math>= \left ( d \vec{r}_1  - d \vec{r}_2 \right ) \cdot \left ( - \vec{\nabla}U(\vec {r}_1 ) \right ) </math>
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::<math>= -\left ( d \vec{r} \right ) \cdot \left (  \vec{\nabla}U(\vec {r}_1 - \vec{r}_2) \right ) </math>
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::<math>= -\left ( d \vec{r} \right ) \cdot \left (  \vec{\nabla}U(\vec {r}) \right ) =-dU</math>
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==Total Mechanical Energy conservation==
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The work done by <math>\vec{F}_{12}</math> as object 1 moves through <math>d\vec{r}_1</math> is given by the work energy theorem as
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: <math>dT_1 = d \vec{r}_1 \cdot \vec{F}_{12} </math>
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similarly for <math>\vec{F}_{21}</math>
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: <math>dT_2 = d \vec{r}_2 \cdot \vec{F}_{21} </math>
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: <math>W_{\mbox{total}} = d \vec{r}_1 \cdot \vec{F}_{12} + d \vec{r}_2 \cdot \vec{F}_{21} = dT_1 + dT_2</math>
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:: <math>= dT = -dU</math>
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Thus
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:<math>dT + dU = 0 </math>
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or
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:<math>E_{\mbox{total}} = T_1 + T_2 + dU =  </math>constant
  
 
=Elastic Collisions=
 
=Elastic Collisions=
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:<math>  v_1^2  =  \left (v_1^{\;\prime} \right )^2 + \left ( v_2^{\;\prime} \right )^2 + 2 \vec{v}_1^{\;\prime} \cdot \vec{v}_2^{\;\prime} </math>
 
:<math>  v_1^2  =  \left (v_1^{\;\prime} \right )^2 + \left ( v_2^{\;\prime} \right )^2 + 2 \vec{v}_1^{\;\prime} \cdot \vec{v}_2^{\;\prime} </math>
  
compare the above conservation of momentum equationwith the conservation of energy equation
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compare the above conservation of momentum equation with the conservation of energy equation
  
 
:<math>  v_1^2 =  \left (v_1^{\;\prime} \right )^2 + \left ( v_2^{\;\prime} \right )^2</math>
 
:<math>  v_1^2 =  \left (v_1^{\;\prime} \right )^2 + \left ( v_2^{\;\prime} \right )^2</math>
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:<math>2 \vec{v}_1^{\;\prime} \cdot \vec{v}_2^{\;\prime} = 0</math>
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:<math>2 \vec{v}_1^{\;\prime} \cdot \vec{v}_2^{\;\prime} = 0 \;\;\;\; \Rightarrow \vec{v}_1^{\;\prime} \perp \vec{v}_2^{\;\prime} </math>  
 
 
 
 
:<math>\Rightarrow \vec{v}_1^{\;\prime} \perp \vec{v}_2^{\;\prime} </math>  
 
  
 
[[Forest_UCM_Energy#Energy_of_Interacting_Particles]]
 
[[Forest_UCM_Energy#Energy_of_Interacting_Particles]]

Latest revision as of 12:39, 29 September 2014

Energy of Interacting particles


Translational invariance

Consider two particles that interact via a conservative force [math]\vec{F}[/math]


Let [math]\vec{r}_1[/math] identify the location of object 1 from an arbitrary reference point and [math]\vec{r}_2[/math] locate the second object.

The vector that points from object 2 to object 1 may be written as

[math]\vec r = \vec{r}_1 - \vec{r}_2[/math]


The distance between the two object is given as the length of the above vector

If the force is a central force

[math]\vec F = \frac{k}{r^3} \vec r = \frac{k}{\left | r_1 - r_2 \right |^3} \left ( \vec{r}_1 - \vec{r}_2 \right )[/math]


Notice
The interparticle force is independent of the coordinate system's position, only the difference betweenthe positions matters


If object 2 was fixed so it is not accelerating and we place the origin of the coordinate system on object 2

Then the force is that of a single object

One potential for Both Particles

Both forces from same potential

If the above force is conservative then a potential exists such that

[math]\vec{F}_{12} = - \vec{\nabla}_1 U(\vec{r}_1) [/math]
[math] = - \vec{\nabla}_1 U(\vec{r}_1) = - \left( \hat i \frac{\partial}{\partial x_1} + \hat j \frac{\partial}{\partial y_1} + \hat k \frac{\partial}{\partial z_1} \right ) U(\vec{r}_1)[/math]
[math]= - \vec{\nabla}_1 U(\vec{r}_1 - \vec{r}_2) [/math]

Newton's 3rd law requries that

[math]\vec{F}_{21} = - \vec{F}_{12} = - \left (- \vec{\nabla}_1 U(\vec{r}_1) \right ) = \vec{\nabla}_1 U(\vec{r}_1 - \vec{r}_2) [/math]
[math]- \left (- \vec{\nabla}_2 U(\vec{r}_2) \right ) = -\vec{\nabla}_2 U(\vec{r}_1 - \vec{r}_2)[/math]

or

[math] \vec{\nabla}_1 U(\vec{r}_1 - \vec{r}_2) = -\vec{\nabla}_2 U(\vec{r}_1 - \vec{r}_2)[/math]


You can find the net external force on a body in the system once you have the potential for the system

Total work given by one potential

The total work is the sum of

the work done by [math]\vec{F}_{12}[/math] as object 1 moves through [math]d\vec{r}_1[/math]

plus

the work done by [math]\vec{F}_{21}[/math] as object 1 moves through [math]d\vec{r}_2[/math]

This NET work can be determine by taking the derivative of the potential energy


Proof:

[math]W_{\mbox{total}} = d \vec{r}_1 \cdot \vec{F}_{12} + d \vec{r}_2 \cdot \vec{F}_{21} [/math]
[math]= d \vec{r}_1 \cdot \vec{F}_{12} + d \vec{r}_2 \cdot (-\vec{F}_{12}) [/math]
[math]= \left ( d \vec{r}_1 - d \vec{r}_2 \right ) \cdot \vec{F}_{12} [/math]
[math]= \left ( d \vec{r}_1 - d \vec{r}_2 \right ) \cdot \left ( - \vec{\nabla}U(\vec {r}_1 ) \right ) [/math]
[math]= -\left ( d \vec{r} \right ) \cdot \left ( \vec{\nabla}U(\vec {r}_1 - \vec{r}_2) \right ) [/math]
[math]= -\left ( d \vec{r} \right ) \cdot \left ( \vec{\nabla}U(\vec {r}) \right ) =-dU[/math]


Total Mechanical Energy conservation

The work done by [math]\vec{F}_{12}[/math] as object 1 moves through [math]d\vec{r}_1[/math] is given by the work energy theorem as

[math]dT_1 = d \vec{r}_1 \cdot \vec{F}_{12} [/math]

similarly for [math]\vec{F}_{21}[/math]

[math]dT_2 = d \vec{r}_2 \cdot \vec{F}_{21} [/math]


[math]W_{\mbox{total}} = d \vec{r}_1 \cdot \vec{F}_{12} + d \vec{r}_2 \cdot \vec{F}_{21} = dT_1 + dT_2[/math]
[math]= dT = -dU[/math]

Thus

[math]dT + dU = 0 [/math]

or

[math]E_{\mbox{total}} = T_1 + T_2 + dU = [/math]constant

Elastic Collisions

Definition

BOTH Momentum and Energy are conserved in an elastic collision

Example


Consider two object that collide elastically

Conservation of Momentum
[math]\left ( p_1 + p_2 \right ) _{\mbox{initial}} = \left ( p_1 + p_2 \right ) _{\mbox{final}}[/math]
Conservation of Energy
[math]\left ( T + U \right ) _{\mbox{initial}} = \left ( T + U \right ) _{\mbox{final}}[/math]

When the initial and final states are far away fromthe collision point

[math]U_{\mbox{initial}} = U_{\mbox{final}} = 0 =[/math] arbitrary constant


Example

Consider an elastic collision between two equal mass objecs one of which is at rest.

Conservation of momentum
[math] m \vec{v}_1 = m \left (\vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime} \right )[/math]
Conservation of Energy
[math] \frac{1}{2} m v_1^2 = \frac{1}{2} m \left (v_1^{\;\prime} \right )^2 + \frac{1}{2} m\left ( v_2^{\;\prime} \right )^2[/math]


Square the conservation of momentum equation
[math] \vec{v}_1 \cdot \vec{v}_1 = \left (\vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime} \right ) \cdot \left (\vec{v}_1^{\;\prime} + \vec{v}_2^{\;\prime} \right )[/math]
[math] v_1^2 = \left (v_1^{\;\prime} \right )^2 + \left ( v_2^{\;\prime} \right )^2 + 2 \vec{v}_1^{\;\prime} \cdot \vec{v}_2^{\;\prime} [/math]

compare the above conservation of momentum equation with the conservation of energy equation

[math] v_1^2 = \left (v_1^{\;\prime} \right )^2 + \left ( v_2^{\;\prime} \right )^2[/math]

and you conclude that


[math]2 \vec{v}_1^{\;\prime} \cdot \vec{v}_2^{\;\prime} = 0 \;\;\;\; \Rightarrow \vec{v}_1^{\;\prime} \perp \vec{v}_2^{\;\prime} [/math]

Forest_UCM_Energy#Energy_of_Interacting_Particles