Difference between revisions of "Forest UCM RBM"

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:<math>\sum \vec R \times m_k  \vec{\dot r}_k^{\;\; \prime} = \vec R \times \sum m_k  \vec{\dot r}_k^{\;\; \prime} 0</math>
 
:<math>\sum \vec R \times m_k  \vec{\dot r}_k^{\;\; \prime} = \vec R \times \sum m_k  \vec{\dot r}_k^{\;\; \prime} 0</math>
  
:: <math>\sum m_k  \vec{\dot r}_k^{\;\; \prime} = 0</math> : The location of the center of mass is at <math>\vec{ r}_k^{\;\; \prime} = 0</math> the derivative is also zero
+
:: <math>\sum m_k  \vec{\dot r}_k^{\;\; \prime} = \sum m_k \left ( \vec {r}_k - \vec R\right ) = \sum m_k \vec {r}_k  - \sum m_k  \vec R = \vec {v}_{cm} - \vec{v}_{cm} 0</math> : The location of the center of mass is at <math>\vec{ r}_k^{\;\; \prime} = 0</math> the derivative is also zero
  
 
: <math>\sum \vec{r}_k^{\;\; \prime} \times m_k \vec \dot R  = \sum m_k \vec{r}_k^{\;\; \prime} \times  \vec \dot R =0  </math>  : The location of the CM is at 0
 
: <math>\sum \vec{r}_k^{\;\; \prime} \times m_k \vec \dot R  = \sum m_k \vec{r}_k^{\;\; \prime} \times  \vec \dot R =0  </math>  : The location of the CM is at 0

Revision as of 16:25, 19 November 2014

Rigid Body Motion

Rigid Body

Rigidy Body
A Rigid Body is a system involving a large number of point masses, called particles, whose distances between pairs of point particles remains constant even when the body is in motion or being acted upon by external force.
Forces of Constraint
The internal forces that maintain the constant distances between the different pairs of point masses.


Consider a rigid body that rotates about a fixed z-axis with the origin at point O.


INSERT PICTURE HERE


let

[math]\vec R[/math] point to the center of mass of the object
[math]\vec {r}_k[/math] points to a mass element [math]m_k[/math]
[math]\vec{r}_k^{\;\;\prime}[/math] points from the center of mass to the mass element [math]m_k[/math]

the angular momentum of mass element [math]m_k[/math] about the point O is given as

[math]\ell_k = \vec {r}_k \times \vec {p}_k = \vec {r}_k \times m \vec {\dot r}_k[/math]

The total angular momentum about the point O is given as

[math] \vec L = \sum \ell_k = \sum \vec {r}_k \times m_k \vec {\dot r}_k[/math]

This can be cast in term of the angular momentum about the center of mass and the angular momentum of the motion

[math]\vec {r}_k = \vec R + \vec{r}_k^{\;\; \prime}[/math]
[math] \vec L = \sum \vec {r}_k \times m_k \vec {\dot r}_k[/math]
[math] = \sum (\vec R + \vec{r}_k^{\;\; \prime}) \times m_k (\vec \dot R + \vec{\dot r}_k^{\;\; \prime})[/math]
[math] = \sum \vec R \times m_k \vec \dot R + \sum \vec R \times m_k \vec{\dot r}_k^{\;\; \prime} + \sum \vec{r}_k^{\;\; \prime} \times m_k \vec \dot R +\sum \vec{r}_k^{\;\; \prime} \times m_k \vec{\dot r}_k^{\;\; \prime} [/math]


[math]\sum \vec R \times m_k \vec \dot R = \vec R \times \sum m_k \vec \dot R = \vec R \times M \vec \dot R = \vec R \times \vec P[/math]
[math]\vec P =[/math] momentum of the center of Mass
[math]\sum \vec R \times m_k \vec{\dot r}_k^{\;\; \prime} = \vec R \times \sum m_k \vec{\dot r}_k^{\;\; \prime} 0[/math]
[math]\sum m_k \vec{\dot r}_k^{\;\; \prime} = \sum m_k \left ( \vec {r}_k - \vec R\right ) = \sum m_k \vec {r}_k - \sum m_k \vec R = \vec {v}_{cm} - \vec{v}_{cm} 0[/math] : The location of the center of mass is at [math]\vec{ r}_k^{\;\; \prime} = 0[/math] the derivative is also zero
[math]\sum \vec{r}_k^{\;\; \prime} \times m_k \vec \dot R = \sum m_k \vec{r}_k^{\;\; \prime} \times \vec \dot R =0 [/math] : The location of the CM is at 0


[math] \vec L = \vec R \times \vec P + \sum \vec{r}_k^{\;\; \prime} \times m_k \vec{\dot r}_k^{\;\; \prime} [/math]


Forest_Ugrad_ClassicalMechanics#Rigid_Body_Motion