
#1
Sep1913, 10:45 AM

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It should for nitpicky convenience's sake, be pointed out that unless the point we choose to calculate the moments around is the C.M of the rigid body, or that the the point moves parallell (including being at rest) to the motion of the C.M, we get an additional term on the RHS.




#2
Sep1913, 10:53 AM

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You get an additional term on the right hand side even if the moments are calculated about the center of mass. Euler was the first to discover these. See http://en.wikipedia.org/wiki/Euler's...body_dynamics), or see any upper level undergraduate classical mechanics text.
Also noteworthy, in many cases engineers do not use rotation about the center of mass, and for very good reasons. It makes a whole lot more sense to represent translation/rotation as being along/about the joint for an object connected to another object at some joint. Things can get mighty hairy when you have multiple objects connected in multiple ways such as in a multijoint robotic arm. Physicists tend not to deal with such problems. Engineers do. 



#3
Sep1913, 11:28 AM

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 Let [itex]m_{i},\vec{r}_{i}, \vec{F}_{i}[/itex] be position vector, mass of the i'th particle relative to an inertial frame, while the F's is the net sum of forces acting at the i'th particle, [itex]\vec{r}_{0}[/itex] be the position vector of our moment point, so that [itex]\vec{r}_{i,0}=\vec{r}_{i}\vec{r}_{0}[/itex] We now sum over all particles in the body, noting that internal forces come in couples, acting along the connecting vector of the two particles, the moments from internal forces disappear, and we get: [tex]\vec{r}_{i,0}\times\vec{F}_{i}^{(ext)}=\vec{r}_{i,0}\times{m}_{i}\vec{a }_{i}[/tex] where superscript (ext) stands for the set of external forces. Now, we get: [tex]\vec{r}_{i,0}\times \vec{F}_{i}^{(ext)} = \frac{d}{dt} (\vec{r}_{i,0}) \times {m}_{i} \vec{v}_{i})(\vec{v}_{i} \vec{v}_{0})\times\vec{v}_{i}{m}_{i}[/tex] In the second expression, the cross product between the v_i's disappear, and summing over all particles, we gain, C.M subscript for Center of Mass: [tex]\vec{r}_{i,0}\times\vec{F}_{i}^{(ext)}=\frac{d}{dt}(\vec{r}_{i,0}\times {m}_{i}\vec{v}_{i})+\vec{v}_{0}\times{M}\vec{v}_{C.M}[/tex] The latter term disappears only when the moment point is moving parallell to the Center of Mass. The first is the rate of change of angular momentum with respect to the moment point. The function of that last term can be appreciated considering the following scenario: Let stuff move with constant, strictly horizontal velocity, while moment point moves strictly vertically. No external forces acts upon stuff, but there sure is calculated rated of change of angular momentum around moment point! The last term brings the RHS side to a strict 0, as it should. 



#4
Sep1913, 11:59 AM

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Nitpicky details of solid body motion[itex]\frac{d}{dt} (I \cdot \vec{\omega})[/itex], which is only equal to [itex]I \cdot \frac{d}{dt} \vec{\omega}[/itex] if the torque doesn't change the moment of inertia. 



#5
Sep1913, 12:04 PM

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#6
Sep1913, 12:26 PM

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That mess is why I split this side discussion from the original thread. 



#7
Sep1913, 12:37 PM

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And, to emphazise:
For full 3D rotational motion, any other representation than with rigid body coordinates is, typically, outright silly (i.e, the Eulerian representation is optimal, in which the twisting of the principal axes relative to the inertial frame is neatly represented). As long as it IS a rigid body then, which not necessarily has much to do with reality.. 


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