
#1
Jun2703, 08:04 PM

P: 364

From Matter and Motion by James Clerk Maxwell, Article 59,
"Let us define a massvector as the operation of carrying a given mass from the origin to the given point. The direction of the massvector is the same as that of the vector of the mass, but its magnitude is the product of the mass into the vector of the mass. Thus, if OA is the vector of the mass A, the massvector is OA*A" I can almost comprehend the idea of a massvector. Simply take the "vector of the mass" and multiply it by the mass. But what is the vector of the mass? What would be the point of the mass vector anyway? Thanks. 



#2
Jun2703, 08:15 PM

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PF Gold
P: 5,540

x_{CM}=(1/M)Σ_{i}m_{i}x_{i} where i=index (1,2,3,...) and M=total mass. 



#3
Jun2803, 11:17 AM

Math
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Thanks
PF Gold
P: 38,904

In modern terms (and not all THAT modern!) multiply the mass and the velocity vector to get the momentum vector.




#4
Jun2803, 01:16 PM

P: 364

Massvectors according to Maxwell 



#5
Jun2803, 01:22 PM

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PF Gold
P: 5,540





#6
Jun2803, 01:36 PM

P: 364

So to find the CM you have to use the displacement vectors of the masses in the system? If you use the momenta, acceleration, velocity, etc, and apply Tom's formula, you'll get the momentum, acceleration, velocity, etc, of the CM.




#7
Jun2803, 01:54 PM

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PF Gold
P: 5,540

(d^{2}/dt^{2})x_{CM}=(1/M)Σ_{i}m_{i}(d^{2}/dt^{2})x_{i} a_{CM}=Σ_{i}m_{i}a_{i} That's the only way I can get the sum of forces from the sum of the mx vectors with the equation for the CM. As you can see, changing from x to a on one side makes it necessary to do it on the other side, so we aren't talking about the location of the CM anymore, but its acceleration. 


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