# Linear momentum translational problem

1. Aug 27, 2008

### Benzoate

1. The problem statement, all variables and given/known data

A number of particles with masses m(1), m(2) , m(3),... are situated at the points with positions vectors r(1),r(2), r(3),... relative to an origin O. The center of mass G of the particles is defined to be the point of space with position vector

R=m(1)r(1)+m(2)r(2)+m(3)r(3)/(m(1)+m(2)+m(3))

Show that if a different Origin O' were used , this definition would still place G at the same point of space

2. Relevant equations
R=m(1)r(1)+m(2)r(2)+m(3)r(3)/(m(1)+m(2)+m(3))

Possibly C+R=R'

3. The attempt at a solution

I think I need to translate R to a new coordinate system , which is O', and essentially show that If a vectors moves into a new coordinate system , calling constant c the distance between the new coordinate system and the old coordinate system, I have to show the magnitude of the vectors don't changed. So here it goes:

R'=(m1)*(r1+c)+(m2)(r2+c)+(m3)(r3+c)/(m1+m2+m3)=m1r1+mc+m2r2+mc+m3r3+mc/(m1+m2+m3)
C=R'-R=c(m1+m2+m3)/(m1+m2+m3)=c; Therefore, R=R'-C. Is that the procedure I would apply to proved That the magnitude of the vectors do not change at all as I move my position vectors into a new coordinate system?

2. Aug 27, 2008

### Dick

I think, if I'm reading what you have written correctly, that you have shown that if you translate all of the ri vectors by c then the center of mass moves by c? If so then doesn't that show that the center of mass is translation independent?

3. Aug 28, 2008

### Benzoate

I am trying to show that the magnitude of the position vectors will not change if I move my positions vectors to a new coordinate system. Isn't that what translation independence is?

4. Aug 28, 2008

### tiny-tim

Hi Benzoate!

Why do you keep going on about magnitude?

Magnitude has nothing to do with it.

As Dick says:
You have proved (very messily, and only for n = 3 … can't you use ∑ notation?) that if R is the average of R1 R2 … Rn then R+C is the average of R1+C R2+C … Rn+C.

In other words: "this definition would still place G at the same point of space".

5. Aug 28, 2008

### Redbelly98

Staff Emeritus
The magnitude of the position vectors does change when you move the origin's location.

|ri + c| is not |ri|

But that is irrelevant to solving this problem.

Moving the origin is equivalent to adding a constant vector c to each position vector. By showing that the center of mass R becomes R + c, you prove that the CM is in the same location after the origin shift. Just as Dick said:

Edit --
Note to self: I owe tiny-tim a beer