How can it be proved that every body has (and only one) CM point?

[Mad_Eye]

how can it be proved that every body has (and only one) CM point?

given the center of the mass is the point that the distances relative to gives:
$$\sum m \cdot r = 0$$

 

[Gerenuk]

You define
$$r_c=\frac{\sum m_i r_i}{\sum m_i}$$
Then you check
$$\sum m_i (r_i-r_c)=\sum m_i r_i-r_c \sum m_i=0$$

 

[Mad_Eye]

oh yeah... didn't know this is the definition...
was able to prov what i wanted nonetheless though :D
good thing i asked anyway :D

 

[arildno]

That is totally wrong, Gerenuk!

SUPPOSE we define ONE C.M as
$$r^{(1}}_{C.M}=\frac{\sum_{i}m_{i}r_{i}}{M}$$
It can then readily be shown that we have:
$$\frac{\sum_{i}m_{i}(r_{i}-r^{(1)}_{C.M})}{M}=0 (*)$$
That is all you did, Gerenuk, NOW follows the proof you should have given:

Assume that there is another point, [itex]r^{(2)}[/tex] that satisfies (*) by taking the place of the defined C.M.<br /> <br /> Then, we have:<br /> $$r^{(1)}_{C.M}-r^{(2)}=\frac{\sum_{i}m_{i}r_{i}}{M}-\frac{M}{M}r^{(2)}=\frac{\sum_{i}m_{i}(r_{i}-r^{(2)})}{M}=0$$<br /> whereby uniqueness of C.M has been proven:<br /> $$r^{(1)}_{C.M}-r^{(2)}=0$$[/itex]

 

[diazona]

"Missing a step" doesn't equate to "totally wrong"

 

[arildno]

"Missing a step" doesn't equate to "totally wrong"

Yes, it does, since that missing step was precisely the proof the asked for.

 

[Mad_Eye]

yeah that what i did but..
if the definition of CM is Gerenuk's first equation, then there is no need to proof there is only one...
(though i did ask to use another given haha)

 

[arildno]

yeah that what i did but..
if the definition of CM is Gerenuk's first equation, then there is no need to proof there is only one...
(though i did ask to use another given haha)

Not at all!

That definition proves that there exists AT LEAST 1 C.M.

It remains to prove that there are no other points having the same property (i.e, uniqueness of the point where the mass-weighted relative positions sums up to 0).

Another argument, ASSUMING the existence of (at least one) point satisfying your equation, can prove that IF such a point exists, then it must be unique.
(You'll need to CONSTRUCT such a point afterwards in order to prove that it does, indeed, exist!)

The uniqueness argument goes then as follows:

Assume that two such points exist. Then we have:
$$\sum_{i}m_{i}(r_{i}-r^{(1)})-\sum_{i}m_{i}(r_{i}-r^{(2)})=0$$
since both terms are, by definition, equal to 0.
The right-hand side is now easily re-written as:
$$(r^{(1)}-r^{(2)})\sum_{i}m_{i}=0$$
and since the total mass is a positive number, uniqueness follows.