1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Dec 18, 2009 #1
    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:
    [tex]\sum m \cdot r = 0[/tex]
     
  2. jcsd
  3. Dec 18, 2009 #2
    You define
    [tex]r_c=\frac{\sum m_i r_i}{\sum m_i}[/tex]
    Then you check
    [tex]\sum m_i (r_i-r_c)=\sum m_i r_i-r_c \sum m_i=0[/tex]
     
  4. Dec 18, 2009 #3
    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
     
  5. Dec 18, 2009 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    That is totally wrong, Gerenuk!

    SUPPOSE we define ONE C.M as
    [tex]r^{(1}}_{C.M}=\frac{\sum_{i}m_{i}r_{i}}{M}[/tex]
    It can then readily be shown that we have:
    [tex]\frac{\sum_{i}m_{i}(r_{i}-r^{(1)}_{C.M})}{M}=0 (*)[/tex]
    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.

    Then, we have:
    [tex]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[/tex]
    whereby uniqueness of C.M has been proven:
    [tex]r^{(1)}_{C.M}-r^{(2)}=0[/tex]
     
  6. Dec 18, 2009 #5

    diazona

    User Avatar
    Homework Helper

    "Missing a step" doesn't equate to "totally wrong"
     
  7. Dec 19, 2009 #6

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Yes, it does, since that missing step was precisely the proof the asked for.
     
  8. Dec 19, 2009 #7
    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)
     
  9. Dec 19, 2009 #8

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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:
    [tex]\sum_{i}m_{i}(r_{i}-r^{(1)})-\sum_{i}m_{i}(r_{i}-r^{(2)})=0[/tex]
    since both terms are, by definition, equal to 0.
    The right-hand side is now easily re-written as:
    [tex](r^{(1)}-r^{(2)})\sum_{i}m_{i}=0[/tex]
    and since the total mass is a positive number, uniqueness follows.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How can it be proved that every body has (and only one) CM point?
Loading...