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Equation for velocity of center of mass

  1. Feb 5, 2005 #1
    What is the equation to find the velocity of the center of mass?

    since...
    [tex]x_{CofM}=\frac{m_1x_1+m_2x_2}{M_{total}}[/tex]
    then...
    [tex]v_{CofM}=\frac{d(m_1x_1+m_2x_2)}{dt(M_{total})}[/tex]
    this means...
    if p=momentum
    [tex]v_{CofM}=\frac{p}{M_{total}}[/tex]

    is this correct?
     
  2. jcsd
  3. Feb 5, 2005 #2

    dextercioby

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    Though i didn't understand very well what your formulas meant (i must be having a bad day :tongue2: ),i can tell you that it's the other way around:
    VCM (velocity of the center of mass) results immediately by computing the total (linear) momentum in 2 ways...

    Daniel.
     
  4. Feb 5, 2005 #3
    I'm not quite sure I undersatnd what you are saying...what 2 ways?

    let me explain.

    [tex]x_{CofM}=\frac{m_1x_1+m_2x_2}{M_{total}}[/tex]
    is the equation to calcualte the distance of center of mass. I took the derivative of it to find the velocity. Therefore, my second equation was the velocity of the center of mass. My third equation went further to say that d(m1v1+m2v2)/dt was really momentum. So... VCM is really momentum over the total mass
     
  5. Feb 5, 2005 #4

    dextercioby

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    I know what u did.It's not wrong at all.I've just given u an alternative approach and i think much more intuitive.

    Daniel.
     
  6. Feb 5, 2005 #5
    does the momentum of the center of mass tell the total linear momentum of the system?

    Or do I have to caculate the momentum of each object and then add them together?
     
  7. Feb 5, 2005 #6

    dextercioby

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    Of course.The total linear momentum of the system is the linear momentum if the CM.

    Daniel.
     
  8. Feb 5, 2005 #7
    in an elastice collision, the VCM is the same before and after to collisions right?
     
  9. Feb 5, 2005 #8

    dextercioby

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    Yes,and that's due to total linear momentum conservation.

    Daniel.
     
  10. Feb 5, 2005 #9
    I have the velocity of the CM to be (3.00i-0.8j)m/s
    would the total linear momentum be...
    [tex]v=\sqrt{3^2+.8^2}[/tex]
    [tex]v=3.1m/s[/tex]
    [tex]p=(m_1+m_2)v[/tex]
    [tex]p=(3kg+2kg)3.1m/s[/tex]
    [tex]15.5Ns[/tex]

    Would 15.5Ns be the total linear momentum?
     
  11. Feb 5, 2005 #10

    dextercioby

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    Yes,of course.

    Daniel.
     
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