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Equation for velocity of center of mass

  • #1
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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?
 

Answers and Replies

  • #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.
 
  • #3
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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
 
  • #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.
 
  • #5
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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?
 
  • #6
dextercioby
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Of course.The total linear momentum of the system is the linear momentum if the CM.

Daniel.
 
  • #7
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in an elastice collision, the VCM is the same before and after to collisions right?
 
  • #8
dextercioby
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Yes,and that's due to total linear momentum conservation.

Daniel.
 
  • #9
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dextercioby said:
Of course.The total linear momentum of the system is the linear momentum if the CM.

Daniel.
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?
 
  • #10
dextercioby
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Yes,of course.

Daniel.
 

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