Equation for velocity of center of mass

[UrbanXrisis]

What is the equation to find the velocity of the center of mass?

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

is this correct?

 

[dextercioby]

Though i didn't understand very well what your formulas meant (i must be having a bad day ),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.

 

[UrbanXrisis]

I'm not quite sure I undersatnd what you are saying...what 2 ways?

let me explain.

x_{CofM}=\frac{m_1x_1+m_2x_2}{M_{total}}
is the equation to calculate 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

 

[dextercioby]

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.

 

[UrbanXrisis]

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?

 

[dextercioby]

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

Daniel.

 

[UrbanXrisis]

in an elastice collision, the VCM is the same before and after to collisions right?

 

[dextercioby]

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

Daniel.

 

[UrbanXrisis]

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...
v=\sqrt{3^2+.8^2}
v=3.1m/s
p=(m_1+m_2)v
p=(3kg+2kg)3.1m/s
15.5Ns

Would 15.5Ns be the total linear momentum?

 

[dextercioby]

Yes,of course.

Daniel.