# Equation for velocity of center of mass

1. Feb 5, 2005

### 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?

2. Feb 5, 2005

### dextercioby

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. Feb 5, 2005

### 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 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. Feb 5, 2005

### 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.

5. Feb 5, 2005

### 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?

6. Feb 5, 2005

### dextercioby

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

Daniel.

7. Feb 5, 2005

### UrbanXrisis

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

8. Feb 5, 2005

### dextercioby

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

Daniel.

9. Feb 5, 2005

### UrbanXrisis

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?

10. Feb 5, 2005

### dextercioby

Yes,of course.

Daniel.