Equation for velocity of center of mass

Click For Summary

Homework Help Overview

The discussion revolves around the equation for the velocity of the center of mass (CofM) in a system of particles, exploring its relationship with momentum and mass. Participants are examining the derivation and implications of the equations related to the center of mass in the context of momentum conservation, particularly in elastic collisions.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the derivation of the velocity of the center of mass from the position equation and its connection to momentum. Questions arise about the interpretation of momentum in relation to the center of mass and whether it can be calculated directly or requires summing individual momenta.

Discussion Status

The discussion is active, with various interpretations being explored. Some participants offer alternative approaches to understanding the relationship between momentum and the center of mass, while others seek clarification on specific points. There is no explicit consensus, but productive dialogue is occurring.

Contextual Notes

Participants are navigating through the definitions and relationships of momentum and center of mass, with some expressing confusion about the formulas and their implications. The context includes considerations of elastic collisions and the conservation of momentum.

UrbanXrisis
Messages
1,192
Reaction score
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?
 
Physics news on Phys.org
Though i didn't understand very well what your formulas meant (i must be having a bad day :-p ),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.
 
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 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
 
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.
 
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?
 
Of course.The total linear momentum of the system is the linear momentum if the CM.

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

Daniel.
 
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?
 

Similar threads

Replies
9
Views
2K
Replies
15
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 27 ·
Replies
27
Views
3K
Replies
18
Views
2K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 12 ·
Replies
12
Views
2K