Center of mass

  • #1
1,197
1
it is said that the posiion vector of the cener of mass of a rigid body can be obtained by:

[tex]r_{CM}= \frac{1}{M} \int r dm[/tex]

I'm not sure I understand this expression. What exactly is dm? and I thought it was the sumation of mass times distance divided by the total mass...

[tex] \frac{m_1x_1+m_2x_2+...}{m_1+m_2+...}[/tex]

that is the integral of r wrt dm?
 

Answers and Replies

  • #2
142
5
mass is the same as density times volume, i.e.

[tex]m=\rho (\bar{x}) * V[/tex]
[tex]dm=\rho (\bar{x}) * dV[/tex]

Then, do the standard volume triple integral. If the density is constant, then it can be taken out of the integral.
 
  • #3
lurflurf
Homework Helper
2,432
132
UrbanXrisis said:
it is said that the posiion vector of the cener of mass of a rigid body can be obtained by:

[tex]r_{CM}= \frac{1}{M} \int r dm[/tex]

I'm not sure I understand this expression. What exactly is dm? and I thought it was the sumation of mass times distance divided by the total mass...

[tex] \frac{m_1x_1+m_2x_2+...}{m_1+m_2+...}[/tex]

that is the integral of r wrt dm?
dm is the mass differential that is density*dV if volume density*dL for length ect.
In word that integral says "center of mass is equal to the average position weighted by mass". It will become the quotient of sums you mention when the mass is concentrated at points. The integral will be useful when the density varies continuously. Say a thin rod has length 1 and the density sin(pi*x).
We approximate the center of mass with a sum. The sum becomes this integral under limit.
[tex]CM=\frac{\int_0^1 x\sin(\pi x) dx}{\int_0^1 \sin(\pi x) dx}=\frac{1}{2}[/tex]
 
  • #4
OlderDan
Science Advisor
Homework Helper
3,021
2
UrbanXrisis said:
it is said that the posiion vector of the cener of mass of a rigid body can be obtained by:

[tex]r_{CM}= \frac{1}{M} \int r dm[/tex]

I'm not sure I understand this expression. What exactly is dm? and I thought it was the sumation of mass times distance divided by the total mass...

[tex] \frac{m_1x_1+m_2x_2+...}{m_1+m_2+...}[/tex]

that is the integral of r wrt dm?
To summarize and expand what others have said, and making explicit that we are talking about a vector position here, the center of mass for an object with distributed mass is

[tex]
\vec r _{CM} = \frac{{\int_V {\vec r \rho (\vec r )dV} }}{{\int_V {\rho (\vec r )dV} }} = \frac{1}{M}\int_V {\vec r \rho (\vec r )dV}
[/tex]

If the mass is uniformly distributed, the density is constant and can be factored out of the integral, or combined with dV into a dm.

[tex]
\vec r _{CM} = \frac{\rho}{M}\int_V {\vec r dV} = \frac{1}{M}\int_V {\vec r dm}
[/tex]

These expressions are all used for center of mass calculations. The first is the most general, but in most cases of interest the mass is uniformly distributed throughout the volume, or at least in sections of the volume that can be integrated separately, so the density can be factored out. For symmetrical objects, the integral of the positon vector can usually be accomplished as one integral, but for more complicated objects it is usually necessary to break the integral into sections and add the contributions together. When you do that you wind up with an equation that looks like your

[tex] \frac{m_1x_1+m_2x_2+...}{m_1+m_2+...}[/tex]

or in three dimensions

[tex] \frac{m_1 \vec r_1+m_2 \vec r_2+...}{m_1+m_2+...}[/tex]

In this case the subscripted position vectors and masses are the centers of mass and mass of each piece of the object. The latter expression can also be used for an assembly of objects, such as finding the center of mass of a system of interacting particles such as colliding objects or planetary systems.
 

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