Centre Of Mass

  • Thread starter andyrk
  • Start date
  • #1
658
4
Why are centre of mass of all uniform symmetric bodies at there geometric centre?
We know the following result:
"Consider a system of point masses m1,m2,m3... located at co-ordinates (x1,y1,z1), (x2,y2,z2)....respectively. The centre of mass of this system of masses is a point whose co-ordinates are ((xcm,ycm,zcm))
Which are given by
xcm=(m1x1+m2x2...)/(m1+m2+m3....)

ycm=(m1y1+m2y2...)/(m1+m2+m3....)

zcm=(m1z1+m2z2...)/(m1+m2+m3....)

How did we get this result?
 
Last edited:

Answers and Replies

  • #2
6,054
391
This is not a result, this is a definition.
 
  • #3
203
6
That comes from the definition of the center of mass.

EDIT : Apologies, already posted above.
 
  • #4
658
4
Why are so many things in Physics simply defined out of nowhere? Like Centre of Mass, Kinetic Energy, Relation between gravitational potential energy and work done, moment of inertia, linear and angular momentum, work etc..?
 
  • #5
6,054
391
They are not defined out of nowhere.

Is you real question "why is CM defined that way"?
 
  • #6
658
4
I had posted another question too. Why is centre of mass of all uniform and symmetric bodies at their geometric centre?
 
  • #7
658
4
They are not defined out of nowhere.

Is you real question "why is CM defined that way"?
Yes. And why are all other quantities defined the way they are?
 
  • #8
6,054
391
When you have many point masses, ##m_1, \ m_2, \ ... \ m_n ##, which interact with some forces from one to another, Newton's third law says that for each force there is one that is exactly the opposite of it, so the sum of all the forces is zero. Because forces are on the right hand side of Newton's second law, the sum of the left hand sides must also be zero, resulting in ##m_1 \vec{a}_1 + m_2 \vec{a}_2 + \ ... \ + m_n \vec{a}_n = 0##. If, in addition to the forces of interaction, there are some external forces, then we have ##m_1 \vec{a}_1 + m_2 \vec{a}_2 + \ ... \ + m_n \vec{a}_n = \vec{F}_e##, where the right hand side is the sum of the external forces.

Using the definition of the CM, we have ## \vec{r}_{CM} = \frac {m_1 \vec{r}_1 + m_2 \vec{r}_2 + \ ... \ + m_n \vec{r}_n} {m_1 + m_2 + \ ... \ + m_n} ##, and its acceleration is then ## \vec{a}_{CM} = \frac {m_1 \vec{a}_1 + m_2 \vec{a}_2 + \ ... \ + m_n \vec{a}_n} {m_1 + m_2 + \ ... \ + m_n}##, and the previous result yields $$ m\vec{a}_{CM} = \vec{F}_e $$

where ##m = m_1 + m_2 + \ ... \ + m_n ##, i.e., the total mass.

So we can replace a system of interacting masses with a single point mass located at the CM of the system, and it will move exactly in the same way. When we do not really care about the motions of the individual point masses, as is the case with solid bodies, this simplifies things quite a bit. And even if we do care, the simple motion of the CM helps, too.
 
  • #9
70
0
that's just a bit of common sense we can apply!
consider the rod
______________________

there are a collection of particles with mass 'dm'

x(cm)= ∫xdx
____
∫dm

solving by taking mass 'dm' at x distance from rod


_________dm_______
distance x

solve the integration or refer anything from webpage on ---- center of mass of continous bodies.
 
  • #10
741
27
If the centre of mass was not at the geometric centre of a uniform body, it would fall over in a gravitational field. We would all be lying on the floor. (Some of us are, actually, and our potential would be fairly limited)
 

Related Threads on Centre Of Mass

  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
862
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
9
Views
2K
Q
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
8
Views
639
  • Last Post
Replies
2
Views
2K
Top