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- Thread starter xxariesxx
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russ_watters

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No need for atoms here - this is just a useful mathematical device for solving problems. You don't have to try to visualize it relating to reality.

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Any compound system is described with its center of inertia coordinates (they do not correspond to a real material point-like heavy particle) and with the relative coordinates.

It is not a replacement of objects, it is a variable change.

When you describe your wooden sphere with only its center of inertia coordinates, you know what a "simplification" you do - you just omit the relative coordinate equations.

Bob.

- #5

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[itex]\frac{dm\mathbf{r_1}+dm\mathbf{r_2}}{2dm}[/itex]

if we have three we have

[itex]\frac{dm\mathbf{r_1}+dm\mathbf{r_2}+dm\mathbf{r_3}}{3}[/itex]

and so on. All we're finding is the average here. Now let's say the 4th mass, as an example, lies at exactly the same place as where the second mass placed was (i.e. there is more mass at that position) then our sum becomes

[itex]\frac{dm\mathbf{r_1}+dm\mathbf{r_2}+dm\mathbf{r_3}+dm\mathbf{r_4}}{4}[/itex]

but [itex]r_4=r_2[/itex], therefore, we can say

[itex]\frac{dm\mathbf{r_1}+2dm\mathbf{r_2}+dm\mathbf{r_3}}{4}[/itex].

Hopefully from here you can see how we arrive at the center of mass concept

- #6

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[itex]\frac{d^2\mathbf{R_{CM}}{dt^2}=\mathbf{A_{CM}}=\frac{\Sigma_i m_i \mathbf{a_i}{M}[/itex]

Now if we times this by our total mass ([itex]M=\Sigma_i m_i[/itex]) we get:

[itex]M\mathbf{A_{CM}}=M\frac{\Sigma_i m_i \mathbf{a_i}}{M}=\Sigma_i m_i \mathbf{a_i}[/itex]

which we see is exactly the equation for the total net force of a system of particles [itex]\Sigma_i m_i \mathbf{a_i} = \Sigma_i \mathbf{F_i} = \mathbf{F_{total}}[/itex]

so we see that by treating the system as a single giant particle of mass M and position [itex]\mathbf{R_{CM}}[/itex] we get the exact same result as if we'd just treated each particle individually. It's just a lot easier.

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