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Steiner theorem demonstration

  • Thread starter Telemachus
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  • #1
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Homework Statement


I have a doubt about the steiner theorem demonstration, its actually trivial, but I can't realize why is this.

Lets see, the demonstration which you can see http://en.wikipedia.org/wiki/Parallel_axis_theorem" [Broken] goes as follows:

[tex]I_{cm} = \int{(x^2 + y^2)} dm[/tex]

[tex]I_z = \int{((x - r)^2 + y^2)} dm[/tex]

[tex]I_z = \int{(x^2 + y^2)} dm + r^2 \int dm - 2r\int{x} dm[/tex]

[tex]I_z = I_{cm} + mr^2[/tex]

What I need to know is why this gives zero:
[tex]2r\int{x} dm[/tex]

Bye there, and thanks for your help :)
 
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Answers and Replies

  • #2
Doc Al
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What I need to know is why this gives zero:
[tex]2r\int{x} dm[/tex]
Because they are using a coordinate system in which the center of mass is at the origin, so:
Xcm ≡ (∫xdm)/M = 0
 
  • #3
832
30
I don't get it. I actually read that explanation before. That integral wouldn't give: [tex]2rxM[/tex]? being M the total mass...? I'm not seeing the "differential" thing and how it works to give zero, I think thats the problem. It must be like integrating over a null area, but I don't see it.
 
  • #4
Doc Al
Mentor
44,910
1,169
I don't get it. I actually read that explanation before. That integral wouldn't give: [tex]2rxM[/tex]? being M the total mass...? I'm not seeing the "differential" thing and how it works to give zero, I think thats the problem. It must be like integrating over a null area, but I don't see it.
Do you agree that the x-coordinate of the center of mass is given by:

[tex]x_{cm} = \frac{1}{M} \int x dm[/tex]

where M is the total mass?
 
  • #5
832
30
Right. I see it now :P thanks.
 

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