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Center of Mass of sheet

  1. Sep 28, 2007 #1
    A uniform sheet of metal is cut in the shape of a semicircle of radius R and lies in the xy plane with its center at the origin and diameter lying along the x axis. Find the position of the CM using polar coordinates. (Center of mass).
    [In this case the sum that defines the CM position becomes a two-D integral of the form [tex]\int[/tex]r[tex]\sigma[/tex]dA where [tex]\sigma[/tex] denotes the surface mass density (mass/area) of the sheet and dA is the element of area dA= rdrd[tex]\phi[/tex].]


    Ok I thought I knew how to start this before I read the bracketed section...
    Could I get a hint on starting this with polar coordinates (never done this actually...) and why is the bracketed section even necessary?

    Thanks alot!
     
  2. jcsd
  3. Sep 28, 2007 #2

    learningphysics

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    so rcenter = [tex]\frac{\int r\sigma dA}{\int \sigma dA}[/tex]

    so rcenter = [tex]\frac{\int_0^{\pi}\int_0^r r\sigma rdrd\phi}{\int \sigma dA}[/tex]

    similarly phicenter = [tex]\frac{\int \phi\sigma dA}{\int \sigma dA}[/tex]

    so phicenter = [tex]\frac{\int_0^{\pi}\int_0^r \phi\sigma rdrd\phi}{\int \sigma dA}[/tex]
     
  4. Sep 28, 2007 #3
    The first thing I'm not following is why density is needed to find the center..
     
  5. Sep 28, 2007 #4
    Can someone explain what this equation is saying?
    Why r AND dr?
     
  6. Sep 28, 2007 #5

    learningphysics

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    center of mass... You don't need the density in this case since the density is uniform... but suppose one part of the semicircle was much denser than the rest... then the center of mass will located closer to that part...
     
  7. Sep 28, 2007 #6

    learningphysics

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    in polar coordinates dA = rdrd[tex]\phi[/tex]... think about it like this for an area dA you need a length times a width... dr is the length... what is the width? it is [tex]rd\phi[/tex]
     
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