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Moment of Inertia of Infinite Rod

  1. Apr 13, 2012 #1
    1. The problem statement, all variables and given/known data

    A thin rod extends along the x-axis from x= +b to infinity. It has a non-uniform linear mass density of A/x[itex]^{4}[/itex] where A is a constant with units of kg m3. Calculate the moment of inertia of the rod for rotation about the origin.

    2. Relevant equations



    3. The attempt at a solution
    Well, I know that you have to put the rod in a coordinate system, break it up into small pieces, take the sum of the those, take the limit as n goes to infinity, and integrate.
    So, I basically want to [itex]\int[/itex]x^2dx because the word "thin" suggest that the y-coordinates do not matter. I understand that by breaking the rod into small pieces and choosing one, I have a [itex]\Delta[/itex]m which I need to relate in terms of dx. I can do this using using the [itex]\frac{\Delta m}{M}[/itex]=[itex]\frac{\Delta x}{A/x^{4}}[/itex], where [itex]\Delta m[/itex] is my piece of mass, M is the total mass, [itex]\Delta x[/itex] is my small width, and then A/x[itex]^{4}[/itex] is my linear mass density. But I don't really know where this b and infinity and limits of integration come into play. :( If any can help me get started, that would be awesome!
     
  2. jcsd
  3. Apr 13, 2012 #2

    tiny-tim

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    hi chromium1387! :smile:
    yes :smile:

    except that's not the way linear density works …

    the mass of a length L is linear density time L, so the mass of [x , x+∆x] is A∆x/x4 :wink:
    you're adding (integrating) the moment of inertia for every [x , x+∆x]

    so you start at one end of the rod, and go to the other … ∫b
     
  4. Apr 13, 2012 #3
    Ohhh.. My bad. Silly algebra mistake.
    And that makes sense. I'm just used to placing one end at the origin.
    Sooo, after I integrate and everything, I get [itex]\frac{MA}{b}[/itex]?
    Thanks for your reply!
     
  5. Apr 13, 2012 #4

    tiny-tim

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    erm :redface: … why do you keep talking about M ? :confused:

    apart from that, yes :smile:
     
  6. Apr 13, 2012 #5
    I don't know.. haha. It's just incorporated into the density, right? :P
     
  7. Apr 13, 2012 #6

    tiny-tim

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    it doesn't exist!! :biggrin:
     
  8. Apr 13, 2012 #7
    okay. :)
     
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