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Moment of inertia of circular flywheel

  1. Dec 5, 2013 #1
    Hi I'm new to this forum so still getting used to it!

    I have to find the moment of inertia of a circular flywheel which has radius, a and mass m! But also had area density (mass per unit area) of ρ0,

    Where ρ0=ρa(1+r/a)

    I know moment of inertia is the integral of a^2 dm and in order to get mass m I just multiply the above expression by area of the flywheel which I think is pi*a^2

    I end up with a rather foolish answer when I have competed the intergratiom!
    Any help would be great

    Thanks guys
     
  2. jcsd
  3. Dec 6, 2013 #2
    Is the answer (9∏[itex]\rho[/itex]a5 /10) ?
    If its' correct pls. reply so that i would post my solution.
     
  4. Dec 6, 2013 #3

    haruspex

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    Your process is not clear from that description. Please post your actual working.
     
  5. Dec 6, 2013 #4

    haruspex

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    That's not how this forum works. We don't just post solutions. The idea is to provides hints and guide the thread originator to the solution.
     
  6. Dec 6, 2013 #5
    Oh! Sorry then.
    The idea is to take a differential strip(a ring of thickness dr) at a general distance r, find its' M.I and then integrate it to get M.I of the whole disc(a flywheel).
     
  7. Dec 7, 2013 #6
    Yea I know sorry!
    Well what I have done is used polar coordinates and defined my moment of inertia as of course

    I= integral of r^2 dm

    And my element of mass is just ρ0+[1+r/α] *2pi a^2 but I am having trouble with integrating the expression! Are my boundaries a -0 or r-0?
     
  8. Dec 7, 2013 #7

    haruspex

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    Seems to me there's some confusion between variables and constants here.
    In the OP you had r as a variable from 0 to a (constant); ρ0 as a variable from ρ a to 2 ρ a, ρ being a constant. That's very unusual - I would have expected ρ and ρ0 to have reversed roles. Assuming you had them crossed over, I'll write ρ(r) = ρ0a(1+r/a). (Maybe you meant ρa, not ρ a, in the OP, but that's just a change to the constant ρ0.)
    You want to integrate wrt r. At radius r the density is ρ(r). To get the element of mass you need to multiply that by the element of area, i.e. the area of an annulus between radius r and radius r+dr. In your above post, you seem to have that as 2pi a^2, which is obviously wrong. Also, is "ρ0+[1+r/α]" a typo? Should it be ρ0[1+r/α], or ρ0a[1+r/α]?
     
  9. Dec 7, 2013 #8
    Why is 2pi a^2 obviously wrong? It is the area of the flywheel is it not?
     
  10. Dec 7, 2013 #9

    haruspex

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    Well, no, the area of the whole flywheel would be pi a2, not 2pi a2. But for the purposes of the integrand we want the area of an element between radius r and radius r+dr, not the whole flywheel, right? That is independent of a.
     
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