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Inertia of a solid vertical disk about it's diameter

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Derive the moment of inertia of a solid, uniform density, disk rotating about a diameter using vertical rods.
    Diagram: http://i.imgur.com/TQIjz.png

    Only the radius of the circle and the inertia of a thin rod is given

    2. Relevant equations
    Inertia of thin vertical rod = m*r^2
    σ= area density
    M= Total mass
    I= inertia
    dm= infinitesimal amount of mass
    dI= infinitesimal amount of inertia
    A= total area
    R= radius of the circle



    3. The attempt at a solution
    This is the method given in my class. Because calculus 1 is the only prerequisite we can not use double integrals.


    I of rod = m*x^2
    dI= dm

    *x^2
    dm=σ*da
    da= 2y*dx
    σ= M/A
    A=pi*R^2
    σ= M/(pi*R^2)
    dm=M/(pi*R^2) * 2y*dx
    dI= M/(pi*R^2) * 2y*x^2 * dx


    From here we have to put y in terms of x which I think must be done using the Pythagorean theorem so:
    y=(R^2-x^2)^(1/2)

    subbing in we get:

    dI= M/(pi*R^2) * 2(R^2-x^2)^(1/2) * x^2 * dx

    This is why I get stuck because I am not quite sure how to integrate that. What is should come out to be is I = 1/4 MR^2. If anyone can offer an explanation it would be very appreciated.
     
  2. jcsd
  3. Apr 14, 2012 #2

    pcm

    User Avatar

    try substituting x=Rsin(theta).
     
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