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Simple Moment of Inertia Question

  1. Dec 1, 2009 #1
    Here is the picture of the problem: http://yfrog.com/3uwtfup
    I understand every other step except for the equation for moment of inertia. In my book the most basic equation for moment is ΔI = (r)^2 Δm but for this problem ΔI = (1/2)(r)^2 Δm. Why is that? Does it have something to do with the method of integration?
     
  2. jcsd
  3. Dec 1, 2009 #2
    When you are integrating to calculate the moment of inertia (MOI), you take the MOI of a tiny mass and sum all of those to get the total. MOI of a point mass is mr^2. In this case the tiny mass is a disk. MOI of a solid cylinder, disk etc is 1/2 mr^2.
     
  4. Dec 2, 2009 #3
    How would you go about deriving the constant that's out front, in this case 1/2?
     
  5. Dec 2, 2009 #4
    Using integration :)
    There is an example or two in almost every physics text that i have seen. Check out some other books if your book doesn't have it.
     
  6. Dec 2, 2009 #5
    The constant comes from the fact that a solid cone is made up of many thin disks (horizontal cross-sections of the cone wrt axis of rotation). So the rotational inertia of the solid cone is equal to the total rotational inertia of all these thin disks. The rotational inertia of a disk of radius R about its center of mass is
    [tex]I_{disk}[/tex] = 1/2 M[tex]R^{2}[/tex]

    Which is derived from the rotational inertia of a ring of radius R about its center of mass
    [tex]I_{ring}[/tex] = M[tex]R^{2}[/tex]

    Which is derived from the rotational inertia of a point mass at a radius R from the axis of rotation.
    [tex]I[/tex] = M[tex]R^{2}[/tex]

    So a small part of the rotational inertia of the cone (d[tex]I_{cone}[/tex]) is equal to the rotational inertia of a thin disk of radius y, 1/2 dm[tex]y^{2}[/tex].
     
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