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Moment of Inertia of a Cone

  1. Nov 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the moment of inertia and center of mass of:
    A uniform cone of mass M, height h, and
    base radius R, spinning about its symmetry
    (x) axis.

    2. Relevant equations

    I = ∫R^2dm

    3. The attempt at a solution

    I tried using I =∫R^2dm, solving for dm I got dm=(M/V)dV, with dV = piR^2*dx. Thus, ∫R^2*(M/V)piR^2*dx V = 1/3*piR^2*h, so ∫R^2*M*(1/(1/3*piR^2*h))piR^2dx. Pi and R^2 cancel, so ∫3(M/h^3)R^2x^2dx, which gets me MR^2, and the right answer is (3/10)MR^2
    1. The problem statement, all variables and given/known data
     
  2. jcsd
  3. Nov 27, 2012 #2

    Simon Bridge

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    The idea is that r is the perpendicular distance from the rotation axis of a small mass dm ... the moment of inertia of that mass is r2dm ... and you add up all the wee masses. It looks to me that you may have attempted to use a method intended for a point mass on a cylindrical one.

    You appear to have adopted the strategy of slicing the cone into disks with x being the rotational axis. So the disks have thickness ##dx## and area ##\pi r^2 ## did you realize that the radius of the disk at x s a function of x?

    You have a better shortcut though - you already know the moment of inertia of a disk ;) So why not just add them up? $$I=\int_0^h I_{disk}(x)dx$$

    The best way to handle these is to write down you reasoning at each step - in words.
    Avoids the need for this kind of guesswork on the part of people checking and/or marking your work ;)
     
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