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Help- parallel axis theorem

  1. Apr 7, 2005 #1
    Could someone please help me with a particular question to this problem?

    Consider a cube of mass m with edges of length a. The moment of inertia I_0 of the cube about an axis through its center of mass and perpendicular to one of its faces is given by I_0 = 1/6ma^2. Find I_edge, the moment of inertia about an axis p through one of the edges of the cube.
    Express I_edge in terms of m and a. Use fractions rather than decimal numbers in your answer.

    You know how the parallel axis theorem is I_edge = I_0 +md^2, what exactly is d in this particular problem. I tried a/2, and I tried using pythagoras' theorem to determine one of the diagonals, getting an answer of d= sqrt(2a^2)/2. Both answers where wrong when i substituted it into the equation.

    A picture is available below.

    Any help would be appreciated.
     

    Attached Files:

  2. jcsd
  3. Apr 7, 2005 #2
    d is the displacement of the axis from the center of mass.

    Now,

    [tex]I_{edge} = I_0 + md^2 = \frac{1}{6}ma^2 + \frac{1}{2}ma^2 = \frac{2}{3}ma^2[/tex].

    That should be the answer.
     
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