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Homework Help: Moment of inertia of a rod at an angle to the axis

  1. Jan 11, 2015 #1
    suppose a uniform rod of mass M and lenght L is at an angle B to the x axis, one end of the the rod touching the axis. wish to find moment of inertia about x axis.

    let the rod touches the axis at x=0. let D=density=M/L, and I will integrate along x axis, that means that at a distance x from the the origin, the little mass dM = D*dx is at a distance x*tanB away from the axis. then the integral I get is I = D*(x^2)*[(tanB)^2] integrated from x=0 to x=L*cosB. the answer I got is (1/3)*m*(L^2)*cosB*(sinB)^2.

    I know this is different from the usual solution where one should integrate along the rod itself,but I dont understand which part of this argument went wrong?

  2. jcsd
  3. Jan 11, 2015 #2


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    Staff: Mentor

    If you integrate dM across the same range that you're using to calculate the moment of inertia, what value do you expect?
  4. Mar 9, 2016 #3
    I know this reply is quite late but I saw the problem and wanted to answer in case anyone else was confused about this.

    Integrating along the x-axis instead of the rod is, of course, absolutely fine, however in this case your infinitesimal mass element dm should be dm = D*(dx/cosB), as each infinitesimal distance element dL along the length of the rod has a projection dx/cosB onto the x-axis.

    With this in mind, the unwanted cosB factor of your result disappears, leaving the correct result of (1/3)*m*(L*sinB)^2
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