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Moment of inertia around displaced and rotated axis

  1. May 3, 2014 #1
    1. The problem statement, all variables and given/known data

    A homogenous disk with radius [itex]R[/itex] and mass [itex]m[/itex] lies in the xy plane so its center matches the origin [itex]O[/itex]. Point [itex]O'[/itex] is on the z axis at a distance [itex]s[/itex] from point [itex]O[/itex]. Axis [itex]y'[/itex] passes through point [itex]O'[/itex] at an angle [itex]\theta[/itex] with the z axis. Find the moment of inertia around axis [itex]y'[/itex].

    2. Relevant equations

    [itex]I_{ij}=\int dm(\delta_{ij}r^2-x_{i}x_{j})[/itex]

    3. The attempt at a solution

    The new coordinates x',y',z' are connected with the old x,y,z through the equations:

    [itex]x'=x[/itex]
    [itex]y'=y\text{sin}\theta+(z-s)\text{cos}\theta[/itex]
    [itex]z'=-y\text{cos}\theta+(z-s)\text{sin}\theta[/itex]

    Since the body has radial symmetry and is 2 dimensional, we can use polar coordinates:

    [itex]x=rcos\varphi[/itex]
    [itex]y=rsin\varphi[/itex]

    and the third coordinate z=0 since its in the xy plane, hence:

    [itex]x'=r\text{cos}\varphi[/itex]
    [itex]y'=r\text{sin}\varphi \text{sin}\theta-s\:\text{cos}\theta[/itex]
    [itex]z'=-r\text{sin}\varphi \text{cos}\theta-s\:\text{sin}\theta[/itex]

    The body is homogenous so:

    [itex]\sigma=\frac{dm}{dA}=const.\rightarrow dm=\sigma dA[/itex]
    [itex]\sigma=\frac{m}{A}=\frac{m}{\pi\:R^{2}}[/itex]

    In polar coordinates:

    [itex]dA=r\:drd\varphi[/itex]

    with limits:

    [itex]0\leq r\leq R[/itex]
    [itex]0\leq \varphi\leq 2\pi[/itex]

    However, I'm not sure whether the "moment of inertia" would in this case mean the [itex]I_{y'\:y'}[/itex] component from the inertia tensor. If that's the case, then:
    [itex]I_{y'\:y'}=\int dm(x'^2+z'^2)=\sigma \int \int r\:drd\varphi[(r\text{cos}\varphi)^2+(-r\text{sin}\varphi \text{cos}\theta-s\:\text{sin}\theta)^2][/itex]
    which Wolfram evaluates as:
    [itex]I_{y'\:y'}=\frac{1}{4} m \left(R^2+R^2 \text{cos}^2\theta+4 s^2 \text{sin}^2\theta\right)[/itex]

    Would this be the correct procedure or not?
     
  2. jcsd
  3. May 3, 2014 #2

    SteamKing

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    Is there a picture associated with the problem description?
     
  4. May 3, 2014 #3
    Not really, but from the description I'm guessing it looks like the attached pic. Also, [itex]y'[/itex] in the pic should lie in the yz plane, and x',y',z' are orthogonal axes (again, that's what I'm assuming as no picture was given).
     

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