1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tensor of inertia

  1. Jan 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the tensor of inertia for a half disk with mass M and then use that to get the angular momentum along the axis in the figure.
    http://img138.imageshack.us/img138/5312/problemr.png [Broken]


    2. Relevant equations
    Moment of inertia for the whole disk (with mass 2M)
    [tex]\begin{bmatrix}
    \frac{MR^{2}}{2} & 0 & 0 \\
    0 & \frac{MR^{2}}{2} & 0 \\
    0 & 0 & MR^{2}
    \end{bmatrix}[/tex]

    3. The attempt at a solution
    For the tensor:
    [tex]\begin{bmatrix}
    \frac{MR^{2}}{2} & 0 & 0 \\
    0 & \frac{MR^{2}}{4} & 0 \\
    0 & 0 & \frac{MR^{2}}{2}
    \end{bmatrix}[/tex]
    And the moment of inertia... I think I should apply Steiner's theorem, but I'm not quite sure how to apply it on a tensor.

    Thanks for your help.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 23, 2010 #2

    jambaugh

    User Avatar
    Science Advisor
    Gold Member

    See:
    http://en.wikipedia.org/wiki/Parallel_axis_theorem" [Broken]

    You take the displacement vector [tex]\mathbf{a}[/tex]

    (written as a column vector) and form the matrix [tex] \mathbf{a}\mathbf{a}^\top[/tex]

    The new moment of inertia tensor is then:
    [tex] \mathbf{I}'=\mathbf{I} + M(|a|^2\mathbf{1} -\mathbf{a}\mathbf{a}^\top)[/tex]

    [edited above, forgot the mass!]
     
    Last edited by a moderator: May 4, 2017
  4. Jan 23, 2010 #3
    Thanks.
    OK, so assuming my "attempt at a solution" tensor was right, I get to the following tensor:

    [tex]
    \begin{bmatrix}
    \frac{MR^{2}}{2} & 0 & 0 \\
    0 & \frac{MR^{2}}{4} & 0 \\
    0 & 0 & \frac{MR^{2}}{2}
    \end{bmatrix}
    +\begin{bmatrix}
    0 & 0 & 0 \\
    0 & MR^{2} & 0 \\
    0 & 0 & MR^{2}
    \end{bmatrix}
    =\begin{bmatrix}
    \frac{MR^{2}}{2} & 0 & 0 \\
    0 & \frac{5MR^{2}}{4} & 0 \\
    0 & 0 & \frac{3MR^{2}}{2}
    \end{bmatrix}
    [/tex]

    I think this can't be right because then when I try to find variation in the angular momentum (at constant rotation speed), I get 0.
     
    Last edited: Jan 23, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Tensor of inertia
  1. Tensor test (Replies: 1)

Loading...