1. Not finding help here? Sign up for a free 30min 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!

Moment of inertia tensor

  1. Dec 12, 2008 #1
    1. The problem statement, all variables and given/known data
    In Cartesian coordinate system are two point masses of mass [tex]m[/tex] connected with massless rod. Masses are in [tex](0,r,0)[/tex] and [tex](0,-r,0)[/tex].
    a) Find the moment of inertia tensor.
    b) Points were rotated in OXY plane such that angle between rod and Y-axis is [tex]\vartheta[/tex] (new cooridnates of masses are [tex](-r \sin \vartheta, r \cos \vartheta, 0)[/tex], [tex](r \sin \vartheta, -r \cos \vartheta, 0)[/tex]

    2. Relevant equations
    [tex]\hat{I} = \left(\begin{matrix}
    I_{xx} & I_{xy} & I_{xz}\\
    I_{yx} & I_{yy} & I_{yz}\\
    I_{zx} & I_{zy} & I_{zz}\\
    [tex]I_{xx} = \sum _{k} m_{k} (y^{2}_{k}+z^{2}_{k}) = \sum _{k} m_{k}(r^{2}_{k} - x^{2}_{k})\![/tex]
    [tex]I_{yy} = \sum _{k} m_{k} (z^{2}_{k}+x^{2}_{k}) = \sum _{k} m_{k}(r^{2}_{k} - y^{2}_{k})\![/tex]
    [tex]I_{zz} = \sum _{k} m_{k} (x^{2}_{k}+y^{2}_{k}) = \sum _{k} m_{k}(r^{2}_{k} - z^{2}_{k})\![/tex]
    [tex]I_{xy} = I_{yx} = - \sum _{k} m_{k} x_{k}y_{k}\![/tex]
    [tex]I_{yz} = I_{zy} = - \sum _{k} m_{k} y_{k}z_{k}\![/tex]
    [tex]I_{zx} = I_{xz} = - \sum _{k} m_{k} z_{k}x_{k}\![/tex]

    3. The attempt at a solution
    a) Using relevant equations I get
    [tex]\hat{I} = \left(\begin{matrix} 2 m r^2 & 0 & -2 m r^2 \\ 0 & 4 m r^2 & 0 \\ -2mr^2 & 0 & 2mr^2 \end{matrix}\right)[/tex]
    is it correct?

    b) I'm not sure, hence I'm asking :) - is it enough (and correct) to calculate:

    [tex]\hat{I}' = \hat{I} \left(\begin{matrix} \cos \varphi & \sin \varphi & 0 \\ -\sin \varphi & \cos \varphi & 0 \\ 0 & 0 & 1 \end{matrix}\right)[/tex] or maybe [tex]\hat{I}' = \left(\begin{matrix} \cos \varphi & \sin \varphi & 0 \\ -\sin \varphi & \cos \varphi & 0 \\ 0 & 0 & 1 \end{matrix}\right) \hat{I}[/tex] or sth else?
  2. jcsd
  3. Dec 12, 2008 #2
    For a) I think there are only contributions in the top left and bottom right components.

    Using your formulae, I_xy, I_xz and I_yz are all zero as they all involve multiplying by the x or z component of either particle's position (both of which are 0). Therefore all the off-diagonal entries must be 0. I agree with you for I_11 and I_33 however - I get both to be 2mr^2. However, the middle component, I_22, must be 0 as it's the sum of the square of the z component and the square of the x component but both x and z are 0 are mentioned previously.

    To sum up, I get 2mr^2 in the top left and bottom right and 0's everywhere else.

    Hope this helps.
  4. Dec 12, 2008 #3
    b) wikipedia tells me the -ve signs should be the other way round in your rotation matrix for a rotation about the z axis (http://en.wikipedia.org/wiki/Rotation_matrix). Wikipedia calls this matrix R_z,theta, however we'll call it L. As your examining the moment of inertia tensor we need to use the transformation law for tensors:


    this implies that in terms of matrices, I'= L I L^T where L^T is the transpose of L

    Just copmute that matrix product or you could repeat the process in a) using the new coordinates but this would be a bit tedious.

    Let me know if this works. Also I'm new-ish here - how do you type in LaTeX?
  5. Dec 13, 2008 #4
    Thanks for your replay.
    But [tex]x_i[/tex] isn't x component of particle position but its distance to X-axis.
    So for example [tex]I_{xz} = -m r \cdot r - m r \cdot r = - 2 mr^2[/tex]. Only [tex]y_i = 0[/tex] because both particles lie on Y-axis.

    PS. To type in LaTeX use [tex] tags.
  6. Dec 13, 2008 #5
    Ok. Well I reckon my method for b) is still right.

    Just reading my notes. I think that I_xx is the copmonent about the x axis.
    r^2 - x^2 represents the perpendicular distance squared betweent he mass and the x axis.
    but here x^2 is just the x component.
    so y^2 + z^2 = r^2 - x^2 is just the sum of the squares of the y and z components and this represents the square of this perpendicular distance.

    To see this, consider I_yy, as you dais both particles lie on the y axis, therefore there can be no moment of inertia about this axis and so I_yy=0
    Last edited: Dec 13, 2008
  7. Dec 13, 2008 #6
    Ok, I was wrong. x_i is x component of mass position.
    But in b) I think if L is a matrix of rotation there should be [tex]I' = L^T I L[/tex].
  8. Dec 13, 2008 #7

    D H

    Staff: Mentor

    Yes, it is. Another way to look at the inertia tensor for a collection of point masses about the origin is

    [tex]\mathbf{I} = (\sum_k m_k\,r_k^2)\mathbf{1} - \sum_k m_k\,\mathbf {r}_k \mathbf {r}_k^T[/tex]

    where [itex]\mathbf{1}[/itex] is the identity matrix, [itex]k[/itex] indexes the point masses, [itex]m_k[/itex] is the mass of the kth point mass, [itex]\mathbf {r}_k[/itex] are the coordinates of the kth point mass written as a column vector, and [itex]\mathbf {r}_k \mathbf {r}_k^T[/itex] is the outer product of that vector with itself. In terms of components, the above expression becomes

    [tex]I_{ij} = \sum_k m_k(r_k^2\delta_{ij} - r_{k,i}r_{k,j})[/tex]

    latentcorpse was exactly right in post #2. The only non-zero components of the inertia tensor for part (a) are Ixx and Izz.
  9. Dec 13, 2008 #8
    Yes. You're correct here. That's what I was getting at but maybe it wasn't clear as I wasn't using LaTeX. However, in your original post, you only appear to have multiplied by one of the transormation matrices.
  10. Dec 13, 2008 #9
    Ok, thank you both! :)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Moment of inertia tensor