Moment of inertia tensor problem

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Homework Statement


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]

Homework 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}\\
\end{matrix}
\right)[/tex]
[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]


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?
 

Answers and Replies

  • #2
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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.
 
  • #3
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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:

I'_i,j=l_i,alpha*l_j,beta*I_alpha,beta

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?
 
  • #4
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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.
 
  • #5
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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:
  • #6
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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].
 
  • #7
D H
Staff Emeritus
Science Advisor
Insights Author
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Thanks for your replay.
But [tex]x_i[/tex] isn't x component of particle position but its distance to X-axis.
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.
 
  • #8
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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].
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.
 
  • #9
19
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Ok, thank you both! :)
 

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