How to Find the Moment of Inertia Tensor for Rotated Point Masses?

[qoqosz]

Homework Statement

In Cartesian coordinate system are two point masses of mass m connected with massless rod. Masses are in (0,r,0) and (0,-r,0).
a) Find the moment of inertia tensor.
b) Points were rotated in OXY plane such that angle between rod and Y-axis is \vartheta (new cooridnates of masses are (-r \sin \vartheta, r \cos \vartheta, 0), (r \sin \vartheta, -r \cos \vartheta, 0)

Homework Equations

\hat{I} = \left(\begin{matrix}<br /> I_{xx} &amp; I_{xy} &amp; I_{xz}\\<br /> I_{yx} &amp; I_{yy} &amp; I_{yz}\\<br /> I_{zx} &amp; I_{zy} &amp; I_{zz}\\<br /> \end{matrix}<br /> \right)
I_{xx} = \sum _{k} m_{k} (y^{2}_{k}+z^{2}_{k}) = \sum _{k} m_{k}(r^{2}_{k} - x^{2}_{k})\!
I_{yy} = \sum _{k} m_{k} (z^{2}_{k}+x^{2}_{k}) = \sum _{k} m_{k}(r^{2}_{k} - y^{2}_{k})\!
I_{zz} = \sum _{k} m_{k} (x^{2}_{k}+y^{2}_{k}) = \sum _{k} m_{k}(r^{2}_{k} - z^{2}_{k})\!
I_{xy} = I_{yx} = - \sum _{k} m_{k} x_{k}y_{k}\!
I_{yz} = I_{zy} = - \sum _{k} m_{k} y_{k}z_{k}\!
I_{zx} = I_{xz} = - \sum _{k} m_{k} z_{k}x_{k}\!

The Attempt at a Solution

a) Using relevant equations I get
\hat{I} = \left(\begin{matrix} 2 m r^2 &amp; 0 &amp; -2 m r^2 \\ 0 &amp; 4 m r^2 &amp; 0 \\ -2mr^2 &amp; 0 &amp; 2mr^2 \end{matrix}\right)
is it correct?

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

\hat{I}&#039; = \hat{I} \left(\begin{matrix} \cos \varphi &amp; \sin \varphi &amp; 0 \\ -\sin \varphi &amp; \cos \varphi &amp; 0 \\ 0 &amp; 0 &amp; 1 \end{matrix}\right) or maybe \hat{I}&#039; = \left(\begin{matrix} \cos \varphi &amp; \sin \varphi &amp; 0 \\ -\sin \varphi &amp; \cos \varphi &amp; 0 \\ 0 &amp; 0 &amp; 1 \end{matrix}\right) \hat{I} or sth else?

 

[latentcorpse]

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.

 

[latentcorpse]

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?

 

[qoqosz]

Thanks for your replay.
But x_i isn't x component of particle position but its distance to X-axis.
So for example I_{xz} = -m r \cdot r - m r \cdot r = - 2 mr^2. Only y_i = 0 because both particles lie on Y-axis.

PS. To type in LaTeX use tags.

 

[latentcorpse]

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

 

[qoqosz]

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 I&#039; = L^T I L.

 

[D H]

Thanks for your replay.
But x_i 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

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

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

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


latentcorpse was exactly right in post #2. The only non-zero components of the inertia tensor for part (a) are I_xx and I_zz.

 

[latentcorpse]

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 I&#039; = L^T I L.

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.

 

[qoqosz]

Ok, thank you both! :)