# Inertial tensor calculation

1. Apr 20, 2005

my homework problem deals w/ rotation in x-y plane. so the tensor is only 2d. inertial tensor still seems obscure to me... my question for now is purely mathematical. assuming the basis are (x,y). I calculated the components of I, is the following correct?
$$I_{xx} = m_i y_{i}^{2}$$
$$I_{yy} = m_i x_{i}^{2}$$
$$I_{xy} = I_{yx} = - m_i x_i y_i$$

for some reason my professor wrote...
$$I_{xx} = m_i x_{i}^{2}$$
and i'm pretty sure its not right

Last edited: Apr 20, 2005
2. Apr 20, 2005

### jamesrc

You're fine (besides the missing summation symbols). Think of it this way: if you are rotating about the x-axis, would your inertia depend on where you are along the x-axis or how far away you are from the x-axis.

3. Apr 20, 2005

what about the off diagonal terms? are they moments about x=y line?... also do you need orthogonal basis for calculating inertial tensor? from the general definition it doesn't look like a requirement.

4. Apr 20, 2005

### jamesrc

Sorry, I don't have a good analogy in mind for the products of inertia; maybe someone else reading the thread can help us out. I guess it comes into play when you have rotation that is not along one of the principal axes.
I suppose you could calculate the inertial tensor using a different basis, but wouldn't that just complicate the algebra(and unnecessarily at that)? I'm sorry if I haven't been much help.

5. Apr 21, 2005

### OlderDan

The full tensor (diagonal and off diagonal terms) is important for rotations about any axis through the origin in the x.y plane, not just the y = x line. The tensor product of I with the angular velocity vector gives the angular momentum vector. When the angular velocity vector is along the x or y axes, the off diagonal terms do not contribute.

6. Apr 21, 2005