- **Advanced Physics Homework**
(*http://www.physicsforums.com/forumdisplay.php?f=154*)

- - **Moment of inertia tensor**
(*http://www.physicsforums.com/showthread.php?t=586895*)

Moment of inertia tensor1 Attachment(s)
1. The problem statement, all variables and given/known dataFind the moment of inertia tensor of the plate attached below 2. Relevant equationsσ = area density 3. The attempt at a solutionSo the main problem I'm having is solving for Ixx and Iyy: 1) Ixx = σ∫∫(y^2 + z^2)dydz, since there are no dz components, I don't see how you end up with 1/3Ma^2. But iIf you approach it like this: 2) Ixx = σ∫∫(y^2 + z^2)dA = (M/a^2)∫(y^2 + z^2)(ady) = 1/3Ma^2, you get the right answer (where after integration you fill in z=0), and it makes sense, but is there a way to get there from 1)? Also, my prof seemed really confused about how the moment of inertia tensor components translate to rotations about different axes. I had a hunch that Ixx would translate to the rotation about the x-axis, Iyy about the y-axis, Izz about the z-axis, Ixy about some diagonal rotation in the xyplane, Ixz about some diagonal rotation in the xzplane, etc. Is there any truth to this? My prof said it doesn't translate that way, but didn't seem sure. I want some closure. Thanks, Ari |

Re: Moment of inertia tensorHi Ari! :smile:
Quote:
_{max} -> 0, σ∫z^{2}dz -> 0 (if we're keeping M, and therefore σz_{max}, constant) :wink:Quote:
symmetry of the body ("principal axes"), the tensor will be diagonal, and the moment of inertia for each axis will give you the ratio between angular momentum and angular velocity for that axis.For any other axes, the tensor isn't actually any use unless the tensor is for part of a composite body, and the axis is a principal axis of the whole body (but not that one part). In no case does I _{xy} etc have any physical significance (other than as part of a calculation). |

All times are GMT -5. The time now is 04:19 AM. |

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.

© 2014 Physics Forums