How Do You Calculate the Moment of Inertia Tensor for a Plate?

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SUMMARY

The moment of inertia tensor for an infinitely thin plate can be calculated using the formula Ixx = σ∫∫(y^2 + z^2)dA, leading to Ixx = 1/3Ma^2 when z is set to zero. The area density σ is defined as mass per unit area. The components of the moment of inertia tensor, such as Ixx and Iyy, correspond to rotations about the principal axes of symmetry, confirming that Ixx relates to rotation about the x-axis and Iyy to the y-axis. However, off-diagonal components like Ixy do not have physical significance unless used in specific composite body calculations.

PREREQUISITES
  • Understanding of moment of inertia and its significance in rotational dynamics
  • Familiarity with calculus, specifically double integrals
  • Knowledge of area density (σ) and its application in physics
  • Concept of principal axes in rigid body dynamics
NEXT STEPS
  • Study the derivation of moment of inertia for various geometric shapes
  • Learn about the application of the parallel axis theorem in calculating moments of inertia
  • Explore the physical significance of off-diagonal components in the inertia tensor
  • Investigate the relationship between angular momentum and angular velocity in rigid body motion
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AriAstronomer
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Homework Statement


Find the moment of inertia tensor of the plate attached below


Homework Equations


σ = area density


The Attempt at a Solution


So 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
 

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Hi Ari! :smile:
AriAstronomer said:
So 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.
… is there a way to get there from 1)?

Yes, this is an "infinitely thin" plate (in the z-direction), so as the thickness zmax -> 0, σ∫z2dz -> 0 (if we're keeping M, and therefore σzmax, constant) :wink:
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.

So long as your z y z axes are axes of 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 Ixy etc have any physical significance (other than as part of a calculation).
 

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