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I have read about moment of inertia in rotational mechanics, and also read about moment of inertia tensor where 9 components are present where Ixx, Iyy, Izz are diagonal elements of moment of inertia tensor and they are called moment of inertia and Ixy, Iyz, Izx are off diagonal elements and they are called product of inertia and we know that the Ixx,Iyy , Izz are the moment of inertia about x,y,z axises So these components tells the rotation of a body about rotation about x,y,z axises.

Questions:

  1. But I don't know what is the axis of rotation of Ixy, Iyz ,Izx I mean about which axis they are rotating?

  2. And in principal moment of inertia those product of inertia terms are become zero ,why they are zero in that case?

  3. By making them zero the moment of inertia tensor is made diagonal matrix, but why they are made zero to make the moment of inertia tensor diagonal?

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The meaning of moment of inertia tensor comes from $$\vec{L}=I\vec{\omega}$$

So for example if you consider an object rotating about the $y$ axis with angular velocity $\omega$, you have

$$\left(\begin{array}{c} L_x\\L_y\\L_z \end{array}\right)=\left(\begin{array}{ccc} I_{xx}&I_{xy}&I_{xz}\\I_{yx}&I_{yy}&I_{yz}\\I_{zx}&I_{zy}&I_{zz}\end{array}\right)\left(\begin{array}{c}0\\ \omega\\0\end{array}\right)$$ $$=\left(\begin{array}{c}I_{xy}\\I_{yy}\\I_{zy}\end{array}\right)\omega$$

So $I_{xy}\omega$ is the $x$-component of angular momentum when the object rotates about the $y$ axis with angular velocity $\omega$, and similarly for the meaning of the other components.

In general $I_{ij}\omega$ is the $i$-th component of the angular momentum of the object rotating with angular velocity $\omega$ about the $j$-th axis.

For axes along which the rotation of an object will give angular momentum along the same axes as well, then the off-diagonal elements are zero if these axes are chosen as basis. Because in that case you have, e.g., for rotation about the $x$-axis $$\vec{L}=I_{xx}\omega \hat{x}$$ for rotation about $y$-axis $$\vec{L}=I_{yy}\omega \hat{y}$$ for rotation about $z$-axis $$\vec{L}=I_{zz}\omega \hat{z}$$ and hence in general for $$\vec{\omega}=\omega_x\hat{x}+\omega_y\hat{y}+\omega_z\hat{z}$$ $$\vec{L}=\left(\begin{array}{ccc}I_{xx}&0&0\\0&I_{yy}&0\\0&0&I_{zz}\end{array}\right)\left(\begin{array}{c}\omega_x\\\omega_y\\\omega_z\end{array}\right)$$

  1. But I don't know what is the axis of rotation of Ixy, Iyz ,Izx I mean about which axis they are rotating?

$I_{ij}$ is talking about the rotation about the $j$-th axis.

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  • $\begingroup$ Yes I think that is more precise $\endgroup$
    – Floris
    Commented Aug 6, 2016 at 15:39
  • $\begingroup$ But Iyy also indecates the rotation about y axis so what is the difference between Iyy and Ixy ? $\endgroup$
    – user101134
    Commented Aug 7, 2016 at 14:41
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    $\begingroup$ In general an object rotating about the $y$ axis with $\omega$ may not have an angular momentum along the $y$ axis, but $\mathbf{L}=I_{xy}\omega\mathbf{\hat{x}}+I_{yy}\omega\mathbf{\hat{y}}+I_{zy} \omega \mathbf{\hat{z}}$. $\endgroup$
    – velut luna
    Commented Aug 7, 2016 at 14:47
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But I don't know what is the axis of rotation of Ixy, Iyz ,Izx I mean about which axis they are rotating?

None. Those nonzero off-diagonal terms, if they exist, mean but one thing:

  • That the object in question lacks symmetry (objects with a spherical mass distribution always have a diagonal inertia tensor), and

  • That one has chosen to use a coordinate system that is not aligned with the principal axes.

That is the only thing that those nonzero off-diagonal terms mean.


Because mass is a scalar quantity, linear momentum is always coaligned with the velocity vector. This is not the case for angular momentum because moment of inertia is a 2nd order tensorial quantity. The inertia tensor of any physical object is symmetric and positive definite. Because the tensor is symmetric and positive definite, one can always find an orthogonal set of axes that make the off-diagonal elements of the inertia tensor vanish. There are however good, solid engineering reasons why one would choose not to do that.

For example, the flight control software for an airplane uses axes pointing along the fuselage, pointing normal to the fuselage along one of the wings, and nominally up or down (normal to both of the first two axes). However, that the plane requires fuel and takes on cargo and passengers almost inevitably makes this choice of axes be not aligned with the principal axes, and thus the inertia tensor will have nonzero off-diagonal elements with this choice of axes.


The only physical significance to these nonzero off-diagonal terms is that the chosen axes are not the principal axes.

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