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# 13.2: The Moment of Inertia Tensor


The angle $$\theta$$ increases in time (if you push hard enough1) in accordance with

$\vec{T}=\underset{\sim}{I}\vec{\alpha},\label{eqn:1}$

in which $$\vec{\alpha}$$ is the angular acceleration and $$\underset{\sim}{I}$$ is a matrix called the moment of inertia. For the simple case shown in Figure 13.1.1, $$\underset{\sim}{I}$$ is proportional to the identity matrix $$\underset{\sim}{\delta}$$, $$\vec{\alpha}$$ is parallel to the axis of rotation (the bolt), and its magnitude $$|\vec{\alpha}|$$ is $$d^2\theta/dt^2$$.

The general definition of the moment of inertia matrix is

$I_{i j}=\int_{V} d V \rho(\vec{x})\left(x_{k} x_{k} \delta_{i j}-x_{i} x_{j}\right),\label{eqn:2}$

where $$\rho(\vec{x})$$ is the density (mass per unit volume). Details can be found in most classical mechanics texts, e.g., Marion (2013).

##### Example $$\PageIndex{1}$$

The particular case illustrated in Figure $$\PageIndex{1}$$ is the rotation of a rectangular prism, with uniform density and edge dimensions $$a$$, $$b$$ and $$c$$, about the $$\hat{e}^{(1)}$$ axis. In this case both torque and angular acceleration are parallel to $$\hat{e}^{(1)}$$, and the only nonzero component of $$\underset{\sim}{I}$$ is $$I_{11}$$, computed as follows:

\begin{aligned} I_{11} &=\int_{V} d V \rho\left(x_{2}^{2}+x_{3}^{2}\right) \\ &=\rho \int_{-a / 2}^{a / 2} d x_{1} \int_{-b / 2}^{b / 2} d x_{2} \int_{-c / 2}^{c / 2} d x_{3}\left(x_{2}^{2}+x_{3}^{2}\right) \\ &=\rho \frac{a b c\left(b^{2}+c^{2}\right)}{12}. \end{aligned} \nonumber

For the simple case of a cube with $$a$$ = $$b$$ = $$c$$ = $$\Delta$$,

$I_{11}=\rho \frac{\Delta^{5}}{6}.\label{eqn:3}$

Is the moment of inertia matrix $$\underset{\sim}{I}$$ a tensor? We would expect so, since it connects two physically real vectors via Equation $$\ref{eqn:1}$$. We can also establish this directly from Equation $$\ref{eqn:2}$$, the general formula for $$\underset{\sim}{I}$$. Like any other integral, $$\underset{\sim}{I}$$ can be written as the limit of a sum:

$I_{i j}=\sum \Delta V \rho\left(x_{k} x_{k} \delta_{i j}-x_{i} x_{j}\right), \nonumber$

where each term in the sum is evaluated at the center of a volume element $$\Delta V$$. Now $$\Delta V$$ and $$\rho$$ are scalars, and so is the dot product $$x_k x_k$$ (section 3.2). Moreover, we know that both $$\delta_{ij}$$ and the dyad $$x_ix_j$$ transform according to Equation 3.3.8. Each term in the sum is therefore a tensor, and so then is the sum itself. Taking the limit as $$\Delta V\rightarrow 0$$, we conclude that $$\underset{\sim}{I}$$ transforms according to Equation 3.3.8. We therefore refer to $$\underset{\sim}{I}$$ as the moment of inertia tensor.

1In the case shown here, $$\vec{F}$$ is really the sum of the force exerted by the person and the opposing force exerted by friction, and similarly for $$\vec{T}$$.

This page titled 13.2: The Moment of Inertia Tensor is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Bill Smyth via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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