Rigid Bodies and the Inertia Tensor
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You should be comfortable with vector cross products, the definition of angular momentum for a single particle, and basic linear algebra, specifically the properties of symmetric matrices.
Why this matters
Rigid bodies do not always rotate simply, and the inertia tensor is the mathematical object that captures this complexity.
For a body spinning about an axis of symmetry, angular momentum is parallel to angular velocity. For a general rigid body this is false. The relationship between these two vectors is governed by a second-rank tensor, the inertia tensor. Understanding this object is essential for analyzing precession, nutation, and the stability of rotating objects, from spinning tops to satellite attitude control. It bridges the gap between simple rotational dynamics and the full complexity of rigid-body motion.
The Core Idea
Imagine a rigid body rotating about a fixed point. Each particle in the body has a specific position and velocity. The total angular momentum is the sum of the angular momenta of all these particles. Because the body is rigid, the distances between particles remain constant, but their positions relative to the rotation axis change as the body spins. This geometric relationship means that the direction of the total angular momentum vector depends not just on the speed of rotation, but on the mass distribution of the body. If the mass is distributed symmetrically, the angular momentum points in the same direction as the angular velocity. If the mass distribution is asymmetric, the angular momentum vector "tilts" away from the angular velocity vector. The inertia tensor is the mathematical bridge that maps the angular velocity vector to the angular momentum vector, accounting for this mass distribution.
Defining the Inertia Tensor
Consider a rigid body with mass elements \( dm \) located at position vectors \( \vec{r} \) relative to a fixed origin. The body rotates with angular velocity \( \vec{\omega} \). The velocity of a mass element is given by \( \vec{v} = \vec{\omega} \times \vec{r} \). The angular momentum \( \vec{L} \) is defined as the integral of the angular momentum of each mass element:
Using the vector triple product identity \( \vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C} \), we can expand this expression. Let \( r^2 = \vec{r} \cdot \vec{r} \). The expression becomes:
This equation shows that \( \vec{L} \) is a linear function of \( \vec{\omega} \). We can write this in component form. Let \( \vec{r} = (x, y, z) \) and \( \vec{\omega} = (\omega_x, \omega_y, \omega_z) \). The \( x \)-component of \( \vec{L} \) is:
Similar expressions hold for \( L_y \) and \( L_z \). We can collect these into a matrix equation \( \vec{L} = \mathbf{I} \vec{\omega} \), where \( \mathbf{I} \) is the inertia tensor. The diagonal elements are the moments of inertia about the coordinate axes, and the off-diagonal elements are the products of inertia.
Properties of the Tensor
The inertia tensor \( \mathbf{I} \) is a real, symmetric \( 3 \times 3 \) matrix. Its components are defined as:
where \( \rho \) is the mass density, \( \delta_{jk} \) is the Kronecker delta, and \( x_j \) are the Cartesian coordinates. The symmetry \( I_{jk} = I_{kj} \) arises naturally from the definition, as the term \( x_j x_k \) is symmetric in \( j \) and \( k \). Because the tensor is real and symmetric, it is orthogonally diagonalizable. This means there exists a set of three mutually perpendicular axes, called the principal axes, along which the tensor is diagonal. In this principal axis frame, the off-diagonal terms (products of inertia) vanish, and the relationship between angular momentum and angular velocity simplifies to:
Here, \( I_1, I_2, I_3 \) are the principal moments of inertia, which are the eigenvalues of the inertia tensor. These values are always non-negative because they represent the integral of a squared distance times mass. The principal axes are the eigenvectors of the tensor. Finding these axes is a standard linear algebra problem involving the characteristic equation \( \det(\mathbf{I} - I \mathbf{1}) = 0 \).
Physical Interpretation
The diagonal elements \( I_{xx}, I_{yy}, I_{zz} \) represent the resistance of the body to rotation about the \( x, y, \) and \( z \) axes, respectively. For example, \( I_{xx} = \int (y^2 + z^2) dm \) measures how far the mass is distributed from the \( x \)-axis. The off-diagonal elements, such as \( I_{xy} = -\int xy \, dm \), measure the asymmetry of the mass distribution with respect to the coordinate planes. If the mass distribution is symmetric under reflection in the \( yz \)-plane (\( x \to -x \)), the products of inertia \( I_{xy} \) and \( I_{xz} \) vanish. In general, for an arbitrary coordinate system, these off-diagonal terms are non-zero, causing the angular momentum vector to be non-parallel to the angular velocity vector. This misalignment is the source of complex rotational dynamics, such as the wobble of a spinning top.
The derivation
The Moment-of-Inertia Tensor
For a rigid body rotating about a fixed reference point with common angular velocity \(\vec{\omega}\), the angular momentum about that point is a linear function of \(\vec{\omega}\), \(L_j=\sum_k I_{jk}\,\omega_k\), where the real symmetric array \(I_{jk}=\sum_i m_i\left(r_i^2\,\delta_{jk}-x_{i,j}x_{i,k}\right)\) is the moment-of-inertia tensor. Being real and symmetric it is orthogonally diagonalizable, possessing three mutually perpendicular principal axes with real non-negative principal moments \(I_1,I_2,I_3\) along which \(\vec{L}\) is parallel to \(\vec{\omega}\).
The inertia tensor depends on the choice of origin. If \( O \) is the centre of mass and the origin is shifted to a point \( O' \) at displacement \( \vec{a} \) from it, the new tensor \( \mathbf{I}' \) is related to the old one by the parallel axis theorem in tensor form. Specifically, \( I'_{jk} = I_{jk} + M(a^2 \delta_{jk} - a_j a_k) \), where \( M \) is the total mass. Separately, note that the principal axes are fixed in the body, but their orientation relative to space changes as the body rotates. Confusing the body frame (where the tensor is constant) with the space frame (where the tensor changes with time) is a common source of error in rigid-body dynamics. C
Worked example
Problem. A uniform rectangular plate of mass \( M = 2.0 \, \mathrm{kg} \) has dimensions \( a = 0.5 \, \mathrm{m} \) (width) and \( b = 0.8 \, \mathrm{m} \) (height). The plate lies in the \( xy \)-plane, centered at the origin. Calculate the moment of inertia about the \( z \)-axis (perpendicular to the plate) and about the \( x \)-axis (along the width).
- Identify the geometry. The plate is a rectangle in the \( xy \)-plane. The \( z \)-axis is perpendicular to the plate, passing through the center. The \( x \)-axis lies in the plane of the plate, along the width \( a \).
- Calculate \( I_z \). For a rectangular plate rotating about an axis perpendicular to its plane through the center, the moment of inertia is given by \( I_z = \frac{1}{12} M (a^2 + b^2) \). Substitute the values: \( I_z = \frac{1}{12} (2.0) (0.5^2 + 0.8^2) \, \mathrm{kg \cdot m^2} \).
- Compute the numerical value. \( 0.5^2 = 0.25 \), \( 0.8^2 = 0.64 \). Sum is \( 0.89 \). \( I_z = \frac{1}{12} (2.0) (0.89) = \frac{1.78}{12} \approx 0.1483 \, \mathrm{kg \cdot m^2} \).
- Calculate \( I_x \). For a rectangular plate rotating about an axis in its plane along the width, the moment of inertia is \( I_x = \frac{1}{12} M b^2 \). Substitute the values: \( I_x = \frac{1}{12} (2.0) (0.8^2) \, \mathrm{kg \cdot m^2} \).
- Compute the numerical value. \( I_x = \frac{1}{12} (2.0) (0.64) = \frac{1.28}{12} \approx 0.1067 \, \mathrm{kg \cdot m^2} \).
Units check. Mass is in \( \mathrm{kg} \), length is in \( \mathrm{m} \). The formula involves mass times length squared, so the units are \( \mathrm{kg \cdot m^2} \), which is correct for moment of inertia.
Three quick tests
- Is the inertia tensor always symmetric?
Answer
Yes, because the definition involves \( x_j x_k \), which is symmetric in \( j \) and \( k \), and the Kronecker delta is symmetric. - What are the principal moments of inertia?
Answer
They are the eigenvalues of the inertia tensor, representing the moments of inertia about the principal axes where the tensor is diagonal. - Can the angular momentum vector be non-parallel to the angular velocity vector?
Answer
Yes, unless the rotation is about a principal axis or the body is spherically symmetric. In general, \( \vec{L} = \mathbf{I} \vec{\omega} \), and if \( \mathbf{I} \) is not a scalar multiple of the identity, \( \vec{L} \) and \( \vec{\omega} \) are not parallel.
Problem set for PU-101 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.