Principal Axes and Rigid-Body Rotation
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You should be comfortable with the definition of the moment-of-inertia tensor and the fact that it is a real, symmetric matrix. Familiarity with the diagonalization of symmetric matrices and the concept of eigenvectors is assumed.
Why this matters
Principal axes simplify the description of rigid-body rotation by aligning the coordinate system with the natural symmetry of the mass distribution.
In the previous lecture, we established that the angular momentum \(\vec{L}\) of a rigid body is related to its angular velocity \(\vec{\omega}\) by the inertia tensor \(\mathbf{I}\). Generally, \(\vec{L}\) and \(\vec{\omega}\) are not parallel, which complicates the equations of motion. However, because \(\mathbf{I}\) is a real symmetric tensor, it possesses a special set of orthogonal axes—the principal axes—along which the tensor is diagonal. In this frame, the components of \(\vec{L}\) are simply proportional to the components of \(\vec{\omega}\), decoupling the rotational dynamics into three independent scalar equations. This simplification is the cornerstone for analyzing complex rotations, such as the tumbling of a satellite or the precession of a top.
The Geometric Intuition
Imagine a rigid body, such as a rectangular block or a dumbbell, rotating about a fixed point. If you choose an arbitrary coordinate system, the mass distribution looks "skewed" relative to the axes, causing the angular momentum vector to point in a direction different from the angular velocity vector. This misalignment arises because mass is distributed asymmetrically with respect to the chosen axes. However, there always exists a set of three mutually perpendicular axes through the center of mass (or the fixed point) for which all the products of inertia vanish — even for a body with no symmetry at all. These are the principal axes. When the body rotates about any one of these axes, the angular momentum is parallel to the angular velocity. The coefficients of proportionality are the principal moments of inertia. Finding these axes is equivalent to finding the eigenvectors of the inertia tensor, and the principal moments are the corresponding eigenvalues. This geometric alignment removes the "cross-coupling" terms in the inertia tensor, making the physics transparent.
Diagonalization of the Inertia Tensor
As established in the derivation of the Moment-of-Inertia Tensor (D-035), the tensor \(\mathbf{I}\) is real and symmetric, satisfying \(\mathbf{I} = \mathbf{I}^{\mathsf T}\). A fundamental theorem of linear algebra states that any real symmetric matrix can be diagonalized by an orthogonal matrix. Let \(\mathbf{Q}\) be an orthogonal matrix whose columns are the normalized eigenvectors of \(\mathbf{I}\). The diagonalization is given by:
Here, \(I_1, I_2, I_3\) are the principal moments of inertia. They are the eigenvalues of \(\mathbf{I}\), found by solving the characteristic equation \(\det(\mathbf{I} - I \mathbf{1}) = 0\). The columns of \(\mathbf{Q}\) define the directions of the principal axes in the original body-fixed coordinate system. In this principal axis frame, the relationship between angular momentum and angular velocity simplifies to:
This decoupling is the primary advantage of working in the principal axis frame. It allows us to treat the rotation about each principal axis as an independent degree of freedom, at least kinematically.
Rotational Kinetic Energy in Principal Axes
The rotational kinetic energy \(T\) of a rigid body is given by \(T = \frac{1}{2} \vec{\omega} \cdot \vec{L}\). Substituting \(\vec{L} = \mathbf{I} \vec{\omega}\), we have \(T = \frac{1}{2} \vec{\omega}^{\mathsf T} \mathbf{I} \vec{\omega}\). In an arbitrary frame, this expression contains cross terms involving products of different components of \(\vec{\omega}\). However, in the principal axis frame, where \(\mathbf{I}\) is diagonal, the expression simplifies significantly:
This form is analogous to the translational kinetic energy \(\frac{1}{2}mv^2\), but generalized to three rotational degrees of freedom. The principal moments of inertia \(I_1, I_2, I_3\) play the role of the "mass" for each rotational component. This diagonal form is crucial for applying conservation of energy in torque-free motion, as it allows us to relate the magnitudes of the angular velocity components directly to the total energy without dealing with coupled terms.
Physical Interpretation of Principal Moments
The principal moments of inertia \(I_1, I_2, I_3\) are non-negative real numbers. Their relative magnitudes determine the stability of rotation. For a general rigid body, the principal moments satisfy the triangle inequalities: \(I_1 + I_2 \geq I_3\), \(I_1 + I_3 \geq I_2\), and \(I_2 + I_3 \geq I_1\). These follow directly from the definitions: for example \(I_1 + I_2 - I_3 = 2\int x_3^2\,dm \ge 0\) in principal axes. If the body is symmetric about one axis (e.g., a cylinder or a sphere), two of the principal moments are equal. Such a body is called a symmetric top. If all three are equal, the body is a spherical top, and the inertia tensor is proportional to the identity matrix, meaning \(\vec{L}\) is parallel to \(\vec{\omega}\) in any frame. For a general asymmetric body, the three principal moments are distinct, and the choice of principal axes is unique up to permutation.
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 \(\vec{\omega}\).
A common misconception is that the principal axes are fixed in space. In reality, they are fixed in the body. As the body rotates, the principal axes rotate with it. Therefore, the components \(\omega_1, \omega_2, \omega_3\) are measured in a rotating frame. This introduces fictitious forces (Coriolis and centrifugal) if one attempts to write Newton's second law directly in this frame. The correct approach is to use Euler's equations, which account for the rotation of the frame itself. Confusing the body-fixed principal frame with an inertial frame leads to incorrect equations of motion. C
Worked example
Problem. A rectangular lamina of mass \(M = 2.0\ \mathrm{kg}\) has dimensions \(a = 0.4\ \mathrm{m}\) and \(b = 0.6\ \mathrm{m}\). It rotates about an axis passing through its center of mass and perpendicular to its plane. Determine the principal moments of inertia and the rotational kinetic energy if the angular speed is \(\omega = 5.0\ \mathrm{rad/s}\).
- Identify the principal axes. For a rectangular lamina, the axes perpendicular to the sides and the axis perpendicular to the plane are principal axes. Let the axis perpendicular to the plane be the 3-axis.
- Calculate \(I_3\). For a rectangular lamina rotating about the axis perpendicular to its plane through the center, \(I_3 = \frac{1}{12} M (a^2 + b^2)\). Substituting the values: \(I_3 = \frac{1}{12} (2.0) (0.4^2 + 0.6^2) = \frac{1}{6} (0.16 + 0.36) = \frac{0.52}{6} \approx 0.0867\ \mathrm{kg \cdot m^2}\).
- Calculate \(I_1\) and \(I_2\). For the in-plane axes through the centre, \(I_1 = \frac{1}{12} M b^2\) and \(I_2 = \frac{1}{12} M a^2\), taking the 1-axis parallel to side \(a\) and the 2-axis parallel to side \(b\) (the mass lies at distances up to \(b/2\) and \(a/2\) from them respectively). As a check, \(I_1 + I_2 = I_3\) (perpendicular-axis theorem). \(I_1 = \frac{1}{12} (2.0) (0.6^2) = \frac{1}{6} (0.36) = 0.060\ \mathrm{kg \cdot m^2}\). \(I_2 = \frac{1}{12} (2.0) (0.4^2) = \frac{1}{6} (0.16) \approx 0.0267\ \mathrm{kg \cdot m^2}\).
- Calculate Kinetic Energy. The rotation is about the 3-axis, so \(\omega_3 = 5.0\ \mathrm{rad/s}\) and \(\omega_1 = \omega_2 = 0\). \(T = \frac{1}{2} I_3 \omega_3^2 = \frac{1}{2} (0.0867) (5.0)^2 = 0.5 \times 0.0867 \times 25 = 1.08375\ \mathrm{J}\).
Units check. \(I\) is in \(\mathrm{kg \cdot m^2}\), \(\omega\) is in \(\mathrm{rad/s}\) (dimensionless). \(T\) is in \(\mathrm{kg \cdot m^2/s^2} = \mathrm{J}\). Correct.
Three quick tests
- Is the inertia tensor always diagonal in a body-fixed frame?
Answer
No, it is only diagonal in the principal axis frame. - What is the relationship between \(\vec{L}\) and \(\vec{\omega}\) when rotation is about a principal axis?
Answer
They are parallel. - Can a rigid body have only two distinct principal moments of inertia?
Answer
Yes, if it is a symmetric top.
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.