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PU-101 / L21 · lecture

Euler's Equations and Torque-Free Precession

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BEFORE YOU START

You must be comfortable with the definition of principal axes of inertia and the expression for angular momentum \(\vec{L} = I_1 \omega_1 \hat{e}_1 + I_2 \omega_2 \hat{e}_2 + I_3 \omega_3 \hat{e}_3\) in the body frame. Familiarity with vector differentiation in rotating frames is assumed.

Why this matters

Euler's equations provide the definitive dynamical description of rigid body rotation.

While Newton's laws describe the motion of particles, Euler's equations extend this framework to extended rigid bodies. By working in the body-fixed frame, we transform the complex vector differential equation of angular momentum balance into a set of three coupled scalar ordinary differential equations. This formulation reveals the intrinsic stability properties of rotation, explaining why a spinning top wobbles or why a tumbling satellite exhibits chaotic motion, phenomena that are counter-intuitive without a rigorous mathematical basis.

The Body-Fixed Perspective

Imagine a rigid body spinning in empty space with no external forces acting on it. If you observe it from a stationary point in the laboratory, the body tumbles and rotates in a complex manner. However, if you attach your coordinate system directly to the body itself, the moments of inertia become constants. The challenge lies in the fact that this body-fixed frame is rotating. When we calculate the rate of change of angular momentum in this rotating frame, we must account for the rotation of the axes themselves. This leads to a term involving the cross product of the angular velocity and the angular momentum. The resulting equations, known as Euler's equations, describe how the components of angular velocity change relative to the principal axes of the body. In the absence of torque, these equations dictate that the angular velocity vector does not generally align with the angular momentum vector, leading to a phenomenon called precession where the axis of rotation sweeps out a cone.

Euler's Equations of Motion

The fundamental law for rotational dynamics is \(\vec{\tau} = \frac{d\vec{L}}{dt}\). When expressed in the body frame, the time derivative includes a correction term for the rotating basis vectors. For a rigid body with principal moments of inertia \(I_1, I_2, I_3\), the components of the angular momentum are \(L_i = I_i \omega_i\). The general form of Euler's equations is given by:

\[ I_1 \dot{\omega}_1 - (I_2 - I_3) \omega_2 \omega_3 = \tau_1 \] \[ I_2 \dot{\omega}_2 - (I_3 - I_1) \omega_3 \omega_1 = \tau_2 \] \[ I_3 \dot{\omega}_3 - (I_1 - I_2) \omega_1 \omega_2 = \tau_3 \]

These equations are cyclic in the indices \(1, 2, 3\). They show that the rate of change of one component of angular velocity depends on the products of the other two components. This coupling is the source of the complex tumbling motions observed in torque-free rigid bodies.

Torque-Free Precession of a Symmetric Top

Consider a symmetric top where \(I_1 = I_2 = I_\perp\) and \(I_3 = I_\parallel\). In the torque-free case (\(\vec{\tau} = 0\)), the third Euler equation simplifies significantly. Since \(I_1 = I_2\), the term \((I_1 - I_2)\omega_1 \omega_2\) vanishes, implying \(\dot{\omega}_3 = 0\). Thus, the component of angular velocity along the symmetry axis is constant. The first two equations become:

\[ I_\perp \dot{\omega}_1 - (I_\perp - I_\parallel) \omega_2 \omega_3 = 0 \] \[ I_\perp \dot{\omega}_2 - (I_\parallel - I_\perp) \omega_3 \omega_1 = 0 \]

Differentiating the first equation with respect to time and substituting the second, we find that \(\omega_1\) and \(\omega_2\) oscillate harmonically. The angular velocity vector \(\vec{\omega}\) precesses about the symmetry axis (the 3-axis) with a precession frequency \(\Omega\). This frequency is given by:

\[ \Omega = \frac{I_\parallel - I_\perp}{I_\perp} \omega_3 \]

This result indicates that the body-fixed angular velocity vector traces a cone around the symmetry axis. The rate of this precession depends on the difference in moments of inertia and the spin rate about the symmetry axis.

Stability of Rotation

A crucial insight from Euler's equations is the stability of rotation about different principal axes. If a rigid body is rotating about its principal axis of maximum moment of inertia, small perturbations result in bounded oscillations, meaning the rotation is stable. Similarly, rotation about the axis of minimum moment of inertia is also stable. However, rotation about the intermediate axis (where \(I_1 < I_2 < I_3\)) is unstable. Any small deviation from this axis will grow exponentially, causing the body to tumble away from the intermediate axis. This is known as the intermediate axis theorem. The stability analysis relies on the conservation of kinetic energy and angular momentum, which constrain the motion to the intersection of an ellipsoid (energy) and a sphere (angular momentum magnitude).

The derivation

◆ DERIVATION EMBED · D-036 · not re-derived here

Euler's Equations and Torque-Free Precession

For a rigid body referred to its principal axes fixed in the body, angular-momentum balance \(\vec{\tau}=\left(\dfrac{d\vec{L}}{dt}\right)_{\text{space}}\) becomes the three Euler equations \(I_i\dot{\omega}_i-(I_j-I_k)\omega_j\omega_k=\tau_i\) (cyclic in \(i,j,k\)). In the torque-free case these predict that a symmetric body's transverse angular velocity precesses about its symmetry axis at a fixed rate \(\Omega=\dfrac{I_\parallel-I_\perp}{I_\perp}\omega_3\), and that steady rotation is stable only about the axes of greatest and least moment of inertia, never the intermediate one.

\[ I_i\dot{\omega}_i-(I_j-I_k)\,\omega_j\omega_k=\tau_i \qquad \Omega=\frac{I_3-I_1}{I_1}\,\omega_3 \qquad K=\frac{(I_3-I_1)(I_1-I_2)}{I_2 I_3}\,\omega_0^{2} \]
Open full derivation →
Tier 3 ยท watch out

A common misconception is that the angular momentum vector \(\vec{L}\) is fixed in the body frame. In reality, \(\vec{L}\) is fixed in the inertial space frame. In the body frame, \(\vec{L}\) appears to precess around the symmetry axis. The distinction between the space-fixed and body-fixed descriptions of \(\vec{L}\) and \(\vec{\omega}\) is critical for correctly interpreting the precession frequency and the geometry of the motion. C

Worked example

Problem. A symmetric top has moments of inertia \(I_1 = I_2 = 2.0\ \mathrm{kg \cdot m^2}\) and \(I_3 = 5.0\ \mathrm{kg \cdot m^2}\). It is spinning freely with an angular velocity component along the symmetry axis of \(\omega_3 = 10\ \mathrm{rad/s}\). Calculate the precession frequency \(\Omega\) of the angular velocity vector about the symmetry axis in the body frame.

  1. Identify the given values: \(I_\perp = 2.0\ \mathrm{kg \cdot m^2}\), \(I_\parallel = 5.0\ \mathrm{kg \cdot m^2}\), \(\omega_3 = 10\ \mathrm{rad/s}\).
  2. Use the formula for the precession frequency: \(\Omega = \frac{I_\parallel - I_\perp}{I_\perp} \omega_3\).
  3. Substitute the values: \(\Omega = \frac{5.0 - 2.0}{2.0} \times 10\).
  4. Calculate the result: \(\Omega = \frac{3.0}{2.0} \times 10 = 1.5 \times 10 = 15\ \mathrm{rad/s}\).
\[ \Omega = 15\ \mathrm{rad/s} \]

Units check. The ratio of moments of inertia is dimensionless. The angular velocity \(\omega_3\) has units of \(\mathrm{rad/s}\). Therefore, \(\Omega\) has units of \(\mathrm{rad/s}\), which is consistent with a frequency.

CHECK YOURSELF

Three quick tests

  1. What is the value of \(\dot{\omega}_3\) for a torque-free symmetric top?
    AnswerIt is zero, meaning the spin about the symmetry axis is constant.
  2. Which principal axis of rotation is unstable for a torque-free rigid body?
    AnswerThe intermediate axis (the axis with the middle moment of inertia).
  3. If \(I_\parallel = I_\perp\), what happens to the precession frequency \(\Omega\)?
    AnswerIt becomes zero, indicating the body is a sphere and does not exhibit this specific precession mode.
PROBLEMS

Problem set for PU-101 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-036 · Threads: force, energy, matter