Rigid Body Dynamics & Euler's Equations
Rigid Body Dynamics & Euler's Equations
A rigid body is a system of particles with fixed inter-particle distances. It has at most 6 degrees of freedom: 3 for the position of the centre of mass and 3 for its orientation. The orientation is specified by three Euler angles $(\phi,\theta,\psi)$ — successive rotations that carry the space frame into the body frame. Alternatively, the orientation is encoded in an orthogonal rotation matrix $R\in SO(3)$ or a unit quaternion. The kinetic energy splits as $T=T_{\text{CM}}+T_{\text{rot}}$, where $T_{\text{rot}}=\frac{1}{2}\boldsymbol{\omega}^\top I\boldsymbol{\omega}$ and $I$ is the inertia tensor.
The inertia tensor $I_{ij}=\sum_a m_a(|\mathbf{r}_a|^2\delta_{ij}-r_{a,i}r_{a,j})$ is a symmetric positive-semi-definite $3\times 3$ matrix. In the principal axis frame (eigenvectors of $I$), it is diagonal: $I=\mathrm{diag}(I_1,I_2,I_3)$, the three principal moments of inertia. The angular momentum in the body frame is $\mathbf{L}=I\boldsymbol{\omega}$. Euler's equations express the torque-free rotation of the body in the body frame:
$$I_1\dot{\omega}_1-(I_2-I_3)\omega_2\omega_3=N_1,$$ $$I_2\dot{\omega}_2-(I_3-I_1)\omega_3\omega_1=N_2,$$ $$I_3\dot{\omega}_3-(I_1-I_2)\omega_1\omega_2=N_3.$$
For torque-free motion ($N_i=0$), both $|\mathbf{L}|^2$ and $T=\frac{1}{2}\boldsymbol{\omega}^\top I\boldsymbol{\omega}$ are conserved. The angular velocity vector traces out an polhode on the inertia ellipsoid $\frac{1}{2}(I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2)=T=\text{const}$. The intermediate-axis theorem states that rotation about the axis with the intermediate principal moment is unstable — the famous "tennis racket theorem" or "Dzhanibekov effect".
Definition — Inertia Tensor & Euler Angles
The inertia tensor: $I_{ij}=\int\rho(\mathbf{r})(r^2\delta_{ij}-r_i r_j)\,d^3r$. Euler angles $(\phi,\theta,\psi)$: first rotate by $\phi$ about $z$ (precession), then by $\theta$ about new $x$ (nutation), then by $\psi$ about new $z$ (spin). The body-frame angular velocity components in terms of Euler angles are $\omega_1=\dot{\phi}\sin\theta\sin\psi+\dot{\theta}\cos\psi$, $\omega_2=\dot{\phi}\sin\theta\cos\psi-\dot{\theta}\sin\psi$, $\omega_3=\dot{\phi}\cos\theta+\dot{\psi}$.
Theorem — Euler's Equations of Motion
In the body (principal axis) frame, the equations of motion for the angular velocity are $I_1\dot{\omega}_1=(I_2-I_3)\omega_2\omega_3+N_1$, (cyclic in $1,2,3$). For torque-free motion, $T=\frac{1}{2}\sum_i I_i\omega_i^2$ and $|\mathbf{L}|^2=\sum_i I_i^2\omega_i^2$ are conserved. Stability: rotation about the axes of largest or smallest $I$ is stable; about the intermediate axis it is unstable.
Example 1: Symmetric Top (Torque-Free)
For $I_1=I_2\neq I_3$ (symmetric top), Euler's equations give $\dot{\omega}_3=0$ (so $\omega_3=\text{const}$) and $\dot{\omega}_1=\Omega\omega_2$, $\dot{\omega}_2=-\Omega\omega_1$ where $\Omega=(I_3-I_1)\omega_3/I_1$. Solutions: $\omega_1+i\omega_2=A e^{i\Omega t}$, so the angular velocity precesses about the symmetry axis at the body-frame precession rate $\Omega$. This torque-free precession is seen in the wobble of a thrown football or in the Earth's Chandler wobble (period $\approx 433$ days).
Example 2: Euler Angle Lagrangian for the Symmetric Top
For a symmetric top ($I_1=I_2$) with one point fixed and subject to gravity, the Lagrangian in Euler angles is $$L=\frac{I_1}{2}(\dot{\theta}^2+\dot{\phi}^2\sin^2\theta)+\frac{I_3}{2}(\dot{\psi}+\dot{\phi}\cos\theta)^2-Mgl\cos\theta.$$ Both $\phi$ and $\psi$ are cyclic: $p_\phi=\text{const}\equiv L_z$ (vertical angular momentum) and $p_\psi=\text{const}\equiv I_3\omega_3$ (body-axis spin). The $\theta$ equation reduces to effective 1D motion in $\theta$ — the basis of the analysis of steady precession and nutation.
Example 3: Steady Precession of the Heavy Top
For a symmetric top ($\theta=\text{const}$, $\dot{\theta}=0$), the $\theta$ equation of motion yields $$I_1\dot{\phi}^2\sin\theta\cos\theta-I_3(\dot{\psi}+\dot{\phi}\cos\theta)\dot{\phi}\sin\theta+Mgl\sin\theta=0.$$ For rapid spin ($\dot{\psi}\gg\dot{\phi}$), this simplifies to the precession rate $$\dot{\phi}\approx\frac{Mgl}{I_3\omega_3}=\frac{Mgl}{p_\psi},$$ inversely proportional to the spin. A gyroscope precesses slowly when spinning fast — a direct consequence of the conservation of $p_\psi$.
Example 4: Intermediate-Axis Instability
For a torque-free asymmetric body ($I_1
Practice Problems
- Compute the inertia tensor of a uniform solid sphere of mass $M$ and radius $R$ about its centre. Identify the principal moments and verify they are all equal (hence any axis is a principal axis).
- A uniform thin rod of mass $M$ and length $L$ rotates about one end. Compute $I$ about the end, the centre, and an axis perpendicular to the rod through one end, and verify the parallel-axis theorem.
- For a torque-free symmetric top ($I_1=I_2$), show that the angular velocity precesses about the symmetry axis at the rate $\Omega=(I_3-I_1)\omega_3/I_1$ in the body frame. What is the precession rate in the space frame?
- A symmetric top with $I_1=I_2=I$, $I_3$, one point fixed, precesses steadily at angle $\theta_0$ and rate $\dot{\phi}$. Derive the exact quadratic equation for $\dot{\phi}$ and show it has two solutions (fast and slow precession).
- Show that for the torque-free asymmetric top, stable rotation about the intermediate principal axis is impossible by constructing the linearised perturbation equations and computing the growth rate explicitly.
Show Answer Key
1. Solid sphere: $I_{ij} = \int\rho(r^2\delta_{ij}-r_ir_j)dV$. By symmetry, $I_{xx}=I_{yy}=I_{zz}$ and off-diagonal $=0$. $I_{xx} = \rho\int(y^2+z^2)dV = \frac{2}{5}MR^2$. All principal moments equal $\frac{2}{5}MR^2$, so any axis through the center is a principal axis (spherical top).
2. About center: $I_c = \frac{1}{12}ML^2$ (standard result from $\int_{-L/2}^{L/2}\rho x^2 dx$). About end: $I_{\text{end}} = \frac{1}{3}ML^2$. Parallel-axis: $I_{\text{end}} = I_c + M(L/2)^2 = \frac{1}{12}ML^2+\frac{1}{4}ML^2 = \frac{1}{3}ML^2$ ✓. Perpendicular through end: same as $I_{\text{end}} = \frac{1}{3}ML^2$ for rotation in the plane.
3. Euler equations (torque-free, $I_1=I_2$): $I_1\dot{\omega}_1 = (I_1-I_3)\omega_2\omega_3$, $I_1\dot{\omega}_2 = (I_3-I_1)\omega_1\omega_3$, $I_3\dot{\omega}_3 = 0$. So $\omega_3 = \text{const}$. Let $\Omega = (I_3-I_1)\omega_3/I_1$. Then $\dot{\omega}_1 = -\Omega\omega_2$, $\dot{\omega}_2 = \Omega\omega_1$: precession at rate $\Omega$ in the body frame. Space-frame rate: $\dot{\psi} = (I_3/I_1)\omega_3\cos\theta/(\cos\theta)$ — depends on geometry.
4. Euler angles $(\phi,\theta,\psi)$: steady precession means $\dot{\theta}=0$, $\dot{\phi}=\text{const}$, $\dot{\psi}=\text{const}$. The $\theta$-equation gives: $I\dot{\phi}^2\cos\theta\sin\theta - I_3(\dot{\phi}\cos\theta+\dot{\psi})\dot{\phi}\sin\theta + Mgl\sin\theta = 0$. Dividing by $\sin\theta$: $I\dot{\phi}^2\cos\theta - I_3\dot{\phi}(\dot{\phi}\cos\theta+\dot{\psi})+Mgl=0$. This is quadratic in $\dot{\phi}$, giving two solutions: fast precession (gyroscopic) and slow precession (gravitational).
5. Let $\omega = (\omega_1^0+\epsilon_1, \omega_2^0+\epsilon_2, \epsilon_3)$ where initially $\omega = (0,\omega_2^0,0)$ (intermediate axis). Euler equations linearized: $I_1\dot{\epsilon}_1 = (I_2-I_3)\omega_2^0\epsilon_3$ and $I_3\dot{\epsilon}_3 = (I_1-I_2)\omega_2^0\epsilon_1$. Combining: $\ddot{\epsilon}_1 = \frac{(I_2-I_3)(I_1-I_2)}{I_1 I_3}\omega_2^{0\,2}\epsilon_1$. Since $I_1 > I_2 > I_3$: $(I_2-I_3)>0$ and $(I_1-I_2)>0$, so the coefficient is positive → exponential growth. Rotation about the intermediate axis is unstable.