Chapter 4 • Theory & Derivations

The Rigid Body Motion & Euler's Equations

Exhaustive treatment of rigid body kinematics and dynamics: degrees of freedom, orthogonal rotation matrices and SO(3) group properties, Eulerian angles (precession, nutation, spin), Euler's rotation theorem, inertia tensor and dyadics, principal axes of inertia, Euler's dynamical equations with fixed points, Poinsot construction for torque-free motion, and the heavy symmetrical top with sleeping top stability.

§4.1 Degrees of Freedom & Orthogonal Transformation Matrices

1. Degrees of Freedom of an Ideal Rigid Body

A rigid body is an idealized assembly of $N$ particles where the inter-particle distances remain invariant under all forces:

|\vec{r}_i - \vec{r}_j| = c_{ij} = \text{constant}, \quad \forall i, j

Specifying the position of 3 non-collinear particles requires 9 coordinates subject to 3 internal distance constraints, yielding 6 independent degrees of freedom:

3 Translational Degrees of Freedom: Specifying the position vector $\vec{R}$ of the center of mass in the space-fixed inertial frame.
3 Rotational Degrees of Freedom: Specifying the orientation of a body-fixed coordinate frame relative to the space-fixed frame.

2. Orthogonal Transformations

Let $\vec{r} = (x, y, z)^T$ be the coordinates of a vector in the space-fixed frame and $\vec{r}' = (x', y', z')^T$ in the rotated frame. The linear transformation is:

\vec{r}' = \mathbf{A} \vec{r}, \quad x_i' = \sum_{j=1}^3 A_{ij} x_j

Invariance of vector length requires $\vec{r}' \cdot \vec{r}' = \vec{r} \cdot \vec{r}$:

(\mathbf{A}\vec{r})^T (\mathbf{A}\vec{r}) = \vec{r}^T (\mathbf{A}^T \mathbf{A}) \vec{r} = \vec{r}^T \mathbf{I} \vec{r}

This necessitates the orthogonality condition:

\mathbf{A}^T \mathbf{A} = \mathbf{A} \mathbf{A}^T = \mathbf{I} \implies \sum_{k=1}^3 A_{ki} A_{kj} = \delta_{ij}

Taking the determinant: $\det(\mathbf{A}^T \mathbf{A}) = (\det \mathbf{A})^2 = 1 \implies \det \mathbf{A} = \pm 1$. For proper physical rotations (preserving coordinate handedness), $\det \mathbf{A} = +1$. The set of all such transformation matrices forms the special orthogonal Lie group $\mathbf{SO(3)}$.

§4.2 The Eulerian Angles & Euler's Rotation Theorem

1. Standard Eulerian Angle Parameterization

To transform an initial space-fixed coordinate system $(x, y, z)$ into an arbitrary body-fixed system $(x', y', z')$, Leonhard Euler defined three successive rotations parameterized by the Eulerian angles $(\phi, \theta, \psi)$:

Precession (Angle $\phi$): Rotation counter-clockwise by angle $\phi$ about the initial $z$-axis ($0 \le \phi < 2\pi$). The resulting line of nodes $\xi$ lies along the intersection of the $(x, y)$ and $(x', y')$ planes: $\mathbf{D}(\phi) = \begin{pmatrix} \cos \phi & \sin \phi & 0 \\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{pmatrix}$
Nutation (Angle $\theta$): Rotation by angle $\theta$ about the intermediate line of nodes $\xi$ ($0 \le \theta \le \pi$): $\mathbf{C}(\theta) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & -\sin \theta & \cos \theta \end{pmatrix}$
Intrinsic Body Spin (Angle $\psi$): Rotation by angle $\psi$ about the final body-fixed $z'$-axis ($0 \le \psi < 2\pi$): $\mathbf{B}(\psi) = \begin{pmatrix} \cos \psi & \sin \psi & 0 \\ -\sin \psi & \cos \psi & 0 \\ 0 & 0 & 1 \end{pmatrix}$

The composite rotation matrix $\mathbf{A} = \mathbf{B}(\psi) \mathbf{C}(\theta) \mathbf{D}(\phi)$ transforms space coordinates into body coordinates.

2. Angular Velocity Vector in Terms of Euler Angles

Projecting the rates of change $(\dot{\phi}, \dot{\theta}, \dot{\psi})$ onto the body-fixed axes $(x', y', z')$ yields the instantaneous angular velocity $\vec{\omega}$:

\omega_{x'} = \dot{\phi} \sin \theta \sin \psi + \dot{\theta} \cos \psi

\omega_{y'} = \dot{\phi} \sin \theta \cos \psi - \dot{\theta} \sin \psi

\omega_{z'} = \dot{\phi} \cos \theta + \dot{\psi}

§4.3 The Inertia Tensor, Dyadics & Principal Axes of Inertia

1. Angular Momentum and the Inertia Tensor

For a rigid body rotating with instantaneous angular velocity $\vec{\omega}$ about a fixed point (or about its center of mass), the velocity of each point mass is $\vec{v}_i = \vec{\omega} \times \vec{r}_i$. The total angular momentum is:

\vec{L} = \sum_{i=1}^N m_i \vec{r}_i \times (\vec{\omega} \times \vec{r}_i) = \sum_{i=1}^N m_i [r_i^2 \vec{\omega} - (\vec{r}_i \cdot \vec{\omega})\vec{r}_i]

In component notation, this linear relationship is written as $\vec{L} = \mathbf{I} \vec{\omega}$:

L_j = \sum_{k=1}^3 I_{jk} \omega_k

where $\mathbf{I}$ is the rank-2 symmetric inertia tensor:

I_{jk} = \int \rho(\vec{r}) (r^2 \delta_{jk} - x_j x_k) dV

The diagonal components are the moments of inertia:

I_{xx} = \int (y^2 + z^2) dm, \quad I_{yy} = \int (x^2 + z^2) dm, \quad I_{zz} = \int (x^2 + y^2) dm

The off-diagonal components are the products of inertia:

I_{xy} = I_{yx} = -\int x y dm, \quad I_{yz} = I_{zy} = -\int y z dm, \quad I_{xz} = I_{zx} = -\int x z dm

2. Rotational Kinetic Energy

The rotational kinetic energy is a quadratic form in angular velocities:

T_{\text{rot}} = \frac{1}{2} \sum_{i=1}^N m_i v_i^2 = \frac{1}{2} \vec{\omega} \cdot \vec{L} = \frac{1}{2} \sum_{j=1}^3 \sum_{k=1}^3 I_{jk} \omega_j \omega_k = \frac{1}{2} \vec{\omega}^T \mathbf{I} \vec{\omega}

3. Principal Axes of Inertia & Diagonalization

Because the inertia matrix $\mathbf{I}$ is real and symmetric, the spectral theorem guarantees the existence of an orthogonal basis of eigenvectors called the principal axes of inertia. In this frame, all products of inertia vanish, and $\mathbf{I}$ becomes purely diagonal:

\mathbf{I} = \begin{pmatrix} I_1 & 0 & 0 \\ 0 & I_2 & 0 \\ 0 & 0 & I_3 \end{pmatrix}

where $I_1, I_2, I_3$ are the principal moments of inertia. The rotational kinetic energy and angular momentum simplify to:

T_{\text{rot}} = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 + \frac{1}{2} I_3 \omega_3^2

\vec{L} = I_1 \omega_1 \hat{e}_1 + I_2 \omega_2 \hat{e}_2 + I_3 \omega_3 \hat{e}_3

§4.4 Euler's Equations of Motion for Rigid Bodies

1. Time Derivatives in Rotating Reference Frames

Let an arbitrary vector $\vec{G}$ be observed in both an inertial space-fixed frame and a body-fixed frame rotating with angular velocity $\vec{\omega}$. The operator relation between time rates of change is:

\left(\frac{d\vec{G}}{dt}\right)_{\text{space}} = \left(\frac{d\vec{G}}{dt}\right)_{\text{body}} + \vec{\omega} \times \vec{G}

2. Derivation of Euler's Dynamical Equations

Newtonian mechanics requires that the rate of change of total angular momentum in the space frame equals the net external torque: $\left(\frac{d\vec{L}}{dt}\right)_{\text{space}} = \vec{N}^{\text{ext}}$. Applying the rotating frame relation:

\left(\frac{d\vec{L}}{dt}\right)_{\text{body}} + \vec{\omega} \times \vec{L} = \vec{N}

Expressing this vector equation along the body's principal axes of inertia where $L_1 = I_1 \omega_1$, $L_2 = I_2 \omega_2$, $L_3 = I_3 \omega_3$:

(\vec{\omega} \times \vec{L})_1 = \omega_2 L_3 - \omega_3 L_2 = (I_3 - I_2) \omega_2 \omega_3

This yields Euler's Equations of Motion:

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

These coupled nonlinear first-order differential equations govern the time evolution of the body-fixed angular velocity components under external torques $N_i$.

§4.5 Torque-Free Motion & The Heavy Symmetrical Top

1. Torque-Free Motion of a Symmetrical Top ($N_i = 0$, $I_1 = I_2 \neq I_3$)

Setting torques to zero and $I_1 = I_2$, Euler's equations become:

I_1 \dot{\omega}_1 = (I_1 - I_3) \omega_2 \omega_3

I_1 \dot{\omega}_2 = -(I_1 - I_3) \omega_1 \omega_3

I_3 \dot{\omega}_3 = 0 \implies \omega_3 = \text{constant}

Defining the constant precession frequency $\Omega_{\text{prec}} = \frac{I_1 - I_3}{I_1} \omega_3$:

\dot{\omega}_1 = \Omega_{\text{prec}} \omega_2, \quad \dot{\omega}_2 = -\Omega_{\text{prec}} \omega_1

Differentiating again yields $\ddot{\omega}_1 + \Omega_{\text{prec}}^2 \omega_1 = 0$. The angular velocity vector $\vec{\omega}$ precesses uniformly around the body symmetry axis ($z'$) with frequency $\Omega_{\text{prec}}$.

2. The Heavy Symmetrical Top with Fixed Pivot

Consider a symmetric top ($I_1 = I_2$) spinning under gravity with its apex supported at a fixed pivot. The Lagrangian in Eulerian angles $(\phi, \theta, \psi)$ is:

L = \frac{1}{2}I_1 (\dot{\theta}^2 + \dot{\phi}^2 \sin^2 \theta) + \frac{1}{2}I_3 (\dot{\phi} \cos \theta + \dot{\psi})^2 - M g l \cos \theta

Coordinates $\phi$ (precession) and $\psi$ (spin) are cyclic:

p_\phi = I_1 \dot{\phi} \sin^2 \theta + I_3 (\dot{\phi} \cos \theta + \dot{\psi}) \cos \theta = \text{const}

p_\psi = I_3 (\dot{\phi} \cos \theta + \dot{\psi}) = I_3 \omega_3 = \text{const}

The nutation angle $\theta(t)$ oscillates between two turning points $\theta_1$ and $\theta_2$ governed by an effective potential.

3. The Sleeping Top Stability Condition

When a top spins vertically upright ($\theta = 0$, the "sleeping top"), small perturbations are stable if and only if the spin angular velocity exceeds the critical threshold:

\omega_3^2 \ge \frac{4 M g l I_1}{I_3^2}

If friction slows $\omega_3$ below this limit, the vertical state undergoes a bifurcation and the top begins to wobble (nutate) wildly.

EXAMINATION PROBLEM SET

Standard University Exam Solved Problems

Solved Problem Example 4.1: Principal Moments of Inertia of a Uniform Solid Cube

Find the inertia tensor of a uniform solid cube of mass $M$ and edge length $a$ about one of its corners as origin, and determine the principal moments of inertia and the orientation of the principal axes.

Step 1: Calculate Moments and Products of Inertia About Corner Origin

$$I_{xx} = \int_0^a dx \int_0^a dy \int_0^a dz \rho (y^2 + z^2) = \frac{M}{a^3} \left( \frac{a^5}{3} + \frac{a^5}{3} \right) = \frac{2}{3} M a^2$$

By symmetry, $I_{xx} = I_{yy} = I_{zz} = \frac{2}{3} M a^2$. The product of inertia is $I_{xy} = -\int_0^a dx \int_0^a dy \int_0^a dz \rho x y = -\frac{M}{a^3} \left(\frac{a^2}{2}\right)\left(\frac{a^2}{2}\right) a = -\frac{1}{4} M a^2$.

Step 2: Construct the Full Inertia Matrix

$$\mathbf{I} = M a^2 \begin{pmatrix} 2/3 & -1/4 & -1/4 \\ -1/4 & 2/3 & -1/4 \\ -1/4 & -1/4 & 2/3 \end{pmatrix}$$

All diagonal elements are $2/3 M a^2$ and all off-diagonal products of inertia are $-1/4 M a^2$.

Step 3: Diagonalize the Matrix to Extract Principal Moments

$$\det(\mathbf{I} - \lambda \mathbf{I}_3) = 0 \implies \lambda_1 = \frac{1}{6} M a^2, \quad \lambda_2 = \lambda_3 = \frac{11}{12} M a^2$$

The eigenvector for $\lambda_1 = \frac{1}{6} M a^2$ points along the main space diagonal $(1, 1, 1)^T$. The other two degenerate eigenvalues $\frac{11}{12} M a^2$ correspond to any orthogonal vectors in the plane perpendicular to the main diagonal.

Final Answer & Physical Insight

Principal moments: I1 = (1/6) M a² (along space diagonal), I2 = I3 = (11/12) M a² (degenerate perpendicular plane).

Solved Problem Example 4.2: Euler's Equations: Intermediate Axis Instability (Tennis Racket Theorem)

An asymmetric rigid body with principal moments of inertia $I_1 < I_2 < I_3$ undergoes torque-free rotation. Use Euler's equations and linear stability analysis to prove that steady rotation about the intermediate axis $I_2$ is dynamically unstable, while rotation about $I_1$ and $I_3$ is stable.

Step 1: Perturb Motion Around the Intermediate Axis 2

$$\vec{\omega} = (\eta_1, \omega_0 + \eta_2, \eta_3), \quad |\eta_i| \ll \omega_0$$

Let the body rotate steadily around principal axis 2 with nominal velocity $\omega_0$, with small perturbations $\eta_1, \eta_2, \eta_3$.

Step 2: Linearize Euler's Equations to First Order in Perturbations

$$I_1 \dot{\eta}_1 = (I_2 - I_3) \omega_0 \eta_3, \quad I_3 \dot{\eta}_3 = (I_1 - I_2) \omega_0 \eta_1$$

To first order, $\dot{\eta}_2 = 0$. Differentiating the first equation: $I_1 \ddot{\eta}_1 = (I_2 - I_3) \omega_0 \dot{\eta}_3 = \frac{(I_2 - I_3)(I_1 - I_2)}{I_3} \omega_0^2 \eta_1$.

Step 3: Evaluate Sign of the Coefficient

$$\ddot{\eta}_1 = \frac{(I_2 - I_3)(I_1 - I_2)}{I_1 I_3} \omega_0^2 \eta_1 = +\Omega^2 \eta_1$$

Since $I_1 < I_2 < I_3$, $(I_2 - I_3) < 0$ and $(I_1 - I_2) < 0$, making their product strictly POSITIVE. The perturbation grows exponentially as $\eta_1(t) \propto e^{+\Omega t}$, proving instability. (For axes 1 or 3, the product is negative, giving stable harmonic oscillation $\ddot{\eta} = -\Omega^2 \eta$).

Final Answer & Physical Insight

Exponential growth rate Ω = ω0 √[(I3 - I2)(I2 - I1) / (I1 I3)] proves the intermediate axis is violently unstable (Dzhanibekov effect).

Solved Problem Example 4.3: Steady Precession Rate of a Fast Heavy Symmetrical Top

A heavy symmetrical top ($I_1 = I_2$, $I_3$) spins with large angular velocity $\omega_3$ about its symmetry axis inclined at angle $\theta$ to the vertical. Derive the slow and fast steady precession rates $\dot{\phi}$ from the torque-free balance equation.

Step 1: Set Up the Nutational Force Balance Equation

$$I_1 \ddot{\theta} = I_1 \dot{\phi}^2 \sin \theta \cos \theta - I_3 \omega_3 \dot{\phi} \sin \theta + M g l \sin \theta = 0$$

For steady precession, nutation is fixed ($\dot{\theta} = 0, \ddot{\theta} = 0$). Dividing by $\sin \theta$ (assuming $\sin \theta \neq 0$) yields a quadratic equation for precession speed $\dot{\phi}$.

Step 2: Solve Quadratic Equation for Precession Speed φ̇

$$I_1 \cos \theta \dot{\phi}^2 - I_3 \omega_3 \dot{\phi} + M g l = 0$$

The roots are $\dot{\phi} = \frac{I_3 \omega_3 \pm \sqrt{I_3^2 \omega_3^2 - 4 I_1 M g l \cos \theta}}{2 I_1 \cos \theta}$.

Step 3: Extract the Fast and Slow Precession Limits for Large Spin ω3

$$\dot{\phi}_{\text{slow}} \approx \frac{M g l}{I_3 \omega_3}, \quad \dot{\phi}_{\text{fast}} \approx \frac{I_3 \omega_3}{I_1 \cos \theta}$$

Expanding the square root for $I_3^2 \omega_3^2 \gg 4 I_1 M g l \cos \theta$: the minus sign gives the ordinary slow gyroscopic precession $\dot{\phi} = \tau / L$, and the plus sign gives the fast inertial precession.

Final Answer & Physical Insight

Slow steady precession: φ̇_slow = M g l / (I3 ω3); Fast precession: φ̇_fast = I3 ω3 / (I1 cos θ).