Rigid-Body Mechanics §

$\vec{\omega}$ is the angular velocity vector. The cross-product form encodes that all points rotate about an instantaneous axis. The acceleration adds a centripetal term:

The term $\vec{\omega} \times (\vec{\omega} \times \vec{r})$ is the centripetal acceleration, always directed toward the rotation axis.

Connecting SO(3) to ω. The rotation matrix $\hat{R}(t)$ is a $3\times 3$ orthogonal matrix with determinant 1. The angular velocity vector is the body-frame representation of the time derivative of the rotation:

where the angular velocity vector $\vec{\omega}$ is the vector corresponding to the skew-symmetric matrix $\hat{R}^{\mathsf{T}}\,\dot{\hat{R}}$ as written below:

A common parameterization of $\hat{R}$ is via three Euler angles (e.g., z-y’-z’’ convention), which reduce the three degrees of freedom of SO(3) to three coordinates.

Step 1 — The inertia tensor as a geometric object. The inertia tensor $\mathbf{I}$ is a symmetric $3\times 3$ matrix. Its eigenvalues are the principal moments of inertia; the eigenvectors define the principal axes. The tensor can be visualized as an ellipsoid (the inertia ellipsoid): the surface $\vec{x}^{\mathsf{T}}\mathbf{I}\,\vec{x} = 1$ has semi-axes of length 1/√I₁, 1/√I₂, and 1/√I₃ along the principal directions.

Step 2 — Off-diagonal terms (products of inertia). The components $I_{ij}$ with $i \neq j$ measure how the mass distribution is tilted relative to the chosen axes: they vanish whenever the body has a symmetry plane containing the corresponding axes. When they are nonzero, $\vec{L} = \mathbf{I}\,\vec{\omega}$ is no longer parallel to $\vec{\omega}$ — the body wobbles as it spins. Choosing the principal axes as the coordinate frame diagonalizes $\mathbf{I}$ and removes this coupling.

Step 3 — Worked example: thin rod about its center. Consider a uniform thin rod of mass $m$ and length $L$, lying along the $x$-axis, rotating about the $y$-axis through its center.

1. The linear density is $\lambda = m/L$.
2. A mass element at position $x$ (with $-L/2 \leq x \leq L/2$) has mass $dm = \lambda\,dx$.
3. Its distance to the y-axis is $|x|$, so it contributes $dI = x^2\,dm = \lambda\,x^2\,dx$.
4. Integrating: $I_y = \lambda\int_{-L/2}^{L/2} x^2\,dx = \frac{m}{L}\cdot\frac{L^3}{12} = \tfrac{1}{12}mL^2$.

Step 4 — Useful theorems.

Parallel-axis theorem (tensor form). Shifting the origin by a vector $\vec{d}$:

The scalar form $I' = I + m\,d^2$ is obtained by projecting onto the axis direction.

Perpendicular axis theorem. For a planar lamina lying in the $xy$-plane:

This follows directly from the definition: for a point in the plane, $z = 0$ so $I_z = \int \rho\,(x^2+y^2)\,dA = I_x + I_y$.

Common values:

• Solid sphere about center: $\tfrac{2}{5}mR^2$
• Solid cylinder about axis: $\tfrac{1}{2}mR^2$
• Thin rod about center: $\tfrac{1}{12}mL^2$
• Thin disk about diameter: $\tfrac{1}{4}mR^2$ (by perpendicular axis theorem)

Authoritative sources

• Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 4

In general $\vec{L}$ and $\vec{\omega}$ are not parallel (they are parallel only along principal axes).

Body frame vs. space frame. The body frame is a coordinate system that rotates with the rigid body, aligned with its principal axes. In this frame the inertia tensor $\mathbf{I}$ is constant (diagonal, with entries $I_1, I_2, I_3$). The space frame is inertial. The time derivative of any vector $\vec{A}$ in the space frame relates to its derivative in the body frame by:

Applying this to $\vec{L}$ and using $\vec{L} = \mathbf{I}\,\vec{\omega}$ with constant $\mathbf{I}$ in the body frame gives Euler’s equations:

The terms $(I_j - I_k)\omega_j\omega_k$ arise from the $\vec{\omega} \times \vec{L}$ correction: they vanish when the body is spherically symmetric ($I_1 = I_2 = I_3$) or when rotation is about a single principal axis.

Authoritative sources

• Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 5

Step 1 — Euler’s equations for a symmetric top. For a body with $I_1 = I_2 \neq I_3$ (e.g., a spinning top, a bicycle wheel), Euler’s equations with $\vec{\tau} = 0$ give:

The component $\omega_3$ along the symmetry axis is constant. In the body frame, $\vec{\omega}$ traces a circle at angular rate $\Omega = (I_3 - I_1)\omega_3 / I_1$ around the symmetry axis.

Step 2 — The two-cone picture. In the space frame, $\vec{L}$ is fixed (no torque). The body’s symmetry axis (the 3-axis) precesses around $\vec{L}$. The angular velocity $\vec{\omega}$ lies between $\vec{L}$ and the symmetry axis. The tip of $\vec{\omega}$ traces:

• a cone on the body (the body cone), with half-angle determined by $\omega_3 / \omega$
• a cone in space (the space cone), with half-angle determined by the angle between $\vec{\omega}$ and $\vec{L}$

The body cone rolls without slipping on the space cone. This is the geometric picture of torque-free precession.

Step 3 — Intermediate axis theorem (tennis racket theorem / Dzhanibekov effect). For a triaxial body ($I_1 < I_2 < I_3$), rotation about the axis of largest moment ($I_3$) or smallest moment ($I_1$) is stable, but rotation about the intermediate axis ($I_2$) is unstable. A small perturbation in $\omega_1$ or $\omega_3$ grows exponentially, causing the body to tumble. This was dramatically demonstrated by cosmonaut Vladimir Dzhanibekov, who observed the tumbling of a wing nut in orbit aboard the space station Salyut 7 in June 1985. The theorem follows directly from the linearized Euler equations: the eigenvalues of the perturbation are real (unstable) for the intermediate axis and imaginary (stable oscillation) for the other two.

Step 4 — Gyroscopic stability. A rapidly spinning symmetric body resists changes to its spin axis because $\vec{L}$ is large. An applied torque $\vec{\tau}$ produces a change $d\vec{L} = \vec{\tau}\,dt$ perpendicular to $\vec{L}$, causing precession rather than tipping. This is the principle behind gyroscopes, spacecraft reaction wheels, and the stability of a spinning bicycle wheel.

Authoritative sources

• Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 5

In principal axes: $T_{\text{rot}} = \tfrac{1}{2}(I_1\omega_1^2 + I_2\omega_2^2 + I_3\omega_3^2)$.

The total kinetic energy of a rigid body is the sum of translational (center of mass) and rotational (about center of mass) parts:

This decomposition is exact for any rigid body.

Lagrangian formulation. The Lagrangian of a rigid body in an inertial frame is

where $V$ is the potential energy (e.g., gravitational). Expressed in body-frame Euler angles $(\phi, \theta, \psi)$ with $\mathbf{I}$ diagonal, the Euler–Lagrange equations $\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{q}_i} - \frac{\partial \mathcal{L}}{\partial q_i} = 0$ reproduce Euler’s equations (see the Angular momentum and torque section). The cyclic nature of $\psi$ (the spin angle) gives the conservation of $L_3 = I_3\omega_3$.

Authoritative sources

• Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 5

Forces: gravity component $mg\sin\theta$ down the ramp, friction $f$ up the ramp (provides torque).

Translation: $ma = mg\sin\theta - f$.

Rotation about center: $I\alpha = fR$, with $I = \tfrac{1}{2}mR^2$.

No-slip constraint: $a = \alpha R$.

Substituting: $\tfrac{1}{2}mR^2 \cdot a/R = fR \Rightarrow f = \tfrac{1}{2}ma$.

Then: $ma = mg\sin\theta - \tfrac{1}{2}ma \Rightarrow \tfrac{3}{2}ma = mg\sin\theta \Rightarrow a = \tfrac{2}{3}g\sin\theta$.

For steady precession, the symmetry axis makes a constant angle $\theta$ with the vertical, and $\dot{\theta} = 0$. The angular velocity decomposes as:

where $\hat{e}_3$ is the unit vector along the symmetry axis.

The torque about the fixed point is $\vec{\tau} = m\vec{g} \times \vec{\ell}$ with magnitude $\tau = mg\ell\sin\theta$.

For steady precession, $\vec{L}$ precesses about the vertical at rate $\Omega$. The rate of change of $\vec{L}$ is:

The component of $\vec{L}$ perpendicular to $\hat{z}$ has magnitude $L\sin\theta$, so:

For a rapidly spinning top, $L \approx I_3\,\omega_3$ (the spin dominates). The steady precession rate is:

This shows that faster spin gives slower precession — the gyroscopic effect. For a complete treatment including nutation (oscillation of $\theta$), see the Lagrangian formulation with cyclic coordinates.