Rigid Body Dynamics

8.1 Rigid Body Kinematics

Definition. A rigid body is a system of particles in which the distance between every pair of particles is fixed.

A rigid body has 6 degrees of freedom: 3 translational (centre of mass position) and 3 rotational (orientation). The orientation is specified by three angles, most commonly the Euler angles $(\phi, \theta, \psi)$ .

8.2 Angular Velocity and Euler Angles

The angular velocity $\boldsymbol{\omega}$ in terms of the Euler angles ( $z$ - $x"$ - $z''$ convention) expressed in the body frame:

$\omega_1 = \dot{\phi}\sin\theta\sin\psi + \dot{\theta}\cos\psi$

$\omega_2 = \dot{\phi}\sin\theta\cos\psi - \dot{\theta}\sin\psi$

$\omega_3 = \dot{\phi}\cos\theta + \dot{\psi}$

8.3 The Inertia Tensor

Definition. The moment of inertia tensor about the centre of mass is:

$I_{jk} = \int \rho(\mathbf{r})\left(\lvert\mathbf{r}\rvert^2 \delta_{jk} - r_j r_k\right) dV$

In matrix form, for a discrete system:

$\mathbf{I} = \begin{pmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} \end{pmatrix}$

Where:

$I_{xx} = \sum_i m_i(y_i^2 + z_i^2), \quad I_{xy} = -\sum_i m_i x_i y_i, \quad \mathrm{etc}.$

Theorem 8.1 (Parallel Axis Theorem). The inertia tensor about a point displaced by $\mathbf{a}$ from the centre of mass is:

$I_{jk}' = I_{jk}^{(\mathrm{CM})} + M(a^2\delta_{jk} - a_j a_k)$

Where $M$ is the total mass.

Proof. Writing $\mathbf{r}_i' = \mathbf{r}_i + \mathbf{a}$ where $\mathbf{r}_i$ is measured from the centre of mass:

$I_{jk}' = \sum_i m_i\left(\lvert\mathbf{r}_i'\rvert^2\delta_{jk} - r_i' r_{i',k}\right) = \sum_i m_i\left(\lvert\mathbf{r}_i + \mathbf{a}\rvert^2\delta_{jk} - (r_{i,j} + a_j)(r_{i,k} + a_k)\right)$

Expanding and using $\sum_i m_i \mathbf{r}_i = \mathbf{0}$ (centre of mass condition):

$I_{jk}' = I_{jk}^{(\mathrm{CM})} + M(a^2\delta_{jk} - a_j a_k)$

$\blacksquare$

8.4 Principal Axes

Definition. The principal axes are the eigenvectors of $\mathbf{I}$ And the principal moments of inertia $I_1, I_2, I_3$ are the corresponding eigenvalues.

In the principal axis frame, the inertia tensor is diagonal:

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

Since $\mathbf{I}$ is a real symmetric matrix, the principal axes are always real and orthogonal.

8.5 Euler’s Equations of Motion

Theorem 8.2 (Euler’s Equations). For torque-free rotation ( $\boldsymbol{\tau} = \mathbf{0}$ ) about the centre of mass, the equations of motion in the body frame are:

$I_1\dot{\omega}_1 = (I_2 - I_3)\omega_2\omega_3$

$I_2\dot{\omega}_2 = (I_3 - I_1)\omega_3\omega_1$

$I_3\dot{\omega}_3 = (I_1 - I_2)\omega_1\omega_2$

Proof. In the body frame, the angular momentum is $\mathbf{L} = \mathbf{I}\,\boldsymbol{\omega}$ . The equation of motion is:

$\left(\frac{d\mathbf{L}}{dt}\right)_{\mathrm{space} = \boldsymbol{\tau}}$

Transforming to the body frame using $\left(\frac{d\mathbf{L}}{dt}\right)_{\mathrm{space} = \left(\frac{d\mathbf{L}}{dt}\right)_{\mathrm{body} + \boldsymbol{\omega} \times \mathbf{L}}}$ :

$\mathbf{I}\,\dot{\boldsymbol{\omega}} + \boldsymbol{\omega} \times (\mathbf{I}\,\boldsymbol{\omega}) = \boldsymbol{\tau}$

In the principal axis frame with $\boldsymbol{\tau} = \mathbf{0}$ This gives Euler’s equations directly. $\blacksquare$

8.6 Stability of Torque-Free Rotation

Theorem 8.3. Torque-free rotation about a principal axis is stable if the axis corresponds to the largest or smallest principal moment of inertia, and unstable for the intermediate axis.

Proof. Consider rotation primarily about the 1-axis: $\boldsymbol{\omega} = (\omega_1, \epsilon_2, \epsilon_3)$ where $\epsilon_2, \epsilon_3$ are small perturbations. From Euler’s equations:

$I_2\dot{\epsilon}_2 = (I_3 - I_1)\omega_1\epsilon_3, \quad I_3\dot{\epsilon}_3 = (I_1 - I_2)\omega_1\epsilon_2$

Combining: $\ddot{\epsilon}_2 = \frac{(I_3 - I_1)(I_1 - I_2)}{I_2 I_3}\omega_1^2\,\epsilon_2$ .

For stability, the coefficient must be negative. This requires $(I_1 - I_3)(I_1 - I_2) \gt 0$ I.e., $I_1$ is either the largest or smallest. If $I_1$ is intermediate, the perturbation grows exponentially. $\blacksquare$

:::caution Common Pitfall The intermediate axis theorem (tennis racket theorem / Dzhanibekov effect) is counterintuitive: a Rigid body spinning about its intermediate axis is unstable. This is not a violation of angular Momentum conservation --- the angular momentum vector remains fixed in space, but the body Tumbles relative to it.

8.7 Worked Example: The Spinning Top

Problem. A symmetric top (moments of inertia $I_1 = I_2 \neq I_3$ ) of mass $M$ spins about its symmetry axis with the tip of the axis fixed. The distance from the fixed point to the centre of mass is $l$ . Find the conditions for steady precession.

Solution

Using Euler angles $(\phi, \theta, \psi)$ where $\theta$ is the tilt from vertical, $\phi$ is the precession angle, and $\psi$ is the spin angle.

The kinetic energy is:

$T = \frac{1}{2}I_1(\dot{\theta}^2 + \dot{\phi}^2\sin^2\theta) + \frac{1}{2}I_3(\dot{\psi} + \dot{\phi}\cos\theta)^2$

The potential energy is $V = Mgl\cos\theta$ .

The Lagrangian is $L = T - V$ . Since $\phi$ and $\psi$ are cyclic:

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

$p_\psi = I_3(\dot{\psi} + \dot{\phi}\cos\theta) = \mathrm{const}$

The quantity $p_\psi = I_3\omega_3$ is the angular momentum component along the symmetry axis. The quantity $p_\phi$ is the angular momentum component along the vertical.

For steady precession at constant $\theta$ and $\dot{\phi} = \Omega$ :

$\dot{\theta} = 0, \quad \dot{\phi} = \Omega = \mathrm{const}, \quad \dot{\psi} = \dot{\psi}_0 = \mathrm{const}$

The equation of motion for $\theta$ gives:

$I_1\Omega^2\sin\theta\cos\theta - I_3(\dot{\psi}_0 + \Omega\cos\theta)\Omega\sin\theta + Mgl\sin\theta = 0$

Dividing by $\sin\theta$ and using $p_\psi = I_3 n$ where $n = \dot{\psi}_0 + \Omega\cos\theta$ :

$I_1\Omega^2\cos\theta - I_3 n\Omega + Mgl = 0$

This is a quadratic in $\Omega$ :

$\Omega_\pm = \frac{I_3 n \pm \sqrt{(I_3 n)^2 - 4I_1 Mgl\cos\theta}}{2I_1\cos\theta}$

Real solutions exist when $(I_3 n)^2 \geq 4I_1 Mgl\cos\theta$ . This is the condition for steady precession. For fast spinning ( $n$ large), $\Omega_\approx \approx I_3 n / (I_1\cos\theta)$ (slow precession) and $\Omega_\approx \approx Mgl / (I_3 n)$ (fast precession). The slow precession is the one observed.

$\blacksquare$

8.8 Worked Example: Gyroscopic Precession

Problem. A bicycle wheel of mass $m$ and radius $R$ is spinning with angular velocity $\dot{\psi}$ about its axle. One end of the axle is supported. Find the precession rate.

Solution

Model the wheel as a symmetric top with $I_3 \approx mR^2$ (thin ring approximation) and $I_1 \approx mR^2/2$ . The axle has length $l$ from pivot to centre of mass.

For a horizontal axle ( $\theta = \pi/2$ ), the steady precession condition becomes:

$I_1\Omega^2 \cdot 0 - I_3 n\Omega + Mgl = 0$

$\Omega = \frac{Mgl}{I_3 n} = \frac{mgl}{mR^2\dot{\psi}} = \frac{gl}{R^2\dot{\psi}}$

This is the gyroscopic precession rate. Notice that it is inversely proportional to the spin rate --- the faster the wheel spins, the slower it precesses.

$\blacksquare$

8.9 Worked Example: Inertia Tensor of a Uniform Rod

Problem. Find the inertia tensor of a uniform thin rod of mass $M$ and length $L$ about one end.

Solution

Place the rod along the $x$ -axis from $x = 0$ to $x = L$ . The linear mass density is $\lambda = M/L$ .

$I_{xx} = \int_0^L \lambda(y^2 + z^2)\, dx = 0$

(since $y = z = 0$ for a thin rod on the $x$ -axis).

$I_{yy} = \int_0^L \lambda(x^2 + z^2)\, dx = \lambda \int_0^L x^2\, dx = \frac{M}{L} \cdot \frac{L^3}{3} = \frac{ML^2}{3}$

$I_{zz} = \int_0^L \lambda(x^2 + y^2)\, dx = \frac{ML^2}{3}$

The products of inertia all vanish by symmetry:

$I_{xy} = I_{xz} = I_{yz} = 0$

The inertia tensor is:

$\mathbf{I} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & ML^2/3 & 0 \\ 0 & 0 & ML^2/3 \end{pmatrix}$

The principal moments about one end are $0$ , $ML^2/3$ , $ML^2/3$ . By the parallel axis theorem, about the centre of mass they would be $0$ , $ML^2/12$ , $ML^2/12$ .

$\blacksquare$

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