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Chapter 3 of Classical Dynamics by David Tong. www.damtp.cam.ac.uk/user/tong/dynamics/dynhtml/S3.html. Section 3.5.2 resolves angular velocity in material-fixed axes; sections 3.6–3.6.3 treat the heavy symmetric top, uniform precession and upright spin stability. This is an authored lecture-note chapter, not an answer set for the assigned exam.

Choose Cartesian coordinates with origin at the upper pivot, $x$ to the right and $y$ upward. The two position vectors areThese expressions satisfy the rod holonomic constraints $|{\mathbf{r}}_1|=\ell$ and $|{\mathbf{r}}_2-{\mathbf{r}}_1|=\ell$ automatically. This is the practical advantage of using the two angles as coordinates: the rod tensions need not be unknowns in the motion calculation.^[ref]

Differentiate these position vectors to obtain the velocities. The dot product between their two rod contributions uses $\cos ({\theta }_1-{\theta }_2)=\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2$, givingAdding the two kinetic energies and measuring gravitational potential energy from the downward equilibrium givesThe coefficient $2$ has a geometric meaning: the upper rod raises both masses, and its motion contributes to both kinetic energies. The coupling in $T$ is present because the lower mass has a moving pivot. The zero chosen for $V$ changes $L$ by a constant and leaves the Euler-Lagrange equations unchanged.^[ref]

The principle of stationary action applies to $S=\int Ldt$. Its fixed-endpoint first variation, after integration by parts, isIndependent allowed variations therefore give one Euler-Lagrange equation for each angle. Stationary action alone does not assert a minimum.^[ref]

A helical symmetry combines an axial rotation and a proportional axial translation. A helix with $z=b\theta$ is unchanged as a set by this transformation. If a mechanical Lagrangian is invariant, the Noether conserved quantity for a mechanical point symmetry is $L_z+b{P}_z$, where the two momenta include all moving components. Pure axial rotation and pure axial translation need not separately be symmetries. The helical symmetry of a potential is a related special case in which the constraint is replaced by an invariant potential.^[ref]

Liouville integrability means having $n$ almost-everywhere independent first integrals in mutual Poisson involution.

For a Hamiltonian phase space of dimension $2n$, Liouville integrability means $n$ functionally independent conserved quantities with pairwise vanishing Poisson brackets. This is the Hamiltonian specialization of integrable systems.

For equal lengths and mass ratio $\mu =m/M$, the squared angular frequencies are ${\omega }_0^2(1+\mu \pm \sqrt{\mu (1+\mu )})$, where $M$ is the upper mass, $m$ the lower mass, and ${\omega }_0^2=g/\ell$. The equal-mass case $\mu =1$ admits particularly simple mode shapes.

Use the ideal model of the double pendulum, with both angles measured from the downward vertical. Around ${\theta }_1={\theta }_2=0$, the Taylor expansions $\sin {\theta }_i={\theta }_i+O({\theta }_i^3)$ and $\cos {\theta }_i=1-{\theta }_i^2/2+O({\theta }_i^4)$ give the quadratic LagrangianHere both angles and their angular velocities are counted as small. Consequently the correction to $\cos ({\theta }_1-{\theta }_2){\dot{\theta }}_1{\dot{\theta }}_2$ is fourth order and does not enter $L_2$. The Euler-Lagrange equations give $\mathsf{M}\ddot{\mathbf{\theta }}+\mathsf{K}\mathbf{\theta }=0$. The mass matrix is not diagonal: an angular motion of the upper rod carries the lower mass too.^[ref]

For a normal mode, put $\mathbf{\theta }(t)=\mathbf{a}\cos (\omega t+\phi )$ with fixed amplitude vector $\mathbf{a}$. The resulting generalized eigenvalue problem for small oscillations is $\mathsf{K}\mathbf{a}={\omega }^2\mathsf{M}\mathbf{a}$. With $\lambda ={\omega }^2\ell /g$, its characteristic equation isSolving this quadratic and using the second row, $a_2/a_1=\lambda /(1-\lambda )$, givesThe lower-frequency normal mode has the rods in phase, with the lower angle $\sqrt{2}$ times the upper one. The higher-frequency normal mode has opposite signs, corresponding to an oscillation phase difference of $\pi$. These are ratios of absolute angles. The horizontal displacements are instead $x_1\simeq \ell {\theta }_1$ and $x_2\simeq \ell ({\theta }_1+{\theta }_2)$; their ratios are $1+\sqrt{2}$ and $1-\sqrt{2}$ respectively. Thus the lower mass moves horizontally opposite to the upper mass in the faster normal mode.

Both squared angular frequencies are positive. Their ratio satisfies ${\omega }_{+}/{\omega }_{-}=1+\sqrt{2}$, independently of $m$, $g$ and $\ell$. For a constructed model with $\ell =1\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the angular frequencies are approximately $2.3972$ and $5.7874{\mathrm{s}}^{-1}$, and the oscillation periods $2\pi /{\omega }_{\pm }$ are $2.6211$ and $1.0857\mathrm{s}$. Doubling both masses leaves these unchanged; multiplying both rod lengths by four doubles both oscillation periods.

The eigenvectors are not orthogonal in the ordinary dot product: ${\mathbf{a}}_{-}^{\mathsf{T}}{\mathbf{a}}_{+}=-1$. They satisfy mass-matrix orthogonality of normal modes, sinceWriting $\mathbf{\theta }={\mathbf{a}}_{-}{q}_{-}+{\mathbf{a}}_{+}{q}_{+}$ therefore removes the cross terms from both quadratic energies. The two normal coordinates satisfy ${\ddot{q}}_{\pm }+{\omega }_{\pm }^2{q}_{\pm }=0$. This is why the normal modes are useful: one coupled system becomes two independently evolving harmonic oscillators, and any small release can be decomposed into them. At larger angles, the discarded terms in the double pendulum equations of motion matter and this exact linear superposition no longer applies.

A constant of motion is a function $I(\mathbf{q},\mathbf{p},t)$ whose total derivative vanishes along every physical trajectory. It may depend explicitly on time; arbitrary functions of conserved quantities are also conserved, so a count normally concerns a specified independent set or symmetry generators.

For a fixed-time infinitesimal variation $\delta q_i=ϵ{\xi }_i(\mathbf{q},t)$, suppose $\delta L=ϵdB/dt$. On a solution of the Euler-Lagrange equations,Therefore Noether theorem givesFor a transformation also changing time by $\delta t=ϵ\tau$, the action variation includes $Ld(\delta t)/dt$, and the transformed velocity varies by $ϵ({\dot{\xi }}_i-{\dot{q}}_i\dot{\tau })$. The same integration by parts gives the general point-symmetry chargeThus a time-translation symmetry gives the energy, a spatial translation gives the corresponding total momentum, and a rotation gives its angular-momentum generator. The required symmetry is a symmetry of the action, allowing a boundary term; demanding that the numerical value of the Lagrangian remain strictly unchanged is unnecessarily restrictive. Conversely, explicit time dependence of a Lagrangian generally prevents conservation of its energy even while other charges survive.^[ref]

For torque-free rotation, angular momentum is fixed in inertial space, but angular velocity need not point along it. In principal body axes,and cyclic permutations. A symmetric top with transverse moment $A$ and axial moment $C$ has constant ${\omega }_3$, while ${\omega }_1+i{\omega }_2$ rotates in body coordinates at signed frequency $(C-A){\omega }_3/A$. The symmetry axis precesses around the fixed angular momentum in space at rate $L/A$. These body-frame and space-frame rates describe different motions and need not coincide.

Poinsot construction expresses the same torque-free motion geometrically. The energy ellipsoid $\sum I_i{\omega }_i^2=2T$ is fixed in the body. Its normal at the angular-velocity point is parallel to $\mathbf{L}$, and $\mathbf{L}\cdot \mathbf{\omega }=2T$ places that point in a fixed plane perpendicular to $\mathbf{L}$. The ellipsoid rolls on this invariable plane without slipping: the instantaneous velocity of its contact point is proportional to $\mathbf{\omega }\times \mathbf{\omega }=0$. The contact curve is the polhode on the body ellipsoid and the herpolhode on the plane.^[ref]

For $I_1<I_2<I_3$, linearization about spin $\Omega$ on the third axis givesThe extreme-moment axes are stable; the intermediate one is unstable. The major-axis relaxation theorem adds weak internal dissipation while retaining negligible external torque. At fixed $L$,Energy loss favours rotation about the axis of largest principal moment. Here “major axis” means maximum inertia, not necessarily the body's longest geometric direction. A perfectly rigid torque-free body does not dissipatively settle; its energy is conserved. Distinguishing stability from relaxation also explains why minimum-inertia rotation can be stable in the ideal rigid model.

Forced precession occurs when an external torque changes $\mathbf{L}$, with $\dot{\mathbf{L}}=\mathbf{\tau }$. For steady precession, $\mathbf{\tau }={\mathbf{\Omega }}_p\times \mathbf{L}$. A rapidly spinning, pivoted heavy gyroscope has $L\simeq C{\omega }_3$ along its symmetry axis, givingThe torque is perpendicular to the nearly axial angular momentum, so it mainly changes its direction. For a steady inclined symmetric top, the more complete condition isIt exhibits slow and fast precession branches when real solutions exist. A generic release excites nutation as well: inclination oscillates instead of staying fixed. Axis constraints and bearing forces must be included when specifying which angular momentum and pivot are being used.

A torque-free rigid body has constant angular momentum $\mathbf{L}$ in inertial space. Its angular velocity need not be parallel to $\mathbf{L}$, because $\mathbf{L}=\mathbf{I}\mathbf{\omega }$. In body principal axes, Euler equations for a torque-free rigid body areFor a symmetric top $I_1=I_2=I_{\perp }$, $I_3=I_{\parallel }$, the component ${\omega }_3$ is constant andThis is the signed rotation frequency of the transverse angular-velocity components in the body frame. In space the symmetry axis instead obeysIts angle to $\mathbf{L}$ is constant, with $\cos \theta =I_{\parallel }{\omega }_3/L$. The body-frame and space-frame frequencies of free symmetric-top precession describe different frames. A spherical body has $\mathbf{\omega }\parallel \mathbf{L}$ and no corresponding wobble. For a triaxial body, rotations about the largest and smallest principal-inertia axes are stable, whereas rotation about the intermediate axis is unstable.^[ref]

For generalized coordinates $q_i$ with Lagrangian $L(q,\dot{q},t)$, the conjugate momentum isA coordinate absent from $L$ is cyclic, and its conjugate momentum is conserved. More generally, Noether's theorem connects continuous symmetries of the action to conserved quantities: spatial translations give linear momentum and rotations give angular momentum. If $L$ has no explicit time dependence, the conserved energy function isThe momentum conjugate to an angle is an angular momentum, not a linear velocity. Canonical momentum can also differ from mechanical momentum: for $L=\frac{1}{2}m{v}^2+q\mathbf{A}\cdot \mathbf{v}-q\varphi$, $\mathbf{p}=m\mathbf{v}+q\mathbf{A}$. Thus a conserved conjugate momentum must be identified from the actual Lagrangian and its symmetry, rather than assumed to equal $m\mathbf{v}$. Gauge-dependent canonical momenta require the corresponding gauge choice or gauge-covariant symmetry interpretation.^[ref]

For positions measured from the chosen origin, the inertia tensor isIt relates angular momentum and rotational kinetic energy to angular velocity:The real symmetric tensor has orthogonal principal axes, in which $\mathbf{I}=\mathrm{diag}(I_1,I_2,I_3)$. Its eigenvalues are nonnegative moments of inertia. Rotation about a principal axis has $\mathbf{L}\parallel \mathbf{\omega }$; a general rotation does not. The moment about a unit axis $\mathbf{n}$ is $I_{\mathbf{n}}=\mathbf{n}\cdot \mathbf{In}$.

Changing from the centre of mass to a parallel origin displaced by $\mathbf{a}$ gives the tensor parallel-axis theorem,For example, a uniform solid sphere has $\mathbf{I}=(2M{R}^2/5)1$, while a thin rod along $x$ has principal moments $0,M{l}^2/12,M{l}^2/12$ in the ideal zero-radius limit. The tensor is needed to describe off-axis rotation and torque-free precession; a single scalar inertia only suffices after an axis is specified.^[ref]

A normal mode is a motion in which all coordinates have a fixed relative pattern and oscillate at one frequency. Near a stable equilibrium, let the quadratic kinetic and potential energies be $T={\dot{\mathbf{q}}}^{\mathsf{T}}\mathbf{M}\dot{\mathbf{q}}/2$ and $V={\mathbf{q}}^{\mathsf{T}}\mathbf{Kq}/2$. With positive-definite mass matrix $\mathbf{M}$, the equations and harmonic ansatz giveThe stiffness and mass matrices are symmetric. For distinct frequencies, their symmetry gives $({\omega }_s^2-{\omega }_r^2){\mathbf{a}}_r^{\mathsf{T}}\mathbf{M}{\mathbf{a}}_s=0$, so the mode vectors are orthogonal in the mass inner product. A basis can also be chosen in a degenerate eigenspace. Expanding $\mathbf{q}={\sum }_s{Q}_s{\mathbf{a}}_s$ decouples the equations into ${\ddot{Q}}_s+{\omega }_s^2{Q}_s=0$; initial displacements and velocities determine the coefficients. Positive stiffness gives real frequencies, a zero eigenvalue can describe a free symmetry motion, and a negative one signals instability rather than an oscillatory mode.^[ref]

For two equal masses between fixed walls with three equal springs, the stiffness matrix is $k\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ and the mass matrix is $m1$. The patterns $(1,1)$ and $(1,-1)$ have frequencies $\sqrt{k/m}$ and $\sqrt{3k/m}$. An initial displacement of one mass excites both modes and can produce beats.

For a uniform taut string fixed at both ends, the wave equation ${\partial }_t^{2}y=c_s^2{\partial }_x^{2}y$, with $c_s=\sqrt{\mathcal{T}/{\mu }_{\ell }}$, has patterns and angular frequenciesThe boundary conditions select the spectrum. Fourier expansion of arbitrary admissible initial data provides the mode amplitudes, illustrating the extension from a finite set of coordinates to a field.

For the normal modes of inductively coupled LC circuits, take two identical circuits with self-inductance $L$, capacitance $C$ and mutual inductance $\mathcal{M}$, where $|\mathcal{M}|<L$. The capacitor charges obey $L{\ddot{q}}_1+\mathcal{M}{\ddot{q}}_2+q_1/C=0$ and the exchanged equation. In-phase and opposite-phase charge patterns haveThe positive inductance matrix for two identical coils is the physical counterpart of positive mass. Here inductance plays the role of the mass matrix. Mechanical, wave and electrical examples therefore share the same eigenvalue structure. Weak forcing excites modes according to its spatial overlap with each pattern; damping broadens the resonances, while strong nonlinearity can invalidate a fixed linear-mode decomposition.

The continuous transformation $\theta \mapsto \theta +ϵ$, ${\theta }_B\mapsto {\theta }_B+ϵ$, $z\mapsto z+bϵ$, $r\mapsto r$ moves both endpoints by the same screw motion and preserves the fixed helix. Both $\Delta$ and $u$ are unchanged, so $\delta L=0$. Noether theorem therefore givesThis is the Noether conserved quantity for a mechanical point symmetry applied to helical symmetry. In physical Cartesian terms it equals total $z$ angular momentum plus $b$ times total $z$ linear momentum: the bead contributes $M{a}^2{\dot{\theta }}_B+b(Mb{\dot{\theta }}_B)$. Neither pure axial translation nor pure axial rotation preserves the fixed coil separately. Their conserved combination is a consequence of the helical symmetry, not two independently conserved momenta. Substitution of the four equations of motion verifies $\dot{\mathcal{Q}}=-kar\sin \Delta +k(ar\sin \Delta +bu)-bku=0$.^[ref]