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In spherical coordinates, the Lagrangian isA central force exerts zero torque, so the angular momentum vector is constant and the orbit lies in a fixed plane. Choose that plane as $\theta =\pi /2$; then $\ell =m{r}^2\dot{\varphi }$ is conserved. Substituting this relation into the energy givesThe effective potential reduces radial motion to a one-dimensional problem. Allowed radii have $E\ge V_{\mathrm{eff}}$, radial turning points satisfy equality, and a circular orbit occurs at $V_{\mathrm{eff}}^{\prime }=0$. A strict local minimum gives stable radial oscillations. Eliminating $\dot{\varphi }$ directly inside the original Lagrangian would require a Routhian; the energy derivation avoids giving the centrifugal term the wrong sign.

For independent generalized coordinates $q_i$, their generalized velocities ${\dot{q}}_i$ and a Lagrangian $L(q,\dot{q},t)$, the Euler-Lagrange equations areHere the partial derivatives treat $q_i,{\dot{q}}_i,t$ as independent arguments, whereas $d/dt$ follows the motion. For a conservative mechanical system, choose coordinates satisfying the mechanical constraints, express the kinetic energy $T$ and potential energy $V$ in those coordinates, set $L=T-V$, and evaluate these equations to obtain the equations of motion. If nonconservative generalized forces $Q_i$ are present, they appear on the right-hand side.

A cyclic coordinate is absent explicitly from $L$: $\partial L/\partial q_i=0$, even though $L$ may depend on ${\dot{q}}_i$. Its canonical momentum is consequently conserved,This does not require ${\dot{q}}_i=0$ or even that ${\dot{q}}_i$ be constant; the relation between canonical momentum and generalized velocity depends on the Lagrangian.

Model the cart as a particle on a fixed plane curve, remaining constrained to the track. Its vertical velocity is $\dot{y}=y^{\prime }(x)\dot{x}$, so its kinetic energy, gravitational potential energy and Lagrangian areConsequently the canonical momentum conjugate to the horizontal coordinate isIt is not simply the horizontal component $m\dot{x}$ of the mechanical momentum, because a change in $x$ also moves the cart vertically. Differentiate carefully in the Euler-Lagrange equation:Subtracting yieldsThe term involving $y^{\prime \prime }$ accounts for the changing track geometry. An independent check is conservation of energy:

Choose zero potential energy at the equilibrium position. For the simple harmonic oscillator, the kinetic energy and elastic potential energy give the LagrangianIts canonical momentum is $p=\partial L/\partial \dot{x}=m\dot{x}$. The Legendre transform defining the Hamiltonian consequently givesThe Euler-Lagrange equation is $m\ddot{x}+kx=0$. Thus along the motionThis proves conservation of energy. Equivalently, the Lagrangian has no explicit time dependence, so $dH/dt=-\partial L/\partial t=0$. The Lagrangian itself need not be constant: the kinetic energy and potential energy exchange periodically.

Treat the ladder as a uniform slender rigid body. While both contacts remain imposed, its centre of mass is $(l\sin \theta ,l\cos \theta )$, with speed $l|\dot{\theta }|$. Its central moment of inertia is $m(2l{)}^2/12=m{l}^2/3$. Adding translational and rotational kinetic energy givesTaking the floor as zero gravitational potential energy, the Lagrangian, conjugate momentum and Euler-Lagrange equation areThe positive sign is appropriate: gravity increases the angle from the vertical. These are the constrained equations of a frictionless sliding ladder. Real wall contact is unilateral: a wall cannot pull the ladder. Once the computed normal force would become negative, the two-contact model ceases to apply; the question explicitly assumes that contact persists.

Write $\omega$ for the imposed constant angular velocity. In inertial polar coordinates, the bead has $v^2={\dot{r}}^2+{\omega }^2{r}^2$. Its constrained Lagrangian isThe reduced Lagrangian has no explicit time dependence, so its Lagrangian energy function is conserved:This is the Jacobi integral $J=E-\omega L_z$, since the bead's actual inertial energy and axial angular momentum areThe motor maintaining the stick's angular velocity supplies torque and does work. Indeed $\dot{E}=2m{\omega }^{2}r\dot{r}=\omega {\dot{L}}_z$, generally nonzero. A time-independent reduced Lagrangian therefore need not imply conservation of the bead's inertial kinetic energy when its imposed constraint is moving.

For the elastic pendulum, set $r=l+x>0$ and measure $\theta$ from the downward vertical. In the vertical plane, Cartesian coordinates are $X=r\sin \theta$ and $Z=-r\cos \theta$, with $Z$ upward. Differentiating gives ${\dot{X}}^2+{\dot{Z}}^2={\dot{x}}^2+r^2{\dot{\theta }}^2$. Thus the Lagrangian isFor $x$, the Euler-Lagrange equation givesFor $\theta$, the canonical momentum is $p_{\theta }=m(l+x{)}^2\dot{\theta }$. Its Euler-Lagrange equation isor, for positive length,At an equilibrium, all time derivatives vanish, so $kx=mg\cos \theta$ and $\sin \theta =0$. Modulo full rotations, the two formal possibilities areThe lower configuration always has positive length. The upper one is a physical straight-spring configuration only if $l>mg/k$. For example, $l=mg/(2k)$ would give the purported upper state a negative length $-mg/(2k)$; at $l=mg/k$ its zero length makes the angular coordinate singular. Also, the straight compressed spring is an idealization; a real unsupported compressed spring can buckle.

The potential energy is $V=k{x}^2/2-mg(l+x)\cos \theta$. At either formal stationary point its mixed second derivative vanishes, whileBoth curvatures are positive at the lower state, making it a stable equilibrium; the upper state, when it has positive length, has negative angular curvature and is unstable.

To linearize about the stable state, define the small displacement $\xi =x-mg/k$, not $x$ itself, and $r_0=l+mg/k$. Products such as ${\dot{\theta }}^2$, $\dot{\xi }\dot{\theta }$ and $\xi \ddot{\theta }$ are second order, while $\cos \theta =1+O({\theta }^2)$ and $\sin \theta =\theta +O({\theta }^3)$. The first-order equations of motion therefore decouple:There are two independent normal modes. The radial one is vertical simple harmonic motion with angular frequency ${\omega }_r=\sqrt{k/m}$. The transverse one is a simple pendulum oscillation at fixed, statically stretched length $r_0$, with angular frequency ${\omega }_{\theta }=\sqrt{g/r_0}$. Their decoupling is a first-order small-oscillation result, not an assertion about the nonlinear motion of the elastic pendulum.

For ordinary conservative classical mechanics, the Lagrangian is $L(q,\dot{q},t)=T-V$, where $T$ is the kinetic energy and $V$ the potential energy. More generally it is the function whose action $\int Ldt$ satisfies the principle of stationary action. The canonical momenta and Hamiltonian areFor a regular Lagrangian, the first equations can be inverted to express the generalized velocities in terms of $q,p,t$; this defines the Legendre transform on phase space.

Along a solution of the Euler-Lagrange equations, ${\dot{p}}_i=\partial L/\partial q_i$. Taking a total time derivative givesThus an autonomous Lagrangian hasFor the usual time-independent mechanical coordinates and a velocity-independent potential energy, the kinetic energy is quadratic in the generalized velocities, so ${\sum }_i{p}_i{\dot{q}}_i=2T$ and $H=T+V$. The conservation proof itself uses the Euler-Lagrange equations, rather than assuming this special expression for energy.

A cyclic coordinate is absent from the Lagrangian itself: $\partial L/\partial q_i=0$. Its conjugate momentum $p_i=\partial L/\partial {\dot{q}}_i$ therefore obeys ${\dot{p}}_i=0$. In the rolling-wedge calculation, the cyclic horizontal translation gives conservation of the system's horizontal momentum.

For a Lagrangian $L(q_i,{\dot{q}}_i,t)$ expressed in independent generalized coordinates, stationary action gives the Euler-Lagrange equationsIdeal constraints can either be incorporated into the independent coordinates or treated using Lagrange multipliers, as in solution.

The massless inextensible string has hanging length $y=l-r$, measured downward from the table. Thus $\dot{y}=-\dot{r}$. In polar coordinates, the tabletop mass has squared speed ${\dot{r}}^2+r^2{\dot{\theta }}^2$. Adding the two kinetic energies and setting the hanging gravitational potential energy to $-Mgy$ givesand the Lagrangian isThe canonical momenta are $p_r=(M+m)\dot{r}$ and $p_{\theta }=m{r}^2\dot{\theta }$. Performing the Legendre transform gives the HamiltonianIt equals the total mechanical energy and is conserved because $L$ has no explicit time dependence. Also $\theta$ is a cyclic coordinate, so $J=p_{\theta }=m{r}^2\dot{\theta }$ is conserved angular momentum about the vertical through the hole. The table and the string guide supply external forces; conservation of the system's total linear momentum does not follow. These are the two independent generic integrals needed for the reduced tabletop mass coupled to a hanging mass.

Choose the vertical $z$ axis downward, and measure the polar angle $\theta$ from it. With azimuth $\varphi$ and fixed rod length $r$, the Cartesian coordinates areThe massless rod has no kinetic energy. Differentiating these coordinates and adding the squared components of velocity yieldsThe gravitational potential energy is $V=-mgz=-mgr\cos \theta$, up to an arbitrary constant. The Lagrangian of this spherical pendulum is thereforeApplying the Euler-Lagrange equation to the two generalized coordinates givesEquivalently, away from the coordinate singularities at the poles,Since $L$ is independent of $\varphi$, the azimuth is a cyclic coordinate. Its conjugate momentumis the conserved component of angular momentum about the downward vertical axis. The polar angle is not a cyclic coordinate. Using an upward $z$ axis instead reverses the sign of the cosine term in $L$ and of the corresponding gravity term; it describes the same motion with the opposite polar angle convention.

Use upward generalized coordinates. Let $h$ be the displacement of the lower pulley and $z$ that of the $4m$ mass. The two inextensible mechanical strings imposeFor ideal massless pulleys and mechanical strings, only the three masses contribute kinetic energy. HenceUp to a constant, their gravitational potential energy isThe Lagrangian of this compound Atwood machine is therefore

A constant displacement proportional to $(1,-3)$ leaves the velocities unchanged and changes $3x+y$ by zero. It is therefore a continuous symmetry of the Lagrangian. The generalized momentum components areNoether's theorem gives the conserved quantityA direct check using the Euler-Lagrange equations isThe normalization of a symmetry generator is arbitrary. One can take the transformation parameter to have length units and $(K_x,K_y)=(1,-3)$ dimensionless, giving $P$ the units of momentum. Alternatively, for the paper's convention with dimensionless parameter and length-valued $K_i$, use $(K_x,K_y)=(\ell ,-3\ell )$; the conserved quantity is then $\ell P$. The conserved content is identical.

For a Lagrangian $\mathcal{L}=T-U$, the Euler-Lagrange equation is $d(\partial \mathcal{L}/\partial \dot{q})/dt=\partial \mathcal{L}/\partial q$ for each unconstrained generalized coordinate. For example, $\mathcal{L}=m{\dot{x}}^2/2-k{x}^2/2$ gives $m\ddot{x}+kx=0$ for a spring oscillator.

For a uniform solid sphere, slicing into disks normal to the rotation axis givesUse total dumb-bell mass $M$ and $\ell =L-r$. Since both centres lie on its spin axis, its moment of inertia about that axis is $2M{R}^2/5$. If $\theta$ measures displacement from the bottom of the track, rolling without slipping imposes spin speed $|\dot{\varphi }|=\ell |\dot{\theta }|/r$. The rolling inside a fixed circular cylinder therefore hasThe equation of motion and its small-amplitude limit areWith $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and the given lengths, $\boxed{T=2\pi /\omega \simeq 80.5\mathrm{s}}$. If the centre rises by height $h$, conservation of energy at the bottom givesThe speed expression uses exact height-energy conversion within the ideal rolling constraint; only the period formula uses the small-angle approximation.

For conservative forces and ideal constraints that do no work, mechanical energy $E=T+U$ is constant. Work-energy balance remains useful with nonconservative forces: their power gives $dE/dt$. In one dimension,locates accessible regions and turning points, and reduces the trajectory to a quadrature $dt=dq/\sqrt{2(E-U)/m}$. A local minimum of potential is a stable equilibrium; a local maximum is unstable. Near a nondegenerate minimum, $U=U_0+K(q-q_0{)}^2/2+\cdots$ gives ${\omega }^2=K/m$.

For a central force, conserved angular momentum reduces radial motion to an effective potential $U_{\mathrm{eff}}=U(r)+L^2/(2m{r}^2)$. Its extrema determine circular orbits and their radial stability. In several coordinates, small oscillations follow from the kinetic-energy mass matrix $M$ and potential Hessian $K$ through $\det (K-{\omega }^{2}M)=0$.

Energy is also central to the Lagrangian $\mathcal{L}=T-U$ and the Euler-Lagrange equations. A time-independent natural mechanical system has conserved energy; cyclic coordinates yield additional momenta. Energy alone does not normally determine every coordinate of a multi-degree-of-freedom motion. Its value is to expose allowed motion, stability and reductions before solving the full differential equations.

Generalized coordinates incorporate holonomic constraints without solving separately for every constraint force. For conservative mechanics use $L=T-V$. The principle of stationary action makes the action $\int Ldt$ stationary for variations vanishing at the endpoints. Integration by parts givesThe canonical momentum is $p_i=\partial L/\partial {\dot{q}}_i$. A cyclic coordinate gives conserved $p_i$; absent explicit time dependence, ${\sum }_i{p}_i{\dot{q}}_i-L$ is conserved. These are symmetry–conservation relationships, not assumptions that all momenta are constant. Nonconservative generalized forces enter the right side.

For example, a pendulum has $L=m{\ell }^2{\dot{\theta }}^2/2+mg\ell \cos \theta$, giving $m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0$. The method extends directly to coupled coordinates and velocity-dependent interactions when the correct Lagrangian is used; $T-V$ with a position-only potential is not the most general form.

In Lagrangian mechanics, choose independent generalized coordinates $q_i$ compatible with the mechanical constraints, write $L=T-U$, and apply the Euler-Lagrange equationEquivalently, the physical path makes the action stationary under variations with fixed endpoints. For ideal holonomic constraints, their unknown reaction forces need not appear in the reduced equations. Direct use of Newton's laws of motion instead resolves vector forces and reactions and enforces the constraints separately. Both approaches describe the same mechanics; the scalar Lagrangian is often shorter in curvilinear coordinates. Nonconservative generalized forces must be added separately. Here a drive maintains the prescribed azimuthal angular velocity $\Omega$.

The mass has polar speed $a\dot{\theta }$ and azimuthal speed $a\Omega \sin \theta$, so its kinetic energy is $m{a}^2({\dot{\theta }}^2+{\Omega }^2{\sin }^2\theta )/2$. The elastic potential energy is $\mu {\theta }^2/2$, and the height is $a\cos \theta$, giving gravitational potential energy $mga\cos \theta$. Thereforeand the Euler-Lagrange equation givesFor $\mu =2mga$, define $q={\Omega }^{2}a/g$ and $F(\theta )=q\sin \theta \cos \theta -2\theta +\sin \theta$, so $\ddot{\theta }=(g/a)F(\theta )$. Near zero,When $q<1$ the upright equilibrium is linearly stable and the small oscillations have ${\omega }_1^2=g/a-{\Omega }^2$.

For $q>1$, the graph of $F$ initially rises, whereas $F(\pi /2)=1-\pi <0$. There must therefore be a zero inside the interval. More strongly, for $0<\theta <\pi /2$,is strictly decreasing: $\cos \theta$ decreases and $\theta /\sin \theta$ increases because $\sin \theta -\theta \cos \theta >0$. The ratio goes from $q-1>0$ to $1-\pi <0$, giving exactly one zero ${\theta }_0$ in that interval. At the zero, $F^{\prime }({\theta }_0)<0$, so it is a stable equilibrium. The rotating torsion-stabilized inverted pendulum also has a symmetric stable angle $-{\theta }_0$ if both signs of tilt are permitted.

At ${\theta }_0=\pi /4$, setting $F({\theta }_0)=0$ gives $q/2-\pi /2+1/\sqrt{2}=0$, soFor $\theta ={\theta }_0+\xi$, the linearized equation is $\ddot{\xi }=(g/a)F^{\prime }({\theta }_0)\xi$. Since $F^{\prime }=q\cos 2\theta -2+\cos \theta$,At the threshold $q=1$ the linear restoring term vanishes, but the cubic term remains:The upright state remains nonlinearly stable, with no nonzero infinitesimal harmonic frequency. Small finite disturbances produce quartic oscillator motion whose frequency is proportional to its small angular amplitude; its period tends to infinity as that amplitude tends to zero. Zero linear stiffness is not the same as an absent restoring torque.

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]

Normal-mode idea. Near a stable equilibrium, expand the potential energy to quadratic order and the kinetic energy as a quadratic form in generalized velocities. The Euler-Lagrange equations then take the linear form $\mathsf{M}\ddot{\mathbf{x}}+\mathsf{K}\mathbf{x}=0$. A normal mode has fixed displacement ratios, $\mathbf{x}=\mathbf{q}\cos (\omega t+\phi )$, with $\mathsf{K}\mathbf{q}={\omega }^2\mathsf{M}\mathbf{q}$. Real systems admit this description for sufficiently small motion, before higher-order nonlinearities or strong damping matter. Superposition permits a general motion to be assembled from the modes. A free symmetry direction may instead be a zero mode, with constant or linearly changing coordinate.

Lagrangian and bead equations. Take each $x_i$ as an arc-length displacement positive in the same sense around the fixed circular wire. Spring length changes are $x_2-x_1$, $x_3-x_2$ and $x_1-x_3$. Equal equilibrium spring tensions give no net tangential force; their linear energy terms sum to zero. Dropping an irrelevant constant, the Lagrangian isFor example, differentiating the potential with respect to $x_1$ gives $k(2x_1-x_2-x_3)$. The three Euler-Lagrange equations therefore areThe wire supplies normal constraint forces but does no tangential work. These are the equal-mass case of the normal modes of three beads connected by springs on a circular hoop.

Frequencies. The stiffness matrix divided by $k$ isIt annihilates $1$. On the two-dimensional subspace $q_1+q_2+q_3=0$, its action is multiplication by three. Thus the angular frequencies of the normal modes areThe repeated frequency is a degeneracy of normal modes, while the zero frequency expresses free rigid rotation of all beads along the wire.

General solution. Two independent vectors spanning the vibrating subspace are ${\mathbf{q}}_A=(1,-1,0{)}^{\mathsf{T}}$ and ${\mathbf{q}}_B=(1,0,-1{)}^{\mathsf{T}}$. Each amplitude satisfies $\ddot{Q}+{\omega }^{2}Q=0$, with $\omega =\sqrt{3k/m}$. The uniform coordinate satisfies ${\ddot{Q}}_0=0$. HenceThe six freely specifiable real constants $A,\varphi ,B,\psi ,C,D$ give the six initial data; $\omega$ is fixed by $k,m$ and is not an additional independent integration constant. The linear zero-mode term must be included to allow nonzero total tangential momentum.

Shapes and uniqueness. In the first term beads 1 and 2 have opposite equal displacements and bead 3 remains at its equilibrium position. In the second term beads 1 and 3 do this while bead 2 stays fixed. The last term moves all three identically, leaving spring extensions unchanged: $D$ is a rigid phase shift and $C$ a uniform drift around the wire. The sketches show positive modal amplitudes; all arrows reverse half a vibration period later.

The two nonzero-frequency eigenvectors are not unique, because any independent linear combinations in their degenerate two-dimensional eigenspace form an equally valid basis. In particular the displayed pair is independent but not orthogonal; one could use $(1,-1,0)$ and $(1,1,-2)$ instead. The zero eigenspace is one-dimensional, so its vector is fixed up to scale and sign.

Specified release. Zero initial velocities permit $\varphi =\psi =0$ and require $C=0$. At $t=0$, the three displacement conditions become $A+B+D=a$, $-A+D=0$, $-B+D=0$. Therefore one convenient set of constants isEquivalently,Their sum remains $a$, as required by zero initial total tangential momentum. Alternative amplitude signs with compensating phase shifts represent the same motion.