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Generalized forces are the coefficients of virtual work in a set of generalized coordinates. A translational coordinate has a force as its coefficient, while an angular coordinate has a torque. A generalized force conjugate to $q_i$ has units of energy divided by the units of $q_i$. Forces not already included in the Lagrangian enter the Euler-Lagrange equations as $d(\partial L/\partial {\dot{q}}_i)/dt-\partial L/\partial q_i=Q_i$.

The time derivative of a generalized coordinate. It is an independent velocity argument of a mechanical Lagrangian, but need not have the units or direction of a Cartesian velocity: for an angular coordinate it is an angular velocity. A regular Legendre transform replaces these variables with the corresponding canonical momenta.

A mechanical configuration specifies the positions and orientations of all parts of a system, subject to its constraints. It is a point of the mechanical configuration space $Q$; generalized coordinates label such points locally. A configuration does not by itself specify velocities, so it generally does not determine the system's full dynamical state. For a bead on a fixed circular hoop, its angle specifies its configuration, while its angular velocity is additional data.

A collection of bodies and specified interactions whose configuration, velocities and evolution are studied in mechanics. Generalized coordinates describe its allowed configurations, while constraints restrict them. A model must specify which bodies and surroundings are included before applying conservation of momentum or conservation of energy.

A nonholonomic constraint is a nonintegrable restriction on allowed velocities: it cannot locally be replaced by relations among generalized coordinates and time. In contrast, a velocity equation $r\dot{\varphi }-R\dot{\theta }=0$ with constant radii integrates to a holonomic constraint $r\varphi -R\theta =C$. Rolling without slipping is therefore not automatically nonholonomic: planar rolling on a prescribed curve can be holonomic, whereas rolling with additional directional freedoms can impose nonintegrable velocity constraints. Unwrapped angles may be needed to record the winding of a rolling trajectory.

Each simple pendulum moves in the vertical plane containing the pivots. Ignoring contact between the ideal rods and point masses, each rod can turn through a full circle. Thus the configuration space is the torusChoose generalized coordinates ${\theta }_1,{\theta }_2$, measured from the downward vertical, positive towards the right and understood modulo $2\pi$. These angles are local coordinates on the torus, rather than a single globally one-to-one real coordinate chart. With upward Cartesian coordinate $y$, the bob positions areTheir velocities satisfy $|{\dot{\mathbf{r}}}_j{|}^2=l^2{\dot{\theta }}_j^2$, so kinetic energy from Cartesian position vectors givesThe spring length is the Euclidean distanceTaking gravitational potential energy zero at pivot height gives $V_g=-mgl(\cos {\theta }_1+\cos {\theta }_2)$. The spring potential energy is $V_s=k(d-r{)}^2/2$, since $r$ is its natural length. Consequently the Lagrangian isIn the paper's notation $t({\theta }_1,{\theta }_2)=V_s$; this function is distinct from the time variable. This is the exact Lagrangian of two spring-coupled pendulums.

One alternative Lagrangian isIt corresponds to setting gravitational potential energy zero at the downward position. The constant disappears from every Euler-Lagrange equation. More generally, total-time-derivative invariance of a Lagrangian allows $\tilde{L}=L+dF({\theta }_1,{\theta }_2,t)/dt$ for any smooth $F$.

Translate the generalized coordinates so that the equilibrium is at zero. Its potential gradient vanishes, and its quadratic potential energy is $q^T\mathsf{V}q/2$, where $\mathsf{V}$ is the symmetric Hessian matrix. The Euler-Lagrange equations for small motion areA normal mode has fixed component ratios, $q(t)=v\cos (\omega t+\varphi )$, withThe angular frequency $\omega$ is its eigenfrequency. A zero mode instead admits a constant or linear time factor. For modes $v_{\alpha },v_{\beta }$, symmetry givesHence mass-matrix orthogonality of normal modes isWithin a degenerate eigenspace, a basis can be chosen orthogonal in this positive weighted inner product.

For three masses connected pairwise, the exact energies areA fixed common translation $x_i\mapsto x_i+a$ preserves all differences and velocities. A fixed rigid rotation $x_i\mapsto Q{x}_i$ preserves their Euclidean norms. Thus both energies have these symmetries. Here the transformation is fixed in time; a moving reference frame would change the kinetic energy. Every equilateral configuration of side $\sqrt{3}$ has zero spring extensions, hence zero force and minimum potential, proving that it is an equilibrium.

For the specified displacement coordinates, writeThe exact potential is $V=(k/2){\sum }_{ij=12,13,23}(d_{ij}-\sqrt{3}{)}^2$. For an equilibrium edge vector $d$ and small perturbation $\delta d$,Thus only each first-order extension is needed for the quadratic potential. These extensions, with an immaterial sign choice for the horizontal edge, areConsequently $V_2=(k/2)(s_1^2+s_2^2+s_3^2)$, with remainder $O(|q{|}^3)$. If $s=Bq$, then $\mathsf{V}=k{B}^{T}B$. Expanding the squares givesThe small-oscillation mass and stiffness matrices are $\mathsf{T}=m{I}_6$ and the displayed $\mathsf{V}$. Hence

The two common translations give zero-mode vectorsAn infinitesimal counterclockwise rotation sends each equilibrium vertex $r_i$ to $J{r}_i$, where $J(x,y)=(-y,x)$. Therefore a third zero-mode vector isEach vector makes all three $s_j$ vanish, so lies in the kernel of a linear map $B$ and therefore in the kernel of $\mathsf{V}$. The rows of $B$ are independent: the $q_4$ and $q_6$ entries first force the coefficients of its last two rows to vanish, then the first coefficient vanishes. Thus $\mathrm{rank}B=3$, and the rank-nullity theorem proves that these are exactly the three zero modes.

For radial motion takeso that $x_i(t)=(1+a(t))r_i$ to first order. The force on mass $i$ is $-ka{\sum }_{j\ne i}(r_i-r_j)=-3ka{r}_i$, since ${\sum }_i{r}_i=0$. Newton's equation therefore gives $m\ddot{a}=-3ka$. Equivalently $\mathsf{V}{v}_b=3k{v}_b$. Hence the breathing normal mode hasAll three masses move radially outward together, then inward together, preserving the triangular shape. The remaining two normal modes of an equilateral spring triangle are shear deformations with squared frequency $3k/(2m)$; together with the breathing mode and three rigid zero modes they account for all six degrees of freedom.

Choose generalized coordinates $q_i$ measured from a static equilibrium point, and assume a conservative system with no velocity-dependent forces. Keeping terms of degree two in displacements and velocities gives the small oscillation LagrangianHere $A$ is the positive-definite symmetric mass matrix evaluated at equilibrium, and $B$ is the symmetric Hessian matrix of the potential energy there. The linear potential term is zero by equilibrium, and an additive constant has been dropped. Coordinate-dependent corrections to the kinetic matrix contribute only at higher order.

The Euler-Lagrange equations are $A\ddot{\mathbf{q}}+B\mathbf{q}=0$. For a normal mode $\mathbf{q}=\mathbf{v}{e}^{i\omega t}$, they give the generalized eigenvalue problem $(B-{\omega }^{2}A)\mathbf{v}=0$. A nonzero mode shape exists exactly whenThis is a degree-$n$ polynomial in ${\omega }^2$. At a stable equilibrium its roots are nonnegative; zero roots correspond to zero modes.

For an ordinary conservative mechanical system, choose generalized coordinates $q_i$ measured from a static equilibrium point. Assume the kinetic energy is quadratic in velocities and there are no velocity-dependent forces. Expanding about equilibrium, the linear term in the potential energy vanishes. Retaining only terms of total degree two gives the small oscillation LagrangianAn irrelevant constant has been dropped. Coordinate dependence of $M$ contributes only terms of third or higher total degree. The matrices $M,K$ are symmetric, $M$ is positive definite for independent coordinates, and $K$ is positive semidefinite at a stable equilibrium. The Euler-Lagrange equations areA normal mode has a fixed mode shape $\mathbf{v}$ and one time factor: $\mathbf{q}=\mathbf{v}Q(t)$. Its angular frequency $\omega$ solves the generalized eigenvalue problem $K\mathbf{v}={\omega }^{2}M\mathbf{v}$, with $\ddot{Q}+{\omega }^{2}Q=0$. Thus $\omega$ is the normal angular frequency, and the frequency in cycles per unit time is $\omega /(2\pi )$. A zero mode satisfies $K\mathbf{v}=0$; its time dependence is $Q=A+Bt$, describing an unopposed displacement or uniform drift rather than a restoring oscillation.

Label the equilibrium vertices in the order $+x,-x,+y,-y,+z,-z$, and write their displacements as ${\mathbf{u}}_i(t)\in {\mathbb{R}}^3$. Their eighteen Cartesian components are generalized coordinates. Linearize the axial Hookean springs about the unstressed equilibrium. For an edge with unit direction ${\mathbf{n}}_{ij}$, its first-order extension is ${\mathbf{n}}_{ij}\cdot ({\mathbf{u}}_i-{\mathbf{u}}_j)$, so the quadratic potential energy isThree zero modes are translations: for each Cartesian unit vector ${\mathbf{e}}_{\alpha }$, take ${\mathbf{u}}_i={\mathbf{e}}_{\alpha }$ at every vertex. Three more are infinitesimal rotations ${\mathbf{u}}_i={\mathbf{e}}_{\alpha }\times {\mathbf{r}}_i$. Explicitly, in the stated vertex order, their displacement vectors areTranslations leave each displacement difference zero. For rotations,because the equilibrium edge vector is parallel to ${\mathbf{n}}_{ij}$. Hence every spring has zero first-order extension in all six patterns, and the stiffness matrix annihilates them.

The translations have nonzero summed displacement, while all the rotations have zero sum. The three rotational patterns are themselves independent because the vertices do not lie on a single line. ThusThese are the rigid-motion zero modes of a spring framework. The rotational displacements describe infinitesimal rigid rotations; an arbitrary finite rotation would require its nonlinear displacement field.

Measure each generalized coordinate ${\theta }_i$ from an unstressed configuration. The diagram has one torsional spring joining the frame to object 1, one joining objects 1 and 2, and one joining objects 2 and 3. The rotational kinetic energy and elastic potential energy areEach spring contribution follows by integrating its restoring torque: ${\int }_0^{\Delta \theta }{k}_{j}sds=k_j(\Delta \theta {)}^2/2$. ThusThe Euler-Lagrange equations giveThese hold for either interpretation of the square's moment of inertia; the geometry-based value from (a) is used first below.

Use the bead angles ${\theta }_1,{\theta }_2,{\theta }_3$ as generalized coordinates. The straight-line spring length between two beads follows from the law of cosines:The hoop is horizontal, so the gravitational potential energy is constant and can be omitted. The kinetic energy is $(m/2){\sum }_i{\dot{\theta }}_i^2$, giving the LagrangianAt ${\theta }_i^0=\Theta +2\pi (i-1)/3$, every spring has its natural length, so the potential energy is zero and its first derivatives vanish. This is an equilibrium for every constant $\Theta$. It is stable against relative displacement because each spring energy is nonnegative; the overall rotation is a neutral direction.

Use the angle $\theta$ from the vertical as the generalized coordinate. For a uniform ladder, its centre of mass is $(l\sin \theta ,l\cos \theta )$, so $v_G^2=l^2{\dot{\theta }}^2$. Its central moment of inertia is $I_G=m(2l{)}^2/12=m{l}^2/3$. With ideal contact constraints that do no work, the Lagrangian, conjugate momentum, and Euler-Lagrange equation areRelease from rest at ${\theta }_0$ also gives ${\dot{\theta }}^2=3g(\cos {\theta }_0-\cos \theta )/(2l)$ by conservation of energy. These results hold while both contacts persist. The ideal wall reaction is normal to the wall; an unspecified tangential wall friction would require an additional generalized force and would prevent determination from the supplied data alone.

Generalized coordinates are variables $q_1,\dots ,q_n$ that specify the allowed instantaneous configuration of a mechanical system. They need not be Cartesian positions: a rigid pendulum's angle is a suitable single coordinate. With constraints solved, the particle positions are functions ${\mathbf{r}}_i(q_1,\dots ,q_n,t)$.

When the block is free, $x$ becomes a dynamical generalized coordinate and its kinetic energy $M{\dot{x}}^2/2$ must be included. Expanding to quadratic order about the downward stationary state and dropping a constant givesThe coordinate $x$ is cyclic. Its conserved canonical momentum is the total horizontal momentum,The two linear Euler-Lagrange equations areEliminating $\ddot{x}$ givesThus the pendulum with a freely sliding support has a zero-frequency translational mode and an oscillatory mode. With four independent constants $X_c,V,C,D$, the general solution isHere $V=P/(M+m)$ and $X_c+Vt$ is the horizontal centre of mass coordinate to first order, since $X_{\mathrm{CM}}=x+[ml/(M+m)]\theta$. The oscillation is faster than that of a pendulum with a fixed pivot; the limit $M\to \infty$ restores $\Omega \to \sqrt{g/l}$.

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.

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.

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

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.