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Beam root rotational compliance is the incremental rotation of a beam's attachment relative to its supporting base per increment of applied root torque. For a linear rotational restraint with stiffness $K_{\theta }>0$,The angle is measured in radians, so the compliance can be reported in $rad/(\mathrm{N}\mathrm{m})$. The root's transverse position is held fixed in this model. The rotational spring boundary condition for a beam implements this finite restraint; a clamped boundary condition is the zero-compliance limit. End moments and section rotations are conjugate quantities in beam virtual work.^[ref]

For an original static derivation, apply a transverse tip force $P$ to a beam of length $L$ and constant beam flexural rigidity $B$. The root moment has magnitude $PL$, giving ${\theta }_0=C_{\theta }PL$. In the small-angle approximation, root rotation contributes $L{\theta }_0$ to tip displacement. Bending relative to that rotated attachment gives the usual cantilever contribution, soThis derivation assumes a linear elastic beam, no root translation and an effectively rigid supporting base. The second term is fixture motion rather than a change in material Young's modulus.

For stipulated $L=0.30\mathrm{m}$, $B=6\mathrm{N}{\mathrm{m}}^2$, $C_{\theta }=0.020rad/(\mathrm{N}\mathrm{m})$ and $P=1\mathrm{N}$, the root rotates through $0.006$ radian. Root motion contributes $1.8\mathrm{mm}$, while bending contributes $1.5\mathrm{mm}$. Attributing the entire $3.3\mathrm{mm}$ to a perfectly clamped beam would infer $B=2.73\mathrm{N}{\mathrm{m}}^2$ instead of the stipulated value six.

A proposed measurement uses a known force and lever arm to apply a root moment, with optical markers on the attachment and on the supporting base to measure their relative angle. A separate displacement channel can detect unwanted root translation. Fit the small-load slope of angle against moment; using only the beam-tip displacement would mix attachment rotation with beam bending. Load reversal tests whether one linear compliance adequately describes the fixture or whether backlash and hysteresis require another model.

Constraint power is the rate of actual work done by constraint forces. For particles with Cartesian velocities ${\mathbf{v}}_i$ and constraint forces ${\mathbf{F}}_i^{\mathrm{c}}$, the total isIt enters the balance of kinetic energy and, after accounting for conservative applied forces, mechanical energy. An ideal mechanical constraint instead imposes zero total virtual work, ${\sum }_i{\mathbf{F}}_i^{\mathrm{c}}\cdot \delta {\mathbf{r}}_i=0$, for allowed virtual displacements at fixed time. Those displacements need not be parallel to actual velocities when the permitted geometry moves.^[ref]

For a smooth ideal restriction $f(\mathbf{r},t)=0$ on one particle, write ${\mathbf{F}}^{\mathrm{c}}=\lambda \nabla f$. Differentiation along its motion gives $\nabla f\cdot \mathbf{v}+{\partial }_{t}f=0$, henceA fixed smooth surface has ${\partial }_{t}f=0$ and zero constraint power. A time-dependent holonomic constraint can have nonzero constraint power even without friction. For example, a bead carried upwards at constant speed $u$ on a horizontal smooth wire receives normal force $mg$ and power $mgu$, increasing its gravitational potential energy. The drive moving the wire supplies that energy.

Zero total constraint power also does not require zero power on every body. Consider two hanging masses connected by a taut, massless, inextensible string over an ideal massless, frictionless pulley. The shared tension is $T$, and the two endpoint speed magnitudes are $v$. If one mass rises while the other falls, the string supplies powers $+Tv$ and $-Tv$: it transfers energy between the bodies while supplying zero net energy to the pair.

With $m_1=0.20\mathrm{kg}$, $m_2=0.30\mathrm{kg}$ and $g=9.81m/{\mathrm{s}}^2$, Newton's second law givesAt $v=0.50m/s$, the two tension powers are $+1.1772\mathrm{W}$ and $-1.1772\mathrm{W}$. Their cancellation leaves gravity's net power, $(m_2-m_1)gv=0.4905\mathrm{W}$, equal to $dK/dt=(m_1+m_2)av$. This constraint power transfer distinguishes an ideal connection that redistributes energy from an externally driven moving support that adds energy to the chosen system.

Constraint power transfer is redistribution of energy between parts of a mechanically connected system through constraint forces. An ideal connection may exert nonzero power on each body while exerting zero total power on the combination. Power on a body is the actual force–velocity scalar product; the zero-virtual work property of an ideal mechanical constraint concerns the sum over admissible virtual displacements.^[ref]

An original ideal rigid link calculation makes the cancellation explicit. Let two endpoint positions be ${\mathbf{r}}_A$ and ${\mathbf{r}}_B$, joined by a massless link of fixed length $\ell$. SetFor axial tension $T$, the link exerts $+T\mathbf{n}$ on A and $-T\mathbf{n}$ on B. The powers supplied to the endpoints areThe endpoint velocities need not be identical: the link can rotate. Only their components along the link must agree.

For stipulated $T=12\mathrm{N}$ and common axial endpoint speed $0.40m/s$, the link supplies $+4.8\mathrm{W}$ to A and $-4.8\mathrm{W}$ to B. It transfers energy without supplying net energy to their combined system. A load cell measuring axial force and tracked endpoint positions measuring velocities would permit these two powers to be compared. External forces determine how the endpoints subsequently accelerate.

Equal and opposite forces alone do not prove power cancellation. If the connector stretches, $\dot{\ell }=\mathbf{n}\cdot ({\mathbf{v}}_B-{\mathbf{v}}_A)$ and the same axial-force calculation givesAn elastic connector can store the corresponding elastic energy; a dissipative connector can convert mechanical energy into heat. Likewise, an externally prescribed moving support can deliver nonzero net constraint power to a subsystem. The location of the system boundary determines whether a measured exchange is internal transfer or externally supplied work.

For a regular material constraint $F(A)=0$, admissible rates of the deformation gradient lie in the kernel of $F_{,A}$. The difference between the nominal stress and the derivative of the strain energy density annihilates that tangent space by virtual work, so it is a Lagrange multiplier times $F_{,A}$. For incompressibility, the derivative of the determinant is its cofactor matrix, and the material-first reaction is $q{A}^{-1}$ when $\det A=1$.

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 $d(\partial L/\partial {\dot{q}}_i)/dt-\partial L/\partial q_i=Q_i$.

An ideal mechanical constraint is a restriction whose constraint forces do zero total virtual work for every admissible virtual displacement at a fixed time:The condition concerns the permitted variations of the whole chosen system. It does not require every constraint force to vanish, and it does not necessarily imply zero actual constraint power. Stewart's treatment distinguishes a displacement along a moving constraint from a virtual displacement made while its time is held fixed.^[ref]

For smooth independent holonomic constraints $f_a({\mathbf{r}}_1,\dots ,{\mathbf{r}}_N,t)=0$, allowed variations obeyConstraint forces of the form ${\mathbf{F}}_i^{\mathrm{c}}={\sum }_a{\lambda }_a{\nabla }_i{f}_a$ therefore satisfy the ideal condition, with Lagrange multipliers ${\lambda }_a$. The argument explains why reaction forces disappear from the virtual-work equation without being absent from the apparatus.

For an original stationary example, constrain a particle of mass $0.20\mathrm{kg}$ to a fixed horizontal circle of radius $0.50\mathrm{m}$. At speed $3m/s$ the required radial force magnitude is $m{v}^2/R=3.6\mathrm{N}$. The allowed virtual displacement is tangential, so its scalar product with that radial normal reaction is zero. The force nevertheless produces the centripetal acceleration needed to maintain the trajectory.

A moving support shows the limitation of equating ideality with energy conservation. For the Cartesian constraint $x=a(t)$, admissible virtual displacements satisfy $\delta x=0$, whereas the actual velocity satisfies $\dot{x}=\dot{a}$. A reaction $F_x$ has zero virtual work but actual power $F_x\dot{a}$. An actuator moving the support can thus supply or remove mechanical energy. A force readout combined with the support's measured speed would detect that exchange.

For a fixed holonomic restriction, actual velocities obey the same homogeneous tangency condition as virtual displacements, giving zero total constraint power. For linked bodies, that total can cancel while individual powers are nonzero, as in constraint power transfer. Friction and compliance require their own model: calling a surface a mechanical constraint does not establish the ideal virtual-work property.

For reference radii $a_0,b_0$, current inner radius $a$ and uniform axial stretch $z>0$, incompressibility integrates to $r^2=a^2+(R^2-a_0^2)/z$. The principal stretches are $r_{,R}=1/(z\lambda )$, $\lambda =r/R$ and $z$. Write $\hat{W}(\lambda ,z)=W((z\lambda {)}^{-1},\lambda ,z)$ and $\mathcal{E}=2\pi {\int }_{a_0}^{b_0}\hat{W}RdR$ per reference length. Virtual work gives $p={\mathcal{E}}_{,a}/(2\pi az)$ and the total axial wall load $N_{\mathrm{wall}}={\mathcal{E}}_{,z}$, both derivatives holding the other displayed parameter fixed. For pressurized closed ends the independently applied axial load is $N_{\mathrm{ext}}=N_{\mathrm{wall}}-\pi a^{2}p$. The distinction follows from the axial pressure power $\pi a^{2}p\dot{z}$.

Mechanical power-port orientation specifies the signs of the conjugate variables used to describe entering power at a mechanical power port. For a translational port with signed force $F$ and velocity $v$, use $p=Fv>0$ for net supply into the chosen system. Reversing the coordinate axis reverses both variables, so $(-F)(-v)=Fv$. Reversing just one variable instead changes the entering-versus-leaving convention and must be stated explicitly. This bookkeeping makes a mechanical supply rate meaningful in a mechanical dissipation inequality.^[ref]

For an element acted on by terminal forces $F,-F$ and terminal velocities $v_1,v_2$,Exchanging terminal labels gives $F^{\prime }=-F$ and $v_1^{\prime }-v_2^{\prime }=-(v_1-v_2)$, preserving power. For $F=100\mathrm{N}$, $v_1=0.030m/s$ and $v_2=0.010m/s$, both descriptions give $2\mathrm{W}$. Taking force from the exchanged convention while retaining the original relative velocity would incorrectly report $-2\mathrm{W}$.

More general coordinate changes require the corresponding conjugate-force change. If physical displacement coordinates satisfy $\mathbf{x}=\mathbf{Tq}$, then $\dot{\mathbf{x}}=\mathbf{T}\dot{\mathbf{q}}$. Equality of virtual work, or instantaneous power, requiresFor the scalar coordinate $q=2x$, the generalized force is $F_q=F_x/2$. Thus $F_x=10\mathrm{N}$ and $\dot{x}=0.10m/s$ correspond to $F_q=5\mathrm{N}$ and $\dot{q}=0.20m/s$, both giving1W. Merely rescaling the motion channel without its conjugate force changes the apparent power. At an ideal connection, equal terminal velocities and opposite entering forces cancel internal port powers, enabling the combined mechanical energy balance.

The moment-free natural boundary condition for a beam is the zero-end-moment condition obtained when an endpoint can rotate freely and no external couple is applied there. For positive beam flexural rigidity $B$, it is $B{w}_{xx}=0$. This differs from prescribing the endpoint beam cross-section rotation.

The simply supported beam gives the full principle of virtual work derivation. At its right support, the admissible virtual displacement $\eta$ obeys $\eta (L)=0$, but ${\eta }^{\prime }(L)$ is arbitrary. The remaining bending boundary contribution is $B(L)w^{\prime \prime }(L){\eta }^{\prime }(L)$. Its vanishing for every permitted end rotation requires $B(L)w^{\prime \prime }(L)=0$. Beam virtual work pairs end moments with rotation variations and forces with displacement variations.^[ref]

An original modified boundary example shows why the missing external couple matters. Define a signed right-end couple $m_L$ by its external virtual work $m_L{\eta }^{\prime }(L)$. The boundary coefficient then becomes $B{w}^{\prime \prime }-m_L$, so the free-rotation condition is $B(L)w^{\prime \prime }(L)=m_L$. With $B=2\mathrm{N}{\mathrm{m}}^2$ and $m_L=0.010\mathrm{N}\mathrm{m}$, the endpoint second derivative is $0.005{\mathrm{m}}^{-1}$, not zero. This sign definition fixes the couple convention explicitly.

Clamping instead sets ${\eta }^{\prime }(L)=0$; that restriction permits an unknown reaction moment rather than forcing it to vanish. Finite rotational restraint leads to a rotational spring boundary condition for a beam. These alternatives distinguish a prescribed motion from its conjugate natural load condition.

Write $B_{i\alpha }={\dot{A}}_{i\alpha }$ and assume that the material constraint is regular, so $F_{,A}\ne 0$. Differentiating the constraint shows that admissible rates of the deformation gradient satisfy $F_{,A_{i\alpha }}{B}_{i\alpha }=0$. The given stress-power identity says that $W_{,A}-N^T$ annihilates this tangent hyperplane. Its orthogonal complement is the one-dimensional span of $F_{,A}$, so a Lagrange multiplier gives the constraint reaction in hyperelastic stress:The sign of $q$ is a convention; the virtual work identity determines no constitutive value for this reaction. Regularity matters: a singular equation describing the same constraint need not have a nonzero gradient.

For incompressibility, take $F=\det A-1$. Differentiating the expression using the Levi-Civita symbol givesThe middle expression is the corresponding entry of the cofactor matrix. Since $\det A=1$, the nominal stress is thereforeFor the neo-Hookean solid, $W_{,A_{i\alpha }}=\mu A_{i\alpha }$, so in the material-first convention $N=\mu A^T+q{A}^{-1}$. The identity matrix satisfies the prescribed dead loading with $q=T-\mu$, proving that the undeformed cube is always an equilibrium.

To obtain the homogeneous tensile bifurcation of a neo-Hookean cube, first seek $A=\mathrm{diag}(\lambda ,\lambda ,{\lambda }^{-2})$, with $\lambda >0$. This has unit determinant. The two distinct diagonal stress equations are $\mu \lambda +q/\lambda =T$ and $\mu {\lambda }^{-2}+q{\lambda }^2=T$. Eliminating $q$ givesThus the nontrivial branch isThe function $h(\lambda )=\lambda +{\lambda }^{-2}$ decreases on $(0,2^{1/3})$ and increases on $(2^{1/3},\infty )$, because $h^{\prime }=1-2{\lambda }^{-3}$. Its minimum is $h(2^{1/3})=3/2^{2/3}$. Above this threshold there are two positive roots ${\lambda }_{-}<2^{1/3}<{\lambda }_{+}$. For $t\ne 2$, neither root is one, and the three choices of the exceptional coordinate for each root give six distinct diagonal deformation gradients, in addition to $I$.

At $t=2$, however, ${\lambda }_{-}=1$ and the three smaller-root matrices coincide with $I$: it would be incorrect to count those as three new deformations. The requested existence of seven distinct homogeneous deformations nevertheless holds even at this value. For any unit vector $n$, putIts positive eigenvalues are ${\lambda }_{+},{\lambda }_{+},{\lambda }_{+}^{-2}$, and its determinant is one. Since $A(n)$ is an orthogonal conjugate of the diagonal solution, the same $q=\mu /{\lambda }_{+}$ gives $\mu A(n{)}^T+qA(n{)}^{-1}=TI$. Choose the six directionsTheir outer products are distinct, and ${\lambda }_{+}>2^{1/3}>1$, so the six $A(n)$ are distinct from one another and from $I$. Constant nominal stress satisfies equilibrium and the prescribed surface tractions. Hence seven distinct homogeneous equilibria exist whenever $T/\mu >3/2^{2/3}$. This is an existence count, not an exhaustive classification: rotated principal axes actually give a continuous family. The conventional seven axis-aligned positive-stretch branches coalesce to four distinct matrices at $T/\mu =2$.

Use $z={\lambda }_z>0$. In cylindrical orthonormal bases the principal stretches are $r_{,R}$, $r/R$ and $z$. Incompressibility givesIntegrate from the inner surface to obtain the inflation and extension of an incompressible tube kinematicsIn particular $b^2=a^2+(b_0^2-a_0^2)/z$. The radial principal stretch is $1/(z\lambda )$, so the reference-volume strain energy density is $\hat{W}(\lambda ,z)$ with the three arguments $(1/(z\lambda ),\lambda ,z)$.

For reference length $L_0$, the stored energy is $L_0\mathcal{E}$, whereLet $N_{\mathrm{wall}}$ denote the total axial resultant acting on the material annulus. The work of the inner pressure on the cylindrical wall is $2\pi az{L}_{0}p\dot{a}$, and axial work is $N_{\mathrm{wall}}{L}_0\dot{z}$. Reversible virtual work therefore saysThe independent derivatives of the circumferential stretch areEquating the coefficients of $\dot{a}$ and $\dot{z}$ provesand the corresponding axial resultant isHere ${\hat{W}}_{,z}$ holds $\lambda$ fixed. If $N$ means the independently applied load on pressurized closed ends, the pressure also does axial end-cap work $\pi a^{2}p{L}_0\dot{z}$. Equivalently its complete work is $pd(\pi a^{2}z{L}_0)/dt$. In that convention,If $N$ includes the end-pressure resultant, it is $N_{\mathrm{wall}}$ itself. Stating this distinction is necessary because the ends are not specified; the pressure formula is the same in both conventions.

For the specified Mooney-Rivlin solid convention, differentiating the constrained strain energy density givesThusPut $c=z{a}^2-a_0^2$. If $c\ne 0$, $z{\lambda }^2-1=c/R^2$ yields $dR/R=-z\lambda d\lambda /(z{\lambda }^2-1)$. The factor cancels becauseThe limits are ${\lambda }_a=a/a_0$ and ${\lambda }_b=b/b_0$, so direct integration givesFor $c=0$, all circumferential stretches equal $z^{-1/2}$ and the original integrand vanishes, so this formula still holds, with $p=0$. This also verifies its continuous limiting value without dividing by zero.

Put $c=a^2-a_0^2$ and use reference cylindrical coordinates $(\rho ,\Theta ,Z)$ and current coordinates $(r,\theta ,z)$. The deformation map is $r=\sqrt{{\rho }^2+c}$, $\theta =\Theta +\alpha Z$, $z=Z$. In the associated orthonormal bases its deformation gradient isIts determinant is one, as required by incompressibility. The radial principal stretch is ${\lambda }_0=\rho /r$. The other two squared principal stretches are the eigenvalues of the tangential-axial block of $F{F}^T$:Writing $K=(r/\rho {)}^2+{\alpha }^2{r}^2+1$, they areThe positive square roots give ${\lambda }_{\pm }$, and ${\lambda }_0{\lambda }_{+}{\lambda }_{-}=1$.

The strain energy density is per reference volume, so the energy per height isThe work rates per height are $M\dot{\alpha }$ from the end torque and $2\pi ap\dot{a}$ from the inner pressure. The outer surface contributes no work because it is traction free, and the fixed height has zero axial work. Thus the virtual work identities for twist and inflation of an incompressible tube are

For a neo-Hookean solid, an irrelevant energy constant can be omitted:Differentiating under the finite integral givesAn antiderivative divided by $\mu$ is $\log \rho -\frac{1}{2}\log ({\rho }^2+c)-c/[2({\rho }^2+c)]+{\alpha }^2{\rho }^2/2$. With $b^2=b_0^2+c$, the neo-Hookean tube pressure under twist and inflation is thereforeIt vanishes for an untwisted undeformed tube, and the twist contributes the last term at fixed radii.

The dynamical method. In Lagrangian mechanics, choose independent generalized coordinates $q_j$ consistent with the constraints, form the Lagrangian $\mathcal{L}=T-V$, and solve the Euler-Lagrange equations,This form applies to a Newtonian system with ideal constraints, so constraint forces do no allowed virtual work, and with the remaining forces derivable from the stated potential. Smooth holonomic constraints can be built into the coordinates. Additional nonconservative generalized forces would appear on the right. Prescribed shaft rotation is a driven constraint; its constant $\Omega$ is a parameter, not a conserved angular momentum.

Energy of the two balls. The diagram labels each ball $m/2$, so their combined mass is $m$. It also shows $\theta$ between an arm and the downward vertical, rather than the full angle between the arms. Each ball has horizontal radius $l\sin \theta$, vertical coordinate $-l\cos \theta$, and squared speed $l^2{\dot{\theta }}^2+l^2{\Omega }^2{\sin }^2\theta$. ThusandThe massless connecting rods and sliding collar enforce the geometry but contribute no energy.

Effective potential. Define ${\omega }_0^2=g/l$. At fixed $\Omega$,This effective potential determines the reduced $\theta$ dynamics. Its conserved effective energy is not the full lab-frame mechanical energy: the shaft drive can exchange work and angular momentum with the balls.

The change of stability. Near $\theta =0$,For $|\Omega |<{\omega }_0$, the downward state is a minimum. At $|\Omega |={\omega }_0$, its quadratic stiffness vanishes but the quartic coefficient is $+{\omega }_0^2/8$. Above that speed, the downward state becomes a maximum and stable nonzero openings appear. This is the gravity-restored centrifugal governor bifurcation.

The physical opening is nonnegative; the two minima in the signed extension represent opposite transverse displacements. The stable opening increases continuously from zero toward a right angle as $|\Omega |$ grows. The apparatus is frictionless in the model, so a displaced system oscillates rather than automatically settling at the equilibrium shown by the minima.

Equilibria. Differentiation givesHence the downward equilibrium is $\theta =0$, and the open equilibria satisfyThe signed extension includes the negative of this angle. Formally $\theta =\pi$ is also stationary, but it is an unstable inverted configuration and lies outside the usual opening range of the mechanism.

Small oscillations. Write $\theta ={\theta }_{\mathrm{eq}}+\eta$. The linear equation is $\ddot{\eta }+{\nu }^2\eta =0$, whereAt the downward state and at a nonzero open state, respectively,These are angular frequencies in ${\mathrm{s}}^{-1}$; ordinary frequencies are $\nu /(2\pi )$. Negative ${\nu }^2$ indicates exponential instability rather than an oscillation. At $\theta =\pi$, ${\nu }^2=-{\omega }_0^2-{\Omega }^2<0$. At the critical speed, the linear frequency is zero and the positive quartic potential gives amplitude-dependent nonlinear restoring motion.

Lagrangian mechanics expresses dynamics through generalized coordinates $q_i$ that respect the constraints. For a system with conservative, velocity-independent forces and standard kinetic energy, the Lagrangian is $L=T-V$. Its equations follow from the principle of stationary action:for variations that vanish at the endpoints. Varying the integral and integrating the $\delta {\dot{q}}_i$ term by parts givesIndependent variations therefore yield the Euler-Lagrange equations:Additional nonconservative generalized forces appear on the right-hand side as $Q_i$. The action is stationary, not necessarily a minimum. For ideal holonomic constraints, using suitable generalized coordinates removes the need to solve explicitly for constraint forces.

The conjugate momentum is $p_i=\partial L/\partial {\dot{q}}_i$. If a coordinate is cyclic, $\partial L/\partial q_i=0$ and its conjugate momentum is conserved. If $L$ has no explicit time dependence, the energy functionis conserved. It equals $T+V$ for the usual quadratic kinetic energy with time-independent geometry, but that identification is not universal for every Lagrangian or constraint.

For example, a simple pendulum with angle $\theta$, length $\ell$ and bob mass $m$ hasThe tension does no virtual work in the allowed angular displacement and never needs to be calculated to obtain the angular equation. The method scales naturally to coupled systems and coordinates with complicated geometry.

A rotational spring boundary condition for a beam lets an endpoint rotate while imposing a restoring torque proportional to that rotation relative to its support. Let $K_{\theta }>0$ be the spring stiffness, with zero rest angle. Its stored elastic energy is $K_{\theta }{\theta }^2/2$, where the small beam cross-section rotation is $\theta =w_x$.

At a right endpoint whose transverse position is fixed, add this energy and the applied-couple potential $-m_L\theta (L)$ to the bending energy. The coefficient of the free rotation variation isHere $B=EI$ is beam flexural rigidity, and $m_L$ is defined by external virtual work $m_L\delta \theta (L)$. End moment and rotation are conjugate variables in the beam principle of virtual work.^[ref] The spring term and the following example are independently derived. At the left endpoint, the bending boundary term has the opposite sign: with no applied end couple, $-B(0)w^{\prime \prime }(0)+K_{\theta }\theta (0)=0$.

For an original fixture model, hold the left endpoint position fixed, restrain its rotation by the spring, and apply a downward end force $P$ at distance $L$. The root moment magnitude is $PL$, giving root rotation $PL/K_{\theta }$. Bending relative to that rotated root contributes $P{L}^3/(3B)$ to the tip displacement; rigid root rotation contributes $P{L}^2/K_{\theta }$. ThusWith $P=1\mathrm{N}$, $L=0.20\mathrm{m}$, $B=2\mathrm{N}{\mathrm{m}}^2$ and $K_{\theta }=20\mathrm{N}m/rad$, these contributions are $1.333$ and $2.000\mathrm{mm}$, giving $3.333\mathrm{mm}$ total. Root rotation is $0.010$ radian, within the stipulated small-rotation model.

As $K_{\theta }$ increases without bound, the root approaches a clamped boundary condition. At zero stiffness this particular end-loaded arrangement becomes a rotational mechanism; it has no finite static deflection for nonzero $PL$. A small-couple fixture check can estimate $K_{\theta }$ from measured moment and angle. The resulting beam root rotational compliance is distinct from deformation of the beam material.

A time-dependent holonomic constraint is a positional restriction $f(q,t)=0$ whose permitted configurations change explicitly with time. Here $q$ denotes the generalized coordinates. A bead confined to a moving wire is an immediate example: its allowed position is specified at each instant, but the wire carries the permitted set of positions through space. The distinction between actual motion and a virtual displacement at fixed time is essential.^[ref]

Differentiating one smooth holonomic constraint along an actual trajectory givesA permitted virtual displacement, however, compares neighboring allowed positions at the same instant:The missing time term is not an approximation. Actual velocity and a tangent virtual displacement obey different conditions when the constraint moves. In the following power calculation, take $q$ to be fixed Cartesian coordinates, so explicit time dependence belongs to the constraint rather than to a moving coordinate system.

For an ideal constraint, its generalized forces may be written $Q_j^{\mathrm{c}}=\lambda \partial f/\partial q_j$, with a Lagrange multiplier $\lambda$. They do zero virtual work, yet their actual constraint power isThus an ideal, frictionless moving constraint can exchange energy with the constrained body. Zero virtual work does not imply zero actual work. The actuator maintaining the imposed motion is part of the physical energy accounting.

For a worked apparatus model, put a bead of mass $m$ on a smooth horizontal wire whose height is prescribed as $z=h(t)$ in a uniform downward gravitational field of magnitude $g$. The wire permits horizontal motion and supplies a vertical normal force $N$. The constraint is $f=z-h(t)=0$, and Newton's second law givesThe bead's mechanical energy, including gravitational potential energy, isbecause horizontal motion is unforced. For $m=0.5\mathrm{kg}$, $g=9.81m/{\mathrm{s}}^2$ and $h(t)=\frac{1}{2}a{t}^2$ with $a=2m/{\mathrm{s}}^2$, the wire exerts $5.905\mathrm{N}$; at $t=1\mathrm{s}$ it delivers $11.81\mathrm{W}$ to the bead. A vertical normal force is perpendicular to every allowed horizontal virtual displacement, but parallel to part of the actual velocity. Measuring wire motion and bead load would expose precisely the energy transfer that is lost by treating the moving wire as a stationary, workless support.

In cylindrical orthonormal bases the deformation gradient is $F=\left(\begin{array}{ccc}\rho /r& 0& 0\\ 0& r/\rho & \alpha r\\ 0& 0& 1\end{array}\right)$ and has unit determinant. One principal stretch is $\rho /r$; the squares of the other two are the roots of $z^2-[(r/\rho {)}^2+{\alpha }^2{r}^2+1]z+(r/\rho {)}^2=0$. The reference-volume energy per height is $E=2\pi \int \rho Wd\rho$. Virtual work gives end torque $M={\partial }_{\alpha }E$ and inner pressure $p=(2\pi a{)}^{-1}{\partial }_{a}E$ at fixed reference radii and height.

A virtual displacement is an infinitesimal change of configuration compatible with the instantaneous mechanical constraints, evaluated at fixed time. It is a variation among admissible configurations, rather than necessarily the displacement made during a time step. In generalized coordinates, writeSubstitution into the applied virtual work identifies the generalized forces conjugate to the coordinates.^[ref]

An original moving-constraint example makes the fixed-time condition visible. Consider a particle constrained to the circleFor a virtual displacement, freeze $t$ and therefore $R$; first-order compatibility gives $x\delta x+y\delta y=0$. For an actual motion, differentiation with respect to time instead givesAt $(x,y)=(R,0)$ with $R>0$, every admissible virtual displacement has $\delta x=0$, while an actual velocity must have $\dot{x}=\dot{R}$. If the circle is expanding at $0.020m/s$, the actual radial velocity is nonzero even though every virtual displacement is tangent to the instantaneous circle.

Consequently, an ideal radial constraint force can have zero virtual work but nonzero instantaneous power during this driven motion. Zero work on admissible variations is not the statement that a moving support never transfers energy. The example is a time-dependent holonomic constraint; the variations satisfy its instantaneous linearized condition. In a beam, the same distinction lets one impose zero endpoint virtual displacement while leaving the virtual endpoint slope free, as in the moment-free natural boundary condition for a beam.

Virtual work is the sum of applied forces dotted with infinitesimal displacements compatible with the instantaneous mechanical constraints. The variation is taken at fixed time; it need not be a displacement the system actually makes during a short time interval. In generalized coordinates it has the formwhich defines the generalized forces $Q_j$.

For a bead on a smooth fixed wire, an admissible virtual displacement is tangent to the wire. The normal reaction is perpendicular to it and contributes no virtual work, allowing that unknown reaction to be eliminated from the equilibrium condition. For such ideal constraints, static equilibrium requires the applied virtual work to vanish for every admissible virtual displacement. The same handling of ideal constraint reactions is useful when deriving the Euler-Lagrange equations.