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Regular constraints and moving frames — Examples

1 · Prerequisites

2 · Summary

Worked checks and model comparisons for cm06-regular-constraints-and-moving-frames. A bead on a uniformly rotating straight wire: the rheonomic ideal constraint, its reaction, and the work it performs declares a bead on a frictionless straight wire that rotates uniformly in a horizontal plane under a uniform downward field, verifies that the constraint datum is regular and rheonomic, computes the unique maximal constrained motion with exponential-radius formula, the multiplier $(2 m Ω r^{'}, m g)$ and the reaction, checks the reaction power $2 m Ω^{2} r r^{'}$ against the mechanical-energy balance, and exhibits on explicit model data that the energy strictly increases by the ideal reaction's work. The planar pendulum as an ideal holonomic model: the multiplier, the tension and the vanishing reaction power of a scleronomic constraint works the planar ideal pendulum: the regular scleronomic circle datum, the multiplier, the reaction read as the tension $T = m (g \cos θ + l θ^{' 2})$ , the tangential projection giving the pendulum equation, the converse check that every solution of that equation is an ideal constrained motion, the identically vanishing reaction power, the conserved energy, the formula $T = 3 m g \cos θ + 2 E / l$ and its evaluation on explicit data. All four inertial terms checked on explicit data, and the exact return to an inertial chart evaluates all four inertial terms on explicit data for a zero-force motion and a chart with quadratic rotation angle and accelerating reference curve, compares their sum with the moving-chart coordinate acceleration, identifies the chart as non-inertial by the criterion, and checks an affine chart with constant rotation against the exact return to an inertial frame. Rolling as distinct models: the planar ideal no-slip disk against frictionless sliding, and the contact capability condition compares two declared reduced models of a rolling disk — the ideal no-slip model with its holonomic constraint and ideal reaction and the frictionless sliding model — with the same kinetic matrix and applied force, computing the accelerations, the multiplier, the required contact force, the vanishing reaction power and the static-friction capability condition, with explicit model data.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passOpen item page →

A bead on a uniformly rotating straight wire: the rheonomic ideal constraint, its reaction, and the work it performs

Physical scope

A worked model check of the rheonomic ideal-constraint branch: one label of mass m constrained without friction to a straight wire through the origin that rotates uniformly in a horizontal plane, under the declared uniform downward field; the verification exhibits the unique maximal constrained motion through consistent data, computes the multiplier and the reaction, verifies the reaction-power identity and the mechanical-energy balance, and shows on explicit model data that an ideal rheonomic constraint performs positive work. The example is a model computation with declared idealizations (frictionless bead, massless rigid wire, prescribed uniform rotation, uniform field with the wire supporting the weight); every number is a model value, not a measurement.

Example

Declare a one-label model of the force branch of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with $N = 1$ , mass $m > 0$ , time window $I_{0} = R$ , configuration domain $Q = R^{3}$ and force map $F : I_{0} \times Q \times R^{3} ⟶ R^{3}, F (t, x, v) : = (0, 0, - m g),$ where $[m] = μ$ and $g > 0$ has dimension $ℓ τ^{- 2}$ (the uniform downward field; conventions of Dimensions and SI unit conventions of the non-relativistic classical mechanics framework). Constrain the label, without friction, to the straight wire through the origin that rotates in the plane $z = 0$ about the $z$ -axis with constant angular speed $Ω > 0$ , and to that plane, by the regular holonomic-constraint datum $O : = R \times R^{3}, k : = 2, G (t, x, y, z) : = (- x \sin Ω t + y \cos Ω t, z)$ of Holonomic-constraint models of the force branch: constraint data, the constrained equation, virtual displacements and the scleronomic/rheonomic distinction, and adopt the ideal-reaction law of Adopted ideal-reaction law for holonomic constraints: the d'Alembert condition, its scope and its limits. Both components of $G$ have length dimension $ℓ$ , and $[Ω] = τ^{- 1}$ .

Let $r_{0} > 0$ be a length and $u_{0} \in R$ the initial radial speed, with $[r_{0}] = ℓ$ and $[u_{0}] = ℓ τ^{- 1}$ . The initial data are $t_{0} : = 0, x_{0} : = (r_{0}, 0, 0), v_{0} : = (u_{0}, Ω r_{0}, 0) .$ Then the constrained model with these data has the unique maximal constrained motion $r (t) : = A e^{Ω t} + B e^{- Ω t}, A : = \frac{r_{0} + u_{0} / Ω}{2}, B : = \frac{r_{0} - u_{0} / Ω}{2}, x (t) : = r (t) (\cos Ω t, \sin Ω t, 0),$ defined for all $t \in R$ , with multiplier, reaction and reaction power $λ (t) = (2 m Ω r^{'} (t), m g), R (t) = 2 m Ω r^{'} (t) (- \sin Ω t, \cos Ω t, 0) + m g e_{3}, R (t) \cdot x^{'} (t) = 2 m Ω^{2} r (t) r^{'} (t),$ kinetic energy $K (t) = \frac{1}{2} m (r^{'} (t)^{2} + Ω^{2} r (t)^{2})$ and mechanical energy $E (t) = K (t) + m g z (t) = K (t)$ satisfying $E^{'} (t) = R (t) \cdot x^{'} (t) = - λ (t) \cdot G_{t} (t, x (t)) = 2 m Ω^{2} r (t) r^{'} (t)$ ; the constraint is rheonomic, and the scleronomic conservation clause does not apply. On the explicit model data $m = 1 k g, Ω = 2 s^{- 1}, r_{0} = 1 m, u_{0} = 0 m s^{- 1}, g = 10 m s^{- 2}$ one has $r (t) = (e^{2 t} + e^{- 2 t}) / 2 m$ , $E (t) = e^{4 t} + e^{- 4 t} J$ and $E^{'} (t) = 4 (e^{4 t} - e^{- 4 t}) W$ ; at $t = \frac{1}{2} s$ , $E = (e^{2} + e^{- 2}) J \approx 7.524391 J$ and $E^{'} = 4 (e^{2} - e^{- 2}) W \approx 29.014883 W$ , so the mechanical energy has strictly increased from its initial value $E (0) = 2 J$ by the work of the ideal reaction. All values are model values under the declared idealizations, not measurements.

Facts & Assumptions

Given: The constants $m, g, Ω, r_{0} > 0$ , the scalar initial radial speed $u_{0} \in R$ , the vector initial velocity $v_{0} = (u_{0}, Ω r_{0}, 0)$ , the model data of the Example, the datum $(O, 2, G)$ , and the curve $x (t) = r (t) (\cos Ω t, \sin Ω t, 0)$ with $r$ as displayed.

[F1]

The force branch and the mass operator: $F : I_{0} \times Q \times R^{3} \to R^{3}$ is the constant map $F (t, x, v) = (0, 0, - m g)$ , hence continuous, state-Lipschitz and of class $C^{1}$ ; the mass operator is $M = m I_{3}$ , symmetric positive definite (Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance, Configuration domain of $N$ labeled particles: open admissible sets, regular holonomic constraints and the mass operator).

[F2]

The holonomic-constraint branch (Holonomic-constraint models of the force branch: constraint data, the constrained equation, virtual displacements and the scleronomic/rheonomic distinction): $G$ is of class $C^{2}$ with $D_{x} G$ of rank $k$ on $Σ$ ; virtual displacements are $\ker D_{x} G (t, x)$ ; the datum is scleronomic exactly when there is $N \subseteq R^{3 N}$ with $Σ = {(t, x) \in O : x \in N}$ , equivalently when $(t, x) \in Σ$ and $(t^{'}, x) \in O$ always imply $(t^{'}, x) \in Σ$ , and then $G_{t} = 0$ on $Σ$ ; a constrained motion satisfies $M x^{''} = F + R$ with a continuous $R$ .

[F3]

The ideal-reaction law and the balance theorem (Adopted ideal-reaction law for holonomic constraints: the d'Alembert condition, its scope and its limits, Ideal holonomic constraints determine the reaction: multiplier form, explicit acceleration, unique constrained motions and the reaction-power identity): along an ideal motion $R = D_{x} G^{T} λ$ with the unique continuous multiplier; with $C = D_{x} G M^{- 1} D_{x} G^{T}$ and $Λ$ of that theorem, the multiplier of a constrained motion is $Λ (t, x, x^{'})$ and the constrained motions with consistent data are precisely the solutions of the resulting first-order system, which has a unique maximal solution through each consistent datum.

[F4]

The rheonomic-power theorem (Reaction work and mechanical-energy balance for ideal holonomic constraints: vanishing power for scleronomic constraints and the rheonomic work term): $R \cdot x^{'} = - λ \cdot G_{t}$ , and for a potential presentation $F_{i} = - \nabla_{x_{i}} U + d_{i}$ , $E^{'} = \partial_{t} U + \sum_{i} ⟨ d_{i}, w_{i} ⟩ - λ \cdot G_{t}$ .

[F5]

Trigonometric and exponential calculus: the derivatives of sine and cosine, the identity $\sin^{2} + \cos^{2} = 1$ , the derivative of the exponential, the chain rule, the algebra of derivatives, and the elementary antiderivative facts used for the explicit solution (Sine and cosine defined by their real power series, The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine, The real exponential function and the number $e$ by a power series, The exponential function is smooth and $(\exp)^{'} = \exp$ , The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral).

[F6]

Kinetic and mechanical energy of a one-label curve family and the potential presentation $U (t, x, y, z) = m g z$ , $d \equiv 0$ (Total mass, total momentum, centre of mass, angular momentum, torque, kinetic energy, power and work of a particle model in an inertial frame clauses (b),(d), Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets clauses (a),(b)); the Euclidean pairing (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ); intervals and constants (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant).

Verification

technique · direct; verify the constraint, the consistency of the data and the equation of motion on the explicit curve, then identify the multiplier, the reaction power and the energy balance, and evaluate the finite model data

1.1F1F2F5algebra

The datum is a regular holonomic-constraint datum with $k = 2 = 3 N - 1$ : $G$ is smooth, and its Jacobian is the constant-rank matrix $D_{x} G (t, x, y, z) = (\begin{array}{ccc} - \sin Ω t & \cos Ω t & 0 \\ 0 & 0 & 1 \end{array});$ the two rows are independent for every $t$ because a combination $a (- \sin Ω t, \cos Ω t, 0) + b (0, 0, 1) = (0, 0, 0)$ has $b = 0$ from the third entry and then $a = 0$ from the first two entries, since $(- \sin Ω t, \cos Ω t) \neq (0, 0)$ by [F5]. The locus is $Σ = {(t, r \cos Ω t, r \sin Ω t, 0) : t, r \in R}$ , nonempty and contained in $I_{0} \times Q$ , and the virtual displacements at $(t, x)$ are the one-dimensional space $R (\cos Ω t, \sin Ω t, 0)$ because $δ \in \ker D_{x} G$ means $δ_{z} = 0$ and $- δ_{x} \sin Ω t + δ_{y} \cos Ω t = 0$ . Finally the datum is not scleronomic: $(0, (1, 0, 0)) \in Σ$ because $G (0, 1, 0, 0) = (0, 0)$ , while for $t_{1} : = π / (2 Ω)$ one has $G_{1} (t_{1}, 1, 0, 0) = - \sin (π / 2) = - 1 \neq 0$ , so $(t_{1}, (1, 0, 0)) \in O = R \times R^{3}$ but $(t_{1}, (1, 0, 0)) \notin Σ$ ; by the membership form of [F2] the datum is rheonomic.

1.2F2givenalgebra

Consistency of the data: $G (0, x_{0}) = (0, 0)$ and, with the declared vector $v_{0} = (u_{0}, Ω r_{0}, 0)$ , $G_{t} (0, x_{0}) + D_{x} G (0, x_{0}) v_{0} = (- Ω r_{0}, 0) + (Ω r_{0}, 0) = (0, 0) .$ Thus $(0, x_{0}, v_{0})$ is consistent.

2.1F1F2F3F5step 1.1algebra

The curve $x$ is a constrained motion with the stated multiplier: for $r (t) = A e^{Ω t} + B e^{- Ω t}$ one has $r^{''} = Ω^{2} r$ by [F5], and the displayed $x$ is $C^{\infty}$ with $x^{'} = r^{'} (\cos Ω t, \sin Ω t, 0) + r Ω (- \sin Ω t, \cos Ω t, 0), x^{''} = (r^{''} - Ω^{2} r) (\cos Ω t, \sin Ω t, 0) + 2 Ω r^{'} (- \sin Ω t, \cos Ω t, 0) = 2 Ω r^{'} (- \sin Ω t, \cos Ω t, 0) .$ Moreover $G (t, x (t)) = (- r \cos Ω t \sin Ω t + r \sin Ω t \cos Ω t, 0) = (0, 0)$ , and $G_{t} (t, x (t)) + D_{x} G (t, x (t)) x^{'} (t) = (- Ω r, 0) + (r Ω, 0) = (0, 0)$ , so $x$ is a constrained curve family with permitted velocity. With $λ (t) : = (2 m Ω r^{'} (t), m g)$ , the reaction $R : = D_{x} G^{T} λ$ is $2 m Ω r^{'} (- \sin Ω t, \cos Ω t, 0) + m g e_{3}$ , and $F + D_{x} G^{T} λ = (0, 0, - m g) + 2 m Ω r^{'} (- \sin Ω t, \cos Ω t, 0) + (0, 0, m g) = m (2 Ω r^{'} (- \sin Ω t, \cos Ω t, 0)) = m x^{''} = M x^{''}$ ; since $R = D_{x} G^{T} λ$ annihilates $\ker D_{x} G$ by [F3], the reaction is ideal and $x$ is a constrained motion of the model with reaction $R$ .

3.1F1F3step 1.1step 1.2step 2.1algebra

The multiplier of step 2.1 is the multiplier $Λ$ of the balance theorem and the motion is the unique maximal one: here $C = D_{x} G M^{- 1} D_{x} G^{T} = (1 / m) I_{2}$ because the two rows of $D_{x} G$ are orthonormal, $D_{x} G M^{- 1} F = (1 / m) (0, - m g) = (0, - g)$ , $G_{t t} = (- Ω^{2} G_{1}, 0) = (0, 0)$ along the motion, $2 G_{t q} [x^{'}] = (- 2 Ω (\cos Ω t x_{1}^{'} + \sin Ω t x_{2}^{'}), 0) = (- 2 Ω r^{'}, 0)$ and $G_{q q} [x^{'}, x^{'}] = 0$ ; hence $Λ = - C^{- 1} [(0, - g) + 0 + (- 2 Ω r^{'}, 0) + 0] = (2 m Ω r^{'}, m g) = λ$ , so by [F3] clause 1 the multiplier of the motion is exactly $λ$ . Here $O^{'} = O = I_{0} \times Q$ , $G$ is smooth with rank $2$ throughout $O$ by step 1.1, and $F$ is $C^{1}$ by [F1], so all hypotheses of [F3] hold. The data of step 1.2 are consistent, the curve solves its first-order system, and a solution defined on all of $R$ is the maximal solution through its datum by uniqueness.

4.1F4F5F6step 1.1step 3.1algebra

Reaction power and energy balance: on the motion, $e_{3} \cdot x^{'} = 0$ and $(- \sin Ω t, \cos Ω t, 0) \cdot x^{'} = r Ω$ , so $R \cdot x^{'} = 2 m Ω r^{'} \cdot r Ω = 2 m Ω^{2} r r^{'}$ , and $G_{t} (t, x (t)) = (- Ω r, 0)$ gives $- λ \cdot G_{t} = - (2 m Ω r^{'}) (- Ω r) = 2 m Ω^{2} r r^{'}$ , in agreement with [F4] clause 1. Since $F = - \nabla U$ for $U (t, x, y, z) = m g z$ with residual forces zero and $\partial_{t} U = 0$ , [F4] clause 2 gives $E^{'} = \partial_{t} U - λ \cdot G_{t} = 2 m Ω^{2} r r^{'}$ . Independently $K^{'} = m (r^{'} r^{''} + Ω^{2} r r^{'}) = m (Ω^{2} r r^{'} + Ω^{2} r r^{'}) = 2 m Ω^{2} r r^{'}$ by [F5], and $z \equiv 0$ , so $E^{'} = K^{'} = R \cdot x^{'}$ . Because the datum is rheonomic (step 1.1), clause 3 of [F4] is not available and the mechanical energy is not asserted constant.

5.1F5F6step 4.1algebra

Finite model data: with $m = 1$ , $Ω = 2$ , $r_{0} = 1$ and $u_{0} = 0$ one has $A = B = \frac{1}{2}$ , so $r (t) = (e^{2 t} + e^{- 2 t}) / 2$ , $r^{'} (t) = (e^{2 t} - e^{- 2 t})$ , $K (t) = \frac{1}{2} (r^{' 2} + 4 r^{2}) = e^{4 t} + e^{- 4 t}$ and $E (t) = K (t) J$ because $z (t) = 0$ ; then $E^{'} (t) = 4 (e^{4 t} - e^{- 4 t}) W$ , while the reaction power at the same time is $2 m Ω^{2} r r^{'} = 8 r (t) r^{'} (t) = \frac{1}{2} \cdot 8 (e^{2 t} + e^{- 2 t}) (e^{2 t} - e^{- 2 t}) = 4 (e^{4 t} - e^{- 4 t})$ , and the two agree. At $t = \frac{1}{2} s$ the values are $E = (e^{2} + e^{- 2}) \approx 7.524391 J$ , $E^{'} = 4 (e^{2} - e^{- 2}) \approx 29.014883 W > 0$ , and $E (0) = 2 J$ ; the energy has increased by the work of the ideal reaction.

6.1step 1.1step 1.2step 2.1step 3.1step 4.1step 5.1∎

Clauses (1)-(4) of the Example are proved: the datum is regular and rheonomic, the data are consistent, the displayed curve is the unique maximal constrained motion with the stated multiplier and reaction, the reaction power is $2 m Ω^{2} r r^{'}$ , the mechanical-energy balance holds with the rheonomic term and no other contribution, and the finite model data exhibit a strict energy increase. The construction is an idealization of a bead threaded on a smooth rigid wire whose rotation is prescribed: friction, wire flexibility, driving details and the ability of the wire to supply the computed reaction are not modelled, and the numbers are model values, not measurements of any apparatus.

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passOpen item page →

The planar pendulum as an ideal holonomic model: the multiplier, the tension and the vanishing reaction power of a scleronomic constraint

Physical scope

A worked model check of the holonomic ideal-constraint branch: a mass point constrained without friction to a circle of radius l in a vertical plane about a fixed pivot under a declared uniform downward field, realized as a bead on a smooth circular wire, with the explicit two-component multiplier (the component tied to the constraining plane vanishing along the motion), the reaction read as the tension along the link, the standard pendulum equation obtained by projecting the constrained equation, the identically vanishing reaction power of the scleronomic constraint, and the conservation of the mechanical energy on explicit model data. The example is a model computation with the declared idealizations (smooth massless wire, equivalently a rigid link or a taut string while it stays taut; frictionless pivot; uniform field); every number is a model value, not a measurement, and the empirical status of a string that can go slack (or of a link that can buckle) is recorded as a limitation of the model.

Example

Declare a one-label model of the force branch of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with $N = 1$ , mass $m > 0$ , time window $I_{0} = R$ , configuration domain $Q = R^{3}$ (coordinates $(x, y, z)$ , dimensions of Dimensions and SI unit conventions of the non-relativistic classical mechanics framework) and force map $F : I_{0} \times Q \times R^{3} ⟶ R^{3}, F (t, x, v) : = (0, - m g, 0),$ where $[m] = μ$ and the constant $g > 0$ has dimension $ℓ τ^{- 2}$ . Let $l > 0$ have dimension $ℓ$ , and use a dimensionless angle $θ$ in radians. Constrain the label, without friction, to the circle of radius $l > 0$ about the origin in the plane $z = 0$ --- the ideal pendulum realized as a bead threaded on a smooth circular wire in a vertical plane --- by the regular holonomic-constraint datum $O : = {(t, x, y, z) \in R^{4} : (x, y) \neq (0, 0)}, k : = 2, G (t, x, y, z) : = (x^{2} + y^{2} - l^{2}, z)$ of Holonomic-constraint models of the force branch: constraint data, the constrained equation, virtual displacements and the scleronomic/rheonomic distinction, and adopt the ideal-reaction law of Adopted ideal-reaction law for holonomic constraints: the d'Alembert condition, its scope and its limits. The two constraint components have dimensions $ℓ^{2}$ and $ℓ$ . The locus is $Σ = {(t, l \sin θ, - l \cos θ, 0) : t, θ \in R}$ ; write $X (θ) : = (l \sin θ, - l \cos θ, 0), X_{θ} (θ) : = \partial_{θ} X (θ) = (l \cos θ, l \sin θ, 0), u (θ) : = (\sin θ, - \cos θ, 0), X (θ) = l u (θ) .$

Then:

the datum is regular and scleronomic, the tangent vector $X_{θ} (θ)$ is permitted for every $θ$ , and the virtual displacements at $X (θ)$ are the line $R X_{θ} (θ)$ ;
along any constrained motion the multiplier is $λ (θ, θ^{'}) = (- \frac{m}{2 l} (g \cos θ + l θ^{' 2}), 0) \in R^{2}$ evaluated at the motion's data, the second component --- tied to the plane constraint $z = 0$ --- vanishing along the motion, and the reaction is $R = - T u (θ), T = m (g \cos θ + l θ^{' 2}),$ a force of magnitude $∣ T ∣$ along the line of the link, toward the pivot when $T > 0$ (the tension of the model); the projection of the constrained equation on the tangent gives $m l^{2} θ^{''} + m g l \sin θ = 0$ , and the projections on $u$ and on the normal $e_{3}$ of the constraining plane give the displayed tension and the vanishing of the second multiplier component;
conversely, for every $C^{2}$ function $θ : J \to R$ with $m l^{2} θ^{''} + m g l \sin θ = 0$ on an interval $J$ with nonempty interior, the curve $x (t) : = X (θ (t))$ is a constrained motion of the model with multiplier and reaction as in clause 2;
the reaction power vanishes identically, $R \cdot x^{'} \equiv 0$ , in agreement with the scleronomic case of the rheonomic-power theorem, and the mechanical energy $E = \frac{1}{2} m l^{2} θ^{' 2} - m g l \cos θ$ is constant along the motion; expressed through the energy, $T = 3 m g \cos θ + 2 E / l$ , and through the initial data $(θ_{0}, θ_{0}^{'})$ , $T = m g (3 \cos θ - 2 \cos θ_{0}) + m l θ_{0}^{' 2}$ ;
for a motion with initial data at $t_{0} = 0$ given by $m = 1 k g$ , $l = 1 m$ , $g = 10 m s^{- 2}$ , $θ_{0} = π / 3$ , $θ_{0}^{'} = 0$ , the initial energy is $E = - 5 J$ and the tension is $T = 30 \cos θ - 10 N$ , with $T = 20 N$ at $θ = 0$ and $T = 5 N$ at $θ = π / 3$ , positive on the reachable arc $∣ θ ∣ \leq π / 3$ .

All values are model values under the declared idealizations; the model is the ideal pendulum realized by a smooth rigid wire (equivalently a massless link, or a taut string while it stays taut), and the empirical qualifications are recorded in the remarks.

Facts & Assumptions

Given: The constants $m, l, g > 0$ , the model data of the Example, the datum $(O, 2, G)$ , and a $C^{2}$ function $θ : J \to R$ on an interval $J$ with nonempty interior.

[F1]

The force branch and mass operator: $F : I_{0} \times Q \times R^{3} \to R^{3}$ is the constant map $F (t, x, v) = (0, - m g, 0)$ , hence has the standing regularity and is of class $C^{1}$ , with potential $U (t, x, y, z) = m g y$ and residual forces zero, $F = - \nabla U$ (Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance, Configuration domain of $N$ labeled particles: open admissible sets, regular holonomic constraints and the mass operator, Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets clause (a), The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[F2]

The holonomic-constraint branch (Holonomic-constraint models of the force branch: constraint data, the constrained equation, virtual displacements and the scleronomic/rheonomic distinction): $G$ is smooth with $D_{x} G = (\begin{array}{ccc} 2 x & 2 y & 0 \\ 0 & 0 & 1 \end{array})$ of rank $2$ on $O$ ; virtual displacements are $\ker D_{x} G$ ; the datum is scleronomic exactly when there is $N \subseteq R^{3}$ with $Σ = {(t, x) \in O : x \in N}$ (equivalently when $(t, x) \in Σ$ and $(t^{'}, x) \in O$ imply $(t^{'}, x) \in Σ$ ), and then $G_{t} = 0$ on $Σ$ ; a constrained motion satisfies $M x^{''} = F + R$ with a continuous ideal reaction.

[F3]

The ideal-reaction law and the balance theorem (Adopted ideal-reaction law for holonomic constraints: the d'Alembert condition, its scope and its limits, Ideal holonomic constraints determine the reaction: multiplier form, explicit acceleration, unique constrained motions and the reaction-power identity, A regular holonomic constraint, its virtual displacements and the ideal-reaction law): $R = D_{x} G^{T} λ$ with the unique continuous multiplier $λ = Λ (t, x, x^{'}) \in R^{2}$ ; hypotheses (h1)-(h2) of the balance theorem hold here because $G$ is smooth with $D_{x} G$ of rank $2$ on all of $O$ by [F2] and $F$ is constant; with consistent data the constrained motions are the solutions of the first-order system and have a unique maximal one through each datum; the power identity $R \cdot x^{'} = - λ \cdot G_{t}$ holds.

[F4]

The rheonomic-power theorem (Reaction work and mechanical-energy balance for ideal holonomic constraints: vanishing power for scleronomic constraints and the rheonomic work term): for a scleronomic constraint the reaction power vanishes identically, and if also $\partial_{t} U \equiv 0$ and the residual forces vanish then $E = K + U$ is constant along the motion.

[F5]

Trigonometric calculus: derivatives of sine and cosine, the identity $\sin^{2} + \cos^{2} = 1$ , the chain rule, the algebra of derivatives, and componentwise differentiation (Sine and cosine defined by their real power series, The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral).

[F6]

Energy of a one-label curve family (Total mass, total momentum, centre of mass, angular momentum, torque, kinetic energy, power and work of a particle model in an inertial frame clauses (b),(d)): $K = \frac{1}{2} m (x^{' 2} + y^{' 2} + z^{' 2})$ and $E = K + U (t, x (t), y (t), z (t))$ ; the Euclidean pairing (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ); a differentiable function with zero derivative on an interval is constant (A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[F7]

Every unit circle point is $(\sin θ, - \cos θ)$ for some real $θ$ ( $t \mapsto (\cos t, \sin t)$ is a bijection from $[0, 2 π)$ onto the real unit circle, applied to the point $(- b / l, a / l)$ ). A continuous scalar function on an interval is integrable on compact subintervals; its oriented basepoint integral has derivative equal to that function, by A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive, For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ and The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ .

Verification

technique · direct; compute the multiplier matrix and the datum quantities, verify the equation of motion on the parametrized curves, and read off the tension, the reaction power and the conserved energy

1.1F2F5algebra

The datum is regular and scleronomic and its virtual displacements are the tangent line: $D_{x} G (t, x, y, z) = (\begin{array}{ccc} 2 x & 2 y & 0 \\ 0 & 0 & 1 \end{array})$ has rank $2$ at every point of $O$ , the first row being nonzero because $(x, y) \neq (0, 0)$ there and the two rows being orthogonal; the locus is $Σ = R \times N$ with $N = {(x, y, z) : x^{2} + y^{2} = l^{2}, z = 0}$ , so it is time-independent and the datum is scleronomic by [F2] (and then $G_{t} = 0$ on $Σ$ ); and $δ \in \ker D_{x} G$ means $2 x δ_{x} + 2 y δ_{y} = 0$ and $δ_{z} = 0$ , which at $X (θ) = l (\sin θ, - \cos θ, 0)$ is the line orthogonal to $u (θ)$ , that is, $R X_{θ} (θ)$ . Moreover $G (X (θ)) = (l^{2} \sin^{2} θ + l^{2} \cos^{2} θ - l^{2}, 0) = (0, 0)$ and $D_{x} G (X (θ)) X_{θ} (θ) = (2 l^{2} (\sin θ \cos θ - \cos θ \sin θ), 0) = (0, 0)$ , so $X_{θ} (θ)$ is permitted for every $θ$ .

1.2F2F5F6F7algebra

Every constrained $C^{2}$ motion admits a $C^{2}$ angular presentation on its interval $J$ . Write its first two coordinates as $(a, b)$ , so $a^{2} + b^{2} = l^{2}$ , and set $ω = (a b^{'} - b a^{'}) / l^{2}$ , a $C^{1}$ function. Differentiating the circle identity gives $(a^{'}, b^{'}) = ω (- b, a)$ . Choose $t_{*} \in J$ and $θ_{*} \in R$ with $(a (t_{*}), b (t_{*})) = l (\sin θ_{*}, - \cos θ_{*})$ by [F7], and put $θ (t) = θ_{*} + \int_{t_{*}}^{t} ω (s) d s$ . The oriented FTC gives $θ^{'} = ω$ and $θ \in C^{2}$ , with domain-relative endpoint derivatives where present. Both $(a, b)$ and $l (\sin θ, - \cos θ)$ satisfy $w^{'} = ω (- w_{2}, w_{1})$ . Their difference has squared norm with derivative zero and value zero at $t_{*}$ , so [F6] makes it zero on the interval. The third coordinate is zero by the constraint. Thus the motion is $X \circ θ$ ; the following calculations apply to every constrained motion.

1.3F2F3F5algebra

The multiplier of the balance theorem on this datum: with $M = m I_{3}$ one has $C = D_{x} G M^{- 1} D_{x} G^{T} = \frac{1}{m} diag ((2 x, 2 y, 0) \cdot (2 x, 2 y, 0), 1) = \frac{1}{m} diag (4 (x^{2} + y^{2}), 1) = \frac{1}{m} diag (4 l^{2}, 1)$ on the locus, $D_{x} G M^{- 1} F = \frac{1}{m} (- 2 m g y, 0) = (- 2 g y, 0)$ , $G_{t t} = 0$ , $G_{t q} = 0$ and $G_{q q} [v, v] = (2 (v_{x}^{2} + v_{y}^{2}), 0)$ ; along $x (t) = X (θ (t))$ the velocity is $\dot{x} = l θ^{'} (\cos θ, \sin θ, 0)$ , so $v_{x}^{2} + v_{y}^{2} = l^{2} θ^{' 2}$ , and hence $Λ = - C^{- 1} [(- 2 g y, 0) + (2 l^{2} θ^{' 2}, 0)] = (\frac{m}{4 l^{2}} (2 g y - 2 l^{2} θ^{' 2}), 0) = (- \frac{m}{2 l} (g \cos θ + l θ^{' 2}), 0)$ at $y = - l \cos θ$ ; this is the displayed multiplier.

2.1F1F2F3F5step 1.1step 1.3algebra

Conversely, a solution of the pendulum equation gives an ideal constrained motion: let $θ : J \to R$ be $C^{2}$ with $m l^{2} θ^{''} + m g l \sin θ = 0$ and put $x : = X (θ)$ , with the model data and potential of [F1]; then $x$ is $C^{2}$ and constraint-preserving by step 1.1, and $\dot{x} = l θ^{'} (\cos θ, \sin θ, 0)$ , $\ddot{x} = l θ^{''} (\cos θ, \sin θ, 0) + l θ^{' 2} (- \sin θ, \cos θ, 0)$ . Set $λ : = (- \frac{m}{2 l} (g \cos θ + l θ^{' 2}), 0)$ and $R : = D_{x} G (x)^{T} λ = (2 x λ_{1}, 2 y λ_{1}, λ_{2}) = - m (g \cos θ + l θ^{' 2}) u (θ)$ ; the triple $(\cos θ, \sin θ, 0)$ , $u (θ)$ , $e_{3}$ is an orthonormal basis of $R^{3}$ , and comparing components of $m \ddot{x}$ and $F + R$ in it: the tangent components are $m l θ^{''}$ and $- m g \sin θ$ , equal by the equation; the $u$ -components are $- m l θ^{' 2}$ and $m g \cos θ - T = - m l θ^{' 2}$ with $T = m (g \cos θ + l θ^{' 2})$ ; and the $e_{3}$ -components are $0$ and $λ_{2} = 0$ . Hence $M \ddot{x} = F + R$ with $R = D_{x} G^{T} λ$ annihilating the kernel, so $x$ is a constrained motion with ideal reaction and multiplier $λ$ .

2.2F3F5step 1.1step 1.3algebra

The tension and the reduced equation along a motion presented as $x = X (θ)$ : projecting $M x^{''} = F + R$ with $R = D_{x} G^{T} λ$ of step 1.3 onto the tangent direction $(\cos θ, \sin θ, 0)$ gives $m l θ^{''} = - m g \sin θ$ , that is $m l^{2} θ^{''} + m g l \sin θ = 0$ ; projecting onto $u (θ)$ gives $- m l θ^{' 2} = m g \cos θ - T$ with $T : = - R \cdot u$ , that is $T = m (g \cos θ + l θ^{' 2})$ , the displayed tension; and projecting onto $e_{3}$ gives $0 = λ_{2}$ , so the plane constraint $z = 0$ exerts no normal force on any constrained motion, whose motion and applied force already lie in the plane. The reaction on the bead is therefore $R = - T u (θ)$ , a force along the radial line toward the origin when $T > 0$ . A load on any support requires additional assumptions about the support and force transmission; the one-particle equation determines only $R$ .

3.1F4F5F6step 2.2algebra

Reaction power and energy: by step 2.2, $R = - T u$ with $T \in R$ , and $x^{'} = l θ^{'} (\cos θ, \sin θ, 0)$ is orthogonal to $u$ , so $R \cdot x^{'} = 0$ identically, in agreement with the scleronomic case of the power identity $R \cdot x^{'} = - λ G_{t}$ of [F3] because $G_{t} = 0$ ; with $U = m g y = - m g l \cos θ$ static and no residual forces, [F4] gives $E = \frac{1}{2} m l^{2} θ^{' 2} - m g l \cos θ$ constant along the motion; independently, $\frac{d}{d t} E = m l^{2} θ^{'} θ^{''} + m g l \sin θ θ^{'} = l θ^{'} (m l θ^{''} + m g \sin θ) = 0$ by the equation of step 2.2.

4.1F5F6step 3.1algebra

Energy expression of the tension and finite model data: substituting $l θ^{' 2} = 2 E / (m l) + 2 g \cos θ$ (equivalent to the definition of $E$ ) into $T = m (g \cos θ + l θ^{' 2})$ gives $T = 3 m g \cos θ + 2 E / l$ ; with $E = \frac{1}{2} m l^{2} θ_{0}^{' 2} - m g l \cos θ_{0}$ this is $T = m g (3 \cos θ - 2 \cos θ_{0}) + m l θ_{0}^{' 2}$ . For $m = 1$ , $l = 1$ , $g = 10$ , $θ_{0} = π / 3$ and $θ_{0}^{'} = 0$ at $t_{0} = 0$ , one has $E = - m g l \cos θ_{0} = - 5 J$ and $T (θ) = 30 \cos θ - 10 N$ , so $T (θ_{0}) = 5 N$ and $T (0) = 20 N$ . Along this solution, conservation of $E$ from step 3.1 gives $θ^{'} (t)^{2} = (20 \cos θ (t) - 10) s^{- 2} \geq 0$ , hence $\cos θ (t) \geq \frac{1}{2}$ and $θ (t) \in ⋃_{k \in Z} [2 π k - π / 3, 2 π k + π / 3]$ for every $t$ in its interval $J ∋ 0$ . Since $J$ is connected and $θ$ is continuous, $θ (J)$ is connected; it contains $θ (0) = π / 3$ , so it lies in the component $[- π / 3, π / 3]$ . Therefore $T \geq 5 N > 0$ throughout this motion. The reaction power is zero at every point of the motion.

5.1step 1.1step 1.2step 1.3step 2.1step 2.2step 3.1step 4.1∎

Clauses 1-5 of the Example are proved: the datum is a regular scleronomic holonomic constraint realizing the circle $x^{2} + y^{2} = l^{2}$ in the plane $z = 0$ , the multiplier has the displayed two components with the second vanishing along every constrained motion, the reaction is the displayed tension, the pendulum equation is the tangential projection of the constrained equation, the ideal reaction performs no work, the mechanical energy is conserved, and the finite data give a positive tension on the reachable arc.

Remarks

Rod versus string. The ideal model supplies the reaction $R = - T u$ with no sign restriction; a smooth wire — the realization declared in the example — and a light rigid link can transmit both tension ( $T > 0$ ) and compression ( $T < 0$ ), while a taut string can pull only. Where the model's formula gives $T < 0$ — for instance for motion near the inverted position — the string assumption fails, the constraint is released, and a slackening model (a different, unilateral model) is required; that regime is outside this example.
Model limits. The pivot is fixed and frictionless, the wire or link is smooth and massless, the field is uniform, and the bob is a point mass; drag, pivot friction, wire or link elasticity, the finite size of the bob and the nonuniformity of a real field are separate model corrections. The example is a conditional computation inside the declared idealizations and reports no measured tension.

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All four inertial terms checked on explicit data, and the exact return to an inertial chart

Physical scope

A worked finite check of the moving-chart equation of the framework: a zero-force one-label model observed in a chart with a uniformly accelerated rotation angle and a uniformly accelerated reference curve; the four inertial terms are evaluated term by term at one time and compared with the moving-chart coordinate acceleration of the motion, a chart with nonzero angular velocity is identified as non-inertial by the criterion, and an explicit chart with affine reference curve and constant rotation is checked to reproduce the inertial-frame description. All values are model values; the example reports no measurement.

Example

Declare a one-label model of the force branch of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with $N = 1$ , mass $m = 1 k g$ , time window $I_{0} = R$ , configuration domain $Q = R^{3}$ and the zero force map $F : I_{0} \times Q \times R^{3} ⟶ R^{3}, F (t, x, v) \equiv 0$ (admissible: constant, hence of the standing regularity), in a fixed inertial frame $F$ with adapted coordinates $(t, x)$ (Inertial frames, stationary observers and the adopted Galilean transformations between frames, Dimensions and SI unit conventions of the non-relativistic classical mechanics framework). For real parameters $ω_{0}, β, γ$ , vectors $a_{0}, v_{c} \in R^{3}$ and Moving spatial charts of an inertial frame: rotating and translating coordinates, angular velocity and inertial terms data, take $θ$ to be a dimensionless angle in radians, with $[ω_{0}] = τ^{- 1}$ , $[β] = τ^{- 2}$ , $[γ] = ℓ τ^{- 2}$ , $[a_{0, j}] = ℓ$ and $[v_{c, j}] = ℓ τ^{- 1}$ for each spatial component $j$ : $θ (t) : = ω_{0} t + \frac{1}{2} β t^{2}, R (t) : = (\begin{array}{ccc} \cos θ (t) & - \sin θ (t) & 0 \\ \sin θ (t) & \cos θ (t) & 0 \\ 0 & 0 & 1 \end{array}), c (t) : = a_{0} + v_{c} t + \frac{1}{2} γ t^{2} e_{1},$ consider the motion $x (t) : = x_{0} + v t$ of the zero-force model with $x_{0} = (1, 0, 0) m$ , $v = (0, 1, 0) m s^{- 1}$ , its moving-chart coordinate curve $y (t) = R (t)^{T} (x (t) - c (t))$ , and the inertial-terms field $τ (t, y, \dot{y}) = - (2 ω (t) \times \dot{y} + ω^{'} (t) \times y + ω (t) \times (ω (t) \times y) + R (t)^{T} c^{''} (t))$ .

Then, for the parameter values $ω_{0} = 2 s^{- 1}$ , $β = 1 s^{- 2}$ , $γ = 3 m s^{- 2}$ and $a_{0} = v_{c} = 0$ :

$R (t) \in S O (3)$ is $C^{\infty}$ and $Ω = R^{T} R^{'}$ has the angular velocity $ω (t) = θ^{'} (t) e_{3} = (2 + t) e_{3}$ and $ω^{'} (t) = e_{3}$ , while $c^{''} (t) = (3, 0, 0)$ ;
$y (0) = e_{1}$ , $\dot{y} (0) = - e_{2}$ , and the moving-chart coordinate acceleration is $\ddot{y} (0) = - 3 e_{1} - e_{2}$ (in units $m s^{- 2}$ ); the inertial-chart acceleration is $\ddot{x} (t) = 0$ for every $t$ ;
the four inertial terms at $t = 0$ are $- 2 ω \times \dot{y} = - 4 e_{1}$ , $- ω^{'} \times y = - e_{2}$ , $- ω \times (ω \times y) = + 4 e_{1}$ and $- R^{T} c^{''} = - 3 e_{1}$ , and their sum is $τ (0, y (0), \dot{y} (0)) = - 3 e_{1} - e_{2} = \ddot{y} (0)$ ; clause 1 of The equation of motion in a moving chart: inertial terms, inertial forces, and when it restricts to an inertial chart therefore holds at $t = 0$ for these data, with $m \ddot{y} = R^{T} F + m τ = m τ$ ;
because $ω \equiv̸ 0$ , clause 3 of that theorem identifies the chart as not inertial; and with the alternative parameters $ω_{0} = β = γ = 0$ , $a_{0} = (1, 0, 0)$ , $v_{c} = (0, 1, 0)$ , the chart has constant rotation $R \equiv I_{3}$ and affine reference curve, hence vanishing inertial terms, and the same motion has $\ddot{y} \equiv 0$ in the chart, which is the inertial-frame description with $y (t) = x (t) - a_{0} - v_{c} t$ .

All values are model values in the declared coordinates; the example reports no measurement and asserts nothing about any apparatus.

Facts & Assumptions

Given: The constants and parameters of the Example, the zero-force model, the moving chart $(I, c, R)$ with $I = R$ , the motion $x (t) = x_{0} + v t$ , and the curve $y (t) = R (t)^{T} (x (t) - c (t))$ .

[F1]

The chart (Moving spatial charts of an inertial frame: rotating and translating coordinates, angular velocity and inertial terms): $R (t) \in S O (3)$ with $Ω = R^{T} R^{'}$ skew and unique representing vector $ω$ , $Ω z = ω \times z$ ; $y$ is the moving-chart coordinate curve; $τ$ is the inertial-terms field of clause (d).

[F2]

The moving-chart equation (The equation of motion in a moving chart: inertial terms, inertial forces, and when it restricts to an inertial chart clauses 1,2,3): $m {\ddot{y}}_{i} = R^{T} F_{i} + m τ$ along a motion; the inertial terms vanish identically exactly when $R$ is constant and $c$ is affine, in which case the chart is an inertial frame's chart and the equation reduces to the covariance law.

[F3]

The zero-force model and its motions: $F \equiv 0$ is of class $C^{1}$ with the standing regularity; $x (t) = x_{0} + v t$ is a motion of the model in the sense of The associated first-order system, motions, maximal motions and the Newtonian flow of a force model clause (c), because $m \ddot{x} = 0 = F$ (Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance, Coordinate velocity and acceleration of a worldline in an inertial frame, and the spatial velocity and acceleration vectors).

[F4]

Rotation and trigonometry: $R (t)$ as displayed lies in $S O (3)$ with $R^{T} R = I_{3}$ ; the derivatives of sine and cosine, the identity $\sin^{2} + \cos^{2} = 1$ , the chain rule and the componentwise algebra of derivatives; the cross product of the standard basis (The orthogonal group $O (3)$ and the rotation group $S O (3)$ , $S O (3)$ is a subgroup of the general linear group, with $R^{- 1} = R^{T}$ , Sine and cosine defined by their real power series, The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The cross product in $R^{3}$ , The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F5]

The kinematic identity of the chart (Angular velocity of a moving frame and the rotating-frame acceleration identity clause 3): $R^{T} x^{''} = y^{''} + 2 ω \times y^{'} + ω^{'} \times y + ω \times (ω \times y) + R^{T} c^{''}$ for the curves of the chart; and $∥ R z ∥_{2} = ∥ z ∥_{2}$ .

[F6]

Intervals and constants (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length); the chart's data are defined for all $t \in R$ .

Verification

technique · direct; compute the chart quantities, evaluate the two sides of the moving-chart equation on explicit data at one time, and check the inertial and non-inertial cases against the criterion

1.1F1F4F6givenalgebra

Chart quantities: $R (t)$ is the rotation about $e_{3}$ by the angle $θ (t)$ , so $R (t)^{T} R (t) = I_{3}$ and $\det R (t) = \cos^{2} θ + \sin^{2} θ = 1$ by [F4], and $R$ is $C^{\infty}$ because sine, cosine and the polynomial $θ$ are on the interval $I = R$ ([F6]); differentiating $Ω = R^{T} R^{'}$ gives $Ω (t) = θ^{'} (t) (\begin{array}{ccc} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}), so ω (t) = θ^{'} (t) e_{3} = (2 + t) e_{3}, ω^{'} (t) = e_{3}$ for the parameter values of the Example, and $c^{''} (t) = γ e_{1} = 3 e_{1}$ .

1.2F3F4givenalgebra

The motion and its chart acceleration at $t = 0$ : the curve $x (t) = x_{0} + v t$ has $x^{''} = 0$ , so it is a motion of the zero-force model by [F3]. At $t = 0$ one has $θ (0) = 0$ , $R (0) = I_{3}$ , $c (0) = 0$ and $c^{'} (0) = 0$ , so $y (0) = x_{0} = e_{1}$ and, differentiating $y = R^{T} (x - c)$ at $t = 0$ with $Ω (0) = 2 [e_{3}]_{\times}$ , $\dot{y} (0) = - Ω (0) y (0) + x^{'} (0) - c^{'} (0) = - 2 e_{3} \times e_{1} + e_{2} = - 2 e_{2} + e_{2} = - e_{2}$ . Differentiating once more at $t = 0$ , $\ddot{y} (0) = - (Ω (0)^{2} + Ω^{'} (0)) y (0) - 2 Ω (0) \dot{y} (0) + x^{''} (0) - c^{''} (0)$ , using $Ω^{'} (0) = θ^{''} (0) [e_{3}]_{\times} = [e_{3}]_{\times}$ and $Ω (0)^{2} = 4 [e_{3}]_{\times}^{2}$ with $[e_{3}]_{\times} e_{1} = e_{2}$ , $[e_{3}]_{\times} e_{2} = - e_{1}$ : $- (Ω^{2} + Ω^{'}) y (0) = 4 e_{1} - e_{2}$ , $- 2 Ω (0) \dot{y} (0) = - 4 [e_{3}]_{\times} (- e_{2}) = - 4 e_{1}$ and $- c^{''} (0) = - 3 e_{1}$ , so $\ddot{y} (0) = - 3 e_{1} - e_{2}$ .

2.1F2F4step 1.1step 1.2algebra

The four inertial terms at $t = 0$ and the identity: with $ω (0) = 2 e_{3}$ , $ω^{'} (0) = e_{3}$ , $R (0) = I_{3}$ , $y (0) = e_{1}$ , $\dot{y} (0) = - e_{2}$ and $c^{''} (0) = 3 e_{1}$ from steps 1.1 and 1.2, the four terms of $τ$ are $- 2 ω \times \dot{y} = - 4 (e_{3} \times (- e_{2})) = - 4 e_{1}$ , $- ω^{'} \times y = - (e_{3} \times e_{1}) = - e_{2}$ , $- ω \times (ω \times y) = - 2 e_{3} \times (2 e_{3} \times e_{1}) = - 4 (e_{3} \times e_{2}) = 4 e_{1}$ and $- R^{T} c^{''} = - 3 e_{1}$ ; their sum is $τ (0, e_{1}, - e_{2}) = - 4 e_{1} - e_{2} + 4 e_{1} - 3 e_{1} = - 3 e_{1} - e_{2}$ , which equals $\ddot{y} (0)$ of step 1.2, so $m \ddot{y} (0) = m τ (0, e_{1}, - e_{2})$ and clause 1 of [F2] holds at $t = 0$ for these data (with $F = 0$ ).

2.2F2F3F5step 1.1algebra

The inertial case checked: with $ω_{0} = β = γ = 0$ , $a_{0} = (1, 0, 0)$ and $v_{c} = (0, 1, 0)$ one has $θ \equiv 0$ , $R \equiv I_{3}$ and $c (t) = a_{0} + v_{c} t$ affine, so $ω \equiv 0$ , $ω^{'} \equiv 0$ and $c^{''} \equiv 0$ ; the inertial terms vanish identically by [F2] clause 3, and the chart is the adapted chart of the inertial frame related to $F$ by the Galilean transformation $y = Q x + v_{*} t + a_{0 *}$ with $Q = I_{3}$ , $v_{*} = - v_{c} = (0, - 1, 0)$ and $a_{0 *} = - a_{0} = (- 1, 0, 0)$ . For the same motion $x (t) = x_{0} + v t$ one gets $y (t) = x_{0} - a_{0} + (v - v_{c}) t = (0, 0, 0)$ and $\ddot{y} \equiv 0$ , consistent with $m \ddot{y} = R^{T} F + m τ = 0$ .

2.3F1F2step 1.1

The chart is not inertial: since $ω (t) = (2 + t) e_{3}$ is not identically zero on $I = R$ , $R$ is not constant, so by clause 3 of [F2] the inertial terms do not vanish identically and the chart is not the adapted chart of an inertial frame; the four-term computation of the preceding check exhibits the nonvanishing explicitly at $t = 0$ .

3.1step 1.1step 1.2step 2.1step 2.2step 2.3∎

The Example's clauses are proved: clause 1 is step 1.1, clauses 2 and 3 are steps 1.2 and 2.1, the non-inertial verdict is step 2.3, and the exact return to an inertial chart is step 2.2.

Remarks

Reading the terms. On this data the Coriolis term and the centrifugal term cancel ( $- 4 e_{1} + 4 e_{1} = 0$ ) and the Euler and translational terms add, so the example is a check of the individual terms rather than of any one of them. A motion with $\dot{y}$ not aligned as here separates the Coriolis contribution; the companion pendulum and bead examples do not use a moving chart at all.
No equivalence principle. The non-inertial chart's terms are computed from its own motion data; nothing in the example compares them with a gravitational field, and the zero-force model remains force-free in every chart.

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Rolling as distinct models: the planar ideal no-slip disk against frictionless sliding, and the contact capability condition

Physical scope

A worked model-discrimination example for the rolling contact: two reduced planar models of a disk with the same declared kinetic matrix and applied force — the ideal no-slip model with its holonomic constraint and ideal reaction, and the frictionless sliding model with no constraint — with the accelerations, the multiplier, the required contact force and a declared Coulomb static-friction capacity model compared on explicit model data. The example uses a reduced two-coordinate description declared as model data, asserts no measured value, and records that general three-dimensional rolling, slip transitions and contact changes are separate models outside it.

Example

Declare a reduced planar disk model on the two coordinates $q = (x, θ) \in R^{2}$ , where $x$ is the centre coordinate along a straight line and $θ$ the oriented spin, with the declared kinetic matrix $M : = (\begin{array}{cc} m & 0 \\ 0 & I \end{array}), m > 0, I > 0,$ of dimension $μ$ and $μ ℓ^{2}$ respectively (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework); the reduction of a rigid disk to these two coordinates, with the spin inertia $I$ as model data, is a declared modelling step and is not derived here. The applied generalized force is $Φ : = (f, 0), f \in R,$ a force $f$ on the centre and no applied spin torque, and the disk radius is $a > 0$ . Here $[x] = [a] = ℓ$ , $[θ] = 1$ in radians, $[m] = μ$ , $[I] = μ ℓ^{2}$ , $[f] = μ ℓ τ^{- 2}$ , and $[g] = ℓ τ^{- 2}$ ; the coefficient $μ_{s}$ below is dimensionless. The two components of $Φ$ have dimensions of force and torque, respectively.

For the contact-capability comparison, additionally declare a horizontal plane with scalar normal load $N = m g$ , and the Coulomb static-contact capacity model in which a signed tangential static reaction $f_{s}$ is available exactly when $∣ f_{s} ∣ \leq μ_{s} N$ . Here $[N] = [f_{s}] = μ ℓ τ^{- 2}$ ; this contact law and its parameters are model assumptions, not measurements.

Consider two models of the planar contact, expressed in the same coordinates and with the same $M$ and $Φ$ :

(M1) ideal no-slip model. The holonomic constraint $G (x, θ) : = x - a θ, G_{q} = (1, - a),$ imposed as $G = 0$ ; the Jacobian has rank one. With the d'Alembert ideal-reaction law of A regular holonomic constraint, its virtual displacements and the ideal-reaction law and virtual displacements $δ \in \ker G_{q} = R (a, 1)$ : the generalized reaction is $G_{q}^{T} λ$ for a multiplier $λ$ , and the motions satisfy $M q^{''} = Φ + G_{q}^{T} λ$ .

(M2) frictionless sliding model. No constraint and no reaction: the motions satisfy $M q^{''} = Φ$ .

Then:

in M1 the ideal motions satisfy $(m + \frac{I}{a^{2}}) x^{''} = f, x^{''} = \frac{f a^{2}}{I + m a^{2}}, θ^{''} = \frac{a f}{I + m a^{2}},$ with the unique multiplier $λ = - \frac{f I}{I + m a^{2}}$ , reaction $G_{q}^{T} λ = (λ, - a λ)$ and reaction power identically zero; the planar straight-line no-slip constraint is holonomic-equivalent, since $x - a θ$ is constant along every admissible curve;
in M2 the motions satisfy $x^{''} = f / m$ and $θ^{''} = 0$ ;
the two models make different predictions for the same applied force: for $f \neq 0$ and $a^{2} m > 0$ , $\frac{x_{M 1}^{''}}{x_{M 2}^{''}} = \frac{m a^{2}}{I + m a^{2}} = \frac{1}{1 + I / (m a^{2})} < 1,$ equal to $\frac{2}{3}$ for a uniform thin disk $I = \frac{1}{2} m a^{2}$ and to $\frac{1}{2}$ for a hoop $I = m a^{2}$ ;
the ideal no-slip model requires the contact to supply the generalized reaction $(λ, - a λ)$ , that is, a tangential contact force $f_{s} = λ$ of magnitude $∣ λ ∣ = ∣ f ∣ I / (I + m a^{2})$ together with its torque. Adopt the Coulomb static-contact capacity model in which, for a declared normal load $N \geq 0$ and dimensionless coefficient $μ_{s}$ , the available tangential static reactions satisfy exactly $∣ f_{s} ∣ \leq μ_{s} N$ ; here declare $N = m g$ for a horizontal plane with weight-only vertical loading. Thus the no-slip idealization is feasible precisely while $∣ λ ∣ \leq μ_{s} N$ ; for the uniform disk this is $∣ f ∣ \leq 3 μ_{s} m g$ . Where the required force exceeds the contact capability, the no-slip constraint cannot be maintained and a slipping model (with kinetic friction as a separately declared force) applies instead;
on the explicit model data $m = 1 k g$ , $a = 1 m$ , $I = \frac{1}{2} k g m^{2}$ , $g = 10 m s^{- 2}$ and $f = 3 N$ : $x_{M 1}^{''} = 2 m s^{- 2}$ , $x_{M 2}^{''} = 3 m s^{- 2}$ , $λ = - 1 N$ and $∣ λ ∣ = 1 N$ , within $μ_{s} N = 5 N$ for the declared $μ_{s} = \frac{1}{2}$ ; with $f = 18 N$ instead, M1 requires $∣ λ ∣ = 6 N > 5 N$ , so under those parameters the declared contact model cannot maintain the no-slip constraint and M1's premises fail.

All values are model values under the declared reduction and contact model; the example reports no measurement.

Facts & Assumptions

Given: The constants $m, a, I > 0$ and $f \in R$ , the reduced model data $(M, Φ)$ , the constraint $G$ , and the two declared models M1 and M2.

[F1]

The ideal-constraint mathematics (A regular holonomic constraint, its virtual displacements and the ideal-reaction law, Reactions annihilating the virtual displacements are transpose multiples, with a positive definite multiplier matrix, The ideal holonomic system in a possibly time-dependent chart: explicit acceleration, constraint preservation, local existence and reaction power): for a $C^{3}$ constraint $G$ with $G_{q}$ of rank $k$ throughout its open ambient domain and a $C^{1}$ applied force map $Φ$ , $M$ symmetric positive definite, $C = G_{q} M^{- 1} G_{q}^{T}$ is symmetric positive definite and invertible, the ideal-constraint motions with permitted velocity are exactly the solutions of the first-order system $q^{'} = v$ , $v^{'} = a (t, q, v)$ with consistent initial data $G (t_{0}, q_{0}) = 0$ and $G_{t} (t_{0}, q_{0}) + G_{q} (t_{0}, q_{0}) v_{0} = 0$ , multiplier $Λ = - C^{- 1} [G_{q} M^{- 1} Φ + G_{t t} + 2 G_{t q} [v] + G_{q q} [v, v]]$ and acceleration $a = M^{- 1} (Φ + G_{q}^{T} Λ)$ , and the reaction power satisfies $R \cdot q^{'} = - Λ \cdot G_{t}$ ; the multiplier is unique, and a vector orthogonal to $\ker G_{q}$ is $G_{q}^{T} λ$ with $λ = C^{- 1} G_{q} M^{- 1} R$ (Reactions annihilating the virtual displacements are transpose multiples, with a positive definite multiplier matrix).

[F2]

Regularity of the constraint data: $G$ is $C^{\infty}$ and $Φ = (f, 0)$ is constant, hence $C^{1}$ , with $G_{q} = (1, - a)$ of rank $1$ ; the kernel is the line $R (a, 1)$ ; $G_{t} = 0$ , $G_{t t} = 0$ , $G_{t q} = 0$ and $G_{q q} = 0$ (Linear map between vector spaces over the same field, $C^{k}$ Euclidean maps and diffeomorphisms, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, The transpose $A^{T}$ of a matrix, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F3]

The sliding model M2: $M q^{''} = Φ$ componentwise is $m x^{''} = f$ and $I θ^{''} = 0$ ; $M$ is invertible with $M^{- 1} = d i a g (m^{- 1}, I^{- 1})$ (Invertible matrices and the general linear group ${GL}_{n} (F)$ ).

[F4]

Calculus and equivalence: on a nondegenerate interval $J$ , a $C^{1}$ curve $q = (x, θ)$ has $\frac{d}{d t} (x - a θ) = x^{'} - a θ^{'}$ . Thus $x^{'} - a θ^{'} = 0$ on $J$ holds if and only if $x - a θ$ is constant on $J$ . For any fixed $t_{0} \in J$ , the zero-level constraint $G (q (t)) = 0$ for all $t \in J$ is equivalent to that velocity law together with $G (q (t_{0})) = 0$ ; the velocity law alone permits any initial constant level (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[F5]

The general nonholonomic definition Velocity-constraint (nonholonomic) models of the force branch: constant-rank constraint laws, the d'Alembert reaction and holonomic equivalence states its factorization criterion on the $3 N$ -dimensional particle configuration domain; this example instead declares reduced coordinates $q = (x, θ) \in U : = R^{2}$ . For this two-coordinate model, a time-independent scalar velocity law $A (q) q^{'} = 0$ on an open $U \subseteq R^{2}$ is holonomic-equivalent when there are $C^{2}$ maps $ψ : U \to R$ and $ϱ : U \to R ∖ {0}$ with $D ψ (q) \neq 0$ and $A (q) = ϱ (q) D ψ (q)$ ; then the chain rule gives $\frac{d}{d t} ψ (q (t)) = ϱ (q (t))^{- 1} A (q (t)) q^{'} (t) = 0$ . Here step 1.1 verifies this reduced-coordinate criterion with $A = (1, - a)$ , $ψ (x, θ) = x - a θ$ and $ϱ = 1$ . Tong's separately described coin rolling without slipping down a slope is a source-reported nonholonomic model, not a classification of all three-dimensional rolling; rank and contact changes are outside this comparison.

Verification

technique · direct; compute the two models' accelerations from their declared equations, compare them, and evaluate the contact capability condition on explicit data

1.1F2F4F5algebra

The M1 constraint data and its holonomic-equivalence: $G_{q} = (1, - a)$ has rank one, $\ker G_{q} = {(a δ, δ) : δ \in R}$ as a set of virtual displacements, and the velocity law $x^{'} - a θ^{'} = 0$ is equivalent to $G (q (t))$ being constant by [F4]. M1 selects its zero level: for any $t_{0}$ in the motion interval, the velocity law and $G (q (t_{0})) = 0$ imply $G = 0$ throughout, and every M1 curve satisfies both. The velocity law is therefore the differentiated family $G =$ constant and is holonomic-equivalent in the sense of [F5].

1.2F1F2algebra

The M1 acceleration, multiplier and reaction: by [F1] with the time-independent data of [F2], $C = G_{q} M^{- 1} G_{q}^{T} = \frac{1}{m} + \frac{a^{2}}{I} = \frac{I + m a^{2}}{m I}$ , $G_{q} M^{- 1} Φ = \frac{f}{m}$ , and $Λ = - C^{- 1} \frac{f}{m} = - \frac{m I}{I + m a^{2}} \cdot \frac{f}{m} = - \frac{f I}{I + m a^{2}} = λ$ . The equations $M q^{''} = Φ + G_{q}^{T} λ$ read $m x^{''} = f + λ$ and $I θ^{''} = - a λ$ ; substituting $λ$ gives $m x^{''} = f - \frac{f I}{I + m a^{2}} = \frac{f m a^{2}}{I + m a^{2}}$ , that is $x^{''} = \frac{f a^{2}}{I + m a^{2}}$ , and $I θ^{''} = \frac{a f I}{I + m a^{2}}$ , that is $θ^{''} = \frac{a f}{I + m a^{2}}$ ; this is the displayed triple, with reaction $G_{q}^{T} λ = (λ, - a λ)$ . The reaction power vanishes because $R \cdot q^{'} = λ x^{'} - a λ θ^{'} = λ (x^{'} - a θ^{'}) = 0$ by the constraint, in agreement with the power identity of [F1] and $G_{t} = 0$ .

1.3F3algebra

The M2 acceleration: by [F3], $x^{''} = f / m$ and $θ^{''} = 0$ .

2.1F3step 1.2step 1.3algebra

The models disagree and the comparison is quantitative: for $f \neq 0$ , subtracting the two predictions gives $x_{M 2}^{''} - x_{M 1}^{''} = \frac{f}{m} - \frac{f a^{2}}{I + m a^{2}} = \frac{f}{m} \cdot \frac{I}{I + m a^{2}} \neq 0$ , so the two declared models are distinct; taking the ratio, $\frac{x_{M 1}^{''}}{x_{M 2}^{''}} = \frac{m a^{2}}{I + m a^{2}} = \frac{1}{1 + I / (m a^{2})}$ , which is $\frac{2}{3}$ at $I = \frac{1}{2} m a^{2}$ and $\frac{1}{2}$ at $I = m a^{2}$ .

2.2givenstep 1.2algebra

The contact capability condition: the no-slip contact can realize the ideal reaction only when the required scalar tangential force $f_{s} = λ$ lies in its declared capacity interval. Step 1.2 gives $∣ λ ∣ = \frac{∣ f ∣ I}{I + m a^{2}}$ ; for the uniform disk $I = \frac{1}{2} m a^{2}$ , this is $\frac{∣ f ∣}{3}$ , so the condition is $∣ f ∣ \leq 3 μ_{s} N$ , or $∣ f ∣ \leq 3 μ_{s} m g$ under the declared load $N = m g$ . If the bound fails, the no-slip idealization is infeasible for that contact model and a separate slipping model is required; transitions between rolling and slipping — like general three-dimensional rolling and rank or contact changes — are outside the present comparison.

3.1givenstep 1.2step 1.3step 2.2algebra

Finite model data: with $m = 1$ , $a = 1$ , $I = \frac{1}{2}$ , $f = 3$ and $g = 10$ one has $x_{M 1}^{''} = \frac{3 \cdot 1}{\frac{1}{2} + 1} = 2 m s^{- 2}$ , $x_{M 2}^{''} = 3 m s^{- 2}$ and $λ = - \frac{3 \cdot \frac{1}{2}}{\frac{1}{2} + 1} = - 1 N$ , so $∣ f_{s} ∣ = 1 N$ ; with the declared $μ_{s} = \frac{1}{2}$ and $N = m g = 10 N$ , the capacity $μ_{s} N = 5 N$ includes the required reaction and M1 is feasible under the declared contact model. With $f = 18$ instead, $x_{M 1}^{''} = 12 m s^{- 2}$ and $λ = - 6 N$ , whose magnitude exceeds $5 N$ , so the required force lies outside the declared contact capability and M1's premises fail.

4.1step 1.1step 1.2step 1.3step 2.1step 2.2step 3.1∎

Clauses 1-5 are proved: the reduced ideal no-slip model and the frictionless sliding model have the displayed accelerations, they are distinct for every nonzero applied force, the ideal model's reaction is the displayed contact pair with vanishing power, and the contact capability condition selects the models on explicit data. The example is a comparison of declared reduced models, not a derivation of the disk's reduction from rigid-body mechanics and not a measurement.

Remarks

The reduced model is declared. The pair $(x, θ)$ with kinetic matrix $d i a g (m, I)$ is a reduced description of the planar disk; the moment of inertia $I$ is model data, not derived here, and the reduction of a rigid body to these coordinates belongs to the rigid-body development of the track. What the example tests is the model discrimination and the capability condition, not the reduction.
Rolling is a family of models. No-slip with sufficient static friction, slipping with kinetic friction, rolling with resistance, and prescribed motion of the contact are different models; their empirical adequacy depends on the contact materials and regime, and transitions between them require contact-change or impact models that are outside this pair.

Regular constraints and moving frames — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

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Example

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Verification

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Remarks

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Facts & Assumptions

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Facts & Assumptions

Verification

Remarks

Sources