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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.

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

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

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