Exercises — Constraints — holonomic vs non-holonomic, rheonomic vs scleronomic

What kind of statement is it? It is an equality in the coordinates $x,y$ only. That is exactly the form $f(\mathbf{r})=0$, so it is holonomic. Does time appear? No $t$ anywhere — the circle of allowed points never moves. So scleronomic. Degrees of freedom. Start with $2$ Cartesian coordinates in the plane. One holonomic equation removes one coordinate: $n=2-1=1$. The single surviving coordinate is the swing angle. See Generalized Coordinates. Answer: holonomic, scleronomic, $n=1$.

Holonomic? It is an equation in coordinate and time — the form $f(\mathbf{r},t)=0$. So holonomic. Time? $t$ appears explicitly and the allowed line $y=vt$ slides upward as time passes. So rheonomic. The moving wire can push the bead vertically and do work on it — a first warning about energy, picked up later in Kinetic Energy and the Hamiltonian. Answer: holonomic, rheonomic.

Holonomic? This is an inequality, not an equation. An inequality cannot be solved to eliminate one coordinate — a molecule deep inside is free in all three directions. So it is non-holonomic. Time? The shell radius $a$ is fixed; no $t$. So scleronomic. Answer: non-holonomic, scleronomic.

Identify $A,B$. Here $A=y$ (multiplies $dx$) and $B=x$ (multiplies $dy$). Test. $\frac{\partial A}{\partial y}=\frac{\partial y}{\partial y}=1$ and $\frac{\partial B}{\partial x}=\frac{\partial x}{\partial x}=1$. They are equal → exact. Integrate. We look for $f$ with $\partial f/\partial x=y$ and $\partial f/\partial y=x$. Both are satisfied by $f=xy$. So the constraint is $d(xy)=0\Rightarrow xy=c$. Verdict: holonomic (it hides the surface $xy=c$), scleronomic (no $t$). Answer: holonomic, $xy=\text{const}$.

Identify. $A=y$, $B=-x$. Test. $\partial A/\partial y=1$, $\partial B/\partial x=-1$. Not equal → not exact. But only two variables. In 2 variables an integrating factor always exists. Try $\mu =1/x^2$: $\frac{y}{x^2}dx-\frac{1}{x}dy=0=-d\left(\frac{y}{x}\right).$ Check: $d(y/x)=\frac{1}{x}dy-\frac{y}{x^2}dx$, and the negative of that is exactly our expression. So it integrates to $y/x=c$, i.e. $y=cx$ — straight lines through the origin. Verdict: holonomic (after the integrating factor), scleronomic. Answer: holonomic, $y/x=\text{const}$.

Why the 2-variable trick fails. Now three variables are coupled, so the "factor always exists" guarantee is gone; we must test properly. Write the one-form $\omega =dz-ydx$. It is integrable to a family of surfaces iff $\omega \wedge d\omega =0$ (the Frobenius condition). Compute. $d\omega =-dy\wedge dx=dx\wedge dy$. Then $\omega \wedge d\omega =(dz-ydx)\wedge (dx\wedge dy)=dz\wedge dx\wedge dy\ne 0,$ because $dx\wedge dx=0$ kills the second piece. Non-zero → not integrable. Verdict: non-holonomic, scleronomic. This is the algebraic skeleton of the rolling disk. Answer: non-holonomic (no surface exists).

List the constraints.

1. Bead 1 on the circle: $x_1^2+y_1^2-R^2=0$.
2. Bead 2 on the circle: $x_2^2+y_2^2-R^2=0$.
3. Fixed rod: $(x_1-x_2{)}^2+(y_1-y_2{)}^2-{\ell }^2=0$. All three are equalities in coordinates only → holonomic, scleronomic. Count. Start $3N=4$ (two particles, plane). Remove one coordinate per independent holonomic constraint: $n=4-3=1$. Sanity check. Once bead 1 sits somewhere on the ring (angle ${\varphi }_1$), the fixed chord forces bead 2 to one of two spots — a discrete choice, not a continuous coordinate. So exactly one continuous freedom survives. $n=1$. ✓ Answer: $3$ holonomic constraints, $n=1$.

Pick a generalized coordinate. Let $q$ = signed distance of the bead along the wire. Then $x=q\cos (\omega t),y=q\sin (\omega t).$ Because $t$ sits inside the map, this is a rheonomic transformation. Differentiate. Using the product rule, $\dot{x}=\dot{q}\cos \omega t-q\omega \sin \omega t,\dot{y}=\dot{q}\sin \omega t+q\omega \cos \omega t.$ Notice each velocity has a $\dot{q}$ piece and a $q\omega$ piece — the second is the $\partial \mathbf{r}/\partial t$ term the wire injects. Kinetic energy. For mass $m$, $T=\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2)$. Expanding (cross terms in $\sin \cos$ cancel): $T=\frac{1}{2}m{\dot{q}}^2+\frac{1}{2}m{q}^2{\omega }^2.$ The first term is the usual quadratic in $\dot{q}$; the second term has no $\dot{q}$ at all — it is a velocity-independent piece supplied by the driven rotation. That extra term is the fingerprint of a rheonomic constraint (see Kinetic Energy and the Hamiltonian). Energy meaning. Because the wire is driven externally at fixed $\omega$, it exerts a sideways force that does work on the bead; mechanical energy of the bead alone need not be conserved. The $q^2{\omega }^2$ term is exactly the energy the wire can pump in or out. Answer: $T=\frac{1}{2}m{\dot{q}}^2+\frac{1}{2}m{q}^2{\omega }^2$; the extra non-quadratic term signals possible non-conservation.

(a) Eliminate $d\theta$. From the two equations, $\frac{dx}{\cos \varphi }=Rd\theta =\frac{dy}{\sin \varphi }$, so $dx\sin \varphi -dy\cos \varphi =0.$ This says the coin may only move along its heading, never sideways. (b) Integrability. Write $\omega =\sin \varphi dx-\cos \varphi dy$ in the three variables $x,y,\varphi$. Frobenius test needs $\omega \wedge d\omega$: $d\omega =\cos \varphi d\varphi \wedge dx+\sin \varphi d\varphi \wedge dy.$ Wedging with $\omega =\sin \varphi dx-\cos \varphi dy$ and keeping only the surviving $dx\wedge dy\wedge d\varphi$ terms: $\omega \wedge d\omega =({\sin }^2\varphi +{\cos }^2\varphi )dx\wedge dy\wedge d\varphi =dx\wedge dy\wedge d\varphi \ne 0.$ Non-zero → not integrable: no surface $f(x,y,\varphi )=c$ exists. (c) Labels: non-holonomic (non-integrable velocity relation), scleronomic (no explicit $t$). (d) Consequence. Even though at each instant one direction is forbidden, by choosing a path (steer, roll forward, steer, roll) you can reach any target $(x,y,\varphi )$. The constraint limits instantaneous velocities, not the set of reachable configurations — the defining feature of non-holonomy. This is why parallel parking works at all. Answer: $dx\sin \varphi -dy\cos \varphi =0$; non-integrable; non-holonomic, scleronomic.

First axis. It is an inequality — it cannot eliminate a coordinate when the particle is in the interior. So non-holonomic. Second axis. Time appears explicitly through $a(t)=a_0+ut$; the boundary moves. So rheonomic. Physical note. When the particle rests against the wall it feels a moving boundary (like L1·2), which can do work — energy of the particle alone need not be conserved. Combining axes: this constraint is non-holonomic and rheonomic — proving the two axes really are independent (compare L1·2 which was holonomic-rheonomic and L1·3 which was non-holonomic-scleronomic). Answer: non-holonomic, rheonomic.

(a) Recognize a total differential. The left side is exactly $d(x^2+y^2+z^2)$. Setting it to zero gives $d(x^2+y^2+z^2)=0$, hence $x^2+y^2+z^2=c.$ Even though it is a velocity/differential relation, it integrated cleanly with no factor needed — so it was holonomic all along. (b) Geometry. $x^2+y^2+z^2=c$ is a sphere of radius $\sqrt{c}$: the particle is pinned to a fixed spherical shell. (c) DOF and labels. Start $3N=3$; one holonomic equation removes one coordinate → $n=2$ (latitude and longitude). No $t$ → scleronomic. Answer: integrable to a sphere $x^2+y^2+z^2=c$; holonomic, scleronomic, $n=2$.

(i) Fixed incline surface $f(\mathbf{r})=0$, no $t$ → holonomic, scleronomic. It's an equality surface that never moves. (ii) $r-(r_0-vt)=0$ is an equation in coordinate and time → holonomic, rheonomic. The allowed radius shrinks with $t$, so time is explicit. (iii) The blade forbids sideways velocity: a relation like $dx\sin \varphi -dy\cos \varphi =0$ in $\ge 3$ variables, non-integrable → non-holonomic, scleronomic (same skeleton as L4·1; rink is fixed, so no $t$). (iv) Rolling without slipping → non-integrable velocity relation → non-holonomic; the bowl is tilted over time so the constraint coefficients depend on $t$ → non-holonomic, rheonomic. Answer: (i) H, S · (ii) H, R · (iii) NH, S · (iv) NH, R — all four boxes of the $2\times 2$ table filled.

The classification decision, one line each. What is the FIRST question you ask of any constraint? ::: Can it be written as an equality $f(\mathbf{r},t)=0$ (possibly after integrating a differential relation)? A differential relation in exactly 2 variables is always...? ::: Integrable (an integrating factor always exists) → holonomic. The second, independent label is decided by...? ::: Whether $t$ appears explicitly — yes → rheonomic, no → scleronomic. Non-holonomic constraints reduce...? ::: Accessible velocity directions, not coordinates or reachable configurations.