Question bank — Constraints using Lagrange multipliers

True — the embedded (reduced-coordinate) method makes constraint forces do zero virtual work and vanish, whereas the multiplier $\lambda \partial f/\partial q_k$ is that force, kept on purpose (see figure s02).

False — once you eliminate the constraint you no longer have an $f$ in those coordinates, so $\partial f/\partial q=0$ and there is no $\lambda$ to read a force from.

False — $\lambda \partial f/\partial q_k$ must have units of generalized force, so $\lambda$'s units are set by $f$; if $f$ has length units, $\lambda$ is a force.

False — $\lambda (t)$ is a function of time; the constraint force changes as the motion proceeds (e.g. $N=mg\sin \theta -mR{\dot{\theta }}^2$ varies with $\theta$ and $\dot{\theta }$).

False — it points along $\partial f/\partial q$ (i.e. $\nabla f$), which is perpendicular to the allowed motion (the surface $f=0$); that is exactly why it does no virtual work under embedding.

True — you get $n$ Euler–Lagrange equations plus $m$ constraint equations, for the $n$ coordinates plus $m$ multipliers: $n+m=n+m$.

True in bookkeeping — the multiplier method keeps all $n$ coordinates (dependent ones included) and adds $\lambda$, trading a smaller coordinate count for direct access to the constraint force. See Generalized Coordinates and Degrees of Freedom.

False — a genuinely non-holonomic constraint $\sum a_k{\dot{q}}_k=0$ is non-integrable, so no position function $f$ exists; you must use the velocity form ${\lambda }_a{a}_{ak}$ directly. See Holonomic vs Non-holonomic Constraints.

True — its sign tells you the direction of the constraint force (e.g. a rope can only pull, so a wrong sign flags the constraint about to be violated / object leaving the surface). See the sign-convention figure s04.

Error — the virtual displacements $\delta q_k$ (imagined time-frozen nudges along the surface) are no longer independent; they must satisfy $\delta f=\sum (\partial f/\partial q_k)\delta q_k=0$, so a weighted sum being zero does not force each term to zero.

Error — after substituting $r=R$ the coordinate is gone and $r-R\equiv 0$ identically; there is no free $r$ to vary, so you cannot recover $N$. Keep $r$ free until after differentiating.

Error — the generalized force on $x$ is $\lambda \partial f/\partial x=\lambda$, which equals $-T$ because the string pulls opposite to the drop-coordinate $x$; the physical tension is $T=|\lambda |$ (see the sign diagram s04).

Error — you forgot the $(n+1)$th equation, the constraint $f=0$ itself; that closes the system and pins down $\lambda$.

Subtle — for rolling, the constraint is $x-R\theta =0$ whose gradient lives in the $(x,\theta )$ coordinate space, not physical space; the resulting $\lambda$ correctly delivers the tangential friction because that is the force enforcing no-slip. See Normal Force and Tension as Constraint Forces.

Error — doing no virtual work means the force is perpendicular to allowed displacements, not that its magnitude is zero; $\lambda$ is precisely the nonzero magnitude of that perpendicular push.

Error — reduced equations have already used up the constraint; $N$ appears only when the redundant coordinate and its $\lambda$ term are retained.

The EL-bracket vector is the collection of left-hand sides $(\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_k}-\frac{\partial L}{\partial q_k})$ stacked into one vector, one entry per coordinate. The allowed nudge $\delta q$ is perpendicular to both this bracket vector and $\nabla f$ (figure s03); by the linear-algebra fact above (two arrows perpendicular to the same line are parallel), bracket $=\lambda \nabla f$ for some scalar $\lambda$.

Adding a zero (since $\delta f=0$) changes nothing physically but lets us choose $\lambda$ to kill the one dependent coordinate's bracket, after which the remaining $\delta q$ are independent. This is the same trick as d'Alembert's Principle and Virtual Work.

The gradient of $f$ points perpendicular to the surface $f=0$, and a constraint force is exactly what pushes a body off/onto that surface — perpendicular to it — so the force must be proportional to $\nabla f$.

Leaving the hoop is the event $N=0$; the embedded method has discarded $N$ entirely, so it has no expression to set to zero.

$n$ EL equations (one per retained coordinate) plus the constraint equation $f=0$, matching the $n$ coordinates plus the extra unknown $\lambda$.

Both enforce a constraint by adding $\lambda \times (\text{constraint})$ and stationarizing; the multiplier measures sensitivity to the constraint. See the inequality-constraint generalization in KKT Conditions — and the concrete mechanical analogue in the edge cases below.

We need $\partial f/\partial q_k$ to exist to form the constraint-force direction; a non-smooth constraint has no well-defined gradient and the method breaks at the kink.

It passes through zero and would go negative if the hoop could only push inward; the sign flip signals the bead physically separating, so the model is only valid up to $\lambda =0$ (this is exactly a mechanical KKT "inactive constraint": the one-sided contact drops out when $\lambda$ would change sign).

The constraint is no longer active — the string goes slack — and you must drop the constraint (set $T=0$) rather than trust a nonphysical negative tension; this is the mechanics version of a KKT constraint switching off (see figure s04).

$T=\frac{mg(I/a^2)}{m+I/a^2}\to 0$; a massless cylinder offers no rotational inertia to react against, so no tension builds and the mass falls at $g$.

$\ddot{x}=\frac{mg}{m+I/a^2}\to 0$ and $T\to mg$; the cylinder won't turn, so the mass hangs essentially in static equilibrium with tension balancing gravity.

The surface itself sweeps in time, so the constraint force $\lambda \nabla f$ now acts on a moving wall and can do real work; energy is not conserved and $\lambda$ tracks a moving push (see figure s05 — the wall displaces the bead as it slides).

It equals whatever transverse force keeps the bead on the line (e.g. supporting its weight component); if no transverse applied force exists, $\lambda =0$ — a degenerate but perfectly valid answer.

The two constraint surfaces touch tangentially so their normals coincide (figure s06); the combination ${\lambda }_1\nabla f_1+{\lambda }_2\nabla f_2$ can be split infinitely many ways, so ${\lambda }_1$ and ${\lambda }_2$ are individually undetermined — a badly posed (redundantly constrained) problem.