Constraints using Lagrange multipliers: Exercises

Prerequisites you may want open: Euler-Lagrange Equations, d'Alembert's Principle and Virtual Work, Generalized Coordinates and Degrees of Freedom, Holonomic vs Non-holonomic Constraints, Normal Force and Tension as Constraint Forces, KKT Conditions.

Level 1 — Recognition

Goal: identify the pieces of the method without heavy algebra.

Exercise L1.1 — Name the gradient

Recall Solution

(a) Move everything to one side: $f (x, y) = y - 3 x + 2 = 0$ . (b) $\frac{\partial f}{\partial x} = - 3$ , $\frac{\partial f}{\partial y} = 1$ . (c) The vector $(- 3, 1)$ is perpendicular (normal) to the line. This matters because a rail/wire pushes a bead sideways off the track — never along it. So the constraint force must point along $\nabla f$ , which is exactly what $λ \nabla f$ delivers. (Check: the line's direction is $(1, 3)$ ; dot product $(- 3) (1) + (1) (3) = 0$ , confirming perpendicular.)

Exercise L1.2 — Count the equations

Recall Solution

Unknowns: the $3$ coordinates $q_{1}, q_{2}, q_{3}$ plus the $2$ multipliers $λ_{1}, λ_{2}$ = $5$ unknown functions of time. Equations: $3$ Lagrange equations (one per coordinate) plus $2$ constraint equations $f_{1} = 0, f_{2} = 0$ = $5$ equations. Balanced: $5 = 5$ . ✓ The constraints "pay the rent" for the extra multipliers.

Level 2 — Application

Goal: run the recipe end-to-end on a clean single-constraint system.

Exercise L2.1 — Bead on a horizontal rotating-free hoop, normal force

Recall Solution

Horizontal plane ⇒ no gravity term. $L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2})$ . $r$ -equation: $\frac{d}{d t} (m \overset{r}{˙}) - m r \dot{θ}^{2} = λ \cdot 1$ . Impose $r = R$ so $\overset{r}{˙} = \overset{r}{¨} = 0$ : $- m R \dot{θ}^{2} = λ$ . With $\dot{θ} = ω$ : $λ = - m R ω^{2}$ . Interpretation: $λ$ is the generalized radial force. Negative means it points toward decreasing $r$ , i.e. inward — the hoop pulls the bead toward the center. Its magnitude $m R ω^{2}$ is exactly the centripetal force. The method reproduces the familiar $m v^{2} / R$ with $v = R ω$ . ✓

Exercise L2.2 — Vertical hoop, derive $N$ then find where the bead leaves

Recall Solution

(a) Derive $N$ from scratch. In plane polar coordinates the position height is $y = r sin θ$ , so $V = m g y = m g r sin θ$ (here $θ$ grows upward against gravity, so the potential carries a plus by our convention). Then $L = T - V = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2}) - m g r sin θ .$ The constraint is $f = r - R = 0$ , so $\partial f / \partial r = 1$ , $\partial f / \partial θ = 0$ . The $r$ -equation of the recipe is

\;\Rightarrow\; \frac{d}{dt}(m\dot r)-\big(mr\dot\theta^2-mg\sin\theta\big)=\lambda\cdot 1.$$ So $m\ddot r-mr\dot\theta^2+mg\sin\theta=\lambda$. Now — *and only now* — impose the constraint $r=R\Rightarrow\dot r=\ddot r=0$: $$-mR\dot\theta^2+mg\sin\theta=\lambda.$$ The physical normal force is $N=\lambda\,\partial f/\partial r=\lambda\cdot 1=\lambda$, so $$\boxed{N=mg\sin\theta-mR\dot\theta^2}.$$ **Sign convention check:** at the very bottom $\theta=-90^\circ$, $\sin\theta=-1$, giving $N=-mg-mR\dot\theta^2<0$; with our convention $\lambda\partial f/\partial r$ negative means the force points toward *decreasing* $r$ (inward), i.e. the hoop pushes the bead toward the center — correct, that is where a floor-like surface pushes. **(b)** The bead presses on the *inside* while the hoop must push inward, i.e. while $N$ (as an inward magnitude) is available; it leaves when the required inward push would reverse. At the top $\theta=+90^\circ$, $\sin\theta=1$, the critical case is $N=0$: $$N=mg(1)-mR\dot\theta^2_{\text{top}}=0\ \Rightarrow\ R\dot\theta^2_{\text{top}}=g\ \Rightarrow\ v_{\text{top}}^2=(R\dot\theta)^2=gR.$$ Energy conservation from bottom ($y=-R$) to top ($y=+R$), height gain $2R$: $$\tfrac12 v_0^2=\tfrac12 v_{\text{top}}^2+g(2R)=\tfrac12 gR+2gR=\tfrac52 gR.$$ $$\boxed{v_0=\sqrt{5gR}}.$$ Having *derived* $N=\lambda$ ourselves is what let us write the "leaves the surface" condition $N=0$ directly — the whole payoff of the multiplier method.

Level 3 — Analysis

Goal: multi-equation systems, extract the physical force, check signs.

Exercise L3.1 — Cylinder on incline, friction as multiplier

The figure below fixes our sign convention: $x$ is measured down the slope (amber arrow), so the disk's center moves in $+ x$ as it descends; friction (cyan arrow) acts up the slope at the contact point, and gravity resolves to $M g sin α$ along $+ x$ . Keep this picture in mind — every sign below is read off it.

Figure — Constraints using Lagrange multipliers

Recall Solution

$x$ -equation ( $\partial f / \partial x = 1$ ): $M \overset{x}{¨} - M g sin α = λ$ . …(i) $θ$ -equation ( $\partial f / \partial θ = - R$ ): $I \ddot{θ} = - R λ$ . …(ii)

Why differentiate the rolling constraint twice? Equations (i) and (ii) contain two unknown accelerations $\overset{x}{¨}, \ddot{θ}$ plus the unknown $λ$ — three unknowns, only two equations. The recipe promises a third equation: the constraint itself, $f = x - R θ = 0$ . But $f = 0$ relates positions, whereas (i)–(ii) are about accelerations. To make the constraint speak the same language, differentiate it twice in time: $f = x - R θ = 0 d / d t \overset{x}{˙} - R \dot{θ} = 0 d / d t \overset{x}{¨} - R \ddot{θ} = 0.$ This gives $\ddot{θ} = \overset{x}{¨} / R$ , the missing third relation that lets us eliminate $\ddot{θ}$ and solve for $λ$ .

Put $\ddot{θ} = \overset{x}{¨} / R$ into (ii): $\frac{I}{R} \cdot \frac{x ¨}{R} = - R λ \Rightarrow λ = - \frac{I x ¨}{R ^{2}}$ . Substitute into (i): $M \overset{x}{¨} - M g sin α = - \frac{I x ¨}{R ^{2}}$ . With $I = \frac{1}{2} M R^{2}$ , so $I / R^{2} = \frac{1}{2} M$ : $M \overset{x}{¨} + \frac{1}{2} M \overset{x}{¨} = M g sin α \Rightarrow \frac{3}{2} \overset{x}{¨} = g sin α \Rightarrow \overset{x}{¨} = \frac{2}{3} g sin α .$ (b) $λ = - \frac{I x ¨}{R ^{2}} = - \frac{1}{2} M \cdot \frac{2}{3} g sin α = - \frac{1}{3} M g sin α .$ Friction magnitude $∣ λ ∣ = \frac{1}{3} M g sin α$ . The physical friction on $x$ is $λ \partial f / \partial x = λ < 0$ , i.e. pointing in $- x$ = up the slope (opposing the descent) — exactly the cyan arrow in the figure.

Exercise L3.2 — Two beads on a rigid rod (shared constraint)

Recall Solution

$L = \frac{1}{2} m \overset{x}{˙}_{1}^{2} + \frac{1}{2} m \overset{x}{˙}_{2}^{2} + F x_{1}$ . $\partial f / \partial x_{1} = - 1, \partial f / \partial x_{2} = + 1$ . $x_{1}$ : $m \overset{x}{¨}_{1} - F = λ (- 1) = - λ$ . …(i) $x_{2}$ : $m \overset{x}{¨}_{2} - 0 = λ (+ 1) = λ$ . …(ii) Constraint twice-differentiated: $\overset{x}{¨}_{2} = \overset{x}{¨}_{1} \equiv a$ . Add (i)+(ii): $ma - F + ma = - λ + λ = 0 \Rightarrow 2 ma = F \Rightarrow a = \frac{F}{2 m} .$ From (ii): $λ = ma = \frac{F}{2}$ . Interpretation: $λ$ is the force in the rod. Bead 2 is dragged forward by the rod with force $+ λ = F /2$ , and bead 1 feels the rod pulling back on it with $- λ = - F /2$ . So the rod is in tension $F /2$ — the rod transmits half the applied force to the second bead, exactly what Newton predicts for two equal masses. $λ = F /2$ .

Level 4 — Synthesis

Goal: combine constraints, degenerate/limiting cases, changing setups.

Exercise L4.1 — Atwood machine with heavy pulley, all limits

The figure shows the geometry: $m_{1}$ (amber block) drops by $x$ while $m_{2}$ rises by $x$ because the string is inextensible, and the pulley of moment $I$ turns by $ϕ$ (white curved arrow) with no slip, giving the constraint $f = x - a ϕ = 0$ . Use it to keep track of which mass moves which way.

Recall Solution

Note $x$ already carries both masses (string is inextensible), so treat $x$ as one coordinate plus $ϕ$ . $\partial f / \partial x = 1, \partial f / \partial ϕ = - a$ . $x$ : $(m_{1} + m_{2}) \overset{x}{¨} - (m_{1} - m_{2}) g = λ$ . …(i) $ϕ$ : $I \ddot{ϕ} = - aλ$ . …(ii) Constraint differentiated twice (same reasoning as L3.1: we need an acceleration-level relation to eliminate $\ddot{ϕ}$ ): $\overset{x}{¨} = a \ddot{ϕ} \Rightarrow \ddot{ϕ} = \overset{x}{¨} / a$ , so (ii) ⇒ $λ = - \frac{I x ¨}{a ^{2}}$ . Into (i): $(m_{1} + m_{2}) \overset{x}{¨} - (m_{1} - m_{2}) g = - \frac{I x ¨}{a ^{2}}$ . $\overset{x}{¨} = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2} + I / a ^{2}}, λ = - \frac{I / a ^{2} ( m _{1} - m _{2} ) g}{m _{1} + m _{2} + I / a ^{2}} .$ (b) $I \to 0$ : $\overset{x}{¨} \to \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$ — the classic Atwood result. ✓ And $λ \to 0$ : a massless pulley needs no net torque, so the tension difference between the two sides vanishes. ✓ (c) $I \to \infty$ : $\overset{x}{¨} \to 0$ . An infinitely heavy pulley cannot be spun, so nothing moves. The system freezes — the degenerate limit behaves sensibly. ✓

Exercise L4.2 — When the constraint force switches off (degenerate case)

Recall Solution

Why an effective gravity? We solve in the elevator's own (accelerating, non-inertial) frame, because in that frame the constraint $y = 0$ is simply "block stays on the floor" with $y$ fixed. Newton's laws don't hold as-is in an accelerating frame; the standard fix is to add a fictitious force $- m a_{frame}$ on every mass. The frame accelerates downward at $a$ , so the fictitious force points upward with magnitude $ma$ . Combined with real gravity $m g$ downward, each mass feels a net downward pull $m (g - a)$ — this is the effective gravity $g_{eff} = g - a$ . It is a genuine derivation, not an assertion: fictitious force + real gravity = effective gravity. With $y$ up, the effective-gravity potential is $V = m g_{eff} y$ (grows as the block rises), so $L = \frac{1}{2} m \overset{y}{˙}^{2} - m g_{eff} y, f = y = 0, \partial f / \partial y = 1.$ $y$ -equation: $m \overset{y}{¨} + m g_{eff} = λ$ . Impose $y = 0 \Rightarrow \overset{y}{¨} = 0$ : $λ = m (g - a) = N .$ (b) Contact requires $N \geq 0 \Rightarrow g - a \geq 0 \Rightarrow a \leq g$ . At $a = g$ (free fall), $λ = 0$ : the multiplier drops to zero, meaning the floor pushes with no force — the block is weightless and on the verge of floating off. For $a > g$ the block would need the floor to pull it ( $λ < 0$ ), which a floor cannot do, so the constraint breaks and $y = 0$ is no longer valid. This is exactly the " $N = 0$ ⇒ leaves the surface" logic. ✓

Level 5 — Mastery

Goal: full research-style problem tying gradient geometry, all cases, and interpretation together.

Exercise L5.1 — Bead on a parabolic wire, full multiplier and the "flying off" analysis

The figure shows the parabola $y = \frac{1}{2} k x^{2}$ (cyan), a bead on it, and the constraint gradient $\nabla f = (- k x, 1)$ (amber arrow) pointing off the wire, normal to it — the only direction a frictionless wire can push. Notice the gradient is not a unit vector: its length $1 + k^{2} x^{2}$ grows as you move away from the vertex, which is exactly why part (c) must multiply $λ$ by $∥\nabla f ∥$ to get the physical force.

Recall Solution

Here $y$ points up and gravity pulls down, so by our convention $V = m g y$ enters $L$ with a minus: $L = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2}) - m g y .$ Gradient of the constraint: $\partial f / \partial x = - k x, \partial f / \partial y = 1$ .

(a) Multiplier equations:

$x$ : $m \overset{x}{¨} = λ (- k x) = - λk x$ . …(i)
$y$ : $m \overset{y}{¨} + m g = λ (1) = λ$ . …(ii)

(b) Constraint $y = \frac{1}{2} k x^{2}$ differentiated twice (again to reach acceleration level, as in L3.1): $\overset{y}{˙} = k x \overset{x}{˙}, \overset{y}{¨} = k \overset{x}{˙}^{2} + k x \overset{x}{¨} .$ From (i): $\overset{x}{¨} = - \frac{λk x}{m}$ . From (ii): $λ = m (\overset{y}{¨} + g) = m (k \overset{x}{˙}^{2} + k x \overset{x}{¨} + g)$ . Substitute this $λ$ into (i): $m \overset{x}{¨} = - k x \cdot m (k \overset{x}{˙}^{2} + k x \overset{x}{¨} + g)$ $\overset{x}{¨} = - k x (k \overset{x}{˙}^{2} + k x \overset{x}{¨} + g)$ $\overset{x}{¨} (1 + k^{2} x^{2}) = - k x (k \overset{x}{˙}^{2} + g)$ $\overset{x}{¨} = - \frac{k x ( g + k x ˙ ^{2} )}{1 + k ^{2} x ^{2}} .$

(c) Back-substitute to get $λ$ : $λ = m (k \overset{x}{˙}^{2} + k x \overset{x}{¨} + g) .$ Using the boxed $\overset{x}{¨}$ : $k x \overset{x}{¨} = - \frac{k ^{2} x ^{2} ( g + k x ˙ ^{2} )}{1 + k ^{2} x ^{2}}$ , so

=m\cdot\dfrac{(k\dot x^2+g)(1+k^2x^2)-k^2x^2(g+k\dot x^2)}{1+k^2x^2}.$$ The $k^2x^2$ terms cancel in the numerator, leaving $$\boxed{\lambda=\dfrac{m(g+k\dot x^2)}{1+k^2x^2}}.$$ The **normal force** magnitude is $N=|\lambda|\cdot\|\nabla f\|=|\lambda|\sqrt{(kx)^2+1}=\dfrac{m(g+k\dot x^2)}{\sqrt{1+k^2x^2}}$, because the *physical* force is the vector $\lambda\nabla f$ and $\|\nabla f\|=\sqrt{k^2x^2+1}$. **Units:** $[k]=\text{m}^{-1}$ (since $\tfrac12 kx^2$ is a length). Then $k\dot x^2$ has units $\text{m}^{-1}(\text{m/s})^2=\text{m/s}^2$, matching $g$. So $m(g+k\dot x^2)$ is $\text{kg}\cdot\text{m/s}^2=$ **newtons**, and the denominator is dimensionless. $\lambda$ is a force. ✓ **(d)** At the vertex $x=0$: $\lambda=\dfrac{m(g+k\dot x^2)}{1}=m(g+k\dot x^2)>0$. More generally the boxed $\lambda=\dfrac{m(g+k\dot x^2)}{1+k^2x^2}$ has a **strictly positive numerator** (both $g>0$ and $k\dot x^2\ge0$) and a **strictly positive denominator** for every $x$ and every $\dot x$. Therefore $\lambda>0$ **always**. Since the physical force is $\lambda\nabla f$ and $\nabla f$ points out of the bowl, the wire always presses the bead **inward/upward**: the bead can **never fly off** the concave-up parabola. This is the complete flying-off analysis — one formula, valid for all $x$, both signs of $\dot x$, and the degenerate vertex. Contrast with the top of a hoop, where the corresponding $\lambda$ *can* reach zero and the bead departs; the parabola is a bowl, so it cannot.

Recap ladder

Recall One-line summary of each level

L1 — $\nabla f$ is normal to the constraint; equations = coordinates + multipliers. L2 — keep the redundant coordinate, differentiate, then impose the constraint; $λ = N$ . L3 — the physical force is $λ \partial f / \partial q$ , sign included; friction/tension pop out. L4 — check $I \to 0, \infty$ limits; a sign change in $λ$ means the (one-sided) constraint breaks. L5 — normal force magnitude is $∣ λ ∣ ∥\nabla f ∥$ , not $∣ λ ∣$ ; one formula covers all cases.