Worked examples — Constraints using Lagrange multipliers

Before we do anything, let us fix the symbols we will reuse everywhere so no notation is ever a surprise:

The gradient $\partial f/\partial q_k$ is just "how fast $f$ changes if I nudge only $q_k$". Picture the constraint as a wall; the gradient is the arrow pointing straight out of the wall, and a wall can only push straight out — which is exactly why the constraint force must be $\lambda$ times that arrow.

The scenario matrix

Every problem this method can throw at you falls into one of these cells. Each row is a distinct case class we must show at least once.

We now walk each example, and each is labelled with the cell(s) it fills.

Example 1 — Bead on a vertical hoop (Cells A, D)

Forecast: guess before reading on — will $N$ be bigger at the bottom of the hoop or at the top? Write down "bigger at ___".

The figure below (s01) sketches the hoop as a violet circle, the bead as a magenta dot at $\theta =55^{\circ }$, the dashed navy radius $r=R$ from centre to bead, the orange arrow that is the radial constraint force $N=\lambda \partial f/\partial r$, and the small magenta $mg$ arrow straight down. Keep your eye on the orange arrow — it is the only direction a frictionless hoop can push.

1. Write the Lagrangian in polar coordinates. $L=T-V=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)-mgR\sin \theta (\text{using }r=R\text{ in }V).$ Why this step? Kinetic energy in the plane is $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$ — the $r^2{\dot{\theta }}^2$ term is the sideways speed of going around. We choose polar so the constraint is the dead-simple $r=R$, giving $\partial f/\partial r=1$.

2. Write the $r$-equation with a multiplier. $\frac{d}{dt}(m\dot{r})-mr{\dot{\theta }}^2+mg\sin \theta =\lambda \cdot \frac{\partial f}{\partial r}=\lambda .$ Why this step? The right-hand side is the constraint force along $r$. Since $\partial f/\partial r=1$, whatever $\lambda$ is, it equals the radial force. In figure s01 the orange arrow points along $r$, straight out of the hoop — exactly where a normal force can point.

3. Impose the constraint $r=R\Rightarrow \dot{r}=0,\ddot{r}=0$. $-mR{\dot{\theta }}^2+mg\sin \theta =\lambda .$ Why this step? On the hoop $r$ never changes, so its time-derivatives die. The radial direction is now pure force balance.

4. Read off the normal force. $N=\lambda =mg\sin \theta -mR{\dot{\theta }}^2.$ Why this step? $\lambda \cdot 1$ is the generalized radial force, and because $r$ is an ordinary length, that generalized force is a genuine force in newtons.

Verify: At the bottom of the hoop $\theta =-90^{\circ }$, so $\sin \theta =-1$: $N=-mg-mR{\dot{\theta }}^2$ (the sign reflects our convention that $+r$ points outward; magnitude $mg+mR{\dot{\theta }}^2$, largest here). At the top $\theta =+90^{\circ }$, $N=mg-mR{\dot{\theta }}^2$, which is smaller. This change of sign of $\lambda$ with position is Cell D in action. Units: $mg\sin \theta$ is $\text{kg}\cdot {\text{m/s}}^2=\text{N}$, and $mR{\dot{\theta }}^2$ is $\text{kg}\cdot \text{m}\cdot {\text{s}}^{-2}=\text{N}$. ✓

Example 2 — Mass on a string wound round a cylinder (Cells B, C)

Forecast: will the tension be less than $mg=20\text{N}$, equal to it, or more? Guess.

1. Lagrangian. $L=T-V=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}I{\dot{\varphi }}^2+mgx.$ Why this step? Two kinetic pieces (falling mass, spinning cylinder) make $T$; the gravitational potential is $V=-mgx$ (dropping distance $x$ lowers height), so $-V=+mgx$.

2. Two multiplier equations. With $\partial f/\partial x=+1$ and $\partial f/\partial \varphi =-a$: $m\ddot{x}-mg=\lambda (\text{i}),I\ddot{\varphi }=-a\lambda (\text{ii}).$ Why this step? This is Cell B: one $\lambda$ shows up in both equations. That single shared unknown is what ties the two motions together.

3. Use the constraint to link accelerations. $x=a\varphi \Rightarrow \ddot{\varphi }=\ddot{x}/a$. Sub into (ii): $\frac{I}{a}\cdot \frac{\ddot{x}}{a}=-a\lambda \Rightarrow \lambda =-\frac{I\ddot{x}}{a^2}.$ Why this step? We now express the mystery $\lambda$ purely through $\ddot{x}$, ready to plug into (i).

4. Solve. Put $\lambda$ into (i): $m\ddot{x}-mg=-\frac{I\ddot{x}}{a^2}\Rightarrow \ddot{x}=\frac{mg}{m+I/a^2}.$ Numbers: $I/a^2=0.02/0.01=2$, so $\ddot{x}=\frac{2\cdot 10}{2+2}=5{\text{m/s}}^2.$

5. Tension. $T=|\lambda |=\frac{I\ddot{x}}{a^2}=2\cdot 5=10\text{N}.$ Why this step? $\lambda$ came out negative (Cell C). By our sign rule, $+x$ points down, so $\lambda <0$ means a force pointing up — a tension. Its magnitude is $T$.

Verify: $\ddot{x}=5<g=10$, correct — the cylinder's inertia holds the mass back. And $T=10\text{N}<mg=20\text{N}$: tension is less than the weight, exactly what lets the mass accelerate downward. Newton cross-check: $mg-T=m\ddot{x}\Rightarrow 20-10=2\cdot 5=10$ ✓. Did you forecast "less than $mg$"?

Example 3 — Solid sphere rolling down an incline (Cells C, F)

Forecast: compared to a frictionless block ($\ddot{x}=g\sin \alpha =5{\text{m/s}}^2$), will the sphere accelerate faster or slower?

The figure below (s02) shows the violet incline at $30^{\circ }$, the magenta sphere resting on it, the orange arrow for friction pointing up the slope (this is $|\lambda |$), and the navy arrow for the gravity component $Mg\sin \alpha$ pointing down the slope. The two opposing arrows are the whole story of why the sphere is slower than a sliding block.

1. Lagrangian. $L=T-V=\frac{1}{2}M{\dot{x}}^2+\frac{1}{2}I{\dot{\theta }}^2+Mgx\sin \alpha .$ Why this step? Same structure as Ex 2, but the "gravity along the coordinate" is only the component down the slope, giving $-V=+Mgx\sin \alpha$.

2. Multiplier equations. $\partial f/\partial x=1$, $\partial f/\partial \theta =-R$: $M\ddot{x}-Mg\sin \alpha =\lambda ,I\ddot{\theta }=-R\lambda .$

3. Constraint $\ddot{x}=R\ddot{\theta }$, eliminate $\lambda$ exactly as before: $\lambda =-\frac{I\ddot{x}}{R^2}=-\frac{(2/5)M{R}^2\ddot{x}}{R^2}=-\frac{2}{5}M\ddot{x}.$ Why this step? $\lambda <0$; since $+x$ points down-slope, the force points up-slope — that is the friction (Cell C/F), pointing exactly as the orange arrow in s02.

4. Solve. $M\ddot{x}-Mg\sin \alpha =-\frac{2}{5}M\ddot{x}\Rightarrow \ddot{x}(1+\frac{2}{5})=g\sin \alpha$, so $\ddot{x}=\frac{5}{7}g\sin \alpha =\frac{5}{7}\cdot 10\cdot \frac{1}{2}=\frac{25}{7}\approx 3.571{\text{m/s}}^2.$

5. Friction. $|\lambda |=\frac{2}{5}M\ddot{x}=\frac{2}{5}\cdot 3\cdot \frac{25}{7}=\frac{30}{7}\approx 4.286\text{N}.$ Cross-check the "$\frac{2}{7}Mg\sin \alpha$" formula: $\frac{2}{7}\cdot 3\cdot 10\cdot \frac{1}{2}=\frac{30}{7}$ ✓.

Verify: $\ddot{x}=3.571<5$: the sphere is slower than a frictionless block, because energy also spins it up — matches forecast "slower". Friction $\frac{30}{7}\text{N}$ points up-slope, which is exactly what supplies the torque to roll it. Units: $Mg\sin \alpha$ is N. ✓

Example 4 — The fly-off: bead leaves the hoop (Cells D, E)

Forecast: a smaller or larger angle from the top? Guess "leaves at ___ degrees below the top".

1. The degenerate condition. The constraint force is a push, never a pull. The bead flies off the instant $N=0$: $mg\sin \theta =mR{\dot{\theta }}^2\Rightarrow g\sin \theta =R{\dot{\theta }}^2.$ Why this step? Cell E is the limiting case $\lambda \to 0$. When the multiplier hits zero, the constraint is doing nothing — the surface has let go. This is the exact question the embedded method could never answer. (And it is Cell D too: just above this angle $\lambda$ was positive, just below it would go negative — the sign flip.)

2. Get ${\dot{\theta }}^2$ from energy. Falling from the top, the bead drops a height $R(1-\sin \theta )$ — using the same $y=R\sin \theta$ relation sketched in figure s01: $\frac{1}{2}m{R}^2{\dot{\theta }}^2=mgR(1-\sin \theta )\Rightarrow R{\dot{\theta }}^2=2g(1-\sin \theta ).$ Why this step? We need ${\dot{\theta }}^2$ in terms of $\theta$ alone to solve. Energy conservation supplies it cleanly.

3. Combine. $g\sin \theta =2g(1-\sin \theta )\Rightarrow \sin \theta =\frac{2}{3}.$ $\theta =\arcsin \frac{2}{3}\approx 41.8^{\circ }.$ Why this step? Set the two expressions for $R{\dot{\theta }}^2$ equal; the $g$ and $R$ cancel — the fly-off angle is universal, independent of size and gravity.

Verify: $\sin \theta =2/3$ gives $R{\dot{\theta }}^2=2g(1-2/3)=\frac{2}{3}g$ and $g\sin \theta =\frac{2}{3}g$ — the two match, so $N=0$ exactly there ✓. Below $\theta =41.8^{\circ }$ the bead has already left. Classic result $\cos (\text{angle from top})=2/3$. ✓

Example 5 — Real-world: two masses over a pulley with rope tension (Cells C, F)

Forecast: which way does the system accelerate, and is $T$ between $m_{2}g$ and $m_{1}g$?

1. Write $L$ with the linking constraint kept. Measuring both coordinates downward: $L=T-V=\frac{1}{2}{m}_1{\dot{x}}_1^2+\frac{1}{2}{m}_2{\dot{x}}_2^2+m_{1}g{x}_1+m_{2}g{x}_2.$ Why this step? Keeping both coordinates lets $\lambda$ appear — that's the tension we want (Cell F, real-world payoff).

2. Multiplier equations. $\partial f/\partial x_1=\partial f/\partial x_2=1$: $m_1{\ddot{x}}_1-m_{1}g=\lambda ,m_2{\ddot{x}}_2-m_{2}g=\lambda .$

3. Constraint ${\ddot{x}}_1+{\ddot{x}}_2=0\Rightarrow {\ddot{x}}_2=-{\ddot{x}}_1$. Subtract the two equations: $m_1{\ddot{x}}_1-m_2{\ddot{x}}_2-(m_1-m_2)g=0\Rightarrow (m_1+m_2){\ddot{x}}_1=(m_1-m_2)g.$ ${\ddot{x}}_1=\frac{m_1-m_2}{m_1+m_2}g=\frac{2}{4}\cdot 10=5{\text{m/s}}^2.$ Why this step? Subtracting cancels the shared $\lambda$ (a common tension acts equally on both sides), and substituting ${\ddot{x}}_2=-{\ddot{x}}_1$ leaves one equation in one unknown — that is why subtraction, not addition, is the right move.

4. Tension. From the first equation, $T=-\lambda =m_{1}g-m_1{\ddot{x}}_1=3\cdot 10-3\cdot 5=15\text{N}.$ Why this step? $\lambda$ is the generalized force from the rope; $+x_1$ points down, so $\lambda <0$ (Cell C) is a force pointing up — the tension. Its magnitude is $15\text{N}$.

Verify: $T=15\text{N}$ lies between $m_{2}g=10$ and $m_{1}g=30$ ✓ (a real rope must). Check $m_2$: $T-m_{2}g=m_2{\ddot{x}}_2\Rightarrow 15-10=1\cdot 5=5$ ✓. Forecast: $m_1$ descends (heavier), $T$ in between.

Example 6 — Two constraints at once (Cell G)

Forecast: will ${\lambda }_1$ (vertical) equal the weight $mg$? Will ${\lambda }_2$ be nonzero even though the drive is purely along $x$?

1. General form for many constraints. With $m$ constraints the RHS sums over them: $\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_k}-\frac{\partial L}{\partial q_k}={\sum }_{a=1}^2{\lambda }_a\frac{\partial f_a}{\partial q_k}.$ Why this step? Cell G: each constraint gets its own honest helper ${\lambda }_a$; the total constraint force on $q_k$ is the sum of their gradients — this is the same stacking of multipliers you meet in the KKT Conditions.

2. $z$-equation. $L=T-V=\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2)-mgz+Fx$. Gradients: $\partial f_1/\partial z=1$, $\partial f_2/\partial z=0$: $m\ddot{z}+mg={\lambda }_1.$ Why this step? Only $f_1$ mentions $z$, so only ${\lambda }_1$ survives here. On the shelf $z=2$ is constant, so $\ddot{z}=0$, giving ${\lambda }_1=mg=10\text{N}.$

3. $x$- and $y$-equations. $\partial f_2/\partial x=-1$, $\partial f_2/\partial y=+1$; $f_1$ contributes nothing: $m\ddot{x}-F=-{\lambda }_2,m\ddot{y}=+{\lambda }_2.$ Why this step? The drive $F$ enters the $x$-equation (from $\partial (Fx)/\partial x=F$); the diagonal wall's gradient splits as $-1$ in $x$ and $+1$ in $y$.

4. Impose the constraint and solve. $y=x\Rightarrow \ddot{y}=\ddot{x}$. Subtract the $y$-equation from the $x$-equation: $m\ddot{x}-F-m\ddot{y}=-{\lambda }_2-{\lambda }_2\Rightarrow -F=-2{\lambda }_2\Rightarrow {\lambda }_2=\frac{F}{2}=2\text{N}.$ Why this step? Subtraction isolates ${\lambda }_2$ because the two equations carry it with opposite signs. Adding them instead cancels ${\lambda }_2$: $m(\ddot{x}+\ddot{y})-F=0\Rightarrow 2\ddot{x}=F\Rightarrow \ddot{x}=2{\text{m/s}}^2$ along the line.

Verify: ${\lambda }_1=mg=10\text{N}$ (the shelf holds up the weight, forecast ✓). ${\lambda }_2=2\text{N}\ne 0$ (the diagonal wall reacts even to an axis-aligned push ✓). Check $x$-equation: $1\cdot 2-4=-2=-{\lambda }_2$ ✓.

Example 7 — The trap: multiplier on an already-eliminated coordinate (Cell H, degenerate)

Forecast: guess what $\partial f/\partial \theta$ equals here.

1. What coordinate survives? After substituting $r=R$, the only coordinate left is $\theta$. The constraint $f=r-R$ no longer mentions any live coordinate. Why this step? This is Cell H, the degenerate case where the constraint has been folded away.

2. Compute the gradient. $\partial f/\partial \theta =0$ (there is no $\theta$ in $r-R$), and there is no live $r$ to differentiate against. $\lambda \frac{\partial f}{\partial \theta }=\lambda \cdot 0=0.$ Why this step? The multiplier term vanishes identically — $\lambda$ can be anything and changes nothing. It is undetermined and meaningless.

3. The lesson. To get a constraint force you must keep the redundant coordinate ($r$ here) so that $\partial f/\partial r=1$ can carry $\lambda$. Eliminating it destroys the very information you wanted.

Verify: With $r$ retained (Ex 1) we found $N=mg\sin \theta -mR{\dot{\theta }}^2$, a real force; with $r$ eliminated we get $\lambda \cdot 0=0$, no information. Same physics, but only the retained-coordinate route yields $N$. This is exactly Error 1 in the parent's mistake list. ✓

Example 8 — Exam twist: a moving constraint (Cell I, time-dependent $f$)

Forecast: with the bead pushed outward and spinning, is $\lambda$ positive (push out) or negative (pull in)?

1. Lagrangian (no gravity). $L=T-V=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)-0.$ Why this step? Only kinetic energy; the wire supplies whatever force is needed via the constraint.

2. $r$-equation with time-dependent constraint. $\partial f/\partial r=1$ (the $t$-part doesn't affect the spatial gradient): $m\ddot{r}-mr{\dot{\theta }}^2=\lambda .$ Why this step? Cell I: even though $f$ contains $t$ explicitly, the gradient in coordinates is still what multiplies $\lambda$. Time enters only through the constraint equation $r=r_0+vt$.

3. Impose the moving constraint. $r=r_0+vt\Rightarrow \dot{r}=v$, $\ddot{r}=0$. At $t=0$, $r=r_0=1$: $\lambda =m\cdot 0-m{r}_0{\omega }^2=-m{r}_0{\omega }^2=-0.5\cdot 1\cdot 9=-4.5\text{N}.$ Why this step? The steady outward push $v$ has $\ddot{r}=0$, so it needs no force; all that remains is the term $-mr{\omega }^2$.

Verify: $\lambda =-4.5\text{N}$ is negative → by our sign rule, $+r$ points outward so $\lambda <0$ means the wire pulls the bead inward (centripetal), because a rotating bead "wants" to fly out and the mechanism resists that. Units: $m{r}_0{\omega }^2={\text{kg⋅m⋅s}}^{-2}=\text{N}$ ✓. Forecast check: negative, pulling inward.

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