Generalized momenta and generalized forces: Worked examples

Prerequisites you can lean on: Lagrangian Mechanics, Euler-Lagrange Equation, Generalized coordinates and constraints, D'Alembert's Principle, Angular momentum, Noether's Theorem.

The scenario matrix

Every problem in this topic is one (or more) of these cells. The 8 worked examples below are labelled by the cell they cover, and together they hit all of them.

Cell	What makes it distinct	Covered by
A. Length coordinate	$q$ is a distance → $p_{j}$ is ordinary momentum, $Q_{j}$ is a real force	Ex 1, Ex 5
B. Angle coordinate	$q$ is an angle → $p_{j}$ is angular momentum, $Q_{j}$ is a torque	Ex 2, Ex 6
C. Sign of $Q_{j}$	force can be positive (driving) or negative (restoring)	Ex 2, Ex 4
D. Cyclic coordinate	$q$ absent from $L$ → $p_{j}$ conserved (Noether)	Ex 3
E. Force hidden in $T$	no potential, yet a generalized force appears from $\partial T / \partial q$	Ex 5
F. Zero / degenerate input	$\dot{θ} = 0$ , $r = 0$ , or a coordinate at an equilibrium	Ex 4, Ex 7
G. Limiting behaviour	small-angle limit, large-mass limit, $ω \to 0$	Ex 6 (small angle), Ex 5 ( $ω \to 0$ )
H. Real-world word problem	figure-skater / merry-go-round story	Ex 7
I. Exam twist	two coupled coordinates, a coupled cyclic momentum	Ex 8

Example 1 — Free particle in a line (Cell A)

The picture below fixes the coordinate: a single number $x$ measured along the rail. Nothing depends on where we put the origin, which is exactly why $x$ is cyclic.

Figure — Generalized momenta and generalized forces

Worked example Statement

A particle of mass $m = 2 kg$ moves freely along the $x$ -axis with velocity $\overset{x}{˙} = 3 m/s$ . No forces act. Find $p_{x}$ and $Q_{x}$ .

Forecast: Is $p_{x}$ just ordinary momentum? Is $Q_{x}$ exactly zero?

Step 1 — Write $L$ . With no potential, $L = T - V = \frac{1}{2} m \overset{x}{˙}^{2} - 0 = \frac{1}{2} m \overset{x}{˙}^{2}$ . Why this step? The Lagrangian is always $L = T - V$ ; here $V = 0$ , so $L$ is pure kinetic energy.

Step 2 — Generalized momentum. $p_{x} = \frac{\partial L}{\partial x ˙} = m \overset{x}{˙} = 2 \cdot 3 = 6 kg\cdotm/s$ . Why this step? This is the definition of $p_{j}$ ; for a length coordinate the "mass factor" is just $m$ , so it reduces to ordinary momentum.

Step 3 — Generalized force. From the boxed identity above, $Q_{x} = \frac{\partial L}{\partial x}$ . Since $T = \frac{1}{2} m \overset{x}{˙}^{2}$ has no $x$ in it, $\frac{\partial T}{\partial x} = 0$ , and $V = 0$ , so $Q_{x} = \frac{\partial L}{\partial x} = \frac{\partial T}{\partial x} - \frac{\partial V}{\partial x} = 0 - 0 = 0$ . Why this step? The generalized force always equals $\partial L / \partial x$ (that is what $\overset{p}{˙}_{x} = \partial L / \partial x$ says). Here the Lagrangian has no $x$ at all, so its derivative vanishes: no force, and $x$ is cyclic.

Verify: Units of $p_{x}$ : $kg \cdot m/s$ ✓ (ordinary momentum). $\overset{p}{˙}_{x} = Q_{x} = 0$ , so momentum is constant → the particle keeps sliding, as Newton demands. ✓

Example 2 — Pendulum at a generic angle (Cells B, C)

Look at the figure: the dashed line at the pivot height is our potential-energy reference level ( $V = 0$ there); the red arrow is the gravity torque trying to pull $θ$ back to zero.

Worked example Solution

Forecast: Will $Q_{θ}$ come out negative (restoring) here? What are its units?

Step 1 — Lagrangian. Combine kinetic and potential energy as $L = T - V = \frac{1}{2} m ℓ^{2} \dot{θ}^{2} - (- m g ℓ cos θ) = \frac{1}{2} m ℓ^{2} \dot{θ}^{2} + m g ℓ cos θ$ . Why this step? $T = \frac{1}{2} m ℓ^{2} \dot{θ}^{2}$ (speed $= ℓ \dot{θ}$ ). For $V$ we measure height $y = - ℓ cos θ$ from the pivot level (the dashed line in the figure, where $V = 0$ ), so $V = m g y = - m g ℓ cos θ$ . Subtracting this in $L = T - V$ flips the sign, which is why the potential enters $L$ with a plus.

Step 2 — Generalized momentum. $p_{θ} = \frac{\partial L}{\partial θ ˙} = m ℓ^{2} \dot{θ} = 1 \cdot 4 \cdot 0.5 = 2 kg\cdotm^{2} / s$ . Why this step? The coordinate is an angle, so its conjugate is angular momentum — units $kg\cdotm^{2} / s$ , not $kg\cdotm/s$ .

Step 3 — Generalized force (a torque). By the boxed identity, $Q_{θ} = \frac{\partial L}{\partial θ} = \frac{\partial T}{\partial θ} - \frac{\partial V}{\partial θ}$ . Here $T = \frac{1}{2} m ℓ^{2} \dot{θ}^{2}$ contains no bare $θ$ , so $\partial T / \partial θ = 0$ and the whole force comes from the potential: $Q_{θ} = - \frac{\partial V}{\partial θ} = - m g ℓ sin θ = - 1 \cdot 9.8 \cdot 2 \cdot sin 3 0^{\circ} = - 9.8 N\cdotm$ . Why this step? We must show why the shortcut $- \partial V / \partial θ$ is legal here: because $\partial T / \partial θ = 0$ , the general $\partial L / \partial θ$ reduces to it. Angle coordinate ⇒ $Q_{θ}$ is a torque; the minus sign says gravity fights the increase of $θ$ — it is restoring (Cell C, negative sign).

Verify: Units: $m ℓ^{2} \dot{θ} = kg\cdotm^{2} / s$ ✓; $m g ℓ sin θ = N\cdotm$ ✓. Sign: at $θ = 3 0^{\circ} > 0$ the bob is to the right, gravity pulls it back left → negative torque ✓.

Example 3 — Central force, cyclic angle (Cell D)

The figure shows the polar coordinates: $r$ (the distance) and $θ$ (the angle swept). Because the force points straight along $r$ , spinning the whole picture about the centre changes nothing — that rotational symmetry is what makes $θ$ cyclic.

Worked example Solution

A planet mass $m$ moves in a plane under a central potential $V (r)$ (depends only on distance $r$ ). Use polar coordinates $(r, θ)$ : $T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2})$ . Show that $p_{θ}$ is conserved and, for $m = 1$ , $r = 2$ , $\dot{θ} = 1.5$ , compute its value.

Forecast: Which coordinate is cyclic — $r$ or $θ$ ? What quantity does the conserved $p_{θ}$ represent physically?

Step 1 — Lagrangian. $L = T - V = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2}) - V (r)$ . Why this step? Standard $L = T - V$ ; note $V$ has no $θ$ in it because the force is central.

Step 2 — Spot the cyclic coordinate. $θ$ appears in $L$ only as $\dot{θ}$ , never as $θ$ itself. So $\frac{\partial L}{\partial θ} = 0$ . Why this step? Noether's Theorem: a coordinate absent from $L$ signals a symmetry (here rotational, visible as "spin the figure"), which forces a conservation law.

Step 3 — Conserved momentum. Euler–Lagrange gives $\overset{p}{˙}_{θ} = \partial L / \partial θ = 0$ , so $p_{θ} = m r^{2} \dot{θ}$ is constant. Numerically $p_{θ} = 1 \cdot 2^{2} \cdot 1.5 = 6 kg\cdotm^{2} / s$ . Why this step? $\overset{p}{˙}_{θ} = 0$ means $p_{θ}$ never changes — this is Angular momentum conservation.

Verify: Units $kg\cdotm^{2} / s$ ✓. Physically $p_{θ} = L_{ang}$ — the planet sweeps equal areas in equal times (Kepler's 2nd law), a direct consequence. ✓

Example 4 — Degenerate input: pendulum at the bottom (Cells C, F)

The figure plots the torque $Q_{θ} (θ) = - m g ℓ sin θ$ . Notice it crosses zero at $θ = 0$ with a downward (negative) slope — the red tangent line — which is the signature of a stable equilibrium.

Worked example Solution

Take the pendulum of Ex 2 but at the equilibrium point $θ = 0$ (hanging straight down), momentarily at rest $\dot{θ} = 0$ . Find $p_{θ}$ and $Q_{θ}$ .

Forecast: Both zero? Or does one of them survive at equilibrium?

Step 1 — Momentum. $p_{θ} = m ℓ^{2} \dot{θ} = m ℓ^{2} \cdot 0 = 0$ . Why this step? $p_{θ} \propto \dot{θ}$ ; with zero angular velocity there is zero angular momentum. This is the degenerate velocity input.

Step 2 — Force. $Q_{θ} = - m g ℓ sin (0) = 0$ . Why this step? $sin 0 = 0$ — at the bottom the gravity torque vanishes. This is exactly the equilibrium condition (Cell F).

Step 3 — Interpret. $\overset{p}{˙}_{θ} = Q_{θ} = 0$ at this instant, so no angular acceleration right now. But it is a stable equilibrium: nudge $θ$ positive and $Q_{θ} = - m g ℓ sin θ < 0$ pushes back (see the negative slope in the figure). Why this step? We check the sign of $Q_{θ}$ near the equilibrium, not just at it, to classify stability.

Verify: At $θ = 0$ : $sin θ = 0 \Rightarrow Q_{θ} = 0$ ✓. Slope $\frac{d Q _{θ}}{d θ}_{0} = - m g ℓ cos 0 = - m g ℓ < 0$ → restoring → stable ✓.

Example 5 — Bead on a rotating wire: force from $T$ (Cells A, E, G)

The figure shows the spinning wire (angular rate $ω$ ) and the bead at distance $r$ . The red arrow is the outward "centrifugal" generalized force — it comes from $\partial T / \partial r$ , not from any potential.

Worked example Solution

A bead mass $m = 0.5 kg$ slides on a frictionless straight wire that is forced to spin about the origin at constant angular rate $ω = 4 rad/s$ ( $θ = ω t$ ). Coordinate: $r$ . Then $T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} ω^{2})$ , $V = 0$ . Find $p_{r}$ and the generalized force $Q_{r}$ at $r = 1 m$ .

Forecast: With $V = 0$ , is $Q_{r}$ zero? Or does a force appear from somewhere else?

Step 1 — Lagrangian. $L = T - V = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} ω^{2}) - 0 = \frac{1}{2} m \overset{r}{˙}^{2} + \frac{1}{2} m r^{2} ω^{2}$ . Why this step? We must write $L$ explicitly before differentiating; $V = 0$ so $L$ is pure kinetic energy, but crucially it contains $r$ (through the $r^{2} ω^{2}$ term).

Step 2 — Momentum. $p_{r} = \frac{\partial L}{\partial r ˙} = m \overset{r}{˙}$ . Why this step? $r$ is a length coordinate (Cell A), so $p_{r}$ is ordinary linear momentum along the wire.

Step 3 — Generalized force. By the boxed identity $Q_{r} = \frac{\partial L}{\partial r} = \frac{\partial T}{\partial r} - \frac{\partial V}{\partial r} = m r ω^{2} - 0 = m r ω^{2}$ . At $r = 1$ : $Q_{r} = 0.5 \cdot 1 \cdot 4^{2} = 8 N$ , pointing outward. Why this step? Even with $V = 0$ , the generalized force $\partial L / \partial r$ is nonzero because $T$ itself depends on $r$ (Cell E). This is the centrifugal effect, and it is legitimately called $Q_{r}$ — not a special " $F_{eff}$ " — because $Q_{r} = \partial L / \partial q$ is the definition.

Step 4 — Equation of motion & limiting check ( $ω \to 0$ ). Euler–Lagrange: $\overset{p}{˙}_{r} = Q_{r} \Rightarrow m \overset{r}{¨} = m r ω^{2}$ . If the wire stops spinning, $Q_{r} = m r \cdot 0 = 0$ — a free bead with no force (Cell G), recovering Ex 1. Why this step? Every formula must reduce to the known simpler case in its limit; $ω \to 0$ recovers the force-free particle.

Verify: $Q_{r} = 0.5 \cdot 1 \cdot 16 = 8 N$ ✓ ; at $ω = 0$ , $Q_{r} = 0$ ✓. Units $kg \cdot m \cdot s^{- 2} = N$ ✓ (length coordinate → real force).

Example 6 — Small-angle pendulum torque (Cells B, G)

The figure overlays the exact torque $- m g ℓ sin θ$ (black) with the linear approximation $- m g ℓ θ$ (red). Near $θ = 0$ they are almost indistinguishable — that is why tiny swings are simple harmonic.

Worked example Solution

For the Ex 2 pendulum ( $m = 1$ , $ℓ = 2$ , $g = 9.8$ ), find the restoring torque at a small angle $θ = 0.05 rad$ , and compare the exact value to the linear approximation $Q_{θ} \approx - m g ℓ θ$ .

Forecast: How close is $sin θ$ to $θ$ for a $0.05$ rad tilt? Under $0.1%$ error?

Step 1 — Exact torque. $Q_{θ} = - m g ℓ sin θ = - 1 \cdot 9.8 \cdot 2 \cdot sin (0.05)$ . Why this step? Angle coordinate ⇒ torque (Cell B); we keep the full $sin$ first.

Step 2 — Linearize. For small $θ$ , $sin θ \approx θ$ , giving $Q_{θ} \approx - m g ℓ θ = - 19.6 \cdot 0.05 = - 0.98 N\cdotm$ . Why this step? The small-angle limit (Cell G) turns the pendulum into a simple harmonic oscillator; this is why "SHM" holds for tiny swings.

Step 3 — Error size. Exact $- 19.6 sin (0.05) = - 0.979837 \dots$ vs approx $- 0.98$ . Relative error $\approx 0.084%$ . Why this step? We must quantify how good the approximation is, not just assert it.

Verify: $∣ Q_{exact} - Q_{approx} ∣/∣ Q_{approx} ∣ \approx 0.00084 < 0.001$ ✓. The linear model is excellent below a few degrees. ✓

Example 7 — Figure skater word problem (Cells F, H)

Look at the figure: the red arrow is angular momentum $p_{θ}$ — the SAME length before and after.

Worked example Solution

Forecast: Does she spin faster or slower? Does $\dot{θ}$ stay constant (a common trap)?

Step 1 — Identify the conserved quantity. $θ$ is cyclic (no external torque, $\partial L / \partial θ = 0$ ), so $p_{θ} = I \dot{θ}$ is conserved. Why this step? It is the momentum $p_{θ}$ that is conserved, not the velocity $\dot{θ}$ — this is the classic mistake.

Step 2 — Apply conservation. $I_{1} \dot{θ}_{1} = I_{2} \dot{θ}_{2} \Rightarrow \dot{θ}_{2} = \frac{I _{1} θ ˙ _{1}}{I _{2}} = \frac{4 \cdot 2}{1} = 8 rad/s$ . Why this step? Equating $p_{θ}$ before and after directly solves for the new rate.

Step 3 — Degenerate check ( $I_{2} = I_{1} = 4$ ). Then $\dot{θ}_{2} = \frac{4 \cdot 2}{4} = 2 rad/s = \dot{θ}_{1}$ : no change of inertia → no change of rate (Cell F). Why this step? A sanity limit — if she does nothing, nothing should change.

Verify: $p_{θ}$ before $= 4 \cdot 2 = 8$ ; after $= 1 \cdot 8 = 8$ ✓ conserved. She spins 4× faster ( $8$ rad/s vs $2$ rad/s) because $I$ dropped 4×; the degenerate case correctly returns $2$ rad/s. ✓ Note the trap: the velocity $\dot{θ}$ changed even though the momentum $p_{θ}$ did not.

Example 8 — Exam twist: coupled cart–pendulum cyclic momentum (Cell I)

The figure shows the cart (coordinate $x$ ) with the hanging bob (angle $θ$ ). Sliding the whole apparatus left or right along the rail changes no energy, so $x$ is cyclic even though the bob feels gravity through $θ$ .

Worked example Solution

A cart mass $M = 3 kg$ slides freely on a horizontal rail (coordinate $x$ ); a bob mass $m = 1 kg$ hangs from it on a light rod, angle $θ$ . No horizontal external force. The horizontal generalized momentum is $p_{x} = (M + m) \overset{x}{˙} + m ℓ \dot{θ} cos θ .$ At an instant $\overset{x}{˙} = 0.4 m/s$ , $ℓ = 2$ , $\dot{θ} = 0.3 rad/s$ , $θ = 6 0^{\circ}$ . Compute $p_{x}$ and state why it is conserved.

Forecast: Is $x$ cyclic even though the pendulum clearly feels gravity? What survives?

Step 1 — Check cyclicity of $x$ . Gravity acts vertically; the potential $V = - m g ℓ cos θ$ depends on $θ$ only, not $x$ . Kinetic energy depends on $\overset{x}{˙}$ but not $x$ . So $\partial L / \partial x = 0$ → $x$ is cyclic. Why this step? Only $x$ -dependence in $L$ matters; the bob's gravity enters through $θ$ , leaving $x$ ignorable. Hence horizontal momentum is conserved.

Step 2 — Evaluate $p_{x}$ . $p_{x} = (3 + 1) \cdot 0.4 + 1 \cdot 2 \cdot 0.3 \cdot cos 6 0^{\circ} = 1.6 + 0.6 \cdot 0.5 = 1.6 + 0.3 = 1.9 kg\cdotm/s$ . Why this step? Just substitute; the $cos θ$ weight is the $\partial r / \partial q$ -style projection of the bob's motion onto the horizontal.

Step 3 — Meaning. $\overset{p}{˙}_{x} = \partial L / \partial x = 0$ , so this $1.9 kg\cdotm/s$ stays fixed while $x$ and $θ$ evolve, exchanging momentum between cart and bob. Why this step? Conservation lets you predict the cart's recoil without solving the full coupled equations.

Verify: $p_{x} = 4 \cdot 0.4 + 2 \cdot 0.3 \cdot 0.5 = 1.6 + 0.3 = 1.9$ ✓. Units $kg\cdotm/s$ ✓ (length coordinate → linear momentum). Conserved because $x$ absent from $L$ . ✓

Recall Quick self-test

Which cell is "no potential yet a force appears"? ::: Cell E — the force hides in $\partial T / \partial q$ (Ex 5, centrifugal term). For a cyclic coordinate, what is constant — velocity or momentum? ::: The momentum $p_{j}$ (Ex 3, Ex 7). Velocity generally changes. What are the units of $Q_{θ}$ for an angle coordinate? ::: Torque, $N\cdotm$ (Ex 2). In the skater problem, why does she spin faster? ::: $p_{θ} = I \dot{θ}$ is conserved; $I$ dropped 4× so $\dot{θ}$ rose 4× (Ex 7).