Kinetic energy in generalized coordinates: Worked examples

Before anything, let me re-anchor the three symbols we will use constantly, in plain words, so nobody is lost from line one.

Two jargon words appear in the table below; define them once here so the matrix reads cleanly.

And the master formula we will apply mechanically each time:

The scenario matrix

Every problem this topic can hand you falls into one of these cells. The last column names the example that covers it. ("Scleronomic" and "rheonomic" are defined just above.)

#	Cell (scenario class)	$\partial_{t} r = 0$ ?	Which pieces survive	Twist to watch	Covered by
A	Flat Cartesian, no constraint	yes (scleronomic)	$T_{2}$ only, $M$ constant	trivial baseline	Ex 1
B	Curvilinear, scleronomic (polar/spherical)	yes	$T_{2}$ only, $M = M (q)$	$M$ depends on position	Ex 2
C	Off-diagonal $M_{ij} \neq = 0$ (skewed coords)	yes	$T_{2}$ with cross term	don't drop the factor-2 pairing	Ex 3
D	Rheonomic, rotation → $T_{0}$ only	no (rheonomic)	$T_{2} + T_{0}$ , $T_{1} = 0$	why does $T_{1}$ vanish here	Ex 4
E	Rheonomic, translation → $T_{1}$ and $T_{0}$	no (rheonomic)	$T_{2} + T_{1} + T_{0}$ all present	linear term is real	Ex 5
F	Curved surface (sphere)	yes	$T_{2}$ , two coupled coords	metric of a 2-sphere	Ex 6
G	Degenerate / limiting input ( $r \to 0$ , $ω \to 0$ )	either	pieces collapse	matrix loses a direction	Ex 7
H	Real-world word problem (elevator + pendulum)	no (rheonomic)	all three pieces	reading physics into the map	Ex 8
I	Exam twist (verify Euler's theorem / find $H$ )	yes	uses $T_{2}$ property	$\sum \overset{q}{˙} \partial_{\overset{q}{˙}} T = 2 T$	Ex 9

Example 1 — Cell A: the flat baseline

Worked example Free particle in the plane, Cartesian coordinates

One particle, mass $m$ , coordinates $q = (x, y)$ , so $r = (x, y)$ , no time in the map. Forecast: guess what $T$ is before reading — you already know it from grade-school physics.

Step 1. Compute the direction vectors $\partial r / \partial q_{i}$ . $\frac{\partial r}{\partial x} = (1, 0), \frac{\partial r}{\partial y} = (0, 1) .$ Why this step? $M_{ij}$ is built by dotting these together, so we need them first. Each says "if only $x$ changes, which way does the point move" — straight along the axis, as the picture shows.

Step 2. Build $M_{ij}$ . $M_{xx} = m (1, 0) \cdot (1, 0) = m, M_{y y} = m, M_{x y} = m (1, 0) \cdot (0, 1) = 0.$ Why this step? We plug the direction vectors into the definition. The zero says the axes are perpendicular — no cross term.

Step 3. Assemble $T$ . $T = \frac{1}{2} m \overset{x}{˙}^{2} + \frac{1}{2} m \overset{y}{˙}^{2} .$ Why this step? No $t$ in $r$ ⟹ $a_{i} = 0$ , $T_{0} = 0$ . Only $T_{2}$ survives.

Verify: This is literally $\frac{1}{2} m ∣ v ∣^{2}$ with $v = (\overset{x}{˙}, \overset{y}{˙})$ . Units: $kg \cdot (m/s)^{2} = J$ . ✓

Example 2 — Cell B: polar, the position-dependent metric

Figure — Kinetic energy in generalized coordinates

Step 1. Direction vectors (look at the figure: the two coloured arrows at the point). $\frac{\partial r}{\partial r} = (cos θ, sin θ), \frac{\partial r}{\partial θ} = (- r sin θ, r cos θ) .$ Why this step? The blue arrow ( $\partial_{r}$ ) points outward — moving $r$ slides you along the ray. The orange arrow ( $\partial_{θ}$ ) points sideways and is $r$ times longer far from the centre, because one radian of angle sweeps a bigger arc when $r$ is big.

Step 2. Dot them for $M_{ij}$ . $M_{r r} = m (cos^{2} θ + sin^{2} θ) = m,$ $M_{θ θ} = m r^{2} (sin^{2} θ + cos^{2} θ) = m r^{2},$ $M_{r θ} = m [cos θ (- r sin θ) + sin θ (r cos θ)] = 0.$ Why this step? $M_{r θ} = 0$ because the blue and orange arrows are perpendicular — the geometry, not luck.

Step 3. $T = \frac{1}{2} m \overset{r}{˙}^{2} + \frac{1}{2} m r^{2} \dot{θ}^{2} .$ Why the $r^{2}$ ? The longer orange arrow means the same $\dot{θ}$ produces more real speed far out — the metric records this.

Verify: At radius $r$ , tangential speed is $r \dot{θ}$ , radial speed $\overset{r}{˙}$ ; perpendicular, so $v^{2} = \overset{r}{˙}^{2} + r^{2} \dot{θ}^{2}$ , giving $T = \frac{1}{2} m v^{2}$ . ✓ Units of $r^{2} \dot{θ}^{2}$ : $m^{2} \cdot s^{- 2}$ ✓.

Example 3 — Cell C: a genuinely off-diagonal metric

Worked example Two masses on a shared coordinate (skewed system)

Two particles on a line: $x_{1} = q_{1}, x_{2} = q_{1} + q_{2}$ , masses $m_{1} = m_{2} = m$ . So a change in $q_{1}$ moves both particles. Forecast: will $M_{12}$ be zero? Think: does moving $q_{1}$ share a direction with moving $q_{2}$ ?

Step 1. Direction vectors, treating the state as $(x_{1}, x_{2})$ : $\frac{\partial r}{\partial q _{1}} = (1, 1), \frac{\partial r}{\partial q _{2}} = (0, 1) .$ Why this step? Moving $q_{1}$ pushes particle 1 by 1 and particle 2 by 1 (both feel it); moving $q_{2}$ only pushes particle 2.

Step 2. Metric (each dot weighted by that particle's mass — here both $m$ ): $M_{11} = m (1)^{2} + m (1)^{2} = 2 m, M_{22} = m (0)^{2} + m (1)^{2} = m, M_{12} = m (1) (0) + m (1) (1) = m .$ Why this step? The shared direction (particle 2) makes $M_{12} = m \neq = 0$ : the coordinates are coupled.

Step 3. Assemble, remembering $M_{12}$ appears for both $(1, 2)$ and $(2, 1)$ : $T = \frac{1}{2} (M_{11} \overset{q}{˙}_{1}^{2} + M_{22} \overset{q}{˙}_{2}^{2}) + M_{12} \overset{q}{˙}_{1} \overset{q}{˙}_{2} = \frac{1}{2} (2 m) \overset{q}{˙}_{1}^{2} + \frac{1}{2} m \overset{q}{˙}_{2}^{2} + m \overset{q}{˙}_{1} \overset{q}{˙}_{2} .$ $T = m \overset{q}{˙}_{1}^{2} + \frac{1}{2} m \overset{q}{˙}_{2}^{2} + m \overset{q}{˙}_{1} \overset{q}{˙}_{2} .$ Why the single (not double) cross term? The $\frac{1}{2} \sum_{i, j}$ counts $(1, 2)$ and $(2, 1)$ separately; together they give $\frac{1}{2} (M_{12} + M_{21}) \overset{q}{˙}_{1} \overset{q}{˙}_{2} = M_{12} \overset{q}{˙}_{1} \overset{q}{˙}_{2}$ . Don't add another factor of 2.

Verify: Directly, $\overset{x}{˙}_{1} = \overset{q}{˙}_{1}, \overset{x}{˙}_{2} = \overset{q}{˙}_{1} + \overset{q}{˙}_{2}$ , so $T = \frac{1}{2} m \overset{q}{˙}_{1}^{2} + \frac{1}{2} m (\overset{q}{˙}_{1} + \overset{q}{˙}_{2})^{2} = m \overset{q}{˙}_{1}^{2} + \frac{1}{2} m \overset{q}{˙}_{2}^{2} + m \overset{q}{˙}_{1} \overset{q}{˙}_{2} . ✓$

Example 4 — Cell D: rotating wire ( $T_{0}$ appears, $T_{1}$ does not)

Step 1. The two kinds of derivative: $\frac{\partial r}{\partial r} = (cos ω t, sin ω t) (radial), \frac{\partial r}{\partial t} = (- r ω sin ω t, r ω cos ω t) (tangential, from the spin) .$ Why this step? $\partial_{r}$ = "how position moves if I slide the bead out"; $\partial_{t}$ = "how position moves because the wire itself rotates while the bead sits still". The figure shows these as perpendicular blue (radial) and orange (tangential) arrows.

Step 2 — the $T_{2}$ piece. $M_{r r} = m (cos^{2} ω t + sin^{2} ω t) = m$ , so $T_{2} = \frac{1}{2} m \overset{r}{˙}^{2}$ . Why this step? $T_{2}$ needs only the $q$ -direction vector dotted with itself; the length of the blue radial arrow is $1$ , giving the plain mass $m$ — the sliding-along-the-wire energy.

Step 3 — the $T_{1}$ piece. $a_{r} = m \partial_{r} r \cdot \partial_{t} r = m [cos ω t (- r ω sin ω t) + sin ω t (r ω cos ω t)] = 0$ . Why this step? $T_{1}$ is the overlap between "sliding out" and "being spun". The radial arrow (blue) and the spin arrow (orange) are perpendicular in the figure, so their dot product — and hence $a_{r}$ — vanishes.

Step 4 — the $T_{0}$ piece. $T_{0} = \frac{1}{2} m ∣ \partial_{t} r ∣^{2} = \frac{1}{2} m (r ω)^{2}$ . Why this step? Even a bead sitting still on the wire ( $\overset{r}{˙} = 0$ ) is dragged in a circle at speed $r ω$ ; $T_{0}$ is exactly that "frozen-but-still-moving" energy.

Step 5 — assemble. $T = \frac{1}{2} m \overset{r}{˙}^{2} + \frac{1}{2} m r^{2} ω^{2} .$ Why $T_{0}$ acts like a potential: it depends on $r$ but not $\overset{r}{˙}$ ; in Euler-Lagrange equations it produces the outward centrifugal push.

Verify: Full speed $v^{2} = \overset{r}{˙}^{2} + (r ω)^{2}$ (radial + tangential, perpendicular), so $T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} ω^{2})$ . ✓

Example 5 — Cell E: sliding frame ( $T_{1}$ and $T_{0}$ both nonzero)

Worked example Coordinate riding a frame that moves at constant

$v$ $x = q + v t, y = w$ ; coordinates $(q, w)$ , mass $m$ , constant speed $v$ . Rheonomic again — the frame is dragged. Forecast: you should now expect all three pieces. Which term carries the $v$ linearly?

Step 1. Direction vectors: $\frac{\partial r}{\partial q} = (1, 0), \frac{\partial r}{\partial w} = (0, 1), \frac{\partial r}{\partial t} = (v, 0) .$ Why this step? The frame drags the $x$ -direction at speed $v$ , so $\partial_{t} r = (v, 0)$ — this time it's parallel to $\partial_{q} r = (1, 0)$ , unlike the rotating case.

Step 2 — the $T_{2}$ piece. $M_{q q} = m (1)^{2} = m, M_{w w} = m (1)^{2} = m, M_{q w} = m (1) (0) + m (0) (1) = 0$ , so $T_{2} = \frac{1}{2} m (\overset{q}{˙}^{2} + \overset{w}{˙}^{2})$ . Why this step? Both coordinate directions are unit vectors along perpendicular axes, so $M$ is just $m$ times the identity — the free-particle quadratic core, unchanged by the drag.

Step 3 — the $T_{1}$ piece. $a_{q} = m \partial_{q} r \cdot \partial_{t} r = m (1) (v) = m v$ , and $a_{w} = m (0) (v) = 0$ , so $T_{1} = m v \overset{q}{˙}$ . Why this step? $T_{1}$ is nonzero now because $\partial_{q} r$ and $\partial_{t} r$ are parallel — full overlap, opposite of the perpendicular rotating case in Ex 4.

Step 4 — the $T_{0}$ piece. $T_{0} = \frac{1}{2} m ∣ \partial_{t} r ∣^{2} = \frac{1}{2} m v^{2}$ . Why this step? Even sitting still in the frame ( $\overset{q}{˙} = \overset{w}{˙} = 0$ ), the particle is carried along at speed $v$ ; that carried energy is $T_{0}$ .

Step 5 — assemble. $T = \frac{1}{2} m (\overset{q}{˙}^{2} + \overset{w}{˙}^{2}) + m v \overset{q}{˙} + \frac{1}{2} m v^{2} = \frac{1}{2} m [(\overset{q}{˙} + v)^{2} + \overset{w}{˙}^{2}] .$ Verify: $\overset{x}{˙} = \overset{q}{˙} + v, \overset{y}{˙} = \overset{w}{˙}$ , so $T = \frac{1}{2} m [(\overset{q}{˙} + v)^{2} + \overset{w}{˙}^{2}]$ . ✓ Pure bookkeeping of $\frac{1}{2} m \overset{x}{˙}^{2}$ , no new physics.

Example 6 — Cell F: particle on a sphere (curved surface)

Step 1. Direction vectors: $\frac{\partial r}{\partial θ} = R (cos θ cos ϕ, cos θ sin ϕ, - sin θ),$ $\frac{\partial r}{\partial ϕ} = R (- sin θ sin ϕ, sin θ cos ϕ, 0) .$ Why this step? $\partial_{θ}$ moves you along a meridian (toward/away from the pole); $\partial_{ϕ}$ moves you along a circle of latitude — and that circle shrinks near the pole, shown by the shorter orange arrow up high in the figure.

Step 2 — diagonal metric entries. $M_{θ θ} = m R^{2} (cos^{2} θ cos^{2} ϕ + cos^{2} θ sin^{2} ϕ + sin^{2} θ) = m R^{2},$ $M_{ϕϕ} = m R^{2} sin^{2} θ (sin^{2} ϕ + cos^{2} ϕ) = m R^{2} sin^{2} θ .$ Why $M_{ϕϕ} = m R^{2} sin^{2} θ$ ? The latitude circle has radius $R sin θ$ , so its speed is $(R sin θ) \dot{ϕ}$ ; squaring gives the $sin^{2} θ$ .

Step 3 — the off-diagonal entry. $M_{θ ϕ} = m R^{2} (- cos θ sin ϕ sin θ cos ϕ + cos θ cos ϕ sin θ sin ϕ + 0) = 0.$ Why $M_{θ ϕ} = 0$ ? The meridian direction ( $\partial_{θ}$ ) and the latitude direction ( $\partial_{ϕ}$ ) are perpendicular everywhere on the sphere — north-south crosses east-west at right angles — so their dot product cancels term by term.

Step 4 — assemble. $T = \frac{1}{2} m R^{2} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2}) .$ Verify: At the equator $θ = π /2$ , $sin^{2} θ = 1$ : full-size latitude circle, term is $\frac{1}{2} m R^{2} \dot{ϕ}^{2}$ . At the pole $θ \to 0$ , $sin^{2} θ \to 0$ : spinning $ϕ$ costs no energy because you're standing on the axis. ✓ (This limiting behaviour is exactly Cell G below.)

Example 7 — Cell G: degenerate and limiting inputs

Worked example What happens as a coordinate collapses

Take the polar result $T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2})$ (Ex 2) and probe the edges. Forecast: what does the metric matrix $M = (m 0 0 m r^{2})$ do as $r \to 0$ ?

Step 1 — limit $r \to 0$ . $M_{θ θ} = m r^{2} \to 0$ . Why this step? We slide the input toward the origin to test where the coordinate system breaks. The energy cost of any $\dot{θ}$ vanishes, and $M$ becomes $(m 0 00)$ — not positive-definite (it has a zero eigenvalue). Geometrically, at the origin the angle $θ$ is undefined: all directions coincide. The coordinate system is degenerate there.

Step 2 — limit $ω \to 0$ in the rotating wire (Ex 4). $T = \frac{1}{2} m \overset{r}{˙}^{2} + \frac{1}{2} m r^{2} ω^{2} \to \frac{1}{2} m \overset{r}{˙}^{2}$ . Why this step? We switch off the spin to check the rheonomic machinery reduces correctly. No spin means no moving constraint: $T_{0}$ disappears and we fall back to a scleronomic (Cell A-like) system. The rheonomic apparatus quietly becomes the simple one — a reassuring consistency check.

Step 3 — bounded scaling in Ex 6, then a determinant test. On the fixed sphere $M_{ϕϕ} = m R^{2} sin^{2} θ$ is bounded by $m R^{2}$ ; nothing blows up as it would in the unbounded plane. For the polar metric, compute the determinant to pin down exactly where degeneracy strikes: $det M = m \cdot m r^{2} = m^{2} r^{2} .$ Why this step? A vanishing determinant is the precise, sign-independent signal that the matrix has lost a direction (a zero eigenvalue). $det M > 0$ for $r > 0$ (invertible, healthy) and $det M = 0$ exactly at $r = 0$ (the coordinate singularity).

Verify: $M$ is positive-definite everywhere except the single point $r = 0$ , matching the geometric story that only the origin is degenerate. ✓

Example 8 — Cell H: real-world word problem

Worked example Pendulum hanging inside an elevator that accelerates

A pendulum of length $ℓ$ , bob mass $m$ , hangs from the ceiling of an elevator whose floor height is $Y (t) = \frac{1}{2} A t^{2}$ (constant upward acceleration $A$ ). Angle $φ$ from vertical is the coordinate. Pivot position: $(0, Y (t))$ . Bob position: $x = ℓ sin φ, y = Y (t) - ℓ cos φ .$ Forecast: all three pieces or not? The elevator moves on its own schedule (rheonomic) — expect $T_{1}$ and $T_{0}$ .

Step 1. Derivatives: $\frac{\partial r}{\partial φ} = (ℓ cos φ, ℓ sin φ), \frac{\partial r}{\partial t} = (0, \dot{Y}) = (0, A t) .$ Why this step? $\partial_{φ}$ swings the bob; $\partial_{t}$ lifts the whole pendulum with the elevator (only a $y$ -component).

Step 2 — the $T_{2}$ piece. $M_{φφ} = m (ℓ^{2} cos^{2} φ + ℓ^{2} sin^{2} φ) = m ℓ^{2}$ , so $T_{2} = \frac{1}{2} m ℓ^{2} \overset{φ}{˙}^{2}$ . Why this step? $T_{2}$ is the swinging energy; the bob traces a circle of radius $ℓ$ , so its speed is $ℓ \overset{φ}{˙}$ and the $cos^{2} + sin^{2} = 1$ collapses the two components into one clean $m ℓ^{2}$ .

Step 3 — the $T_{1}$ piece. $a_{φ} = m \partial_{φ} r \cdot \partial_{t} r = m (ℓ cos φ) (0) + m (ℓ sin φ) (A t) = m A t ℓ sin φ$ , so $T_{1} = m A t ℓ \overset{φ}{˙} sin φ$ . Why $T_{1} \neq = 0$ here but was $0$ for the rotating wire? The lift direction $(0, A t)$ overlaps the swing direction through its $y$ -part $ℓ sin φ$ — a nonzero dot product whenever the bob is off-vertical.

Step 4 — the $T_{0}$ piece. $T_{0} = \frac{1}{2} m ∣ \partial_{t} r ∣^{2} = \frac{1}{2} m (A t)^{2}$ . Why this step? Even a bob hanging straight and still is carried up by the elevator at speed $A t$ ; that carried energy is $T_{0}$ .

Step 5 — assemble. $T = \frac{1}{2} m ℓ^{2} \overset{φ}{˙}^{2} + m A t ℓ \overset{φ}{˙} sin φ + \frac{1}{2} m A^{2} t^{2} .$ Verify: Directly $\overset{x}{˙} = ℓ \overset{φ}{˙} cos φ, \overset{y}{˙} = A t + ℓ \overset{φ}{˙} sin φ$ . $T = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2}) = \frac{1}{2} m [ℓ^{2} \overset{φ}{˙}^{2} cos^{2} φ + (A t + ℓ \overset{φ}{˙} sin φ)^{2}] .$ Expanding: $\frac{1}{2} m ℓ^{2} \overset{φ}{˙}^{2} + m A t ℓ \overset{φ}{˙} sin φ + \frac{1}{2} m A^{2} t^{2}$ . ✓ (uses $cos^{2} + sin^{2} = 1$ ).

Example 9 — Cell I: the exam twist (Euler's theorem, and $H$ )

Worked example Confirm

$\sum_{i} \overset{q}{˙}_{i} \partial T / \partial \overset{q}{˙}_{i} = 2 T_{2} + T_{1}$ and find the energy function Use the elevator pendulum's $T$ from Ex 8. See Energy function and Hamiltonian and Generalized momentum and conserved quantities. Forecast: For a pure quadratic $T_{2}$ , Euler's theorem gives exactly $2 T$ . With $T_{1}$ present, guess the correction.

Step 1. Differentiate $T$ w.r.t. $\overset{φ}{˙}$ : $\frac{\partial T}{\partial φ ˙} = m ℓ^{2} \overset{φ}{˙} + m A t ℓ sin φ .$ Why this step? This is the generalized momentum $p_{φ}$ ; multiplying by $\overset{φ}{˙}$ builds the sum in Euler's theorem.

Step 2. Form $\overset{φ}{˙} \partial T / \partial \overset{φ}{˙}$ : $\overset{φ}{˙} \frac{\partial T}{\partial φ ˙} = m ℓ^{2} \overset{φ}{˙}^{2} + m A t ℓ \overset{φ}{˙} sin φ = 2 T_{2} + T_{1} .$ Why this step? Euler's theorem for a homogeneous function of degree $k$ gives $k \times (that piece)$ : degree 2 → factor 2 on $T_{2}$ , degree 1 → factor 1 on $T_{1}$ , degree 0 → nothing from $T_{0}$ .

Step 3. The energy function $h = \sum_{i} \overset{q}{˙}_{i} \partial T / \partial \overset{q}{˙}_{i} - L$ with $L = T - V$ becomes $h = (2 T_{2} + T_{1}) - (T_{2} + T_{1} + T_{0}) + V = T_{2} - T_{0} + V .$ Why this step? Only the pure-quadratic $T_{2}$ contributes its full weight; the $- T_{0}$ shows the energy function is not $T + V$ when constraints move (rheonomic).

Verify: For a scleronomic system ( $T_{1} = T_{0} = 0$ ), $h = T_{2} + V = T + V$ — the familiar total energy. ✓ For our elevator, $h = \frac{1}{2} m ℓ^{2} \overset{φ}{˙}^{2} - \frac{1}{2} m A^{2} t^{2} + V$ , which is explicitly time-dependent, so energy is not conserved — physically correct, the elevator motor injects energy.

Recall Which cell does each surviving-term pattern signal?

Only $T_{2}$ ::: scleronomic (no $t$ in the map) — Cells A, B, C, F, G(ω→0). $T_{2} + T_{0}$ , no $T_{1}$ ::: moving constraint whose motion is perpendicular to every coordinate direction (rotating wire) — Cell D. $T_{2} + T_{1} + T_{0}$ ::: moving constraint with a component along a coordinate direction (sliding frame, lifting elevator) — Cells E, H. Energy function $h = T + V$ ::: only when scleronomic ( $T_{1} = T_{0} = 0$ ) — Cell I.