Torque and Angular Momentum about a Fixed Axis

Push a heavy door at its handle and it swings open with two fingers. Push with the same force a hand-width from the hinge and you can lean your shoulder into it for nothing. Force alone evidently does not decide rotation — where and which way the force acts matter just as much. The quantity that packages all three is torque, and it plays the role for spinning that force plays for pushing: it is the thing that changes rotational motion. The previous article built the description of rotation — $\theta$ , $\omega$ , $\alpha$ , and their suvat algebra — while carefully postponing the question of cause. This article supplies it, and the payoff is the rotational half of Newtonian dynamics arriving in one sweep by the dictionary substitution: force becomes torque, mass becomes moment of inertia $I = \sum m_i r_i^2$ , and the second law reappears as $\tau = I\alpha$ . Crucially, this is not a new postulate. A rigid body is a swarm of ordinary particles each obeying plain $F = ma$ ; multiply each particle's equation by its radius, add up the swarm, and $\tau = I\alpha$ falls out — with the internal forces cancelling and the $r^2$ in the moment of inertia manufactured before your eyes. The same sweep delivers rotation's energy bookkeeping ( $W = \int\tau\,d\theta$ , $P = \tau\omega$ — the equation on every engine's spec sheet) and its most consequential new character: the angular momentum $L = I\omega$ , whose rate of change is the torque and whose conservation, when torques vanish, explains the spinning skater and half of astrophysics. Wrenches, flywheels, and figure skaters in one article — all of it Newton's laws, summed around an axis.

Key takeaways

Torque measures a force's effectiveness at turning: $\tau = rF\sin\phi = r_\perp F = rF_\perp$ — force times lever arm; only the component perpendicular to the radius turns anything.
Newton's second law for a rigid body about a fixed axis is $\tau_{\rm net} = I\alpha$ , with the moment of inertia $I = \sum_i m_i r_i^2$ playing the role of mass — derived by summing $F = ma$ over the body's particles, not decreed.
$I$ depends on the axis, not just the body: mass far from the axis costs quadratically more to spin up (hoop $MR^2$ vs disk $\tfrac12 MR^2$ ).
Work and power translate by the dictionary: $W = \int \tau\,d\theta$ , $\ P = \tau\omega$ , and the work–energy theorem becomes $W_{\rm net} = \Delta\!\left(\tfrac12 I\omega^2\right)$ .
The angular momentum about the axis is $L = I\omega$ , and the deepest form of the law is $\tau_{\rm net} = dL/dt$ : no net torque, $L$ frozen — the skater's spin-up and the flywheel's constancy.
Key idea: torque is to rotation what force is to translation — $\tau = I\alpha = dL/dt$ is $F = ma = dp/dt$ summed around an axis.

The core insight

About a fixed axis, the turning effectiveness of a force is $\tau = rF_\perp$ , and applying $F = ma$ to every particle of a rigid body and summing yields $\tau_{\rm net} = I\alpha = dL/dt$ with $I = \sum m_i r_i^2$ and $L = I\omega$ — the complete rotational translation of Newton's second law, energy theorem included ( $W = \int\tau\,d\theta$ , $P = \tau\omega$ ).

Where this lives today

$P = \tau\omega$ is the sentence the machine world is written in. An electric vehicle's spec sheet is a torque-versus-rpm curve; the motor delivers near-constant torque from zero speed, which is why EVs launch so hard, and the gearbox is a torque-exchange machine trading $\omega$ for $\tau$ at almost constant power. Wind turbines run the trade in reverse — enormous blade torque at a stately 10 rpm, geared up to generator speed — and every robot joint is a motor plus harmonic drive sized in newton-metres. Dynamometers, the instruments every engine and motor is certified on, measure exactly the two factors of equation (3) and multiply. The conservation law (6) runs the more spectacular applications. Reaction wheels aboard Hubble and JWST steer without propellant: motor torques a flywheel one way, spacecraft turns the other, total $L$ frozen. The same bookkeeping in reverse spins up neutron stars — a collapsing stellar core shrinks its $I$ by a factor of billions, and conservation turns a leisurely stellar rotation into a millisecond pulsar, the skater's trick at cosmic stakes. And at the human scale, a falling cat, a diver mid-somersault, and a gymnast twisting off the beam are all solving $L = I\omega =$ constant with reshapeable bodies — trading moment of inertia against spin rate on live television.

The physical picture

Feel the door again, and notice that three separate changes kill your effort. Slide your push toward the hinge: nothing turns — small $r$ . Keep the distance but angle your push along the door, toward the hinge line: nothing turns — you are pulling the door through its own plane, not around. Push perpendicular, at the far edge: maximum effect. Torque is the product that scores all three at once, $\tau = rF\sin\phi$ , and the geometry can be read two equivalent ways: either the full force times the lever arm $r_\perp = r\sin\phi$ (the perpendicular distance from the axis to the force's line of action — slide the force along its own line and nothing changes), or the full distance times the tangential component $F_\perp = F\sin\phi$ (the only part of the force with any interest in turning).

Figure

Torque geometry — lever arm and tangential component, two readings of one product

Why does inertia pick up a factor of radius squared? Because the radius taxes rotation twice. Take one particle of the body, mass $m$ at radius $r$ , and try to give the body angular acceleration $\alpha$ : the particle must gain tangential acceleration $a_t = r\alpha$ (farther means faster), which needs force $mr\alpha$ ; but your handle on it is torque, and its lever arm is $r$ again — so the torque bill is $mr^2\alpha$ . Once converting demand into linear terms and once converting your means into angular ones: $r$ enters twice, and the swarm's total, $I = \sum m_i r_i^2$ , weights every particle by the square of its distance. That is why a hoop (all mass at the rim) is twice as hard to spin up as a disk of equal mass and radius, why a figure skater's arms are her throttle, and why flywheels store their mass at the rim on purpose.

Figure

The swarm view — one shared alpha, torque bills summed over particles

The math: the dictionary's dynamic rows, earned

First the definition, in its full vector dress and its working fixed-axis form. About a chosen origin, the torque of a force $\vec F$ applied at position $\vec r$ is

\vec\tau = \vec r \times \vec F, \qquad \tau = rF\sin\phi = r_\perp F = rF_\perp \qquad (1)

where: $\phi$ — the angle between $\vec r$ and $\vec F$ ; $r_\perp = r\sin\phi$ — the lever arm; $F_\perp = F\sin\phi$ — the tangential component; units $\mathrm{N\,m}$ (dimensionally joules, but never called that — torque is not energy, and the distinction is exactly the $\sin\phi$ geometry). For rotation about a fixed axis, the working quantity is the component of $\vec\tau$ along that axis, with a sign: counterclockwise positive by convention. The vector form's direction (right-hand rule, perpendicular to both $\vec r$ and $\vec F$ ) becomes essential in the rigid-body chapters where the axis itself moves; here it politely reduces to " $+$ or $-$ about the axis."

Derivation ANewton's second law for rotation

We are deriving $\tau_{\rm net} = I\alpha$ — the rotational second law — from nothing but $F = ma$ applied to each particle of a rigid body, thereby earning the dictionary rows $F \to \tau$ and $m \to I$ rather than assuming them. The strategy is the swarm sum of the physical picture: write the tangential second law for one particle, trade its private linear acceleration for the shared $\alpha$ , convert force to torque by multiplying with the radius, and add over the body. Two things to watch: where the $r^2$ is manufactured (Steps 2–3, one $r$ from each trade), and why the fearsome sum of internal forces vanishes (Step 4) — without that cancellation the law would be useless, since we can never catalogue the interatomic forces.

Step 1. Newton's second law, tangentially, for one particle. A mass element $m_i$ at radius $r_i$ experiences some net tangential force $F_{t,i}$ (external plus internal) and obeys $F_{t,i} = m_i a_{t,i}$ . Why tangential only: radial forces point through the axis, have zero lever arm, and cannot influence the turning; they are the hinge's business.

Step 2. Trade the private acceleration for the shared one. Rigidity gives every particle the same $\alpha$ , with $a_{t,i} = r_i\alpha$ from the kinematics article: $F_{t,i} = m_i r_i \alpha$ . Why this is the pivotal move: one unknown $\alpha$ now serves the entire swarm — the summing becomes possible.

Step 3. Convert to torque: multiply by $r_i$ . $\ \tau_i = r_i F_{t,i} = m_i r_i^2\,\alpha$ . Why: torque is the honest currency of turning — and this multiplication mints the second factor of $r_i$ , completing the $r^2$ .

Step 4. Sum over the body and watch the internal forces die. Particles exert forces on each other, but by Newton's third law these come in equal-and-opposite pairs acting along the line joining the pair — so each pair's two torques about the axis share one line of action, hence one lever arm, and cancel exactly (the internal forces considered here being central forces). Only external torques survive:

\tau_{\rm net}^{\rm ext} = \Big(\sum_i m_i r_i^2\Big)\alpha \equiv I\,\alpha \qquad (2)

Why the cancellation matters: it is the rotational twin of internal forces cancelling in $F = Ma_{\rm cm}$ — the reason a body cannot spin itself up by its own interatomic tugs, and the reason (2) contains only forces you can list.

Step 5. Read the definition the sum forced on us: $I = \sum_i m_i r_i^2$ , or for continuous bodies $I = \int r^2\,dm$ — the moment of inertia about this axis. Standard results worth memorizing: point mass $mr^2$ ; hoop $MR^2$ ; solid disk or cylinder $\tfrac12 MR^2$ ; solid sphere $\tfrac25 MR^2$ ; thin rod $\tfrac1{12}ML^2$ about its centre, $\tfrac13 ML^2$ about an end. Why $I$ is body-and-axis, not body alone: the weights $r_i^2$ are distances from a particular line — the same rod is four times lazier about its end than about its centre, and the parallel-axis theorem article systematizes exactly this. *The catalogue values follow from $\int r^2\,dm$ ; for the disk: $\int_0^R r^2\,(2\pi r\sigma)\,dr = \tfrac12 MR^2$ with $\sigma = M/\pi R^2$ ✓. Worked example 2 closes the loop numerically, confirming $(T_1 - T_2)R = I\alpha$ to three figures.*

Go deeper · Derivation B · 1 minWork, power, and kinetic energy of rotation

Derivation BWork, power, and kinetic energy of rotation

The goal is rotation's energy bookkeeping: the work of a torque $W = \int\tau\,d\theta$ , the power $P = \tau\omega$ , and the work–energy theorem $W_{\rm net} = \Delta(\tfrac12 I\omega^2)$ . The strategy is again translation-by-summation: work is force through distance for each particle, and every particle's distance is an arc. Watch how the same sum $\sum m_i r_i^2$ that resisted angular acceleration reappears as the warehouse of kinetic energy — one quantity, two jobs, exactly like mass in the linear story.

Step 1. Work on one particle, through a body turn $d\theta$ . The particle rides its arc $ds_i = r_i\,d\theta$ ; only the tangential force component works: $dW_i = F_{t,i}\,r_i\,d\theta = \tau_i\,d\theta$ . Why: force through distance, with the distance measured along the arc — the radial components do no work, being perpendicular to the motion.

Step 2. Sum: $dW = \tau_{\rm net}\,d\theta$ , so over a finite turn

W = \int \tau\,d\theta \qquad \text{and, dividing by } dt, \qquad P = \tau\omega \qquad (3)

Why $P = \tau\omega$ runs the machine world: an engine's shaft delivers energy at torque times spin rate — this is why spec sheets quote torque curves against rpm, and why gearboxes, trading $\omega$ for $\tau$ at nearly constant $P$ , exist at all.

Step 3. Kinetic energy of the spinning swarm: $K = \sum_i \tfrac12 m_i v_i^2 = \tfrac12 \sum_i m_i r_i^2\,\omega^2$ , since $v_i = r_i\omega$ with $\omega$ shared. Hence

K = \tfrac12 I\omega^2 \qquad (4)

Why the same $I$ : the sum $\sum m_i r_i^2$ never asked which job it was doing — resisting $\alpha$ in (2), storing energy in (4). The rotational-kinetic-energy article develops (4)'s consequences at length; here it completes the theorem.

Step 4. Assemble the work–energy theorem: integrate $\tau = I\alpha = I\,\omega\,\dfrac{d\omega}{d\theta}$ over angle to get $W_{\rm net} = \tfrac12 I\omega_f^2 - \tfrac12 I\omega_0^2$ . Why the chain-rule step: it is the same $a = v\,dv/dx$ manipulation that produced the linear theorem — the dictionary at work in the calculus itself. *Worked example 1 checks $W = \tau\theta$ against $\Delta(\tfrac12 I\omega^2)$ to three figures (790 J vs 789 J, rounding). Dimensional check: $P = \tau\omega$ carries units $\mathrm{N\,m \cdot s^{-1}} = \mathrm{W}$ ✓.*

Go deeper · Derivation C · 2 minAngular momentum about the axis, and the deepest form of the law

Derivation CAngular momentum about the axis, and the deepest form of the law

We are introducing rotation's conserved currency: the angular momentum about the axis, $L = I\omega$ , and upgrading the second law to $\tau_{\rm net} = dL/dt$ — the form that survives even when $I$ itself changes, which $\tau = I\alpha$ quietly does not. The strategy: define each particle's contribution as lever arm times linear momentum (mirroring torque as lever arm times force), sum the swarm, then differentiate. The payoff to watch is Step 4: when the net torque vanishes, $L$ is frozen — and a body that can reshape itself must then trade $I$ against $\omega$ , which is the entire physics of the spinning skater.

Step 1. Define per particle, mirroring torque. A particle of momentum $p_i = m_i v_i$ moving on its circle has angular momentum about the axis $L_i = r_i p_i = m_i r_i v_i$ . Why this definition: lever arm times momentum is to "quantity of rotation" what lever arm times force is to "turning effort" — the dictionary's $p \to L$ row.

Step 2. Sum with the shared $\omega$ : $v_i = r_i\omega$ , so $L = \sum_i m_i r_i^2\,\omega$ , i.e.

L = I\omega \qquad (5)

Why: the same $I$ a third time — resister of $\alpha$ , warehouse of $K$ , and now the exchange rate between spin and angular momentum.

Step 3. Differentiate: for a rigid body ( $I$ fixed), $\dfrac{dL}{dt} = I\dfrac{d\omega}{dt} = I\alpha = \tau_{\rm net}$ by (2). So

\tau_{\rm net} = \frac{dL}{dt} \qquad (6)

Why this form is the deeper one: (6) remains true even when the body reshapes and $I$ varies (the honest general proof, via $\vec L = \sum \vec r_i \times \vec p_i$ , is the angular-momentum article's business in the rigid-body chapters); $\tau = I\alpha$ is its rigid special case. Integrated over time it gives the angular impulse–momentum theorem, $\int \tau\,dt = \Delta L$ — the rotational twin of impulse.

Step 4. Set the torque to zero and collect the physics: $L = I\omega =$ constant. A skater pulling her arms in slashes her $I$ ; conservation demands $\omega$ rise in exact proportion — she spins faster without anyone applying a torque. Note the energy is not conserved in the process: $K = \tfrac12 I\omega^2 = L^2/2I$ rises as $I$ falls, the difference supplied by the muscular work of pulling her arms inward against the centrifugal tendency. Why flag the energy: "conserved $L$ " and "conserved $K$ " are independent claims, and conflating them is a classic exam wound. *Skater check with numbers: $I$ halved gives $\omega$ doubled and $K$ doubled, consistent with $K = L^2/2I$ ✓. The conservation law's full scope — central forces, Kepler's second law, stellar collapse — is developed in the angular-momentum-conservation articles of the orbits chapter.*

Which equation answers which question

you want	use	notes
turning effect of a force	$\tau = rF\sin\phi = r_\perp F$ , eq. (1)	slide the force along its line freely
spin-up rate	$\alpha = \tau_{\rm net}/I$ , eq. (2)	the rotational analogue of $a = F/m$
the body's laziness	$I = \int r^2 dm$ + catalogue	body and axis; Steiner shifts axes
energy delivered by a shaft	$W = \int\tau\,d\theta$ , $\ P = \tau\omega$ , eq. (3)	the engine-spec equation
speed after a spin-up	$W_{\rm net} = \Delta(\tfrac12 I\omega^2)$	or suvat if $\alpha$ is constant
reshaping bodies, collisions	$L = I\omega$ , $\ \int\tau\,dt = \Delta L$ , eqs. (5)–(6)	conservation when $\tau_{\rm net} = 0$

Limiting cases

Case	Result	Meaning
$\phi = 0$ or $\pi$ (force along the radius)	$\tau = 0$	pushing through the axis turns nothing
$\phi = \pi/2$	$\tau = rF$ , maximum	push perpendicular, at the far edge
all mass at the axis ( $r \to 0$ )	$I \to 0$	no lever arms to tax — spins up freely
same body, axis moved to the rim	$I$ jumps by $MR^2$	parallel-axis theorem: $I = I_{\rm cm} + Md^2$
constant $\tau$	constant $\alpha$ — suvat applies	the kinematics article takes over
$\tau_{\rm net} = 0$ , rigid body	$\omega$ constant	rotational first law: flywheels coast
$\tau_{\rm net} = 0$ , body reshapes	$L$ constant, $\omega \propto 1/I$	the skater; $K = L^2/2I$ changes
point mass on a string	$I = mr^2$ , $\tau = I\alpha$ trivially	the swarm sum's atomic case

Common trap"A bigger force always out-turns a smaller one"

Turning contests are decided by torque, not force, and the lever arm can overwhelm any strength deficit: two fingers at a door's handle ( $r = 0.9$ m) beat a shoulder shove a hand-width from the hinge ( $r = 0.1$ m) nine to one, and a child at a seesaw's end lifts the adult sitting near the pivot. Three sibling errors ride with this one. First, force along the radius counts for nothing — only $F\sin\phi$ turns, which is why pulling a wrench toward the bolt achieves zero. Second, torque is not energy despite sharing the units $\mathrm{N\,m}$ : a torque does work only through an angle ( $W = \tau\theta$ ), and a stuck bolt under enormous torque receives no energy at all. Third, a nonzero net torque can coexist with zero net force (two opposite forces on opposite rims — a "couple": pure spin-up, no push) and vice versa (a single force through the axis: pure push, no spin) — translation and rotation keep separate books, and every rigid-body problem must balance both.

Worked examples

1 · Spinning up a grindstone (confidence builder)

A grindstone is a solid disk of mass $M = 20\ \mathrm{kg}$ and radius $R = 0.40\ \mathrm{m}$ . A motor belt applies a steady tangential force $F = 25\ \mathrm{N}$ at the rim. Find the angular acceleration, the time to reach 300 rpm, and verify the energy books at that moment.

Step 1. Torque: tangential force at the rim has lever arm $R$ , so $\tau = FR = 25 \times 0.40 = 10\ \mathrm{N\,m}$ .

Step 2. Moment of inertia (solid disk): $I = \tfrac12 MR^2 = \tfrac12 \times 20 \times 0.16 = 1.6\ \mathrm{kg\,m^2}$ .

Step 3. Second law (2): $\alpha = \tau/I = 10/1.6 = \boxed{6.25\ \mathrm{rad/s^2}}$ .

Step 4. Time to 300 rpm $= 31.4\ \mathrm{rad/s}$ : from the suvat set, $t = \omega/\alpha = 31.4/6.25 = \boxed{5.0\ \mathrm{s}}$ , during which the stone turns $\theta = \tfrac12\alpha t^2 = \tfrac12 \times 6.25 \times 25.24 = 79\ \mathrm{rad}$ (about 12.6 turns).

Step 5. Energy audit: work in, $W = \tau\theta = 10 \times 79.0 = 790\ \mathrm{J}$ ; kinetic energy held, $K = \tfrac12 I\omega^2 = \tfrac12 \times 1.6 \times (31.4)^2 = 789\ \mathrm{J}$ ✓ (rounding). Every joule the belt delivered through its torque is on deposit in the spin — the theorem of equation (3)–(4) closing to three figures.

2 · Atwood's machine with a real pulley (exam level)

Masses $m_1 = 4.0\ \mathrm{kg}$ and $m_2 = 2.0\ \mathrm{kg}$ hang from a light cord over a solid-disk pulley of mass $M = 2.0\ \mathrm{kg}$ and radius $R = 0.25\ \mathrm{m}$ ; the cord does not slip. Find the acceleration and both tensions, and verify the pulley's torque equation. (Take $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Write one law per body — the frictionless-pulley shortcut " $T_1 = T_2$ " is dead on arrival here, because the pulley needs a net torque to spin up. Falling side: $m_1 g - T_1 = m_1 a$ . Rising side: $T_2 - m_2 g = m_2 a$ . Pulley, eq. (2): $(T_1 - T_2)R = I\alpha$ with $I = \tfrac12 MR^2$ .

Step 2. Add the bridge (no slipping): the cord's linear acceleration matches the rim's tangential one, $a = R\alpha$ .

Step 3. Reduce the pulley equation: $(T_1 - T_2) = \tfrac12 MR^2 \cdot \dfrac{a}{R}\cdot\dfrac1R = \tfrac12 Ma$ . The radius cancels — the pulley acts like an extra $M/2$ of ordinary inertia hidden in the cord.

Step 4. Add all three equations (the tensions telescope): $\ (m_1 - m_2)g = \left(m_1 + m_2 + \tfrac{M}{2}\right)a$ , so $a = \dfrac{2.0 \times 9.8}{4.0 + 2.0 + 1.0} = \boxed{2.8\ \mathrm{m/s^2}}$ .

Step 5. Tensions: $T_1 = m_1(g - a) = 4.0 \times 7.0 = \boxed{28.0\ \mathrm{N}}$ ; $\ T_2 = m_2(g + a) = 2.0 \times 12.6 = \boxed{25.2\ \mathrm{N}}$ . They differ by $2.8$ N — the pulley's fee.

Step 6. Verify the torque books: $(T_1 - T_2)R = 2.8 \times 0.25 = 0.70\ \mathrm{N\,m}$ , while $I\alpha = (\tfrac12 \times 2.0 \times 0.0625)(2.8/0.25) = 0.0625 \times 11.2 = 0.70\ \mathrm{N\,m}$ ✓. (Massless-pulley check: set $M = 0$ and $a \to 2g/6 = 3.27\ \mathrm{m/s^2}$ , $T_1 = T_2$ — the old idealization recovered as a limit.)

Origins · 1 minArchimedes and how this was discovered

Archimedes

c. 287–212 BC · Syracuse

Torque's founding document is two millennia older than Newton: Archimedes' On the Equilibrium of Planes states the law of the lever — magnitudes balance at distances inversely proportional to their weights — and proves it from symmetry postulates, making $rF$ the first quantitative law of mechanics. "Give me a place to stand and I will move the Earth" is the lever-arm scaling read as a boast, and his compound pulleys and war engines at the siege of Syracuse were the scaling read as engineering. What Archimedes had was the statics of torque, the balance of turning tendencies; what he lacked was the dynamics — that an unbalanced $rF$ produces angular acceleration in proportion to a mass-weighted $r^2$ sum. That closure waited for the seventeenth century's calculus and for Euler, who named the moment of inertia and wrote rotation's equations of motion. The word "torque" itself is the newest thing in this article: James Thomson (Lord Kelvin's brother) proposed it only in 1884, from the Latin torquere, to twist.

History

1638

— Galileo's Two New Sciences analyses levers and beams, carrying Archimedes' moment reasoning into the new mechanics.

1673

— Huygens's Horologium Oscillatorium solves the compound pendulum — the first dynamical calculation with what we now call a moment of inertia.

1687

— Newton's Principia provides the particle laws whose swarm sum is this article's derivation.

1765

— Euler's Theoria motus corporum solidorum names the moment of inertia and casts rigid-body rotation as differential equations:

\tau = I\alpha

in its mature form.

1884

— James Thomson coins the word "torque"; engineering adopts it as the shaft-transmission currency of the machine age.

Going further

Texts. Halliday, Resnick & Walker, Fundamentals of Physics, Ch. 10–11 · Kleppner & Kolenkow, An Introduction to Mechanics, Ch. 7 · Morin, Introduction to Classical Mechanics, Ch. 8 · Archimedes, On the Equilibrium of Planes, for the lever's founding document. Key numerical results: disk integral $\int_0^R r^2(2\pi r\sigma)dr = \tfrac12 MR^2$ ; worked example 1 gives $\tau = 10\ \mathrm{N\,m}$ , $I = 1.6\ \mathrm{kg\,m^2}$ , $\alpha = 6.25\ \mathrm{rad/s^2}$ , $t = 5.0$ s, $W = 790$ J vs $K = 789$ J; worked example 2 gives $a = 19.6/7 = 2.8\ \mathrm{m/s^2}$ , $T_1 = 28.0$ N, $T_2 = 25.2$ N, torque check $0.70 = 0.70\ \mathrm{N\,m}$ ; skater energy bookkeeping: $K = L^2/2I$ .

Builds on: Rotational Kinematics (θ, ω, α) · Newton's Second Law · Newton's Third Law · Kinetic Energy. Leads to: Rolling Motion · Rotational Kinetic Energy · Angular Momentum of a Rigid Body · Moment of Inertia Tensor · Parallel Axis (Steiner) Theorem.

Torque has units of newton-metres, which is joules. Is torque a kind of energy?

No — and the units coincidence is worth defusing properly. Energy is force times distance along the motion; torque is force times distance perpendicular to the force's line. The two products combine different geometric partners, and no angle of squinting turns a lever arm into a displacement. The cleanest way to see the difference: a torque produces energy only by acting through an angle, $W = \tau\,\theta$ — and there the radian, being dimensionless, silently converts $\mathrm{N\,m}$ of torque into honest joules of work. A stuck bolt under 200 N·m receives zero energy; the same torque turning a shaft through $100$ rad has delivered 20 kJ. Convention keeps the books straight by never writing torque in joules: newton-metres for twist, joules for energy, and $P = \tau\omega$ (watts) as the bridge between them.

Why do the internal forces really cancel in the big sum? Atoms pull on each other constantly.

They cancel in pairs, by Newton's third law plus one geometric observation. Take atoms A and B: A pulls B with some force, B pulls A with the exact opposite force, and both forces act along the same line — the line through the two atoms (interatomic forces are central). Two opposite forces on one line of action share a single lever arm about any axis, so their torques are equal and opposite: each pair contributes exactly zero net torque, and the sum over all pairs is zero regardless of how ferocious the individual forces are. This is why a rigid body cannot spin itself up from inside (no isolated system can — that is conservation of $L$ in embryo) and why equation (2) is usable at all: the millions of unknowable internal forces exit the problem, leaving only the external torques you can actually list. The same pairwise logic, run on forces instead of torques, is what gives $F_{\rm net}^{\rm ext} = Ma_{\rm cm}$ in the centre-of-mass article.

Does torque depend on where I put the axis? That seems to make it arbitrary.

It does depend on the reference point, and that is a feature to manage, not a bug. Torque is defined about a point (or axis): move the reference and every lever arm changes, so every torque changes. For this article's rigid bodies the physics chooses the axis for you — the hinge, the axle, the fixed rotation line — and all torques are taken about it, full stop. The freedom becomes a tool in statics: when a body is in equilibrium you may take torques about any point (all choices give valid equations), so you choose the point that makes unknown forces disappear — put the axis where the unknown reaction acts, and it vanishes from the torque equation. One clean theorem to keep: if the net force on a body is zero (as for a couple), the net torque is the same about every point — a couple is an unambiguous, reference-free twist.

Why is I defined with r squared and not just r? Momentum L = mvr seems to use one power.

Count the trades. Dynamics: to give the body $\alpha$ , a particle at radius $r$ needs tangential force $m(r\alpha)$ — one $r$ for the demand — and your torque handle on that force has lever arm $r$ — a second $r$ for the means. Hence $mr^2\alpha$ per particle. Energy tells the same story: $v = r\omega$ , and kinetic energy squares the speed, so $\tfrac12 mv^2 = \tfrac12 m r^2\omega^2$ . Angular momentum looks like one power — $L_i = r_i p_i = m_i r_i v_i$ — but the momentum itself hides another: $v_i = r_i\omega$ , so $L_i = m_i r_i^2\omega$ . All three roads mint $r^2$ , which is why one quantity $I = \sum m_i r_i^2$ serves as resister of $\alpha$ , warehouse of $K$ , and exchange rate of $L$ . Physically: far-flung mass must be moved faster for the same body turn, and your leverage on it (or its momentum's lever arm) is longer — every rotational role compounds the radius twice.

The skater speeds up "for free" — where does her extra kinetic energy come from, if no torque acts?

From her muscles, honestly paid. With no external torque, $L = I\omega$ is locked, so pulling her arms in ( $I$ down) forces $\omega$ up. But the kinetic energy is $K = L^2/2I$ : at fixed $L$ , halving $I$ doubles $K$ . The books balance because pulling her arms inward is hard work — her hands are moving in circles, and dragging them toward the axis means hauling them against their own inertia along a spiral path; the work her muscles do is exactly the $\Delta K$ . Run the film backward for the same lesson: letting her arms fly outward, she slows and her muscles absorb energy braking the outward rush. Conserved $L$ and conserved $K$ are independent claims: torque-free means $L$ fixed always, but $K$ fixed only if the body is truly rigid. A reshaping body under zero torque changes its kinetic energy through precisely the internal work that reshaping costs.

Can there be torque without rotation, and rotation without torque?

Both, and each direction of the mismatch is its own subject. Torque without rotation is statics: a wrench straining at a seized bolt, a beam pinned in a wall, your forearm holding a coffee cup — in each case some other torque balances the applied one, the net is zero, and nothing turns despite large individual torques. (Equilibrium demands both $F_{\rm net} = 0$ and $\tau_{\rm net} = 0$ ; the second condition is what keeps ladders from spinning into the floor.) Rotation without torque is the first law of the spinning world: a flywheel in good bearings, a gyroscope in vacuum, the Earth itself — all coast at constant $L$ because nothing twists them. The pairing to internalize: torque correlates with angular acceleration, never with angular velocity — asking "what torque keeps it spinning?" is the rotational version of Aristotle's error about force and motion.

Is L = Iω always true? I've heard angular momentum is more complicated than this.

It is exactly true for what this article treats — rotation about a fixed axis, with $L$ meaning the component of angular momentum along that axis — and that covers flywheels, doors, pulleys, and most of engineering. The complication you have heard of is real and arrives when the axis is freed: for a general rigid body, $\vec L$ and $\vec\omega$ are both vectors, and they need not be parallel — spin a badly balanced wheel and $\vec L$ swings around $\vec\omega$ , shaking the bearings. The honest general relation replaces the number $I$ with a $3 \times 3$ object, the inertia tensor, and $\vec L = \mathbf{I}\vec\omega$ ; the fixed-axis formula is its diagonal slice. That machinery — principal axes, why exactly three special directions exist where $\vec L \parallel \vec\omega$ , and the strange dynamics of the free rigid body — is the moment-of-inertia-tensor article's opening act in the rigid-body chapters. Everything you learn here survives; it becomes one component of a richer vector story.