Rotational Kinematics (θ, ω, α)

Here is the best bargain in classical mechanics: you already know rotational kinematics. You learned it the semester you learned position, velocity, and acceleration — you just learned it under different names. Every kinematic law you own has a rotational twin, produced by a systematic substitution: position becomes angle, velocity becomes angular velocity $\omega$ , acceleration becomes angular acceleration $\alpha$ . The suvat equations survive with every symbol swapped, solved by exactly the moves you already trust. One dictionary, and the entire description of spinning things unlocks at the cost of learning three Greek letters. Why bother with new variables at all? Watch a vinyl record: a dust speck near the spindle and one at the rim complete each revolution together — same angle swept, same rate — yet the rim speck moves through space several times faster. Linear language forces you to track every point separately; angular language describes the whole rigid body with one $\theta$ , one $\omega$ , one $\alpha$ , and hands back any individual point's speed on demand through the exchange rate $v = r\omega$ . That single conversion — radius as the price of translating between the two languages — is the working core of this article, and the reason it must be spoken in radians. This article builds the kinematic layer completely: the angular variables, the rotational suvat set, the bridges to linear motion, and the two-component acceleration of a point on a spinning body. The dynamics — what torque is, why $I$ replaces $m$ , what makes $\omega$ change — is the next article's business; here we master the description that it will need.

Key takeaways

A rigid body rotating about a fixed axis is described by one angle $\theta(t)$ ; its rates are the angular velocity $\omega = d\theta/dt$ and angular acceleration $\alpha = d\omega/dt$ , shared by every point of the body.
For constant $\alpha$ , the suvat equations translate verbatim: $\omega = \omega_0 + \alpha t$ , $\ \theta = \omega_0 t + \tfrac12\alpha t^2$ , $\ \omega^2 = \omega_0^2 + 2\alpha\theta$ .
The radius is the exchange rate to linear language: $s = r\theta$ , $v = r\omega$ , $a_t = r\alpha$ — valid only in radians.
A point on the body carries two perpendicular acceleration components: tangential $a_t = r\alpha$ (changes speed) and centripetal $a_c = \omega^2 r$ (bends the path) — total $a = \sqrt{a_t^2 + a_c^2}$ .
$\vec\omega$ is a vector along the rotation axis (right-hand rule); the full dictionary continues into dynamics with $m \to I$ , $F \to \tau$ in the next article.
Key idea: one shared set of angular variables describes the whole spinning body; each point's private linear motion is bought back at the exchange rate $r$ .

The core insight

Rotation about a fixed axis is one-dimensional kinematics in disguise: the angle $\theta$ , angular velocity $\omega = \dot\theta$ , and angular acceleration $\alpha = \dot\omega$ obey the same equations as $x$ , $v$ , $a$ — with the radius converting between the shared angular description and any point's private linear motion via $s = r\theta$ , $v = r\omega$ , $a_t = r\alpha$ .

Where this lives today

Rotational kinematics is the dial-reading language of the machine age. Motor and turbine engineering lives in rpm and $\mathrm{rad/s}$ : a wind-turbine controller schedules blade pitch against measured $\omega$ ; an electric-vehicle inverter commands torque from a motor whose speed it tracks tens of thousands of times per second; hard-drive servo firmware once counted every one of the platter's 7200 rpm from the pattern of passing sectors. Your phone contains the inverse instrument: MEMS gyroscopes report the device's $\vec\omega$ around three axes several hundred times per second, and the screen's rotation, the camera's stabilization, and a drone's entire attitude-control loop are integrations of $\theta = \int\omega\,dt$ performed in real time — rotational suvat run live, with all the drift pitfalls that integrating a noisy $\omega$ implies. The extremes are instructive. Ultracentrifuges reach the $10^5\,g$ of worked example 2 and are the workhorse of biochemistry; spin-stabilized projectiles and satellites exploit the constancy of $\vec\omega$ that the next articles will explain dynamically. And the most spectacular $\omega$ in nature belongs to millisecond pulsars — city-sized stars at $\omega \approx 4000\ \mathrm{rad/s}$ , whose rim speeds $v = r\omega$ approach a tenth of light speed: the record-player fact, promoted to relativistic astrophysics.

The physical picture

Watch the record spin. A dust speck near the spindle and one at the rim complete each revolution together — same angle swept, same angular velocity $\omega$ , same angular acceleration when the turntable spins up. But the rim speck races through a far bigger circle each turn: its speed is proportional to its radius, $v = r\omega$ . This is why the angular variables are the natural language of rotation: one $\theta$ , one $\omega$ , one $\alpha$ describe the whole rigid body at once, while every point has its own private $v$ and $a$ , each recoverable by multiplying by that point's radius. Rigidity is the licence for the shared description — if the body could deform, points could slip to different angles and the single- $\theta$ bookkeeping would collapse.

Figure

One spinning disk, two points — shared angular language, private linear speeds

The second half of the picture is the shape of the time story. Plot $\omega$ against $t$ for a spinning-up turntable and you get exactly the graphs of one-dimensional kinematics with the axes relabelled: constant $\alpha$ is a straight line, the swept angle is the area under it, and every graphical trick you learned for $v$ – $t$ diagrams — slopes, areas, average values — transfers untouched. This is not an analogy to be checked case by case; both subjects are the calculus of one variable and its derivatives, so they must share all their mathematics.

Figure

The omega-t graph — 1D kinematics with the axes relabelled

The math: angular variables and the rotational suvat set

Definitions first, because the whole subject is three of them. Fix an axis and a reference direction; the body's orientation is the angle $\theta$ (in radians) between the reference and a line painted on the body. Its rates are

\omega = \frac{d\theta}{dt}, \qquad \alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2} \qquad (1)

where: $\theta$ — orientation angle in rad, unwrapped: a wheel that has made ten and a quarter turns is at $\theta = 20.5\pi$ , not $\pi/2$ , because $\theta$ is a running odometer, not a compass reading; $\omega$ — angular velocity in $\mathrm{rad/s}$ (positive = counterclockwise by the usual convention); $\alpha$ — angular acceleration in $\mathrm{rad/s^2}$ . Everyday rotation rates arrive in other units — rpm, revolutions per second, degrees per second — and the first move of every problem is conversion: multiply rpm by $2\pi/60 \approx 0.1047$ to get $\mathrm{rad/s}$ (300 rpm $\approx 31.4\ \mathrm{rad/s}$ ). One full turn is $2\pi$ rad; the period and frequency connect through $\omega = 2\pi/T = 2\pi f$ .

Derivation AThe rotational suvat equations

We are deriving the constant- $\alpha$ kinematic set — $\omega = \omega_0 + \alpha t$ , $\theta = \omega_0 t + \tfrac12 \alpha t^2$ , and $\omega^2 = \omega_0^2 + 2\alpha\theta$ — the exact rotational twins of the linear suvat equations. The strategy is deliberate: rather than quoting the analogy, we integrate the definitions (1) directly, so the equations are earned and the analogy emerges as a theorem about identical mathematics rather than a mnemonic. Watch what is assumed and where: constant $\alpha$ enters at the first integration, and everything downstream inherits that restriction — for varying $\alpha$ you must return to the integrals themselves.

Step 1. Integrate $\alpha = d\omega/dt$ once, with $\alpha$ constant, from $0$ to $t$ :

\omega(t) = \omega_0 + \alpha t \qquad (2)

Why: a constant rate of change accumulates linearly — this is the straight line of the $\omega$ – $t$ graph, slope $\alpha$ , intercept $\omega_0$ .

Step 2. Integrate again, using $\omega = d\theta/dt$ with (2) as the integrand, taking $\theta(0) = 0$ :

\theta(t) = \omega_0 t + \tfrac12 \alpha t^2 \qquad (3)

Why: the two terms are the two pieces of area under the $\omega$ – $t$ line — the rectangle $\omega_0 t$ and the triangle $\tfrac12(\alpha t)(t)$ . Angle is the area under angular velocity, exactly as displacement is the area under velocity.

Step 3. Eliminate time between (2) and (3). Solve (2) for $t = (\omega - \omega_0)/\alpha$ and substitute into (3); the algebra collapses to

\omega^2 = \omega_0^2 + 2\alpha\theta \qquad (4)

Why: this is the timeless member of the set — reach for it whenever the question links spin rate to swept angle with no clock in sight, e.g. "how many turns does the flywheel need to reach 3000 rpm?"

Step 4. Note the two useful corollaries that come along free: the average angular velocity over the interval is $\bar\omega = (\omega_0 + \omega)/2$ (average of endpoints, straight-line graph), so $\theta = \bar\omega t$ — often the fastest route to turn-counting. Why: the same shortcuts you trust in linear kinematics hold verbatim, because the mathematics is identical symbol for symbol. *Self-consistency check: differentiating (3) returns (2), and differentiating (2) returns $\alpha$ — as required; every term of (3) carries units of radians, confirming dimensional consistency. The set is the exact rotational twin of the linear suvat equations under the substitution $x \to \theta$ , $v \to \omega$ , $a \to \alpha$ .*

Go deeper · Derivation B · 2 minThe bridges — arc, speed, and the two accelerations of a spinning point

Derivation BThe bridges — arc, speed, and the two accelerations of a spinning point

We are deriving the exchange-rate relations $s = r\theta$ , $v = r\omega$ , $a_t = r\alpha$ , and assembling the total acceleration of a point on a rotating body, $a = \sqrt{a_t^2 + a_c^2}$ . The strategy is two differentiations of the arc-length relation at fixed radius, then an honest reckoning with the fact that a point on a circle accelerates even at constant speed — the centripetal component from the circular-motion article must be re-imported, or the acceleration budget is only half done. The step to watch is the very first: $s = r\theta$ is the definition of the radian, which is why every bridge silently fails in degrees.

Step 1. Start from the definition of the radian. A point at radius $r$ on a body that has turned through $\theta$ has travelled the arc

s = r\theta \qquad (5)

Why: the radian is defined as arc over radius — this is a definition, not a result, and it is the reason radians are non-negotiable everywhere downstream.

Step 2. Differentiate at fixed $r$ (the body is rigid, so each point keeps its radius): $\dot s = r\dot\theta$ , i.e.

v = r\omega \qquad (6)

Why: rigidity freezes $r$ ; only the angle changes, so the point's speed inherits the angle's rate, scaled by the radius. This is the record-player fact: same $\omega$ , bigger circle, faster ride.

Step 3. Differentiate once more: $\dot v = r\dot\omega$ , giving the tangential acceleration

a_t = r\alpha \qquad (7)

Why: this is the component along the direction of motion — the piece that changes the point's speed. It exists only while the spin rate is changing.

Step 4. Re-import the centripetal component. Even at constant $\omega$ , the point's velocity direction turns, demanding the inward acceleration derived in the circular-motion article: $a_c = \omega^2 r = v^2/r$ . The two components are perpendicular by construction — $a_t$ along the tangent, $a_c$ toward the axis — so the total is

a = \sqrt{a_t^2 + a_c^2} = r\sqrt{\alpha^2 + \omega^4} \qquad (8)

Why: forgetting $a_c$ is the classic budget error; note from the second form that as a spin-up proceeds, $\omega^4$ swiftly buries $\alpha^2$ — within a fraction of a second of most spin-ups the acceleration is essentially all centripetal, tilted barely off the inward radial direction.

Step 5. Record the direction bookkeeping for $\omega$ itself: for fixed-axis work, counterclockwise-positive suffices, but $\vec\omega$ is properly a vector along the rotation axis, oriented by the right-hand rule (curl your fingers with the spin; the thumb gives $\vec\omega$ ), and the point-velocity bridge generalizes to $\vec v = \vec\omega \times \vec r$ . Why mention it now: the cross-product form is the doorway to the rigid-body chapters, where the axis itself moves — and it explains a curiosity: finite rotations do not commute (rotate a book 90° about two axes in both orders and compare), so finite angles are not vectors, but infinitesimal ones are, which is what lets their rate $\vec\omega$ be one. *Self-consistency check: equation (8)'s two components are the $\dot r = 0$ specialization of the general polar-coordinate acceleration — setting the radial-velocity term to zero recovers exactly $a_t = r\alpha$ and $a_c = \omega^2 r$ . Dimensional check: $r\sqrt{\alpha^2 + \omega^4}$ carries $\mathrm{m}\cdot\mathrm{s^{-2}}$ ✓.*

The dictionary — kinematic rows earned, dynamic rows previewed

Linear	Rotational	Bridge / status
position $x$	angle $\theta$	$s = r\theta$ — earned here
velocity $v = \dot x$	angular velocity $\omega = \dot\theta$	$v = r\omega$ — earned here
acceleration $a = \dot v$	angular acceleration $\alpha = \dot\omega$	$a_t = r\alpha$ , plus $a_c = \omega^2 r$ — earned here
suvat set	eqs. (2)–(4)	translate verbatim — earned here
mass $m$	moment of inertia $I$	next article: $I = \sum m_i r_i^2$
force $F$	torque $\tau$	next article: $\tau = rF_\perp$ , $\ \tau = I\alpha$
momentum $p$	angular momentum $L$	next article: $L = I\omega$
$K = \tfrac12 mv^2$	$K = \tfrac12 I\omega^2$	rotational-energy article

The lower rows are the reason this article exists: once the description is solid, the dynamics arrives by the same substitution, and $F = ma$ will reappear as $\tau = I\alpha$ with a two-line derivation. Learn the table's upper half cold and the lower half is free.

Limiting cases

Case	Result	Meaning
$\alpha = 0$	$\omega$ constant, $\theta = \omega t$	uniform rotation — the circular-motion article's regime
angles in degrees	(5)–(7) fail	$s = r\theta$ presupposes radians; degrees smuggle in $\pi/180$
$r \to 0$ (on the axis)	$v = a_t = a_c = 0$	axis points share the angles but do not move
constant $\alpha$ assumed, $\alpha$ actually varying	(2)–(4) invalid	return to the integrals $\omega = \int \alpha\,dt$ , $\theta = \int \omega\,dt$
just after spin-up starts	$a \approx a_t$	at $\omega \approx 0$ the centripetal piece hasn't grown yet
a moment later	$a \approx a_c$	$\omega^4$ buries $\alpha^2$ ; the acceleration swings inward
one revolution	$\theta = 2\pi$ rad	turns convert via $\theta/2\pi$ ; $\omega = 2\pi/T = 2\pi f$

Common trap"Every point on a rigid wheel has the same velocity and acceleration"

Half of this is true and the halves get swapped constantly. What every point shares are the angular quantities: one $\theta$ , one $\omega$ , one $\alpha$ for the whole rigid body — that shared description is the entire point of the subject. What every point does not share are the linear quantities: speed $v = r\omega$ , tangential acceleration $a_t = r\alpha$ , and centripetal acceleration $a_c = \omega^2 r$ all scale with the point's radius, so the rim of a wheel moves and accelerates strictly more than points near the hub, and points on the axis do not move at all. The error's favourite disguise is the exam phrase "the acceleration of the wheel" — always ask which point, and remember even a rim point's acceleration has two perpendicular components, of which the centripetal one usually dominates within moments of any spin-up.

Worked examples

1 · Braking a bicycle wheel (confidence builder)

A bicycle rolls at $7.0\ \mathrm{m/s}$ on wheels of radius $0.35\ \mathrm{m}$ . The brakes bring it uniformly to rest in $4.0\ \mathrm{s}$ . Find the initial angular velocity, the angular acceleration, the tangential acceleration of a rim point, and the number of turns each wheel makes while stopping.

Step 1. Initial angular velocity from bridge (6): $\omega_0 = v/r = 7.0/0.35 = 20\ \mathrm{rad/s}$ .

Step 2. Angular acceleration from suvat (2): $0 = 20 + \alpha(4.0)$ , so $\alpha = \boxed{-5.0\ \mathrm{rad/s^2}}$ — negative because it opposes the positive spin.

Step 3. Tangential acceleration of a rim point from bridge (7): $a_t = r\alpha = 0.35 \times (-5.0) = -1.75\ \mathrm{m/s^2}$ — exactly the bicycle's linear deceleration, as rolling demands (the rolling-motion article makes that connection law).

Step 4. Angle swept from suvat (3): $\theta = 20(4.0) - \tfrac12(5.0)(16) = 80 - 40 = 40\ \mathrm{rad}$ , i.e. $40/2\pi = \boxed{6.4\ \mathrm{turns}}$ . (Faster route: $\bar\omega = 10\ \mathrm{rad/s}$ , times $4.0$ s = $40$ rad ✓.)

Step 5. Cross-check through the linear language: distance rolled $= r\theta = 0.35 \times 40 = 14\ \mathrm{m}$ , which matches average speed times time, $3.5 \times 4.0 = 14\ \mathrm{m}$ ✓. Two dictionaries, one story.

2 · The ultracentrifuge (exam level)

A laboratory ultracentrifuge spins up from rest to $60{,}000$ rpm in $30$ s at constant angular acceleration. For a sample sitting at $r = 8.0\ \mathrm{cm}$ : (a) find $\alpha$ and the number of revolutions during spin-up; (b) find the sample's centripetal acceleration at full speed, in units of $g$ ; (c) compare $a_t$ and $a_c$ during the spin-up, and find the moment they are equal.

Step 1. Convert first, always: $\omega_f = 60{,}000 \times \dfrac{2\pi}{60} = 6283\ \mathrm{rad/s}$ . Then $\alpha = \omega_f/t = 6283/30 = \boxed{209\ \mathrm{rad/s^2}}$ .

Step 2. Revolutions from suvat (3): $\theta = \tfrac12 \alpha t^2 = \tfrac12 (209.4)(900) = 9.42 \times 10^4\ \mathrm{rad}$ , i.e. $9.42 \times 10^4/2\pi = \boxed{15{,}000\ \mathrm{rev}}$ . Sanity: average rate $30{,}000$ rpm for half a minute is $15{,}000$ revolutions ✓.

Step 3. Part (b), centripetal demand at full speed: $a_c = \omega_f^2 r = (6283)^2 \times 0.080 = 3.95 \times 10^7 \times 0.080 = 3.16 \times 10^6\ \mathrm{m/s^2}$ , which is $3.16 \times 10^6/9.81 = \boxed{3.2 \times 10^5\ g}$ — three hundred thousand gravities, the number that lets a centrifuge separate molecules by weight.

Step 4. Part (c), the tangential piece: $a_t = r\alpha = 0.080 \times 209.4 = 16.8\ \mathrm{m/s^2}$ — fixed throughout the spin-up, and utterly dwarfed by $a_c$ within moments.

Step 5. They are equal when $\omega^2 r = r\alpha$ , i.e. $\omega = \sqrt{\alpha} = 14.5\ \mathrm{rad/s}$ , which (2) says happens at $t = \omega/\alpha = 0.069\ \mathrm{s}$ . For the remaining $29.9$ seconds the acceleration is effectively all centripetal: the total-acceleration vector swings from tangent to radially inward within the first tenth of a second — the generic fate of every spin-up, by the $\omega^4$ -versus- $\alpha^2$ anatomy of equation (8).

Origins · 1 minLeonhard Euler and how this was discovered

Leonhard Euler

1707–1783 · Basel, St Petersburg, Berlin

The angular language of this article is largely Euler's handwriting. Huygens had computed swing rates of compound pendulums, but it was Euler's Theoria motus corporum solidorum seu rigidorum (1765) that treated the orientation angle of a rigid body as a dynamical variable in its own right, introduced the angular velocity of a body as a vector, and wrote rotation as differential equations in these variables. He proved that any rigid body's most general motion decomposes into a translation plus a rotation about an axis — the theorem that licenses "one $\theta$ for the whole body" — and his three orientation angles remain the standard coordinates when the axis itself is freed to move, in the rigid-body chapters ahead. Euler treated rotation not as new physics but as the calculus of Newton's mechanics faithfully rewritten in angular variables — precisely the spirit of the dictionary.

History

1673

— Huygens's Horologium Oscillatorium analyses compound pendulums — rotational motion quantified before its variables had names.

1743

— d'Alembert's Traité de dynamique systematizes constrained systems of particles, groundwork for the rigid body as a single summed object.

1765

— Euler's Theoria motus corporum solidorum makes orientation, angular velocity, and angular acceleration first-class dynamical variables and proves the translation-plus-rotation decomposition.

1834

— Poinsot's Théorie nouvelle de la rotation des corps gives rotation its geometric picture, completing the kinematic toolkit.

1852

— Foucault names the gyroscope and uses a spinning rotor to make the Earth's own

\omega

— all of

7.3 \times 10^{-5}

rad/s — visible on a lab bench.

Going further

Texts. Halliday, Resnick & Walker, Fundamentals of Physics, Ch. 10 · Kleppner & Kolenkow, An Introduction to Mechanics, Ch. 6–7 · Morin, Introduction to Classical Mechanics, Ch. 8 for where the kinematics graduates to vectors and tensors · Euler, Theoria motus corporum solidorum (1765), for the historical source. Key numerical results for quick reference: worked example 1 gives $\omega_0 = 20\ \mathrm{rad/s}$ , $\alpha = -5.0\ \mathrm{rad/s^2}$ , 6.4 turns, and a stopping distance of 14 m (consistent in both angular and linear language); worked example 2 gives $\omega_f = 6283\ \mathrm{rad/s}$ , $\alpha = 209\ \mathrm{rad/s^2}$ , $1.5 \times 10^4$ rev during spin-up, $a_c = 3.16 \times 10^6\ \mathrm{m/s^2} = 3.2 \times 10^5\,g$ , $a_t = 16.8\ \mathrm{m/s^2}$ , and a tangential–centripetal crossover at $t = 0.069$ s. The rpm-to-rad/s conversion factor is $2\pi/60$ .

Builds on: One-Dimensional Kinematics · Velocity and Acceleration · Circular Motion and Centripetal Acceleration. Leads to: Torque and Angular Momentum about a Fixed Axis · Rolling Motion · Rotational Kinetic Energy · Moment of Inertia Tensor (the graduate layer).

Why are radians "natural"? Degrees describe the same rotation.

Both label the same physical turn, but only the radian makes the bridge equations clean, because the radian is defined as arc length over radius: $\theta = s/r$ . With that definition, $s = r\theta$ is exact by construction, and its derivatives $v = r\omega$ , $a_t = r\alpha$ follow with no constants. Write the same relations in degrees and every one sprouts a factor $\pi/180$ — survivable, but a guaranteed source of silent errors, especially where the angle hides inside a derivative or an energy formula. The deeper point: the radian is a ratio of lengths, so it is dimensionless — which is why $\omega^2 r$ comes out in honest $\mathrm{m/s^2}$ without unit surgery. Convert everything (rpm, degrees, revolutions) to radians as step zero of every problem; convert back to friendly units only in the final line.

Is the angle θ limited to one turn, like a compass heading?

No — and treating it that way breaks the kinematics. The $\theta$ of the suvat set is an odometer, not a compass: it accumulates without wrapping, so a flywheel that has made ten and a quarter counterclockwise turns sits at $\theta = 20.5\pi\ \mathrm{rad}$ , not at $\pi/2$ . You can see why unwrapping is mandatory from equation (4): $\omega^2 = \omega_0^2 + 2\alpha\theta$ links the spin rate to the total angle swept — wrap $\theta$ back into $[0, 2\pi)$ and the equation returns nonsense after the first revolution. The same convention makes "number of revolutions" trivial: $N = \theta/2\pi$ . Orientation-modulo- $2\pi$ has its uses (where is the valve stem right now?), but for kinematics, let the odometer run.

Which way does the vector ω point, and why along the axis of all places?

Along the rotation axis, by the right-hand rule: curl the fingers of your right hand in the direction of spin, and your thumb points along $\vec\omega$ . The axis is the only honest choice, by elimination: every direction in the plane of rotation is constantly changing and none is special, while the axis is the unique direction the motion leaves fixed. The choice of which way along the axis (right hand rather than left) is pure convention, but once adopted it makes the point-velocity formula $\vec v = \vec\omega \times \vec r$ work with the standard cross product. There is one honest subtlety behind the vector status: finite rotations do not commute — rotate a book 90° about vertical then horizontal axes, then repeat in the other order, and it ends up differently — so finite angles are not vectors. Infinitesimal rotations do commute, and $\vec\omega$ , being a rate of infinitesimal rotations, inherits clean vectorhood.

If a point on a wheel moves at constant speed, how can it be accelerating?

Because velocity is a vector, and turning counts as changing. At constant $\omega$ , a rim point's speed never varies, but its direction of travel swings continuously — and the rate of that swing is an honest acceleration, $a_c = \omega^2 r$ , pointed at the axis. This is the founding fact of the circular-motion article, re-imported here because rotational problems make it dangerously easy to forget: the suvat set tracks only the angular story, in which constant $\omega$ looks like "nothing happening," while the rim of that placid flywheel may be pulling hundreds of thousands of $g$ (worked example 2). The full budget for any point is always two perpendicular pieces — $a_t = r\alpha$ for speed changes, $a_c = \omega^2 r$ for direction changes — and only their vector sum is the acceleration.

The suvat equations look identical to the linear ones. Is that a coincidence I need to double-check case by case?

It is a theorem, not a coincidence, and it needs checking exactly once. Both sets are statements about one variable and its time derivatives: given any quantity $q(t)$ with constant second derivative, integrating twice yields $\dot q = \dot q_0 + \ddot q\,t$ and $q = \dot q_0 t + \tfrac12 \ddot q\,t^2$ , and eliminating $t$ gives $\dot q^2 = \dot q_0^2 + 2\ddot q\,q$ . Feed in $q = x$ and you have linear suvat; feed in $q = \theta$ and you have equations (2)–(4). The calculus never asked what $q$ means. The same immunity extends to every graphical technique (slopes, areas, average-of-endpoints) — and it is why the dictionary keeps paying off in the dynamics articles: wherever the equations have the same shape, all their solutions transfer wholesale.

What does negative α mean — slowing down?

Not by itself. $\alpha$ is the rate of change of a signed $\omega$ , so negative $\alpha$ means the angular velocity is becoming more negative — which is "slowing down" only if the body currently spins in the positive sense. A wheel with $\omega < 0$ and $\alpha < 0$ is speeding up (clockwise, ever faster), exactly as a car with negative velocity and negative acceleration gains speed while reversing. The reliable test is the product: $\omega\alpha > 0$ means the spin rate's magnitude is growing; $\omega\alpha < 0$ means genuine braking. Watch also the moment a braking wheel passes through $\omega = 0$ under a persistent $\alpha$ : it does not stay stopped, it reverses — the rotational twin of a ball at the top of its vertical flight.

Bottom line: when do I work in angular variables and when in linear ones?

Let the rigid body decide. The moment several points move together — a wheel, a rotor, a swinging door — the angular description is strictly more economical: one $\theta(t)$ replaces a velocity field, and questions about "the machine" (spin rate, turns, time to speed) live entirely in equations (1)–(4). Switch to linear language only when the question asks about a specific point — how fast is the rim moving, what acceleration does the sample feel — and cross the bridge at that point's radius: $v = r\omega$ , $a_t = r\alpha$ , $a_c = \omega^2 r$ . Composite problems (a cable unwinding from a drum, a wheel rolling on a road) use both languages plus the bridge as a constraint tying them together; the rolling-motion article is built on exactly that pattern. What you should not do is solve a whole-body problem point by point in Cartesian coordinates — it works, and it is how you discover why angular variables were invented.