Relation to linear quantities - v = rω, a_t = rα, a_c = rω²: Worked examples

This is a practice companion to the parent topic. There we derived the three bridge formulas. Here we hunt down every kind of situation they can be thrown into — every sign, every zero, every degenerate case, and the sneaky exam twists — and solve one example for each.

Prerequisites you may want open: Angular velocity and angular acceleration, Radian measure and arc length, Uniform vs non-uniform circular motion, Centripetal force and circular motion.

Definition Quick recall of the symbols on this page

Before any numbers, here is exactly what each letter means (fuller versions live in Angular velocity and angular acceleration):

$r$ — the radius: distance in metres of a point from the spin axis. A picture: how far the bug sits from the middle of the record.
$θ$ — the angle turned, always measured in radians here (one full turn $= 2 π$ rad).
$t$ — time in seconds; $Δ t$ means a change in time (an elapsed interval), used inside $α = Δ ω /Δ t$ .
$ω$ (Greek "omega") — the angular velocity: how many radians the body sweeps per second, i.e. $ω = \frac{Δ θ}{Δ t}$ , unit rad/s. Same for every point on a rigid body. Its sign shows the spin direction: positive = counter-clockwise, negative = clockwise.
$α$ (Greek "alpha") — the angular acceleration: how fast $ω$ itself is changing, $α = \frac{Δ ω}{Δ t}$ , unit rad/s². Positive = speeding up (in the $+$ sense), negative = slowing down.
$v$ — the linear (tangential) speed, m/s, along the circle. When we keep $ω$ 's sign, $v = r ω$ is a signed velocity along the tangent; its magnitude (how fast, no direction) is $∣ v ∣ = r ∣ ω ∣$ .
$a_{t}$ — tangential acceleration (changes how fast), m/s²; $a_{c}$ — centripetal acceleration (points to the centre, changes direction), m/s².
$ϕ$ (Greek "phi") — an angle we will compute in Ex 6: the small tilt of the total acceleration vector away from the pure inward (radial) direction.

The scenario matrix

Every problem this topic throws at you is one of these cells. The columns are what is changing; the rows are what special value is involved.

Cell	What makes it special	Which formula is stressed	Example
A	Plain uniform spin ( $α = 0$ )	$v = r ω$ , $a_{c} = r ω^{2}$	Ex 1
B	Same $ω$ , different $r$	$v = r ω$ scaling	Ex 2
C	Degenerate: point on the axis, $r = 0$	all $\to 0$	Ex 3
D	Speeding up ( $α > 0$ )	$a_{t} = r α$	Ex 4
E	Slowing down ( $α < 0$ , sign matters)	$a_{t} = r α$ negative	Ex 5
F	Both accelerations at once (vector total)	$a = a_{t}^{2} + a_{c}^{2}$ + angle	Ex 6
G	Unit trap: degrees / rpm given, must convert	radian discipline	Ex 7
H	Limiting behaviour: $ω \to$ large, what dominates?	$a_{c} \sim ω^{2}$ growth	Ex 8
I	Real-world word problem	pick the right bridge	Ex 9
J	Exam twist: work backwards (given $v$ , find $ω$ or $r$ )	invert the formula	Ex 10
K	Negative $ω$ (clockwise spin): sign of $v$	signed $v = r ω$	Ex 11

Each cell below is filled. Let's go.

Ex 1 — Cell A: plain uniform spin

Worked example A grinding wheel spins steadily at

$ω = 8 rad/s$ . A point sits at $r = 0.15 m$ . Find its speed $v$ and its acceleration.

Forecast: Steady spin means $α = 0$ . So is the point accelerating at all? Guess yes/no before reading.

Speed. $v = r ω = 0.15 \times 8 = 1.2 m/s$ . Why this step? $ω$ is already in rad/s, so we multiply directly — the bridge $v = r ω$ needs no conversion.
Tangential acceleration. Steady spin $\Rightarrow α = 0 \Rightarrow a_{t} = r α = 0$ . Why this step? $a_{t}$ only exists when the rate of turning changes. It doesn't here.
Centripetal acceleration. $a_{c} = r ω^{2} = 0.15 \times 8^{2} = 0.15 \times 64 = 9.6 m/s^{2}$ . Why this step? Even at constant speed the velocity direction keeps swinging, so there is always a pull toward the center.

Verify: Cross-check with $a_{c} = v^{2} / r = 1. 2^{2} /0.15 = 1.44/0.15 = 9.6 m/s^{2}$ . ✓ The two formulas for $a_{c}$ agree, so the answer is consistent. Units: $m \cdot (rad/s)^{2} = m/s^{2}$ (radian is dimensionless). ✓

Ex 2 — Cell B: same $ω$ , different radius

The picture makes the point unmistakable: same angular speed, but the outer bug sweeps a bigger circle in the same time, so it must move faster.

Ex 3 — Cell C: degenerate case, a point right on the axis

Ex 4 — Cell D: speeding up ( $α > 0$ )

Worked example A potter's wheel of radius

$r = 0.20 m$ starts from rest and reaches $ω = 12 rad/s$ in $6 s$ (uniform angular acceleration). Find $α$ and the rim's tangential acceleration $a_{t}$ .

Forecast: Will $a_{t}$ change during the spin-up, or stay constant?

Angular acceleration. $α = \frac{Δ ω}{Δ t} = \frac{12 - 0}{6} = 2 rad/s^{2}$ . Why this step? $α$ is the rate of change of $ω$ ; with a uniform speed-up it's just the total change over the total time $Δ t$ .
Tangential accel. $a_{t} = r α = 0.20 \times 2 = 0.4 m/s^{2}$ . Why this step? $a_{t}$ turns angular "getting faster" into linear "getting faster along the tangent."

Verify: Because $r$ and $α$ are both constant, $a_{t} = 0.4 m/s^{2}$ is constant the whole time — it does not depend on the current $ω$ . Sanity check with kinematics: rim speed at $t = 6$ is $v = r ω = 0.20 \times 12 = 2.4 m/s$ , and $v = a_{t} t = 0.4 \times 6 = 2.4 m/s$ . ✓ They match.

Ex 5 — Cell E: slowing down ( $α < 0$ , sign matters)

Worked example A bicycle wheel, radius

$r = 0.35 m$ , is spinning at $ω_{0} = 15 rad/s$ and brakes to a stop in $Δ t = 5 s$ . Find $α$ , the tangential acceleration $a_{t}$ , and interpret the sign.

Forecast: Will $a_{t}$ come out negative, and what does a negative $a_{t}$ mean physically?

Angular acceleration. $α = \frac{ω _{final} - ω _{0}}{Δ t} = \frac{0 - 15}{5} = - 3 rad/s^{2}$ . Why this step? Slowing down means $ω$ decreases, so $Δ ω < 0$ and $α$ is negative. The sign is not decoration — it encodes direction.
Tangential accel. $a_{t} = r α = 0.35 \times (- 3) = - 1.05 m/s^{2}$ . Why this step? The minus sign tells us $a_{t}$ points against the motion (backward along the tangent) — it is a braking acceleration.

Verify: Magnitude of speed loss: $∣ a_{t} ∣ \times Δ t = 1.05 \times 5 = 5.25 m/s$ . Initial rim speed was $v_{0} = r ω_{0} = 0.35 \times 15 = 5.25 m/s$ , which must drop to $0$ . ✓ Exactly consumed. The sign correctly delivered a full stop, no more no less.

Recall Does the centripetal acceleration also point backward while braking?

No — $a_{c} = r ω^{2}$ is always $\geq 0$ and always points toward the center, no matter the sign of $α$ . Only the tangential piece flips sign.

Ex 6 — Cell F: both accelerations, the vector total (geometry)

Worked example At the instant the braking wheel of Ex 5 is at

$ω = 15 rad/s$ with $α = - 3 rad/s^{2}$ , $r = 0.35 m$ : find $a_{c}$ , $a_{t}$ , the total acceleration magnitude, and the angle $ϕ$ it makes with the radius.

Forecast: Which is bigger at this instant — the center-pull $a_{c}$ or the braking $a_{t}$ ?

Centripetal. $a_{c} = r ω^{2} = 0.35 \times 1 5^{2} = 0.35 \times 225 = 78.75 m/s^{2}$ . Why this step? $a_{c}$ handles the direction-change part of the motion; it dominates at high $ω$ .
Tangential (magnitude). $∣ a_{t} ∣ = ∣ r α ∣ = 0.35 \times 3 = 1.05 m/s^{2}$ . Why this step? This is the speed-change part; small here because braking is gentle.
Total magnitude. The two are perpendicular ( $a_{c}$ along the radius, $a_{t}$ along the tangent), so use Pythagoras: $a = a_{c}^{2} + a_{t}^{2} = 78.7 5^{2} + 1.0 5^{2} \approx 78.75 m/s^{2} .$ Why this step? Perpendicular vectors add like the legs of a right triangle — that is exactly what Pythagoras is for. Here $a_{t}$ is so tiny that, to the same 2-decimal precision as $a_{c}$ , the total is indistinguishable from $a_{c}$ .
Angle $ϕ$ from the (inward) radius. $ϕ = arctan (\frac{∣ a _{t} ∣}{a _{c}}) = arctan (\frac{1.05}{78.75}) \approx 0.76 4^{\circ}$ . Why this step? We ask "which angle has this tangent-to-radial ratio?" — that's what $arctan$ answers, and it tilts the total vector slightly off the pure inward direction. $ϕ$ is that tilt.

Verify: Since $a_{t} ≪ a_{c}$ , the total should be a hair above $a_{c}$ : $78.7 5^{2} + 1.0 5^{2} = 78.757 \dots$ , which rounds to $78.75 m/s^{2}$ at 2 d.p. ✓ And the tilt angle should be tiny: $0.7 6^{\circ}$ is nearly straight to the center. ✓

The figure shows the right triangle: a long inward $a_{c}$ leg, a short tangential $a_{t}$ leg, and the slightly-tilted total arrow (the hypotenuse).

Ex 7 — Cell G: the unit trap (rpm and degrees)

Ex 8 — Cell H: limiting behaviour, what dominates as $ω$ grows

Worked example A rotor at

$r = 0.3 m$ has a fixed spin-up $α = 4 rad/s^{2}$ . Compute $a_{t}$ and $a_{c}$ at $ω = 2$ , then at $ω = 20 rad/s$ . Which acceleration wins as $ω$ grows?

Forecast: As $ω$ climbs, do $a_{t}$ and $a_{c}$ grow at the same pace?

Tangential (unchanging). $a_{t} = r α = 0.3 \times 4 = 1.2 m/s^{2}$ at both speeds. Why this step? $a_{t}$ has no $ω$ in it — it stays put no matter how fast we spin.
Centripetal at $ω = 2$ . $a_{c} = r ω^{2} = 0.3 \times 4 = 1.2 m/s^{2}$ .
Centripetal at $ω = 20$ . $a_{c} = 0.3 \times 400 = 120 m/s^{2}$ . Why this step? $a_{c}$ grows with $ω^{2}$ — the square — so a $10 \times$ faster spin gives $100 \times$ the center-pull.

Verify: Ratio $a_{c} (20) / a_{c} (2) = 120/1.2 = 100 = 1 0^{2}$ . ✓ The square law is confirmed. Limiting conclusion: as $ω \to$ large, $a_{c}$ ( $\propto ω^{2}$ ) utterly overwhelms the constant $a_{t}$ — this is why fast rotors fly apart at the rim, not because of tangential push.

The graph makes it visual: the flat $a_{t}$ line versus the parabola of $a_{c}$ that leaves it far behind.

Ex 9 — Cell I: real-world word problem

Worked example A car's tyre has radius

$r = 0.31 m$ . The car drives (without skidding) at $v = 25 m/s$ on a straight road. How fast is the tyre spinning ( $ω$ ), and what centripetal acceleration does a stone stuck in the tread feel?

Forecast: A stone in a tyre on a straight road — is it accelerating even though the car goes straight?

Angular speed from rolling. No skidding means the tread speed equals road speed $v$ , so invert the bridge: $ω = \frac{v}{r} = \frac{25}{0.31} \approx 80.6 rad/s$ . Why this step? Rolling without slipping links road speed to spin via the same bridge $v = r ω$ , just solved for $ω$ . See Rolling motion.
Centripetal accel on the stone. In the wheel's frame the stone circles the axle: $a_{c} = r ω^{2} = 0.31 \times 80. 6^{2} \approx 0.31 \times 6503 \approx 2016 m/s^{2}$ . Why this step? Relative to the axle the stone still traces a circle, so it feels the center-pull even while the car travels in a straight line.

Verify: Cross-check via $a_{c} = v^{2} / r = 2 5^{2} /0.31 = 625/0.31 \approx 2016 m/s^{2}$ . ✓ Both routes agree. That's over $200 g$ — which is exactly why stones and mud get flung out of tyres.

Ex 10 — Cell J: the exam twist, work backwards

Ex 11 — Cell K: negative $ω$ (clockwise spin)

Worked example A disk spins

clockwise, which by our sign convention means $ω = - 8 rad/s$ . A rim point sits at $r = 0.15 m$ . Find the signed tangential velocity $v$ , its magnitude (how fast), and the centripetal acceleration $a_{c}$ .

Forecast: Does the minus sign make the point move slower, or does it survive only as a direction? And is $a_{c}$ negative now?

Signed velocity. $v = r ω = 0.15 \times (- 8) = - 1.2 m/s$ . Why this step? Keeping $ω$ 's sign, $v = r ω$ inherits it: the minus means the point moves in the clockwise tangential sense (opposite to the counter-clockwise $+$ direction). The number is negative but that is a direction flag, not a slower speed.
Magnitude (how fast). $∣ v ∣ = r ∣ ω ∣ = 0.15 \times 8 = 1.2 m/s$ . Why this step? "How fast" strips the sign: $∣ v ∣ = r ∣ ω ∣$ . Same speed as the $ω = + 8$ case in Ex 1 — only the direction differs.
Centripetal accel. $a_{c} = r ω^{2} = 0.15 \times (- 8)^{2} = 0.15 \times 64 = 9.6 m/s^{2}$ . Why this step? $ω$ is squared, so $(- 8)^{2} = 64$ : the sign vanishes. Centripetal acceleration always points inward and its magnitude does not care which way you spin.

Verify: $a_{c}$ matches Ex 1 exactly ( $9.6 m/s^{2}$ ), confirming the square kills the sign; and $∣ v ∣$ matches Ex 1's speed ( $1.2 m/s$ ), confirming only the direction of $v$ flipped. ✓

Recap: which cell needs which move

Recall Quick self-test across the matrix

Doubling $ω$ (fixed $r$ , fixed $α$ ) changes $v$ , $a_{t}$ , $a_{c}$ by what factors? ::: $v$ doubles, $a_{t}$ unchanged, $a_{c}$ quadruples. At $r = 0$ , what are $v$ , $a_{t}$ , $a_{c}$ ? ::: All zero. A negative $α$ makes which acceleration negative? ::: Only $a_{t}$ ; $a_{c}$ stays positive (inward). A negative $ω$ makes which quantities negative? ::: Only the signed $v$ (its direction flips); $a_{c} = r ω^{2}$ stays positive because $ω$ is squared. To convert $210$ rpm to rad/s you multiply by? ::: $\frac{2 π}{60} = \frac{π}{30}$ .