Angular displacement θ, angular velocity ω, angular acceleration α: Worked examples

Before anything, one habit that saves you every time:

The scenario matrix

Every rotational-kinematics problem is one (or a blend) of these cells. Each example below is tagged with the cell it covers.

Cell	What makes it distinct	Trap it hides	Example
A. Speed-up from rest	$ω_{0} = 0$ , $α > 0$	forgetting rad vs rev	Ex 1
B. Slow-down to rest, no time	know $ω, ω_{0}, θ$ , want $α$	which equation (time-free)	Ex 2
C. Reversal	$α$ opposite to $ω$ , motion crosses zero	net vs total angle	Ex 3
D. Angular → linear at a point	need $v, a_{t}, a_{c}$ at radius $r$	$a_{t}$ vs $a_{c}$ confusion	Ex 4
E. Two radii, one body	same $ω, α$ , different $v$	who shares what	Ex 5
F. Degenerate: $α = 0$	uniform spin, period/frequency	can't use $α$ equations blindly	Ex 6
G. Real-world word problem	rpm units, "how many turns"	unit conversion rpm→rad/s	Ex 7
H. Exam twist: coupled wheels	belt/gear links two rims	equal rim speed, not equal $ω$	Ex 8

Example 1 — Cell A: speed-up from rest

Worked example Statement

A potter's wheel starts from rest and reaches $ω = 20 rad/s$ after $4 s$ of constant angular acceleration. Find $α$ and the total angle turned. How many full revolutions is that?

Forecast: guess before reading — will the angle be more or less than $40$ rad? (Average speed is half of $20$ , over $4$ s...)

Find $α$ : $α = \frac{ω - ω _{0}}{t} = \frac{20 - 0}{4} = 5 rad/s^{2} .$ Why this step? With constant $α$ the definition $α = Δ ω /Δ t$ is exact — average equals instantaneous, so it's just rise over run.
Find $θ$ : $θ = ω_{0} t + \frac{1}{2} α t^{2} = 0 + \frac{1}{2} (5) (4)^{2} = 40 rad .$ Why this step? We know $ω_{0}, α, t$ and want $θ$ — the displacement equation uses exactly those.
Convert to revolutions: $N = \frac{θ}{2 π} = \frac{40}{2 π} \approx 6.37 rev .$ Why this step? One full turn is $2 π$ rad; dividing counts turns.

Verify: Average angular speed $= (0 + 20) /2 = 10 rad/s$ , times $4 s = 40 rad$ . ✓ Matches. Units: $(rad/s) \cdot s = rad$ . ✓

Example 2 — Cell B: slow to rest, no time given

Worked example Statement

A ceiling fan spinning at $ω_{0} = 30 rad/s$ is switched off and coasts to a stop after turning through $θ = 10 rad$ . Find its angular acceleration.

Forecast: the sign of $α$ — will it be positive or negative? (It's slowing, and we chose its spin as $+$ ...)

Choose the right equation. We know $ω = 0, ω_{0} = 30, θ = 10$ , but not the time. The equation with no $t$ is $ω^{2} = ω_{0}^{2} + 2 α θ$ . Why this step? Reaching for $ω = ω_{0} + α t$ would force us to invent $t$ . Match knowns to the equation that omits the unknown.
Solve: $0 = 3 0^{2} + 2 α (10) \Rightarrow 20 α = - 900 \Rightarrow α = - 45 rad/s^{2} .$ Why this step? Pure algebra once the equation is chosen.

Verify: The minus sign is correct — $α$ is opposite to the (positive) spin, which is exactly what "slowing down" means by our sign rule. Sanity: $∣ α ∣$ is large because a lot of speed ( $30$ ) is killed in a small angle ( $10$ rad). ✓

Example 3 — Cell C: reversal (net vs total angle)

Worked example Statement

A disc spins at $ω_{0} = + 6 rad/s$ (counter-clockwise) while a constant $α = - 3 rad/s^{2}$ acts. Find (a) the time to momentarily stop, (b) its net angular displacement at $t = 4 s$ , and (c) the total angle actually swept in those $4 s$ .

Forecast: at $t = 4 s$ is the disc spinning forward, stopped, or backward? Will net and total angle be equal?

Time to stop: set $ω = 0$ in $ω = ω_{0} + α t$ : $0 = 6 + (- 3) t \Rightarrow t = 2 s .$ Why this step? "Momentarily stops" means $ω = 0$ ; the velocity equation ties $ω$ to $t$ .
Net displacement at $4$ s: $θ = ω_{0} t + \frac{1}{2} α t^{2} = 6 (4) + \frac{1}{2} (- 3) (4)^{2} = 24 - 24 = 0 rad .$ Why this step? The displacement equation gives signed position — forward and backward sweeps cancel.
Total angle swept. Forward phase ( $0$ to $2$ s): $θ_{1} = 6 (2) + \frac{1}{2} (- 3) (2)^{2} = 12 - 6 = 6 rad$ . Backward phase ( $2$ to $4$ s): by symmetry it sweeps $6 rad$ the other way. Total distance $= 6 + 6 = 12 rad .$ Why this step? Distance ignores direction — you must split at the turning point $t = 2$ s and add magnitudes, never subtract.

Verify: At $t = 4$ : $ω = 6 + (- 3) (4) = - 6 rad/s$ — exactly reversed from the start, mirroring $θ_{net} = 0$ (it's back where it began, now spinning backward). ✓ Net $0 \neq =$ total $12$ : the classic reversal signature. ✓

Example 4 — Cell D: angular converted to linear at a point

Worked example Statement

Take the potter's wheel of Example 1 ( $ω = 20 rad/s$ , $α = 5 rad/s^{2}$ ) at radius $r = 0.30 m$ , at the instant it reaches $20 rad/s$ . Find the rim's linear speed $v$ , tangential acceleration $a_{t}$ , centripetal acceleration $a_{c}$ , and the total linear acceleration magnitude $a$ .

Forecast: which is bigger at high speed — $a_{t}$ or $a_{c}$ ?

Linear speed: $v = r ω = 0.30 \times 20 = 6 m/s .$ Why this step? $v = r ω$ is the bridge from turning rate to how fast the rim point actually travels.
Tangential acceleration: $a_{t} = r α = 0.30 \times 5 = 1.5 m/s^{2} .$ Why this step? $a_{t}$ is the speed-changing part; it comes from $α$ , the changing spin rate.
Centripetal acceleration: $a_{c} = ω^{2} r = 2 0^{2} \times 0.30 = 120 m/s^{2} .$ Why this step? Even at constant speed a point on a circle accelerates inward to keep turning; that's $a_{c}$ , direction-changing only.
Total: $a = a_{t}^{2} + a_{c}^{2} = 1. 5^{2} + 12 0^{2} \approx 120.0 m/s^{2} .$ Why this step? $a_{t}$ (along path) and $a_{c}$ (inward) are perpendicular, so combine by Pythagoras.

Verify: $a_{c} ≫ a_{t}$ — at high $ω$ the turning dominates, so the total is almost pure centripetal, as expected. Units: $(rad/s)^{2} \cdot m = m/s^{2}$ (radian is dimensionless). ✓

Example 5 — Cell E: two radii on one rigid body

Worked example Statement

On the same wheel ( $ω = 20 rad/s$ , $α = 5 rad/s^{2}$ ), a bug sits at the hub $r_{1} = 0.10 m$ and a chip of clay at the rim $r_{2} = 0.30 m$ . Compare their $ω$ , their linear speeds, and their tangential accelerations.

Forecast: which of $ω, v, a_{t}$ is the same for both?

Angular quantities: both share $ω = 20 rad/s$ and $α = 5 rad/s^{2}$ . Why this step? A rigid body turns as one piece — every point sweeps the same angle in the same time, so $θ, ω, α$ are common to all.
Linear speeds: $v_{1} = r_{1} ω = 0.10 \times 20 = 2 m/s$ ; $v_{2} = r_{2} ω = 0.30 \times 20 = 6 m/s .$ Why this step? $v = r ω$ : farther out means faster, even with identical $ω$ .
Tangential accelerations: $a_{t 1} = r_{1} α = 0.5 m/s^{2}$ ; $a_{t 2} = r_{2} α = 1.5 m/s^{2} .$ Why this step? $a_{t} = r α$ scales with $r$ the same way.

Verify: Ratios: $v_{2} / v_{1} = 6/2 = 3 = r_{2} / r_{1} = 0.30/0.10$ . ✓ The angular quantities matched exactly; only the linear ones scaled with radius. That's the whole reason we invented angular variables. ✓

Example 6 — Cell F: degenerate case $α = 0$ (uniform spin)

Worked example Statement

A hard drive platter spins steadily (no speed-up) at $ω = 7200$ rpm (revolutions per minute). Find its period $T$ , frequency $f$ , and angular velocity in rad/s. What is its angular acceleration?

Forecast: can we use $θ = ω_{0} t + \frac{1}{2} α t^{2}$ here? What is $α$ ?

Angular acceleration is zero: "spins steadily" ⇒ $ω$ constant ⇒ $α = 0$ . Why this step? The three kinematic equations still hold, but with $α = 0$ they collapse to $ω = const$ and $θ = ω t$ . Don't misuse the $\frac{1}{2} α t^{2}$ term — it vanishes.
Frequency: $f = 7200 rev/min = 7200/60 = 120 Hz (rev/s) .$ Why this step? Frequency is turns per second; divide rpm by $60$ .
Period: $T = 1/ f = 1/120 \approx 0.00833 s .$ Why this step? Period is seconds per turn — the reciprocal of frequency.
Angular velocity: $ω = 2 π f = 2 π (120) \approx 753.98 rad/s .$ Why this step? Each turn is $2 π$ rad; $ω = 2 π / T = 2 π f$ .

Verify: $ω \cdot T = 753.98 \times 0.00833 \approx 6.283 \approx 2 π$ — one full turn per period, as it must be. ✓ Units: $rev/s \to \times 2 π rad/rev = rad/s$ . ✓

Example 7 — Cell G: real-world word problem (rpm, "how long")

Worked example Statement

A car engine idles at $800 rpm$ and is revved up to $3000 rpm$ in $2.5 s$ at constant angular acceleration. Find $α$ (in rad/s²) and the number of revolutions the crankshaft makes during this rev-up.

Forecast: more or fewer than $60$ revolutions in $2.5$ s?

Convert both speeds to rad/s: $ω_{0} = 800 \cdot \frac{2 π}{60} \approx 83.78 rad/s$ , $ω = 3000 \cdot \frac{2 π}{60} \approx 314.16 rad/s .$ Why this step? Every kinematics formula needs rad/s; rpm must be converted first ( $\times 2 π /60$ ).
Angular acceleration: $α = \frac{ω - ω _{0}}{t} = \frac{314.16 - 83.78}{2.5} \approx 92.15 rad/s^{2} .$ Why this step? Constant $α$ ⇒ rise over run of angular velocity.
Angle turned: use average speed × time: $θ = \frac{ω _{0} + ω}{2} t = \frac{83.78 + 314.16}{2} (2.5) \approx 497.4 rad .$ Why this step? With constant $α$ , average angular speed is the midpoint; this avoids re-plugging $α$ .
Revolutions: $N = θ /2 π \approx 497.4/6.283 \approx 79.2 rev .$ Why this step? Convert rad to turns.

Verify: Cross-check $θ$ with $θ = ω_{0} t + \frac{1}{2} α t^{2} = 83.78 (2.5) + \frac{1}{2} (92.15) (2.5)^{2} \approx 209.4 + 288.0 = 497.4 rad$ . ✓ Both methods agree. ✓

Example 8 — Cell H: exam twist, two wheels linked by a belt

Worked example Statement

A small pulley of radius $r_{A} = 0.05 m$ drives a large pulley of radius $r_{B} = 0.20 m$ through a non-slipping belt. The small pulley spins at $ω_{A} = 40 rad/s$ . Find the belt speed, the large pulley's $ω_{B}$ , and the ratio of their revolution rates.

Forecast: which pulley spins faster (bigger $ω$ ) — the small or the large one?

Belt speed = rim speed of the driver: $v = r_{A} ω_{A} = 0.05 \times 40 = 2 m/s .$ Why this step? A non-slipping belt moves at the linear speed of the rim it wraps — this is the shared quantity, not $ω$ .
Same belt on pulley B: $v = r_{B} ω_{B} \Rightarrow ω_{B} = \frac{v}{r _{B}} = \frac{2}{0.20} = 10 rad/s .$ Why this step? The belt's linear speed is common to both rims; invert $v = r ω$ for B.
Ratio of spins: $\frac{ω _{A}}{ω _{B}} = \frac{40}{10} = 4 = \frac{r _{B}}{r _{A}} = \frac{0.20}{0.05} .$ Why this step? Equal rim speed forces $ω \propto 1/ r$ — the smaller wheel spins faster.

Verify: Belt speed from B: $r_{B} ω_{B} = 0.20 \times 10 = 2 m/s$ ✓ matches the driver. The small pulley spins $4 \times$ faster, inverse to its $4 \times$ -smaller radius. ✓

Recall Quick self-test on the cells

Which equation for "know $ω, ω_{0}, θ$ , want $α$ , no time"? ::: Time-free: $ω^{2} = ω_{0}^{2} + 2 α θ$ . After a reversal, is net displacement equal to total angle swept? ::: No — net is signed and can cancel; total adds magnitudes of each phase, split at the stop time. Two clay chips at different radii on one platter: what do they share? ::: $θ, ω, α$ (same); $v, a_{t}, a_{c}$ differ, scaling with $r$ . Belt-linked pulleys share what? ::: Rim (linear) speed $v$ , so $r_{A} ω_{A} = r_{B} ω_{B}$ ; not $ω$ . Convert $3000$ rpm to rad/s. ::: $3000 \times 2 π /60 \approx 314.16$ rad/s.

Connections

Parent topic (Hinglish) — the definitions and derivations these examples exercise.
Uniform Circular Motion — Cell F ( $α = 0$ ) is exactly this case.
Centripetal Acceleration and Force — the $a_{c} = ω^{2} r$ used in Example 4.
Linear Kinematics Equations — same algebra, swap $x, v, a \to θ, ω, α$ .
Rolling Motion — Cell H's $v = r ω$ logic extends to a wheel rolling on the ground.
Torque and Angular Acceleration — what causes the $α$ in these speed-up examples.
Moment of Inertia — the rotational "mass" behind that $α$ .