Angular displacement θ, angular velocity ω, angular acceleration α: Visual walkthrough

We will earn, in order:

what an angle even is once we insist on a clean measuring stick (the radian),
how a turning angle drags a point along an arc,
how "turning per second" ( $ω$ ) becomes "metres per second" ( $v$ ),
how "turning-speed changing" ( $α$ ) becomes "speeding up along the path" ( $a_{t}$ ),
the edge cases: the centre point, zero spin, and reversing spin.

Step 1 — Paint one dot and ask the right question

Figure — Angular displacement θ, angular velocity ω, angular acceleration α

WHY this setup. We could try to track the dot in plain $x, y$ metres, but that mixes "how far out" with "how far around" into an ugly changing pair. Rotation has one thing in common for every dot on the disc: they all swing through the same amount of turn each second. So we invent a number that captures only the "around" part and throws away the "out" part. That number is the angle.

WHAT IT LOOKS LIKE. In the figure the fixed length is the cyan spoke of length $r$ . The amber wedge is the "amount of turn" — the thing we are about to measure properly.

Step 2 — Choose a fair measuring stick for the angle: the radian

Term by term, right where they sit:

$s$ — the arc length: the curved distance the dot actually rode along the rim (metres).
$r$ — the spoke: straight distance centre-to-dot (metres).
$θ$ — their ratio. Both are metres, so metres cancel: $θ$ is a pure number. We name that unit the radian.

WHY a ratio, and why not degrees? A degree is an arbitrary human choice — chop a full circle into 360 slices "because ancient astronomers liked 360". Nothing in the geometry cares about 360. But the ratio $s / r$ is forced on us by the picture itself: it asks "how many spoke-lengths of curved edge did we sweep?" When the arc equals exactly one spoke ( $s = r$ ), the angle is exactly $1$ radian. No conversion factor is ever born. This is the whole reason radians make the later formulas clean.

WHAT IT LOOKS LIKE. In the figure, the amber arc and the cyan spoke are drawn the same length — that pictures the angle "1 radian" directly.

Step 3 — A full trip round the circle checks the definition

WHY do this. It's a sanity check: a definition is only trustworthy if a known case comes out right. One full turn should give one clean number.

$θ_{full} = \frac{s}{r} = \frac{2 π r}{r} = 2 π rad$

$2 π r$ — the circumference (arc for one full loop),
$r$ — cancels top and bottom,
$2 π$ — the answer: one full turn is $2 π$ radians ( $\approx 6.283$ ), which we also call $360°$ .

WHAT IT LOOKS LIKE. The figure marks the quarter-turns: $\frac{π}{2}$ , $π$ , $\frac{3 π}{2}$ , $2 π$ around the loop — so you can see radians laid out like clock positions.

Step 4 — Add a clock: turning-per-second becomes angular velocity $ω$

Term by term:

$Δ θ$ — the extra angle swept (radians) — the amber wedge growing between two snapshots.
$Δ t$ — the time between those two snapshots (seconds).
$ω$ — radians per second: how much turn packs into each second.

WHY the ratio $d θ / d t$ and not something else? We want rate: "how much angle per unit time". A rate is always a change-in-thing over change-in-time — that's exactly what dividing $Δ θ$ by $Δ t$ does. Shrinking $Δ t$ toward zero (the derivative $d θ / d t$ ) gives the instantaneous spin at one moment, not just the average over a chunk.

WHAT IT LOOKS LIKE. Two dotted spokes in the figure — the position at $t$ and at $t + Δ t$ — with the amber wedge between them labelled $Δ θ$ . Same wedge every second means constant $ω$ .

Step 5 — Turn $ω$ into a real speed: derive $v = r ω$

WHY differentiate $s = r θ$ . Linear speed means "metres of arc per second", i.e. $v = d s / d t$ . We already have $s = r θ$ from Step 2. Since $r$ is fixed for our dot, we just watch how $s$ grows as $θ$ grows:

$v = \frac{d s}{d t} = \frac{d}{d t} (r θ) = r \frac{d θ}{d t} = r ω$

$d s / d t$ — metres of arc per second = the linear speed $v$ ,
$r$ — pulled out because it's constant (the dot doesn't drift in or out),
$d θ / d t$ — recognised from Step 4 as $ω$ .

$v = r ω$

WHAT IT LOOKS LIKE. The figure shows an inner cyan dot and an outer amber dot. Same wedge (same $ω$ ), but the outer arrow (its velocity) is longer — longer $r$ scales up $v$ . The velocity arrow points tangent to the circle (along the direction of travel), not along the spoke.

Recall Why is

$v$ tangent, not along the spoke? Because the dot's next instant of motion is along the rim it's riding — the spoke direction ( $r$ ) never changes for our dot, so no motion happens along it. ::: The velocity is tangent to the circle at the dot.

Step 6 — Let the spin itself change: derive $a_{t} = r α$

WHY differentiate $v = r ω$ . "Speeding up along the path" means "rate of change of the linear speed", $a_{t} = d v / d t$ . We have $v = r ω$ with $r$ fixed:

$a_{t} = \frac{d v}{d t} = \frac{d}{d t} (r ω) = r \frac{d ω}{d t} = r α$

$d v / d t$ — metres-per-second per second: the tangential acceleration $a_{t}$ ,
$r$ — constant, pulled out,
$d ω / d t$ — recognised as $α$ .

$a_{t} = r α$

WHAT IT LOOKS LIKE. The figure shows the velocity arrow at time $t$ (short) and at $t + Δ t$ (longer), both tangent. The extra amber arrow between the tips is $a_{t}$ , pointing along the motion — it lengthens the speed.

Step 7 — Edge and degenerate cases (never let the reader fall off the map)

Case A — the centre dot ( $r = 0$ ). Put the dot on the pin. Then $v = r ω = 0 \cdot ω = 0$ and $a_{t} = r α = 0$ . WHY it makes sense: the centre never rides any arc — it only pivots in place. It still shares the same $θ, ω, α$ as every other dot, but its linear speed is zero. The figure shows this dot fixed while an outer dot sweeps around it.

Case B — steady spin ( $α = 0$ ). If the spin rate never changes, $a_{t} = r \cdot 0 = 0$ : no speeding up along the path. But the dot is still accelerating — inward, via $a_{c}$ — because its direction keeps turning. This steady-spin case is exactly Uniform Circular Motion.

Case C — reversing spin ( $α$ opposite to $ω$ ). If $α$ points against the current spin (negative $α$ while $ω > 0$ ), the disc slows, stops, then spins backward. The signs, not the pictures, carry this: $ω$ shrinks to $0$ , passes through, and grows negative. The figure shows the wedge shrinking, then reopening the other way. (This is the fan of parent-note Example 3, where $α = - 45 rad/s^{2}$ .)

The one-picture summary

This single blueprint compresses the whole walkthrough: the angle $θ = s / r$ (Step 2), its rate $ω = d θ / d t$ (Step 4), and the two bridges $v = r ω$ (Step 5) and $a_{t} = r α$ (Step 6), with the tangential $a_{t}$ and inward $a_{c}$ arrows shown perpendicular at the dot.

Recall Feynman retelling of the whole walkthrough

I paint one dot on a spinning disc, a distance $r$ from the pin. To measure how far around it went, I don't count degrees — I measure the curved trail it left and divide by the spoke length. That ratio is the angle in radians, and rearranged it says the trail length is just the spoke times the angle: $s = r θ$ . Now I start a clock. How fast the angle grows is $ω$ . Because the trail length is $r$ times the angle, and $r$ never changes, the trail grows at $r$ times the angle's growth rate — so the dot's real speed is $v = r ω$ . An outer dot and an inner dot turn the same amount each second, but the outer one has more $r$ , so it zooms. Finally, if the spin itself speeds up (that's $α$ ), the dot's speed climbs along its path at $a_{t} = r α$ . Special dots keep me honest: the centre dot has $r = 0$ so it never moves in metres; a steady spin has $α = 0$ so no path-speeding but still an inward turning pull; and a spin fighting itself ( $α$ against $ω$ ) slows, stops, and reverses. Three bridges, one radius, the same picture the whole way down.

Connections

Parent topic note — the summary this page derives visually.
Uniform Circular Motion — the $α = 0$ case of Step 7B.
Centripetal Acceleration and Force — the inward $a_{c}$ arrow perpendicular to $a_{t}$ .
Linear Kinematics Equations — the linear world the bridges connect to.
Torque and Angular Acceleration — what causes the $α$ we used in Step 6.
Moment of Inertia — the rotational analogue of mass.
Rolling Motion — where $v = r ω$ meets straight-line travel.