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

WHAT the unit tells us. Radians live in the angle world. A "per second" means "rate of change with time." A "per second squared" means "rate of change of a rate."

• $\text{rad}$ alone → an angle → $\theta$.
• $\text{rad/s}$ → angle per time → $\omega$.
• ${\text{rad/s}}^2$ → (angle per time) per time → $\alpha$, angular acceleration.

So $4.5{\text{rad/s}}^2$ is an angular acceleration: it says the spin rate $\omega$ grows by $4.5\text{rad/s}$ every second.

WHY radians here. One full turn sweeps an arc equal to the whole circumference $2\pi r$, and $\theta =s/r=2\pi r/r=2\pi$ rad. So one revolution $=2\pi$ rad — no calculator needed for the idea, just multiply. $\theta =3\times 2\pi =6\pi \approx 18.85\text{rad}.$

WHAT is shared vs scaled. For any rigid body, every point turns through the same angle in the same time — that is the whole reason we invented angular language.

• Angle: both ants sweep $\frac{\pi }{2}$ rad — equal.
• Arc (distance): $s=r\theta$, so ant B travels $s_B=0.3\times \frac{\pi }{2}\approx 0.471$ m while ant A travels $s_A=0.1\times \frac{\pi }{2}\approx 0.157$ m.

Ant B covers 3× the arc, purely because it is 3× farther out. (See the figure below — same wedge angle, longer outer arc.)

WHY this equation. We know start ${\omega }_0=0$, end $\omega$, and time $t$ — that is exactly the trio in $\omega ={\omega }_0+\alpha t$. $\alpha =\frac{\omega -{\omega }_0}{t}=\frac{12-0}{3}=4{\text{rad/s}}^2.$ Check the meaning: $+4$ means $\omega$ climbs by $4\text{rad/s}$ each second: $0\to 4\to 8\to 12$. ✓

WHY $v=r\omega$ and $a_t=r\alpha$. These are the two "bridges" from the angular world to the linear world at a point sitting at radius $r$. $v=r\omega =0.25\times 12=3\text{m/s}.$ $a_t=r\alpha =0.25\times 4=1.0{\text{m/s}}^2.$ $a_t$ is the part of acceleration along the path — it is what makes the rim's speed grow.

WHY the displacement equation. We know ${\omega }_0=0$, $\alpha =4$, $t=3$ and want $\theta$ — that is $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$. $\theta =0+\frac{1}{2}(4)(3^2)=\frac{1}{2}(4)(9)=18\text{rad}.$ Turns: $18/(2\pi )\approx 2.86$ revolutions.

WHY the third equation. We know ${\omega }_0$, final $\omega =0$, and $\theta$, but not $t$. The only kinematic equation with no $t$ is ${\omega }^2={\omega }_0^2+2\alpha \theta$. $0=40^2+2\alpha (100)\Rightarrow 0=1600+200\alpha .$ $\alpha =-\frac{1600}{200}=-8{\text{rad/s}}^2.$ The minus sign is the physics: $\alpha$ opposes $\omega$, i.e. it is a deceleration.

WHY two accelerations. A rim point does two things at once: its speed grows (tangential, along the path) and its direction bends (centripetal, toward the centre). These are perpendicular, so we combine by Pythagoras. $a_t=r\alpha =0.2\times 6=1.2{\text{m/s}}^2(\text{along the path}).$ $a_c={\omega }^{2}r=10^2\times 0.2=20{\text{m/s}}^2(\text{toward centre}).$ $a=\sqrt{a_t^2+a_c^2}=\sqrt{1.2^2+20^2}=\sqrt{1.44+400}=\sqrt{401.44}\approx 20.04{\text{m/s}}^2.$ Notice $a_c\gg a_t$: at high spin the turning dominates the speeding-up. The figure shows the two perpendicular arrows and their resultant.

WHY differentiate for (b). Average asks "total angle ÷ total time"; instantaneous asks "the rate right now", which is the derivative $\omega =d\theta /dt$.

• (a) Average: $\theta (2)=2(8)=16$, $\theta (1)=2(1)=2$. ${\omega }_{avg}=\frac{16-2}{2-1}=14\text{rad/s}.$
• (b) Instantaneous: $\omega =\frac{d}{dt}(2t^3)=6t^2$, so at $t=2$: $\omega =6(4)=24\text{rad/s}.$

Why different: the motor is accelerating ($\alpha =d\omega /dt=12t\ne 0$), so its rate at the end ($24$) exceeds its rate averaged over the interval ($14$). For constant $\alpha$ they'd tie only at the midpoint of the interval.

WHY convert to angular first. Rolling without slipping links road-distance $s_{road}$ to wheel-angle: each turn advances the car by one circumference. So the car's forward speed $v$ ties to the wheel's spin through $v=r\omega$, its forward acceleration through $a=r\alpha$, and the road distance through $s_{road}=r\theta$.

Step 1 — angular speeds: ${\omega }_0=\frac{v_0}{r}=\frac{5}{0.35}\approx 14.286\text{rad/s},\omega =\frac{20}{0.35}\approx 57.143\text{rad/s}.$ Step 2 — angle turned: the wheel rolls $150$ m, so $\theta =s_{road}/r=150/0.35\approx 428.57$ rad. Step 3 — α (time-free, we don't know the time): ${\omega }^2={\omega }_0^2+2\alpha \theta \Rightarrow 57.143^2=14.286^2+2\alpha (428.57).$ $3265.3=204.1+857.14\alpha \Rightarrow \alpha =\frac{3061.2}{857.14}\approx 3.57{\text{rad/s}}^2.$ Cross-check via linear kinematics: $a=(v^2-v_0^2)/(2s)=(400-25)/300=1.25{\text{m/s}}^2$, and $a/r=1.25/0.35\approx 3.57{\text{rad/s}}^2$. ✓ Same answer, two languages. Revolutions: $428.57/(2\pi )\approx 68.2$ rev.

WHY split into phases. The equations of constant-$\alpha$ kinematics only apply within a phase where $\alpha$ is constant. Two different $\alpha$ values → two separate calculations, then add.

Phase 1 (accelerating, $0$–$5$ s): ${\omega }_1={\omega }_0+{\alpha }_{1}t=0+8(5)=40\text{rad/s}.$ ${\theta }_1=\frac{1}{2}{\alpha }_1{t}^2=\frac{1}{2}(8)(25)=100\text{rad}.$ Phase 2 (constant speed, $5$–$10$ s): $\alpha =0$, so ${\theta }_2={\omega }_{1}t=40\times 5=200$ rad. Total: $\theta ={\theta }_1+{\theta }_2=100+200=300\text{rad}.$

WHY watch the sign. $\omega$ and $\alpha$ point opposite ways at first (wheel slowing), but $\alpha$ never quits — so after the stop, $\omega$ goes negative (reverse spin). The equations handle all of this automatically if you keep signs.

(a) Momentary stop: set $\omega =0$ in $\omega ={\omega }_0+\alpha t$: $0=6+(-3)t\Rightarrow t=2\text{s}.$ (b) Displacement at $t=4$ s: $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2=6(4)+\frac{1}{2}(-3)(16)=24-24=0\text{rad}.$ Net angle is zero — the wheel wound forward, stopped, then unwound back to its start. (c) At $t=4$ s: $\omega =6+(-3)(4)=-6\text{rad/s}$: it is now spinning backward at the same rate it began forward. (See the θ-vs-t parabola — it rises, peaks at $t=2$, and returns to $0$ at $t=4$.)

WHY distance ≠ displacement. Displacement is net (with sign); the total angle turned adds up the forward part and the backward part as positive amounts, because the wheel physically moved both ways.

Forward stretch (0 → 2 s): peak angle at $t=2$: ${\theta }_{peak}=6(2)+\frac{1}{2}(-3)(4)=12-6=6\text{rad}.$ Backward stretch (2 → 4 s): from $+6$ rad back to $0$ rad → another $6$ rad of turning (in reverse). Total angle turned: $|{\theta }_{forward}|+|{\theta }_{back}|=6+6=12\text{rad}.$ So the wheel swept $12$ rad of arc even though its net displacement is $0$.

WHY test a limit. A good general formula must collapse to the simpler special case when the extra ingredient (here $\alpha$) is switched off — this is a sanity check on the whole framework, and it tells us exactly which familiar motion sits inside the general one.

Step 1 — kill the $\alpha$ term in the displacement equation: $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2\stackrel{\alpha \to 0}{\to }\theta ={\omega }_{0}t.$ The $\frac{1}{2}\alpha t^2$ term vanishes because it is multiplied by $\alpha =0$. What remains, $\theta ={\omega }_{0}t$, is angle growing linearly with time — the hallmark of a constant spin rate.

Step 2 — check the other two equations agree: $\omega ={\omega }_0+\alpha t\stackrel{\alpha \to 0}{\to }\omega ={\omega }_0(\text{constant spin}),$ ${\omega }^2={\omega }_0^2+2\alpha \theta \stackrel{\alpha \to 0}{\to }{\omega }^2={\omega }_0^2(\text{speed never changes}).$ All three collapse consistently — the framework is self-consistent.

Step 3 — the physical picture (connect to Uniform Circular Motion). With $\alpha =0$ the tangential acceleration $a_t=r\alpha \to 0$, so the rim's speed never changes. What is left is the centripetal acceleration $a_c={\omega }^{2}r$, which only bends the direction. A body at constant $\omega$ tracing a circle at constant speed while its velocity continually turns is exactly uniform circular motion. So uniform circular motion is not a separate theory — it is the $\alpha \to 0$ corner of angular kinematics, the same way constant-velocity motion is the $a\to 0$ corner of linear kinematics.

Can you now, without looking:

• state which of θ/ω/α a unit belongs to? ::: Yes — count the "per seconds": none→θ, one→ω, two→α.
• choose the time-free equation when $t$ is missing? ::: Yes — use ${\omega }^2={\omega }_0^2+2\alpha \theta$.
• split a two-phase spin at the point where α changes? ::: Yes — solve each phase, carry ω forward, add θ.
• keep signs so a reversing wheel gives net θ=0 but path 12 rad? ::: Yes — fix a positive direction and never drop a minus.