CLASSICAL-MECHANICS · Interactive Practice — Unit 5 · Video 6

Spin as an Arrow: Angular Velocity and Acceleration as Vectors

IKey Formulas

Formula	Name	Description
$\vec{\omega} = \dfrac{d\theta}{dt}\,\hat{k} = \omega_z\,\hat{k}$	Angular velocity	Lies along the rotation ( $z$ ) axis; sign of $\omega_z$ sets the sense
$\omega \equiv \left\lvert \omega_z \right\rvert = \left\lvert \dfrac{d\theta}{dt}\right\rvert$	Angular speed	Magnitude only — always positive
$\vec{v} = \vec{\omega}\times\vec{r} = r\,\dfrac{d\theta}{dt}\,\hat{\theta}$	Tangential velocity	From $\hat{k}\times\hat{r}=\hat{\theta}$
$\vec{\alpha} = \dfrac{d^2\theta}{dt^2}\,\hat{k} = \alpha_z\,\hat{k}$	Angular acceleration	Also along the rotation axis

Sense of rotation: $\omega_z > 0 \Rightarrow$ counterclockwise, $\vec{\omega}$ points up $(+\hat{k})$ ; $\omega_z < 0 \Rightarrow$ clockwise, $\vec{\omega}$ points down $(-\hat{k})$ .

Speeding up vs. slowing down: when $\vec{\omega}$ and $\vec{\alpha}$ point the same way the spin speeds up; when they point opposite ways it slows down.

IIInteractive Visualizations

Visualization 1 — Spin direction as an arrow

The sign of $\omega_z$ fixes both the sense of rotation and which way $\vec{\omega}$ points along the axis.

ω_z (rad/s) 1.5

💡 Right-hand rule: curl your right fingers along the motion and your thumb points the way $\vec{\omega}$ points.

Visualization 2 — When the spin reverses

With $\theta(t) = A\,t - B\,t^3$ , the rotation pauses and reverses exactly when $\omega_z = A - 3Bt^2$ reaches zero.

t (s) 0.00 A (rad/s) 4 B (rad/s³) 1

💡 The particle's angle $\theta$ is largest at the reversal instant $t_1 = \sqrt{A/(3B)}$ — the moment $\omega_z$ changes sign.

Visualization 3 — Speeding up or slowing down

Whether the spin speeds up depends on how $\vec{\alpha}$ compares to $\vec{\omega}$ , not on which way $\vec{\alpha}$ points.

Rotation ω Angular accel α

💡 An upward $\vec{\alpha}$ speeds up a CCW spin but slows a CW one: same direction speeds up, opposite directions slow down.

IIIQuiz Questions

Question 1

A wheel spins counterclockwise in the $xy$ -plane (viewed from the $+z$ side), and its angular speed is steady. In which direction does the angular velocity vector $\vec{\omega}$ point?

✅ Correct! CCW means ω_z > 0, so ω points up along +k̂.

❌ Not quite. ω always lies along the rotation (z) axis, never tangent or radial. For CCW, ω_z > 0.

Show solution

Solution:

For counterclockwise motion the angle $\theta$ is increasing, so $\omega_z = d\theta/dt > 0$ . Since $\vec{\omega} = \omega_z\,\hat{k}$ , a positive component means $\vec{\omega}$ points up, in the $+\hat{k}$ direction (out of the plane toward the viewer).

Right-hand rule: curl your right fingers counterclockwise and your thumb points up.

Question 2

A particle on a circle has angle $\theta(t) = A\,t - B\,t^{3}$ with $A, B > 0$ . At what time $t_1$ does the angular velocity momentarily vanish ( $\omega_z = 0$ )?

✅ Correct! Set A − 3B t² = 0 and solve for t.

❌ Not quite. Differentiate first: ω_z = A − 3B t². Setting it to zero gives t² = A/(3B), then take the square root.

Show solution

Solution:

Differentiate to get the angular velocity component: $\omega_z = \frac{d\theta}{dt} = \frac{d}{dt}\left(A t - B t^3\right) = A - 3B t^2.$

Set it to zero and solve: $A - 3B t_1^2 = 0 \;\Rightarrow\; t_1^2 = \frac{A}{3B} \;\Rightarrow\; t_1 = \sqrt{\frac{A}{3B}}.$

Question 3

True or False: If the angular acceleration vector $\vec{\alpha}$ points in the $+\hat{k}$ direction (up), then the object must be speeding up.

✅ Correct! With ω down (CW) and α up, the spin slows down — so α up does not guarantee speeding up.

❌ Not quite. Compare α to ω: if ω points down (clockwise) while α points up, the object slows down.

Show solution

Solution: False.

The direction of $\vec{\alpha}$ alone does not decide speeding up vs. slowing down — you must compare it to $\vec{\omega}$ .

If the object turns counterclockwise ( $\vec{\omega}$ up) and $\vec{\alpha}$ is also up, the vectors agree → speeds up.
If the object turns clockwise ( $\vec{\omega}$ down) while $\vec{\alpha}$ is up, the vectors oppose → slows down.

The rule: same direction speeds up, opposite directions slows down.

Question 4

A point object starts from rest and has angular acceleration $\alpha_z(t) = b\left(1 - \dfrac{t}{t_1}\right)$ for $0 \le t \le t_1$ (with $b > 0$ ). What is its angular velocity $\omega_z$ at time $t_1$ ?

✅ Correct! Integrating the triangular α_z over [0, t₁] gives b·t₁/2.

❌ Not quite. Integrate α_z from 0 to t₁: ∫b(1 − t'/t₁)dt' = b(t₁ − t₁/2) = b·t₁/2. (The b·t₁²/3 result is the angle, not the angular velocity.)

Show solution

Solution:

The change in angular velocity is the integral of the angular acceleration. Starting from rest, $\omega_z(0) = 0$ , so: $\omega_z(t_1) = \int_0^{t_1} b\left(1 - \frac{t'}{t_1}\right) dt' = b\left(t_1 - \frac{t_1^2}{2 t_1}\right) = b\left(t_1 - \frac{t_1}{2}\right) = \frac{b\,t_1}{2}.$

(Geometrically: the area under the straight line from $b$ down to $0$ over the interval $[0, t_1]$ is the triangle $\tfrac{1}{2} b\, t_1$ .)

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