CLASSICAL-MECHANICS · Interactive Practice | Unit 5 · Video 4

Tangential and Radial Acceleration

IKey Formulas

For a particle on a circle of radius $R$ with angle $\theta(t)$ and angular velocity $\omega \equiv \dfrac{d\theta}{dt}$ :

Formula	Name	Description
$\vec{a} = a_r\,\hat{r} + a_\theta\,\hat{\theta}$	Acceleration in polar form	Radial + tangential components
$\vec{v} = R\dfrac{d\theta}{dt}\,\hat{\theta}$	Velocity on a circle	Purely tangential
$a_\theta = R\dfrac{d^2\theta}{dt^2}$	Tangential acceleration	Nonzero only when speed changes
$a_r = -R\omega^2$	Radial (centripetal) acceleration	Always $\le 0$ — points inward

Key identity from the derivation: $\dfrac{d\hat{\theta}}{dt} = -\dfrac{d\theta}{dt}\,\hat{r}$ , because $\hat{\theta} = -\sin\theta\,\hat{i} + \cos\theta\,\hat{j}$ and $\hat{r} = \cos\theta\,\hat{i} + \sin\theta\,\hat{j}$ .

IIInteractive Visualizations

Visualization 1 — The rotating polar basis

Unlike the fixed axes $\hat{i},\hat{j}$ , the polar basis $\hat{r},\hat{\theta}$ reorients at every point of the circle.

Angle θ 45°

💡 Because this frame is not constant, differentiating $\vec{v} = R\dot{\theta}\,\hat{\theta}$ needs the product rule — and the extra term $-R\dot{\theta}^2\,\hat{r}$ is exactly the inward acceleration.

Visualization 2 — Tangential vs. radial acceleration

Because $a_r = -R\omega^2$ , the inward arrow survives either sign of $\omega$ , while the tangential arrow flips with $\alpha$ .

ω (rad/s) 1.00 α (rad/s²) 0.50 position θ 60°

Visualization 3 — Where centripetal acceleration vanishes

For $\theta(t) = At^3 - Bt$ , the radial curve kisses zero exactly at $t_1 = \sqrt{B/3A}$ while the tangential keeps climbing.

A 1.0 B 3.0 R 1.0 time t 0.00

💡 At $t_1$ the angular velocity passes through zero — the particle pauses and reverses its circulation, yet $a_\theta \neq 0$ means it is still angularly accelerating.

IIIQuiz Questions

Question 1

For a particle on a circle of radius $R$ with angular velocity $\omega = \dfrac{d\theta}{dt}$ , what is the radial (centripetal) component of the acceleration?

✅ Correct! The radial component is $a_r = -R\omega^2$ .

❌ Not quite. The radial component comes from the term with $\hat{r}$ , which is $-R\omega^2$ (note the minus sign and the square of ω, not θ¨).

Show solution

Solution:

Differentiating $\vec{v} = R\dot{\theta}\,\hat{\theta}$ with the product rule and using $\dfrac{d\hat{\theta}}{dt} = -\dot{\theta}\,\hat{r}$ gives

$\vec{a} = R\ddot{\theta}\,\hat{\theta} - R\dot{\theta}^2\,\hat{r}.$

Reading off the radial component (the coefficient of $\hat{r}$ ):

$a_r = -R\dot{\theta}^2 = -R\omega^2.$

The minus sign and the square are what make it always point inward.

Question 2

True or False: If the particle reverses its direction of circulation (so the sign of $\omega$ flips from positive to negative), the radial acceleration $a_r$ switches from pointing inward to pointing outward.

✅ Correct! Since $a_r = -R\omega^2$ depends on $\omega^2$ , it stays negative (inward) no matter the sign of ω.

❌ Not quite. $a_r = -R\omega^2$ depends on $\omega^2$ , not ω, so flipping the sign of ω leaves $a_r$ unchanged — still inward.

Show solution

Solution:

$a_r = -R\omega^2.$

Here $R > 0$ and $\omega^2 \ge 0$ , so $a_r \le 0$ regardless of the sign of $\omega$ . A negative radial component means the acceleration points opposite to $\hat{r}$ (which points outward), i.e. inward, toward the center. Reversing the circulation direction changes the sign of $\omega$ but not of $\omega^2$ , so $a_r$ stays inward.

The statement is therefore False.

Question 3

In the derivation, we need the time derivative of the unit vector $\hat{\theta} = -\sin\theta\,\hat{i} + \cos\theta\,\hat{j}$ . Using the chain rule (since $\theta$ depends on $t$ ), what is $\dfrac{d\hat{\theta}}{dt}$ ?

✅ Correct! The bracket collapses to $-\hat{r}$ , giving $d\hat{\theta}/dt = -(d\theta/dt)\,\hat{r}$ .

❌ Not quite. Differentiating gives a factor $(d\theta/dt)$ times $(-\cos\theta\,\hat{i} - \sin\theta\,\hat{j}) = -\hat{r}$ . Watch the sign.

Show solution

Solution:

Differentiate term by term with the chain rule:

$\frac{d\hat{\theta}}{dt} = -\cos\theta\,\frac{d\theta}{dt}\,\hat{i} - \sin\theta\,\frac{d\theta}{dt}\,\hat{j} = \frac{d\theta}{dt}\big(-\cos\theta\,\hat{i} - \sin\theta\,\hat{j}\big).$

The bracket is $-(\cos\theta\,\hat{i} + \sin\theta\,\hat{j}) = -\hat{r}$ , so

$\frac{d\hat{\theta}}{dt} = -\frac{d\theta}{dt}\,\hat{r}.$

This is the term that injects the inward (radial) acceleration.

Question 4

A particle on a circle of radius $R$ has $\theta(t) = A t^3 - B t$ with $A, B > 0$ . At what time $t_1 > 0$ does the centripetal (radial) acceleration first equal zero?

✅ Correct! Set $3At^2 - B = 0$ and solve: $t_1 = \sqrt{B/(3A)}$ .

❌ Not quite. $a_r = 0$ needs $3At^2 - B = 0$ , so $t^2 = B/(3A)$ and $t_1 = \sqrt{B/(3A)}$ — keep the 3A in the denominator under the root.

Show solution

Solution:

The radial acceleration is $a_r = -R\left(\dfrac{d\theta}{dt}\right)^2$ with $\dfrac{d\theta}{dt} = 3A t^2 - B$ . Since $R \neq 0$ , $a_r = 0$ requires the squared quantity itself to vanish:

$3A t^2 - B = 0 \;\Rightarrow\; 3A t^2 = B \;\Rightarrow\; t^2 = \frac{B}{3A}.$

Taking the positive root (time is positive):

$t_1 = \sqrt{\frac{B}{3A}}.$

At this instant $\omega = \dfrac{d\theta}{dt} = 0$ — the particle momentarily stops rotating and reverses. Note the tangential acceleration $a_\theta = 6ARt_1 \neq 0$ there.

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