CLASSICAL-MECHANICS · Interactive Practice — Unit 5 · Video 7

When the Radius Moves: Velocity, Acceleration, and the Coriolis Term

IKey Formulas

For a particle in a plane with polar coordinates $(r, \theta)$ where both $r$ and $\theta$ change in time:

Formula	Name	Description
$\vec{v} = \dot{r}\,\hat{r} + r\dot{\theta}\,\hat{\theta}$	Velocity	Radial part $\dot{r}$ is the new piece vs. circular motion
$a_r = \ddot{r} - r\dot{\theta}^2$	Radial acceleration	Outward $\ddot{r}$ minus centripetal $r\dot{\theta}^2$
$a_\theta = 2\dot{r}\dot{\theta} + r\ddot{\theta}$	Tangential acceleration	Coriolis term $+$ angular-acceleration term
$a_{\text{cor}} = 2\dot{r}\dot{\theta}$	Coriolis acceleration	The cross term — appears only when the radius moves

Here $\dot{r} = \dfrac{dr}{dt}$ , $\ddot{r} = \dfrac{d^2r}{dt^2}$ , and likewise for $\theta$ .

IIInteractive Visualizations

Visualization 1 — Velocity Gains a Radial Component

Once the radius grows, the velocity leaves the tangent line and picks up the radial component $v_r = \dot{r}$ .

Radial rate ṙ 1 Angular rate θ̇ 1.5 Radius r 1.5

Visualization 2 — The Coriolis Term in a Spiral

On the spiral $r = bct^2$ , how do the Coriolis and angular-acceleration pieces of $a_\theta$ grow as the particle winds outward?

Spiral b 1 Angle c 0.5 Time t 1.0

💡 $a_\theta = 2\dot{r}\dot{\theta} + r\ddot{\theta} = 8bc^2t^2 + 2bc^2t^2 = 10bc^2t^2$ — the Coriolis term alone supplies $8/10$ of the tangential acceleration.

Visualization 3 — When the Radial Acceleration Vanishes

On the same spiral $a_r = 2bc - 4bc^3 t^4$ begins outward and flips to centripetal — where does it cross zero?

Spiral b 1 Angle c 1

IIIQuiz Questions

Question 1

For a particle whose radius changes in time, the velocity in polar coordinates is $\vec{v} = \dot{r}\,\hat{r} + r\dot{\theta}\,\hat{\theta}$ .

Which term is the new radial component that does not appear in pure circular motion?

✅ Correct! $v_r = \dot{r}$ is the radial component that circular motion was missing.

❌ Not quite. Look for the piece along $\hat{r}$ . The tangential term $r\dot\theta$ was already in circular motion.

Show solution

Solution:

In pure circular motion $r$ is constant, so $\dot{r} = 0$ and the velocity is purely tangential: $\vec{v} = r\dot{\theta}\,\hat{\theta}$ .

Once the radius is allowed to change, the product rule on $\vec{r} = r\,\hat{r}$ gives

$\vec{v} = \dot{r}\,\hat{r} + r\dot{\theta}\,\hat{\theta}.$

The brand-new piece is the radial component $v_r = \dot{r}$ . The tangential component $r\dot{\theta}$ was already present in circular motion.

Question 2

The radial acceleration in polar coordinates is $a_r = \ddot{r} - r\dot{\theta}^2$ .

A student writes the radial acceleration as simply $a_r = \ddot{r}$ , forgetting the second term. True or False: for a particle moving on a circle of fixed radius ( $\dot{r} = \ddot{r} = 0$ ) with constant $\dot{\theta}$ , that student's formula would give the correct nonzero centripetal acceleration.

✅ Correct! The dropped $-r\dot\theta^2$ term IS the centripetal acceleration; without it $a_r = 0$ , which is wrong.

❌ Not quite. With $\ddot{r}=0$ , the formula $a_r=\ddot r$ gives zero — but a circling particle has nonzero centripetal acceleration.

Show solution

Solution:

For uniform circular motion $\dot{r} = \ddot{r} = 0$ , so the full formula gives

$a_r = \ddot{r} - r\dot{\theta}^2 = 0 - r\dot{\theta}^2 = -r\dot{\theta}^2,$

the familiar centripetal acceleration (pointing inward, hence the minus sign).

The student's truncated formula $a_r = \ddot{r} = 0$ would predict zero radial acceleration — completely wrong. The essential centripetal term is exactly the $-r\dot{\theta}^2$ piece they dropped, so the statement is False.

Question 3

A particle follows the spiral $r = bc t^2$ , so that $\dot{r} = 2bct$ and $\dot{\theta} = 2ct$ .

Using $a_{\text{cor}} = 2\dot{r}\dot{\theta}$ , what is the Coriolis acceleration as a function of time?

✅ Correct! 2·(2bct)·(2ct) = 8bc²t².

❌ Not quite. Use only a_cor = 2ṙθ̇ = 2·(2bct)·(2ct); don't add the rθ̈ term.

Show solution

Solution:

The Coriolis acceleration is the cross term

$a_{\text{cor}} = 2\dot{r}\dot{\theta} = 2\,(2bct)\,(2ct) = 8bc^2 t^2.$

Common traps:

$10bc^2t^2$ is the total tangential acceleration $a_\theta = 2\dot r\dot\theta + r\ddot\theta = 8bc^2t^2 + 2bc^2t^2 $— it includes the extra$ r\ddot\theta$ term.
4bc²t² drops the factor of 2 in the Coriolis definition.
2bc²t has the wrong power of $t$ — it loses a factor of $t$ when multiplying $\dot r\dot\theta$ .

The Coriolis term alone is 8bc²t².

Question 4

For the spiral example the radial acceleration is $a_r = 2bc - 4bc^3 t^4$ . Setting $a_r = 0$ and solving gives the time $t_1$ at which the radial acceleration vanishes, and the corresponding angle is $\theta_1 = c\,t_1^2$ .

What is the value of $\theta_1$ , and what does it depend on?

✅ Correct! Both b and c cancel, leaving θ₁ = √2/2 ≈ 0.71 rad.

❌ Not quite. Substitute t₁ = (1/2c²)^(1/4) into θ₁ = c·t₁² and watch every constant cancel.

Show solution

Solution:

Set $a_r = 0$ :

$2bc = 4bc^3 t^4 \;\Rightarrow\; t^4 = \frac{1}{2c^2} \;\Rightarrow\; t_1 = \left(\frac{1}{2c^2}\right)^{1/4}.$

The factor $bc$ cancels, so $t_1$ depends only on $c$ . Now the angle:

$\theta_1 = c\,t_1^2 = c\left(\frac{1}{2c^2}\right)^{1/2} = c\cdot\frac{1}{c\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \approx 0.71\ \text{rad}.$

Both $b$ and $c$ cancel entirely, so the radial acceleration always vanishes at the same angle $\theta_1 = \sqrt{2}/2 \approx 0.71$ rad — independent of both constants.

(Note $\sqrt{2}/2 \approx 0.707$ , distinct from $\pi/4 \approx 0.785$ .)

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