Classical-Mechanics · Unit 5 · Video 2 · Interactive Practice

Velocity in Circular Motion

IKey Formulas

Formula	Name	Description
$\hat{r} = \cos\theta\,\hat{\imath} + \sin\theta\,\hat{\jmath}$	Radial unit vector	Points outward, along the position vector
$\hat{\theta} = -\sin\theta\,\hat{\imath} + \cos\theta\,\hat{\jmath}$	Tangential unit vector	$\hat{r}$ rotated $90^\circ$ counterclockwise
$\dfrac{d\hat{r}}{dt} = \dfrac{d\theta}{dt}\,\hat{\theta}$	Derivative of $\hat{r}$	Radial direction turns into tangential
$\dfrac{d\hat{\theta}}{dt} = -\dfrac{d\theta}{dt}\,\hat{r}$	Derivative of $\hat{\theta}$	Tangential direction turns back inward
$\vec{v} = R\,\dfrac{d\theta}{dt}\,\hat{\theta} = v_\theta\,\hat{\theta}$	Velocity	Purely tangential; no radial part
$v = R\,\omega,\quad \omega = \left\lvert\dfrac{d\theta}{dt}\right\rvert$	Speed and angular speed	Radius times angular speed

IIInteractive Visualizations

Visualization 1 — Rotating Polar Unit Vectors

As the object circles, $\hat{r}$ and $\hat{\theta}$ spin together and stay perpendicular, while $\hat{\imath}$ and $\hat{\jmath}$ at the origin stay fixed.

Polar angle θ (°) 50°

💡 The length $|\vec{r}| = R$ never changes — only its direction turns, and that turning of $\hat{r}$ is the entire source of the motion.

Visualization 2 — Velocity Is Purely Tangential

Does an object on a fixed circle ever have velocity pointing toward or away from the center?

Object angle θ (°) 130° Angular rate dθ/dt (rad/s) 1.00

Visualization 3 — Speed Is Radius Times Angular Speed

How does the speed $v = R\,\omega$ respond as you scale the radius or the angular speed?

Radius R (m) 2.0 Angular speed ω (rad/s) 1.50

💡 Challenge: double $R$ with $\omega$ fixed, then double $\omega$ with $R$ fixed — confirm the speed doubles each time.

IIIQuiz Questions

Question 1

Differentiating the radial unit vector with respect to time gives $\dfrac{d\hat{r}}{dt}$ . Which vector does this point along?

✅ Correct! dr̂/dt = (dθ/dt) θ̂ — the radial direction turns into the tangential one.

❌ Not quite. Factor out dθ/dt; the leftover vector (-sinθ î + cosθ ĵ) is θ̂, not r̂.

Show solution

Solution:

Start from $\hat{r} = \cos\theta\,\hat{\imath} + \sin\theta\,\hat{\jmath}$ with $\hat{\imath},\hat{\jmath}$ constant. By the chain rule:

$\frac{d\hat{r}}{dt} = -\sin\theta\,\frac{d\theta}{dt}\,\hat{\imath} + \cos\theta\,\frac{d\theta}{dt}\,\hat{\jmath} = \frac{d\theta}{dt}\left(-\sin\theta\,\hat{\imath} + \cos\theta\,\hat{\jmath}\right).$

The vector in parentheses is exactly $\hat{\theta}$ , so

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

Differentiating the radial direction gives the tangential direction.

Question 2

Carrying out the same differentiation for the tangential unit vector, what is $\dfrac{d\hat{\theta}}{dt}$ ?

✅ Correct! dθ̂/dt = -(dθ/dt) r̂ — the tangential direction turns back toward the center.

❌ Not quite. Watch the sign: the leftover vector is -(cosθ î + sinθ ĵ) = -r̂, giving a minus sign.

Show solution

Solution:

From $\hat{\theta} = -\sin\theta\,\hat{\imath} + \cos\theta\,\hat{\jmath}$ , differentiate term by term with the chain rule:

$\frac{d\hat{\theta}}{dt} = -\cos\theta\,\frac{d\theta}{dt}\,\hat{\imath} - \sin\theta\,\frac{d\theta}{dt}\,\hat{\jmath} = \frac{d\theta}{dt}\left(-\cos\theta\,\hat{\imath} - \sin\theta\,\hat{\jmath}\right).$

The parenthesized vector is $-(\cos\theta\,\hat{\imath} + \sin\theta\,\hat{\jmath}) = -\hat{r}$ , so

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

Differentiating the tangential direction points you back inward, along $-\hat{r}$ .

Question 3

For an object moving on a circle of fixed radius $R$ , the velocity is $\vec{v} = \dfrac{d}{dt}(R\,\hat{r})$ .

True or False: The velocity has both a radial ( $\hat{r}$ ) component and a tangential ( $\hat{\theta}$ ) component.

✅ Correct! With R constant, v = R(dθ/dt)θ̂ — purely tangential, no radial component.

❌ Not quite. Since R is constant, dr̂/dt = (dθ/dt)θ̂ leaves only a tangential term.

Show solution

Solution:

Because $R$ is constant on a fixed circle, it pulls straight out of the derivative:

$\vec{v} = \frac{d}{dt}(R\,\hat{r}) = R\,\frac{d\hat{r}}{dt} = R\,\frac{d\theta}{dt}\,\hat{\theta}.$

There is no $\hat{r}$ term at all — the velocity is purely tangential. This matches intuition: an object on a circle moves along the circle, never toward or away from the center. So the statement is False.

Question 4

A satellite moves on a circle of radius $R = 3$ m with angular speed $\omega = 4$ rad/s, but it travels clockwise, so $\dfrac{d\theta}{dt} = -4$ rad/s. What is its speed $v$ ?

✅ Correct! v = Rω = 3 × 4 = 12 m/s. Speed uses ω = |dθ/dt|, always positive.

❌ Not quite. Speed is a magnitude — it can't be negative. The sign of dθ/dt only sets direction, not speed.

❌ Not quite. Use v = Rω: multiply R = 3 by ω = |dθ/dt| = 4, don't add or divide.

Show solution

Solution:

Speed is the magnitude of velocity and uses the angular speed $\omega = \left\lvert d\theta/dt\right\rvert = 4$ rad/s:

$v = R\,\omega = (3)(4) = 12 \text{ m/s}.$

The clockwise direction makes $d\theta/dt = -4$ rad/s, which flips the velocity vector to point clockwise, but speed is always positive — the sign drops out under the absolute value. So $v = 12$ m/s.

IVSummary

The whole story of velocity in circular motion comes down to one move: differentiating the rotating radial direction $\hat{r}$ turns it into the tangential direction $\hat{\theta}$ .

$\dfrac{d\hat{r}}{dt} = \dfrac{d\theta}{dt}\,\hat{\theta}$ and $\dfrac{d\hat{\theta}}{dt} = -\dfrac{d\theta}{dt}\,\hat{r}$ — the two unit vectors feed into each other as the object turns.
$\vec{v} = R\,\dfrac{d\theta}{dt}\,\hat{\theta} = v_\theta\,\hat{\theta}$ — purely tangential, no radial part.
$v = R\,\omega$ with $\omega = \left\lvert\dfrac{d\theta}{dt}\right\rvert$ — speed is radius times angular speed.

The motion was never in a stretching vector. It lived entirely in a spinning one.

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