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

Constant Speed, Constant Acceleration: The Geometry of Uniform Circular Motion

IKey Formulas

Formula	Name	Description
$v = r\,\lvert\omega\rvert$	Speed	Constant in uniform circular motion
$T = \dfrac{2\pi}{\omega}$	Period	Time for one full orbit
$f = \dfrac{1}{T} = \dfrac{\omega}{2\pi}$	Frequency	Orbits per second, in hertz (Hz)
$\lvert a_r\rvert = r\omega^2 = \dfrac{v^2}{r} = 4\pi^2 r f^2 = \dfrac{4\pi^2 r}{T^2} = v\,\lvert\omega\rvert$	Centripetal acceleration	Always points inward $(-\hat{r})$

IIVisualization 1 — Velocity and acceleration vectors

Constant speed still means constant acceleration: the velocity stays tangent while a steady pull points inward.

Position θ (deg) 45° Radius r (m) 2.0 Angular speed ω (rad/s) 1.5

💡 Because $v = r\omega$ grows linearly but $a_r = r\omega^2$ grows with the square of $\omega$ , doubling the spin rate doubles the speed yet quadruples the inward pull.

IIIVisualization 2 — Period and frequency

Angular speed alone fixes both the period and the frequency — and they move in opposite directions as $\omega$ grows.

Angular speed ω (rad/s) 2.0

IVVisualization 3 — Deriving the inward acceleration

Where does the inward acceleration come from? Subtract two equal-length velocities a small angle apart.

Angle Δθ (deg) 60°

💡 Dividing $\lvert\Delta\vec{v}\rvert \approx v\,\Delta\theta$ by $\Delta t$ and letting $\Delta\theta \to 0$ gives $\lvert a_r\rvert = v\lvert\omega\rvert = r\omega^2$ .

VQuiz Questions

Question 1

An object moves in uniform circular motion. Which statement is true about its velocity and acceleration?

✅ Correct! Constant speed, but a constant inward (centripetal) acceleration that bends the path into a circle.

❌ Not quite. Changing direction counts as acceleration even when the speed is fixed — and that acceleration points inward, not along the motion.

Show solution

Solution:

In uniform circular motion the tangential force is zero, so the speed $v = r\lvert\omega\rvert$ stays constant. But the direction of the velocity changes continuously, so the velocity vector changes — that change is an acceleration. The surviving acceleration is purely radial:

$\vec{a}_r = -r\omega^2\,\hat{r}$

The minus sign means it points inward, toward the center. "Uniform" means constant speed, not zero acceleration.

Question 2

An object orbits with angular speed $\omega = 4\ \text{rad/s}$ . What is its period $T$ ?

(Use $T = \dfrac{2\pi}{\omega}$ .)

✅ Correct! T = 2π/4 = π/2 ≈ 1.57 s.

❌ Not quite. Use T = 2π/ω. The other options either multiply by ω instead of dividing, or drop the factor of 2π.

Show solution

Solution:

$T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \approx 1.57\ \text{s}$

The period depends only on the angular speed — the radius does not enter. The faster the object sweeps angle, the shorter the time for one trip around.

Question 3

A car rounds a circular track of radius $r = 50\ \text{m}$ at a constant speed $v = 20\ \text{m/s}$ . What is the magnitude of its centripetal acceleration?

(Use $\lvert a_r\rvert = \dfrac{v^2}{r}$ .)

✅ Correct! v²/r = 400/50 = 8 m/s².

❌ Not quite. Remember to square the speed: a_r = v²/r = (20²)/50.

Show solution

Solution:

$\lvert a_r\rvert = \frac{v^2}{r} = \frac{(20)^2}{50} = \frac{400}{50} = 8\ \text{m/s}^2$

This is the speed-and-radius form of the centripetal acceleration. The "0.4" answer comes from forgetting to square the speed ( $v/r = 20/50$ ), and "2.5" comes from inverting the ratio ( $r/v$ pieces). Always square the speed.

Question 4

True or False: If you double the angular speed ω of an object in uniform circular motion (keeping the radius fixed), its centripetal acceleration also exactly doubles.

✅ Correct! Since a_r = rω², doubling ω quadruples the acceleration (×4).

❌ Not quite. The speed v = rω doubles, but a_r = rω² depends on the square of ω, so it grows by a factor of 4.

Show solution

Solution: The statement is False.

The centripetal acceleration is $\lvert a_r\rvert = r\omega^2$ . Doubling ω replaces $\omega^2$ with $(2\omega)^2 = 4\omega^2$ , so the acceleration becomes four times larger, not twice.

$\frac{\lvert a_r\rvert_{\text{new}}}{\lvert a_r\rvert_{\text{old}}} = \frac{r(2\omega)^2}{r\omega^2} = 4$

(The speed $v = r\omega$ does double — but acceleration depends on ω squared.)

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