Centripetal Acceleration

Equivalent forms

a_c = \omega^2 r

a_c = \frac{4\pi^2 r}{T^2}

Reveals that uniform circular motion is constantly accelerating — a beautiful paradox of constant speed with changing velocity.

Symbol	Name	SI unit	Dimension	Range
$a_c$	Centripetal Accelerationoutput Acceleration directed toward the center of the circular path	m/s²	LT⁻²	0 – 200
$v$	Tangential Velocity Speed of the object along the circular path	m/s	LT⁻¹	0.1 – 100
$r$	Radius Radius of the circular path	m	L	0.1 – 1000

Unit systems

Symbol	SI	CGS	Imperial
$a_c$	m/s²	cm/s²	ft/s²
$v$	m/s	cm/s	ft/s
$r$	m	cm	ft

Where it holds

Valid for uniform circular motion at non-relativistic speeds. For relativistic circular motion, Lorentz factor corrections apply.

Dimensional analysis

a_

c \to [LT^{-2}]

v^{2}/r \to [LT^{-1}]^{2}/[L] = [L^{2}T^{-2}]/[L] = [LT^{-2}]

Both sides are

[LT^{-2}] = m/s^{2}

.

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Discovery

Christiaan Huygens · 1659

Huygens derived the formula while studying pendulum clocks, publishing it in Horologium Oscillatorium (1673).

Try this

At what speed does a car on a 50 m roundabout feel 1g of lateral acceleration?

Solve a = v²/r for v: v = √(ar) = √(9.8 × 50) ≈ 22.1 m/s ≈ 80 km/h. That's the speed where you feel pressed sideways as hard as gravity pulls you down.

Research status: stable

Real-world applications

Banked road design
Centrifuge separation in labs
Satellite orbit mechanics
Amusement park ride safety limits

Common misconceptions

Centripetal force is not a new force — it's provided by tension, gravity, friction, etc.
Centrifugal 'force' is fictitious — it only appears in rotating reference frames
Objects in circular motion are constantly accelerating even at constant speed

Experimental verification

Conical pendulum measurements, centrifuge g-force readings, satellite orbital velocity analysis.

Derivation

Consider an object moving at constant speed v on a circle of radius r.

In time

\Delta t

, the velocity vector rotates

by \Delta \theta = v\Delta t/r

.

The change in velocity |

\Delta v| = v\Delta \theta = v^{2}\Delta t/r

, so

a = |\Delta v|/\Delta t = v^{2}/r

, directed inward.

Limiting cases

v \to 0

⟶ a_

c \to 0

No speed means no circular motion and no centripetal acceleration.

r \to \infty

⟶ a_

c \to 0

An infinitely large circle approximates straight-line motion.

r \to 0

⟶ a_

c \to \infty

Infinitely tight turns require infinite acceleration — physically impossible.

What if…

What if you double the speed on the same curve?

Centripetal acceleration quadruples $(v^{2}$ dependence). At 160 km/h on a 50 m roundabout: a_ $c = 39.5\,\mathrm{m/s}^{2} \approx 4g$ . The car would skid.

What if the road is banked?

Banking angle $\theta$ provides a gravity component toward center: $\tan (\theta ) = v^{2}/(rg)$ . At the design speed, no friction is needed.

What happens at the top of a loop-the-loop?

Both gravity and normal force point inward. Minimum speed: $v = \sqrt{rg}$ so that gravity alone provides the centripetal force at the top.

1

Car on a roundabout

Given ·

v:: 22.1
r:: 50

Find · a_c

Steps

Car speed: $80\,\mathrm{km/h} = 22.2\,\mathrm{m/s}$ , radius $= 50\,\mathrm{m}$
a_ $c = v^{2}/r = (22.2)^{2}/50 = 493/50 = 9.86\,\mathrm{m/s}^{2}$
This is approximately 1g — you'd feel pushed sideways as hard as gravity pulls you down

Result · a_

c = v^{2}/r = (22.1)^{2}/50 = 488.4/50 = 9.77\,\mathrm{m/s}^{2} \approx 1g

2

ISS orbital acceleration

Given ·

v:: 7660
r:: 6779000

Find · a_c

Steps

ISS orbits at 7660 m/s, altitude $408\,\mathrm{km} \to r = 6371 + 408 = 6779\,\mathrm{km}$
a_ $c = (7660)^{2} / (6.779 \times 10^{6}) = 5.868 \times 10^{7} / 6.779 \times 10^{6}$
a_ $c = 8.65\,\mathrm{m/s}^{2}$ — this equals $GM/r^{2} at$ that altitude, confirming gravity provides the centripetal force

Result · a_

c = v^{2}/r = (7660)^{2}/(6.779e_{6}) = 8.65\,\mathrm{m/s}^{2}

— matching gravitational g at that altitude

Newton's Second Law→Newton's Law of Universal Gravitation→Simple Harmonic Motion Period→