Playground
Object orbits in a circle with adjustable speed and radius. Centripetal acceleration vector points inward.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Centripetal Accelerationoutput Acceleration directed toward the center of the circular path | m/s² | LT⁻² | 0 – 200 | |
| Tangential Velocity Speed of the object along the circular path | m/s | LT⁻¹ | 0.1 – 100 | |
| Radius Radius of the circular path | m | L | 0.1 – 1000 |
Deep dive
Derivation
Consider an object moving at constant speed v on a circle of radius r. In time Δt, the velocity vector rotates by Δθ = vΔt/r. The change in velocity |Δv| = vΔθ = v²Δt/r, so a = |Δv|/Δt = v²/r, directed inward.
Experimental verification
Conical pendulum measurements, centrifuge g-force readings, satellite orbital velocity analysis.
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
Real-world applications
- Banked road design
- Centrifuge separation in labs
- Satellite orbit mechanics
- Amusement park ride safety limits
Worked examples
Car on a roundabout
Given:
- v:
- 22.1
- r:
- 50
Find: a_c
Solution
a_c = v²/r = (22.1)²/50 = 488.4/50 = 9.77 m/s² ≈ 1g
ISS orbital acceleration
Given:
- v:
- 7660
- r:
- 6779000
Find: a_c
Solution
a_c = v²/r = (7660)²/(6.779e6) = 8.65 m/s² — matching gravitational g at that altitude
Scenarios
What if…
- scenario:
- What if you double the speed on the same curve?
- answer:
- Centripetal acceleration quadruples (v² dependence). At 160 km/h on a 50 m roundabout: a_c = 39.5 m/s² ≈ 4g. The car would skid.
- scenario:
- What if the road is banked?
- answer:
- Banking angle θ provides a gravity component toward center: tan(θ) = v²/(rg). At the design speed, no friction is needed.
- scenario:
- What happens at the top of a loop-the-loop?
- answer:
- Both gravity and normal force point inward. Minimum speed: v = √(rg) so that gravity alone provides the centripetal force at the top.
Limiting cases
- condition:
- v → 0
- result:
- a_c → 0
- explanation:
- No speed means no circular motion and no centripetal acceleration.
- condition:
- r → ∞
- result:
- a_c → 0
- explanation:
- An infinitely large circle approximates straight-line motion.
- condition:
- r → 0
- result:
- a_c → ∞
- explanation:
- Infinitely tight turns require infinite acceleration — physically impossible.
Context
Christiaan Huygens · 1659
Huygens derived the formula while studying pendulum clocks, publishing it in Horologium Oscillatorium (1673).
Hook
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.
Dimensions:
- lhs:
- a_c → [LT⁻²]
- rhs:
- v²/r → [LT⁻¹]²/[L] = [L²T⁻²]/[L] = [LT⁻²]
- check:
- Both sides are [LT⁻²] = m/s². ✓
Validity: Valid for uniform circular motion at non-relativistic speeds. For relativistic circular motion, Lorentz factor corrections apply.