Centripetal force
In uniform circular motion the speed stays the same but the direction keeps changing. A change in velocity is an acceleration, so there must be a net force. This acceleration points towards the centre:
ac = v²/r = ω² r
By the second law the net force towards the centre is
Fc = m v²/r = m ω² r
This is the centripetal force ("centre-seeking"). It is not a new force; it is the name for whatever real force (or sum of forces) points to the centre:
- Stone on a string: tension.
- Moon round the Earth: gravity.
- Car turning on a level road: static friction.
- Car on a banked road: part of the normal force (plus friction).
If this force suddenly stops, the body moves off along the tangent (first law). Centripetal force is always perpendicular to velocity, so it changes only the direction, not the speed.
Motion of a car on a level road
Forces on the car: weight mg down, normal force N up, and static friction f along the road towards the centre.
Vertical: N = mg. Towards the centre: f = m v²/r.
Friction cannot be larger than μs N, so m v²/r ≤ μs m g, which gives
vmax = √(μs r g)
The safe speed does not depend on the mass of the car. On a wet road μs is smaller, so vmax falls. The friction here is static, because the tyres are not sliding sideways.
Motion of a car on a banked road
Banking means raising the outer edge of a curved road above the inner edge by an angle θ. Then the normal force tilts towards the centre.
Without friction
Vertical: N cos θ = mg. Towards the centre: N sin θ = m v²/r. Dividing,
tan θ = v²/(r g), so the ideal (optimum) speed is v0 = √(r g tan θ). At this speed no friction is needed and tyre wear is least.
With friction (maximum safe speed)
If the car goes faster, it tends to slide outward, so friction f acts down the slope, towards the inside. Taking f = μs N at the limit:
Vertical: N cos θ = mg + f sin θ. Towards the centre: N sin θ + f cos θ = m v²/r.
So v² = r g (sin θ + μs cos θ)/(cos θ − μs sin θ), which gives
vmax = √[ r g (μs + tan θ)/(1 − μs tan θ) ]
If the car goes too slowly it tends to slide down, friction acts up the slope, and vmin = √[ r g (tan θ − μs)/(1 + μs tan θ) ].
Put θ = 0 and you get back vmax = √(μs r g) for a level road. Put μs = 0 and you get v0.
Cyclist leaning
A cyclist on a turn leans inward by θ with tan θ = v²/(r g), the same relation, so that the ground's force passes through the centre of gravity.
Try it at home
Bucket swing: half-fill a small bucket with water and swing it fast in a vertical circle. The water stays in because the bucket's push supplies the centripetal force. Do it outdoors!
Coin on a turntable: put a coin near the edge of a lazy Susan and spin it faster. It slides off when friction can no longer give mv²/r. A coin nearer the centre stays longer.
In the 3D: on the last step set μ = 0 and find the one speed at which the car does not skid (it equals v₀).
Key formulas and definitions
- a_c = v²/r = ω²r
- F_c = m v²/r = m ω² r
- Level road: v_max = √(μs r g)
- Banked, no friction: tanθ = v₀²/(r g)
- Banked with friction: v_max = √[r g (μs + tanθ)/(1 − μs tanθ)]
- v_min = √[r g (tanθ − μs)/(1 + μs tanθ)]
Worked examples
1. A 0.5 kg stone moves in a circle of radius 1 m at 2 m/s. Find the tension in the string (ignore gravity).
T = mv²/r = 0.5 × 4/1 = 2 N.
2. The stone in the example above is whirled at 4 m/s. Find the new tension.
T = 0.5 × 16/1 = 8 N. Doubling v makes the force 4 times.
3. A 1000 kg car turns on a level road of radius 50 m at 10 m/s. Find the friction needed.
f = mv²/r = 1000 × 100/50 = 2000 N, towards the centre.
4. Find the maximum safe speed on a level curve of radius 50 m if μs = 0.5 (g = 10 m/s²).
vmax = √(μs r g) = √(0.5 × 50 × 10) = √250 ≈ 15.8 m/s ≈ 57 km/h.
5. A curve of radius 50 m is banked at 15°. Find the speed at which no friction is needed (tan 15° = 0.268, g = 10).
v₀ = √(r g tanθ) = √(50 × 10 × 0.268) = √134 ≈ 11.6 m/s.
6. At what angle should a curve of radius 90 m be banked for a speed of 15 m/s? (g = 10)
tanθ = v²/(rg) = 225/900 = 0.25 → θ = tan⁻¹(0.25) ≈ 14°.
7. The 15° banked curve of radius 50 m has μs = 0.3. Find the maximum safe speed.
vmax² = r g (μs + tanθ)/(1 − μs tanθ) = 500 × (0.3 + 0.268)/(1 − 0.0804) = 500 × 0.618 = 309. vmax ≈ 17.6 m/s. On a level road with μs = 0.3 it would be only √150 ≈ 12.2 m/s.
Common mistakes
- Drawing a separate "centripetal force" on the free-body diagram in addition to the real forces. It is the net of the real forces towards the centre.
- Thinking there is an outward force pushing the car off the road. In the ground frame the car skids out because the inward force is not enough (inertia).
- Using kinetic friction μk for a car taking a turn without skidding. The tyres are not sliding sideways, so it is static friction.
- Mixing up the level-road and banked-road formulas, or using degrees in a calculator set to radians.